NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

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Introduction 

In earlier classes, you have learnt about congruence of two geometric figures, and also some basic theorems and results on the congruence of triangle. Two geometric figures having same shape and size are congruent to each other but two geometric figures having same shape are called similar. Two congruent geometric figures are always similar but the converse may or may not be true.

 

All regular polygons of same number of sides such as equilateral triangle, squares, etc. are similar. All circles are similar.

In some cases, we can easily notice that two geometric figures are not similar. For example, a triangle and a rectangle can never be similar. In case, we are given two triangles, they may appear to be similar but actually they may not be similar. So, we need some criteria to determine the similarity of two geometric figures. In particular, we shall discuss similar triangles.

~ Historical Facts

Euclid was a very great Greek mathematician born about 2400 years ago. He is called the father of geometry because he was the first to establish a school of mathematics in Alexandria. He wrote a book on geometry called "The Elements" which has 13 volumes and has been used as a text book for over 2000 years. This book was further systematized by the great mathematician of Greece like Thales, Pythagoras, Pluto and Aristotle.

Abraham Lincoln, as a young lawyer was of the view that this greek book was a splendid sharpner of human mind and improves his power of logic and language.

A king once asked Euclid, "Isn't there an easier way to understand geometry" 

Euclid replied : "There is no royal-road way to geometry. Every one has to think for himself when studying."

Thales (640-546 B.C.) a Greek mathematician was the first who initiated and formulated the theoretical study of geometry to make astronomy a more exact science. He is said to have introduced geometry in Greece. He is believed to have found the heights of the pyramids in Egypt, using shadows and the principle of similar triangles. The use of similar triangles has made possible the measurements of heights and distances. He proved the well-known and very useful theorem credited after his name : Thales Theorem. 

~ Congruent Figures 

Two geometrical figures are said to be congruent, provided they must have same shape and same size.

Congruent figures are alike in every respect.

Ex. 1. Two squares of the same length.

2. Two circle of the same radii.

3. Two rectangles of the same dimensions.

4. Two wings of a fan.

5. Two equilateral triangles of same length.

~ Similar Figures 

Two figures are said to be similar, if they have the same shape. Similar figures may differ in size. Thus, two congruent figures are always similar, but two similar figures need not be congruent. NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRevNCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Ex. 1. Any two line segments are similar.

2. Any two equilateral triangles are similar

3. Any two squares are similar.

4. Any two circles are similar.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRevNCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRevNCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRevNCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRevNCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRevNCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRevNCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

We use the symbol '~' to indicate similarity of figures.

~ Similar Triangles 

DABC and DDEF are said to be similar, if their corresponding angles are equal and the corresponding sides are proportional.

i.e., when ÐA = ÐD, ÐB = ÐE, ÐC = ÐF

and NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev.

And, we write DABC ~ DDEF.

The sign '~' is read as 'is similar to'.

Theorem-1 (Thales Theorem or Basic Proportionality Theorem) : If a line is drawn parallel to one side of a triangle intersecting the other two sides, then the other two sides are divided in the same ratio. 

Given : A DABC in which line l parallel to BC (DEBC) intersecting AB at D and AC at E.

To prove :NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Construction : Join D to C and E to B. Through E drawn EF perpendicular to AB i.e., EF ^ AB and through D draw DG ^ AC.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Proof :

STATEMENT REASON 

1. Area of (DADE) = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev(AD × EF) Area of D = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Area of (DBDE) = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev(BD × EF)

2.NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev By 1.

3.NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Similarly

4. Area (DBDE) = Area (DCDE) Ds BDE and CDE are on the same base BC and between the same parallel lines DE and BC.

5.NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev By 3. & 4.

6.NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev By 1. & 5.

Hence proved.

Theorem-2 (Converse of Basic Proportionality Theorem) : If a line divides any two sides of a triangle proportionally, the line is parallel to the third side. 

Given : A DABC and DE is a line meeting AB and AC at D and E respectively such that =

To prove : DEBC

Proof :

STATEMENT REASON 

1. If possible, let DE be not parallel to BC.

Then, draw DFBC

2.NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev By Basic Proportionality Theorem.

3.NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Given

4.NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev=NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev From 2 and 3.

Þ +1 =NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev + 1 Adding 1 on both sides.

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev By addition.

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev AF + FC = AC and AE + EC = AC.

Þ FC = EC Þ E and F coincide.

But, DFNCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRevBC. Hence DEBC.

Hence, proved.

Ex.1 In the adjoining figure, DEBC.

(i) If AD = 3.4 cm, AB = 8.5 cm and AC = 13.5 cm, find AE.

(ii) If NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev and AC = 9.6 cm, find AE.

Sol. (i) Since DEBC, we have NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev = 5.4

Hence, AE = 5.4 cm.

(ii) Since DEBC, we have NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRevNCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Let AE = x cm. Then, EC = (AC _ AE) = (9.6 _ x) cm.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Þ 5x = 3(9.6 _ x)

Þ 5x = 28.8 _ 3x Þ 8x = 28.8 Þ x = 3.6.

AE = 3.6 cm.

Ex.2 In the adjoining figure, AD = 5.6 cm, AB = 8.4 cm, AE = 3.8 cm and AC = 5.7 cm. Show that DEBC.

Sol. We have, AD = 5.6 cm, DB = (AB _ AD) = (8.4 _ 5.6) cm = 2.8 cm.

AE = 3.8 cm, EC = (AC _ AE) = (5.7 _ 3.8) cm = 1.9 cm.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev and NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Thus, NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

DE divides AB and AC proportionally.

Hence, DEBC

Ex.3 In fig, NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev and ÐPST = ÐPRQ. Prove that PQR is an isosceles triangle.  [NCERT] 

Sol. It is given that NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

So, STNCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRevQR [Theorem]

Therefore, ÐPST = ÐPQR [Corresponding angles] - (1)

Also, it is given that

ÐPST = ÐPRQ (2)

So, ÐPRQ = ÐPQR [From 1 and 2]

Therefore PQ = PR [Sides opposite the equal angles]

i.e., PQR is an isosceles triangle.

Ex.4 Prove that any line parallel to parallel sides of a trapezium divides the non-parallel sides proportionally (i.e., in the same ratio).

or

ABCD is a trapezium with DCAB. E and F are points on AD and BC respectively such that EFAB. Show that NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev [NCERT] 

Sol. We are given trapezium ABCD.

CDNCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRevBA

EFAB and CD both

We join AC.

It mets EF at O.

In DACD, OECD

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev ...(i)

(Basic Proportionality Theorem)

In DCAB, OFAB

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev [B.P.T]

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev ...(ii)

From (i) and (ii)

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Hence, proved.

Ex.5 Prove that the internal bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle.

(Internal Angle Bisector Theorem) 

Sol. Given :  A DABC in which AD is the internal bisector of ÐA.

To Prove : NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Construction : Draw CEDA, meeting BA produced at E.

Proof :

STATEMENT REASON 

1. Ð1 = Ð2 AD is the bisector of ÐA

2. Ð2 = Ð3 Alt. Ðs are equal, as CEDA and AC is the transversal

3. Ð1 = Ð4 Corres. Ðs are equal, as CEDA and BE is the transversal

4. Ð3 = Ð4 From 1, 2 and 3.

5. AE = AC Sides opposite to equal angles are equal

6. In DBCE, DACE

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev By B.P.T.

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Using 5

Hence, Proved.

Remark : The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle. i.e., if in a DABC, AD is the bisector of the exterior of angle ÐA and intersect BC produced in D, NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev.

 ~ Axioms of similarity of triangles 

1.  AA (Angle-Angle) Axiom of Similarity : 

If two triangles have two pairs of corresponding angles equal, then the triangles are similar. In the given figure, DABC and DDEF are such that

ÐA = ÐD and ÐB = ÐE.

DABC ~ DDEF

2.  SAS (Side-Angle-Side) Axiom of Similarity : 

If two triangles have a pair of corresponding angles equal and the sides including them proportional, then the triangles are similar.

In the given fig, DABC and DDEF are such that

ÐA = ÐD and NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

DABC ~ DDEF.

3.  SSS (Side-Side-Side) Axiom of Similarity : 

If two triangles have three pairs of corresponding sides proportional, then the triangles are similar.

If in DABC and DDEF we have :

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev, then DABC ~ DDEF.

Ex.6 In figure, find ÐL.

Sol. In DABC and DLMN,

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev and NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Þ DABC ~ DLMN (SSS Similarity)

Þ ÐL = ÐA = 180° _ ÐB _ ÐC

= 180° _ 50° _ 70° = 60°

ÐL = 60°

Ex.7 In the figure, AB ^ BC, DE ^ AC, and GF ^ BC. Prove that DADE ~ DGCF.

Sol. Ð1 + Ð4 = Ð1 + Ð2 (each side = 90°)

Þ Ð4 = Ð2

Þ ÐA = ÐG ...(i)

Also ÐE = ÐF ...(ii) (each equal to 90°)

From (i) and (ii), we get AA similarity for triangles ADE and GCF.

Þ DADE ~ DGCF

Ex.8 In fig, NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev and Ð1 = Ð2. Prove that DPQS ~ DTQR.

Sol. Ð1 = Ð2 (Given)

Þ PR = PQ ...(i)

(Sides opposite to equal angles in DQRP)

Also NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev (Given) ...(ii)

Form (i) and (ii), we have

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev ...(iii)

Now, in triangles PQR and TQR, we have

ÐPQS = ÐTQR (each = Ð1)

and (from (3))

Þ DPQS ~ DTQR (SAS Similarity)

Ex.9 In fig, CD and GH are respectively, the medians of DABC and DFEG, If DABC ~ DFEG, prove that

(i) DADC ~ DFHG

(ii) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev(NCERT) 

Sol. DABC ~ DFEG (given)

Þ ÐA = ÐF, ...(i) (Q the corresponding angles of the similar triangles are equal)

Also, NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev (Corresponding sides are proportional)

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRevNCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev ...(ii)

Now, in triangles ADC and FHG, we have

ÐA = ÐF and NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev (By (i) and (ii))

Þ DADC ~ DFHG  (SAS similarity)

(ii) DADC ~ DFHG

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev (Corresponding sides proportional)

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Ex.10 ABC is a right triangle, right angled at B. If BD is the length of the perpendicular drawn from B to AC. Prove that:

(i) DADB ~ DABC and hence AB2 = AD × AC (ii) DBDC ~ DABC and hence BC2 = CD × AC

(iii) DADB ~ DBDC and hence BD2 = AD × DC (iv) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Sol. Given : ABC is right angled triangle at B and BD ^ AC

To prove : 

(i) DADB ~ DABC and hence AB2 = AD × AC

(ii) DBDC ~ DABC and hence BC2 = CD × AC

(iii) DADB ~ DBDC and hence BD2 = AD × DC

(iv) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Proof : (i) In two triangles ADB and ABC, we have :

ÐBAD = ÐBAC (Common)

ÐADB = ÐABC (Each is right angle)

ÐABD = ÐACB (Third angle)

ÐADB ~ ÐABC (AAA Similarity)

Triangle ADB and ABC are similar and so their corresponding sides must be proportional.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev ÞNCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Þ AB × AB = AC × AD Þ AB2 = AD × AC This proves (a).

(ii) Again consider two triangles BDC and ABC, we have

ÐBCD = ÐACB (Common)

ÐBDC = ÐABC (Each is right angle)

ÐDBC = ÐBAC (Third angle)

Triangle are similar and their corresponding sides must be proportional.

i.e., DBDC ~ DABC NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Þ BC × BC = DC × AC Þ BC2 = CD × AC This proves (ii)

(iii) In two triangles ADB and BDC, we have :

ÐBDA = ÐBDC = 90°

Ð3 = Ð2 = 90° _ Ð1 [Q Ð1 + Ð2 = 90°, Ð1 + Ð3 = 90°]

Ð1 = Ð4 = 90° _ Ð2 [Q Ð1 + Ð2 = 90°, Ð2 + Ð4 = 90°]

DADB ~ DBDC (AAA criterion of similarity)

Þ Their corresponding sides must be proportional.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Þ BD × BD = AD × DC

BD2 = AD × DC

Þ BD is the mean proportional of AD and DC

(iv) From (i), we have : AB2 = AD × AC

(ii), we have : BC2 = CD × AC

(iii), we have : BD2 = AD × DC

Consider NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

= NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev (from (iii))

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Thus we have proved the following :

If a perpendicular is drawn from the vertex containing the right angle of a right triangle to the hypotenuse then: 

(a) The triangle on each side of the perpendicular are similar to each other and also similar to the original triangle. 

i.e., DADB ~ DBDC, DADB ~ DABC, DBDC ~ DABC 

(b) The square of the perpendicular is equal to the product of the length of two parts into which the hypotenuse is divided by the perpendicular i.e., BD2 = AD × DC. 

~ RESULTS ON AREA OF SIMILAR TRIANGLES 

Theorem-3 : the areas of two similar triangles are proportional to the squares on their corresponding sides.  

Given : DABC ~ DDEF

To prove :NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Construction : Draw AL ^ BC and DM ^ EF.

Proof :

STATEMENT REASON 

1.NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Area of D = × Base × Height

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

2. In DALB and DDME, we have

(i) ÐALB = ÐDME Each equal to 90°

(ii) ÐABL = ÐDEM DABC ~ DDEF Þ ÐB = ÐE

DALB ~ DDME AA-axiom

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Corresponding sides of similar Ds are proportional.

3. DABC ~ DDEF Given.

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Corresponding sides of similar Ds are proportional.

4.NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev From 2 and 3.

5. Substituting in 1, we get :

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

6. Combining 3 and 5, we get :

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Corollary-1 : the areas of two similar triangles are proportional to the squares on their corresponding altitude.  

Given : DABC ~ DDEF, AL ^ BC and DM ^ EF.

To prove :NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Proof :

STATEMENT REASON 

1.NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Area of D = × Base × Height

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

2. In DALB and DDME, we have

(i) ÐALB = ÐDME Each equal to 90°

(ii) ÐABL = ÐDEM DABC ~ DDEF Þ ÐB = ÐE

Þ DALB ~ DDME AA-axiom

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Corresponding sides of similar Ds are proportional.

3. DABC ~ DDEF Given.

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Corresponding sides of similar Ds are proportional.

4.NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev From 2 and 3.

5. Substituting in 1, we get :

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev.

Hence, proved.

Corollary-2 : the areas of two similar triangles are proportional to the squares on their corresponding medians.  

Given : DABC ~ DDEF and AP, DQ are their medians.

To prove :NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Proof :

STATEMENT REASON 

1. DABC ~ DDEF Given

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev.......I. Areas of two similar Ds are proportional to the

squares on their corresponding sides.

2. DABC ~ DDEF

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev......II Corresponding sides of similar Ds are proportional.

3.NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev and ÐA = ÐD From II and the fact the DABC ~ DDEF

Þ DAPB ~ DDQE By SAS-similarity axiom

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev .......III

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev From II and III.

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev..........IV

4.NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev From I and IV.

Hence, proved.

Corollary-3 : The areas of two similar triangles are proportional to the squares on their corresponding angle bisector segments. 

Given : DABC ~ DDEF and AX, DY are their

bisectors of ÐA and ÐD respectively.

To prove :NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Proof :

STATEMENT REASON 

1.NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Areas of two similar Ds are proportional to the squares of the corresponding sides.

2. DABC ~ DDEF Given

Þ ÐA = ÐD

Þ ÐA = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRevÐD

Þ ÐBAX = ÐEDY ÐBAX = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRevÐA and ÐEDY = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRevÐD

3. In DABX and DDEY, we have Given

ÐBAX = ÐEDY From 2.

ÐB = ÐE DABC ~ DDEF

DABX ~ DDEY By AA similarity axiom

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

4.NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev From 1 and 3.

Hence, proved.

Ex.11 It is given that DABC ~ DPQR, area (DABC) = 36 cm2 and area (DPQR) = 25 cm2. If QR = 6 cm, find the length of BC.

Sol. We know that the areas of similar triangles are proportional to the squares of their corresponding sides.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Let BC = x cm. Then,

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRevÛ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRevÛ x2 = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Û x = = = 7.2.

Hence BC = 7.2 cm

Ex.12 P and Q are points on the sides AB and AC respectively of DABC such that PQBC and divides DABC into two parts, equal in area. Find PB : AB.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev 

Sol. Area (DAPQ) = Area (trap. PBCQ) [Given]

Þ Area (DAPQ) = [Area (DABC) _ Area (DAPQ)]

Þ 2 Area (DAPQ) = Area (DABC)

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev ...(i)

Now, in DAPQ and DABC, we have

ÐPAQ = ÐBAC [Common ÐA]

ÐAPQ = ÐABC [PQBC, corresponding Ðs are equal]

DAPQ ~ DABC.

We known that the areas of similar Ds are proportional to the squares of their corresponding sides.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev [Using (i)]

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev i.e., AB = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev · AP

Þ AB = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev (AB _ PB) Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev PB = (NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev _ 1) AB

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev.

PB : AB = ( _ 1) : NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Ex.13 Two isosceles triangles have equal vertical angles and their areas are in the ratio 16 : 25. Find the ratio of their corresponding heights.

Sol. Let DABC and DDEF be the given triangles in which AB = AC, DE = DF, ÐA = ÐD and

Draw AL ^ BC and DM ^ EF

Now, NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev = 1 and NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev = 1 [Q AB = AC and DE = DF]

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev.

In DABC and DDEF, we have

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev and ÐA = ÐD

Þ DABC ~ DDEF [By SAS similarity axiom]

But, the ratio of the areas of two similar Ds is the same as the ratio of the squares of their corresponding heights.

Þ Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

AL : DM = 4 : 5, i.e., the ratio of their corresponding heights = 4 : 5.

Ex.14 If the areas of two similar triangles are equal, prove that they are congruent.

Sol. Let DABC ~ DDEF and area (DABC) = area (DDEF).

Since the ratio of the areas of two similar Ds is equal to the ratio of the squares on their corresponding sides, we have

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev= NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev = 1 [Q Area (DABC) = Area (DDEF)]

Þ AB2 = DE2, AC2 = DF2 and BC2 = EF2

Þ AB = DE, AC = DF and BC = EF

DABC NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev DDEF [By SSS congruence]

Ex.15 In fig, the line segment XY is parallel to side AC of DABC and it divides the triangle into two parts of equal areas. Find the ratio NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev.  [NCERT] 

Sol. We are given that XYAC.

Þ Ð1 = Ð3 and Ð2 = Ð4 [Corresponding angles]

Þ DBXY ~ DBAC [AA similarity]

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev [By theorem] ...(i)

Also, we are given that

ar (DBXY) = × ar (DBAC) Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev ...(ii)

From (i) and (ii), we have Þ = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev ...(iii)

Now, NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev = 1 _ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev = 1 _ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev = 1 _ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev [By (iii)]

= NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Hence, NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Theorem-4 [Pythagoras Theorem] : In a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. 

Given : A DABC in which ÐB = 90°.

To prove : AC2 = AB2 + BC2.

Construction : From B, Draw BD ^ AC.

Proof :

STATEMENT REASON 

1. In DADB and DABC, we have :

ÐBAD = ÐCAB = ÐA Common

ÐADB = ÐABC Each = 90°

DADB ~ DABC By AA axiom of similarity

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Corr. sides of similar Ds are proportional

Þ AB2 = AD × AC ..(i)

2. In DCDB and DCBA, we have :ÐCDB = ÐCBA Each = 90°

ÐBCD = ÐACB = ÐC Common

DCDB ~ DCBA By AA axiom of similarity

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Corr. sides of similar Ds are proportional

Þ BC2 = DC × AC ..(ii)

3. Adding (i) and (ii), we get

AB2 + BC2 = AD × AC + DC × AC

= (AD + DC) × AC = ACQ AD + DC = AC

Hence, AB2 + BC2 = AC2.

Theorem-5 [Converse of pythagoras Theorem] : In a triangle if the square of one side is equal to the sum of the squares of the other two sides, then the triangle is right angled. 

Given : A DABC in which AB2 + BC2 = AC2

To prove : ÐB = 90°

Construction : Draw a DDEF in which

DE = AB, EF = BC and ÐE = 90°

Proof :

STATEMENT REASON 

1. In DDEF, we have: ÐE = 90°

DE2 + EF2 = DF2 By Pythagoras Theorem

Þ AB2 + BC2 = DF2 Q DE = AB and EF = BC

Þ AC2 = DF2 Q AB2 + BC2 = AC2 (Given)

Þ AC = DF

2. In DABC and DDEF, we have :

AB = DE By construction

BC = EF By construction

AC = DF Proved above

DABC DDEF By SSS congruence

Þ ÐB = ÐE c.p.c.t

Þ ÐE = 90° Q ÐE = 90°

Hence, ÐB = 90°

Ex.16 If ABC is an equilateral triangle of side a, prove that its altitude = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Sol. DABC is an equilateral triangle.

We are given that AB = BC = CA = a. AD is the altitude, i.e., AD ^ BC.

Now, in right angled triangles ABD and ACD, we have

AB = AC [Given]

and AD = AD [Common side]

Þ DABD NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev DACD [By RHS congruence]

Þ BD = CD Þ BD = DC = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRevBC = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

From right triangle ABD,

AB2 = AD2 + BD2 Þ a= AD2 + NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Þ AD2 = a2 _ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Þ AD = a

Ex.17 In a DABC, obtuse angled at B, if AD is perpendicular to CB produced, prove that :

AC2 = AB2 + BC2 + 2BC × BD

Sol. In DADB, ÐD = 90°.

AD2 + DB2 = AB2 ... (i) [By Pythagoras Theorem]

In DADC, ÐD = 90°

AC2 = AD2 + DC2 [By Pythagoras Theorem]

= AD2 + (DB + BC)2

= AD2 + DB2 + BC2 + 2DB × BC

= AB2 + BC2 + 2BC × BD [Using (i)]

Hence, AC2 = AB2 + BC2 + 2BC × BD.

Ex.18 In the given figure, ÐB = 90°. D and E are any points on AB and BC respectively. Prove that :

AE2 + CD2 = AC2 + DE2.  [NCERT] 

Sol. In DABE, ÐB = 90°

AE2 = AB2 + BE2 ...(i)

In DDBC, ÐB = 90°.

CD2 = BD2 + BC2 ...(ii)

Adding (i) and (ii), we get :

AE2 + CD2 = (AB2 + BC2) + (BE2 + BD2)

= AC2 + DE2 [By Pythagoras Theorem]

Hence, AE2 + CD2 = AC2 + DE2.

Ex.19 A point O in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D.

Prove that: OA2 + OC2 = OB2 + OD2.

Sol. Through O, draw EOFAB. Then, ABFE is a rectangle.

In right triangles OEA and OFC, we have:

OA2 = OE2 + AE2

OC2 = OF2 + CF2

OA2 + OC2 = OE2 + OF2 + AE2 + CF2 ... (i)

Again, in right triangles OFB and OED, we have :

OB2 = OF2 + BF2

OD2 = OE2 + DE2

OB2 + OD2 = OF2+ OE2 + BF2 + DE2

= OE2 + OF2 + AE2 + CF2 ...(ii) [Q BF = AE & DE = CF]

From (i) and (ii), we get

OA2 + OC2 = OB2 + OD2.

Ex20 In the given figure, DABC is right-angled at C.

Let BC = a, CA = b, AB = c and CD = p, where CD ^ AB.

Prove that: (i) cp = ab (ii) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev 

Sol. (i) Area of DABC = AB × CD = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev cp.

Also, area of DABC = BC × AC = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev ab.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev cp = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev ab. Þ cp = ab

(ii) cp = ab Þ p = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Þ p2 = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev [Q c2 = a2 + b2]

Þ NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev.

Ex.21 Prove that in any triangle, the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median which bisects the third side. (Appollonius Theorem) 

Sol. Given: A DABC in which AD is a median.

To prove : AB2 + AC2 = 2AD2 + 2NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev or AB2 + AC2 = 2(AD2 + BD2)

Construction : Draw AE ^ BC.

Proof : Q AD is median

BD = DC

Now, AB2 + AC2 = (AE2 + BE2) + (AE2 + CE2) = 2AE2 + BE2 + CE

= 2[AD2 _ DE2] + BE2 + CE

= 2AD2 _ 2DE2 + (BD + DE)2 + (DC _ DE)

= 2AD2 _ 2DE2 + (BD + DE)2 + (BD _ DE)

= 2(AD2 + BD2) = = 2AD2 + 2NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Hence, Proved.

~ Synopsis 

8 SIMILAR TRIANGLES. Two triangles are said to be similar if

(i) Their corresponding angles are equal and (ii) Their corresponding sides are proportional.

All congruent triangles are similar but the similar triangles need not be congruent.

Two polygons of the same numbers of sides are similar, if

(i) their corresponding angles are equal and

(ii) their corresponding sides are in the same ratio.

8 BASIC PROPORTIONALITY THEOREM. In a triangle, a line drawn parallel to one side, to intersect the other sides in distinct points, divides the two sides in the same ratio.

8 CONVERSE OF BASIC PROPORTIONALITY THEOREM. If a line divides any two sides of a triangle in the same ratio, the line must be parallel to the third side.

8 AAA-SIMILARITY. If in two triangles, corresponding angles are equal, i.e., the two corresponding angles are equal, then the triangles are similar.

8 SSS-SIMILARITY. If the corresporiding sides of two triangles are proportional, then they are similar.

8 SAS-SIMILARITY. If in two triangles one pair of corresponding sides are proportional and the included angles are equal then the two triangles are similar.

8 The ratio of the areas of similar triangles is equal to the ratio of the squares of their corresponding sides.

8 PYTHAGORAS THEOREM. In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

8 CONVERSE OF PYTHAGORAS THEOREM. In a triangle, if the square of one side is equal to the sum of the squares of the other two sides then the angle opposite to the first side is a right angle.

Exercise-1 (for school/board exams) 

Objective type Questions 

Choose The Correct One 

1. Triangle ABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. Triangle DEF is similar to DABC. If EF = 4 cm, then the perimeter of DDEF is :

(A) 7.5 cm (B) 15 cm (C) 22.5 cm (D) 30 cm

2. In DABC, AB = 3 cm, AC = 4 cm and AD is the bisector of ÐA. Then, BD : DC is :

(A) 9 : 16 (B) 16 : 9 (C) 3 : 4 (D) 4 : 3

3. In an equilateral triangle ABC, if AD ^ BC, then:

(A) 2AB2 = 3AD2 (B) 4AB2 = 3AD2 (C) 3AB2 = 4AD2 (D) 3AB2 = 2AD2

4. ABC is a triangle and DE is drawn parallel to BC cutting the other sides at D and E. If AB = 3.6 cm, AC = 2.4 cm and AD = 2.1 cm, then AE is equal to :

(A) 1.4 cm (B) 1.8 cm (C) 1.2 cm (D) 1.05 cm

5. The line segments joining the mid points of the sides of a triangle form four triangles each of which is :

(A) similar to the original triangle. (B) congruent to the original triangle.

(C) an equilateral triangle. (D) an isosceles triangle.

6. In DABC and DDEF , ÐA = 50°, ÐB = 70°, ÐC = 60°, ÐD = 60°, ÐE = 70°, ÐF = 50°, then DABC is similar to:

(A) DDEF (B) DEDF (C) DDFE (D) DFED

7. D, E, F are the mid points of the sides BC, CA and AB respectively of DABC. Then DDEF is congruent to triangle

(A) ABC (B) AEF (C) BFD, CDE (D) AFE, BFD, CDE

8. If in the triangles ABC and DEF, angle A is equal to angle E, both are equal to 40°, AB : ED = AC : EF and angle F is 65°, then angle B is :-

(A) 35° (B) 65° (C) 75° (D) 85°

9. In a right angled DABC, right angled at A, if AD ^ BC such that AD = p, If BC = a, CA = b and AB = c, then:

(A) p2 = b2 + c2 (B)

(C) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev (D) p2 = bc2

10. In the adjoining figure, XY is parallel to AC. If XY divides the triangle into equal parts, then the value of NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev=

(A) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev (B) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

(C) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev (D) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

11. The ratio of the corresponding sides of two similar triangles is 1 : 3. The ratio of their corresponding heights is :

(A) 1 : 3 (B) 3 : 1 (C) 1 : 9 (D) 9 : 1

12. The areas of two similar triangles are 49 cm2 and 64 cm2 respectively. The ratio of their corresponding sides is:

(A) 49: 64 (B) 7: 8 (C) 64: 49 (D) None of these

13. The areas of two similar triangles are 12 cm2 and 48 cm2. If the height of the smaller one is 2.1 cm, then the corresponding height of the bigger one is:

(A) 4.41 cm (B) 8.4 cm (C) 4.2 cm (D) 0.525 cm

14. In the adjoining figure, ABC and DBC are two triangles on the same base BC, AL ^ BC and DM.^ BC. Then, NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev is equal to :

(A) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev (B) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

(C) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev (D) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev.

15. In the adjoining figure, AD : DC = 2 : 3, then ÐABC is equal to :

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

(A) 30° (B) 40° (C) 45° (D) 110°

16. In DABC, D and E are points on AB and AC respectively such that DE ||BC. If AE = 2 cm, EC = 3 cm and BC = 10 cm, then DE is equal to :

(A) 5 cm (B) 4 cm (C) 15 cm (D) cm

17. In the given figure, ÐABC = 90° and BM is a median, AB = 8 cm and BC = 6 cm. Then, length BM is equal to:

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

(A) 3 cm (B) 4 cm (C) 5 cm (D) 7cm

18. If D, E, F are respectively the mid points of the sides BC, CA and AB of DABC and the area of DABC is 24sq. cm, then the area of DDEF is :-

(A) 24 cm2 (B) 12 cm2 (C) 8 cm2 (D) 6 cm2

19. In a right angled triangle, if the square of the hypotenuse is twice the product of the other two sides, then one of the angles of the triangle is :-

(A) 15° (B) 30° (C) 45° (D) 60°

20. Consider the following statements :

1. If three sides of a triangle are equal to three sides of another triangle, then the triangles are congruent.

2. If three angles of a triangle are respectively equal to three angles of another triangle, then the two triangles are congruent.

Of these statements,

(A) 1 is correct and 2 is false (B) both 1 and 2 are false

(C) both 1 and 2 are correct (D) 1 is false and 2 is correct

(objective) Exercise

ANSWER KEY

Exercise-2 (for school/board exams)

Subjective type Questions 

Very Short Answer Type Questions 

1. In the given figure, XYBC.

Given that AX = 3 cm, XB = 1.5 cm and BC = 6 cm.

Calculate :

(i) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev (ii) XY.

2. D and E are points on the sides AB and AC respectively of DABC. For each of the following cases, state whether DEBC:

(i) AD = 5.7 cm, BD = 9.5 cm, AE = 3.6 cm and EC = 6 cm

(ii) AB = 5.6 cm, AD = 1.4 cm, AC = 9.6 cm and EC = 2.4 cm.

(iii) AB = 11.7 cm, BD = 5.2 cm, AE = 4.4 cm and AC = 9.9 cm.

(iv) AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm and AE = 2.8 cm.

3. In DABC, AD is the bisector of ÐA. If BC = 10 cm, BD = 6 cm and AC = 6 cm, find AB.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

4. AB and CD are two vertical poles of height 6 m and 11 m respectively. If the distance between their feet is 12 m, find the distance between their tops.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

5. DABC and DPQR are similar triangles such that area (DABC) = 49 cm2 and area (DPQR) = 25 cm2. If AB = 5.6 cm, find the length of PQ .

6. DABC and DPQR are similar triangles such that area (DABC) = 28 cm2 and area (DPQR) = 63 cm2. If PR = 8.4 cm, find the length of AC.

7. DABC ~ DDEF. If BC = 4 cm, EF = 5 cm and area (DABC) = 32 cm2, determine the area of DDEF.

8. The areas of two similar triangles are 48 cm2 and 75 cm2 respectively. If the altitude of the first triangle be 3.6 cm, find the corresponding altitude of the other.

9. A rectangular field is 40 m long and 30 m broad. Find the length of its diagonal.

10. A man goes 15 m due west and then 8 m due north. How far is he from the starting point?

11. A ladder 17 m long reaches the window of a building 15 m above the ground. Find the distance of the foot of the ladder from the building.

Short Answer Type Questions 

1. In the given fig, DEBC.

(i) If AD = 3.6 cm, AB = 9 cm and AE = 2.4 cm, find EC.

(ii) If NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev = and AC = 5.6 cm, find AE. NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

(iii) If AD = x cm, DB = (x_2) cm, AE = (x+2) cm and EC = (x_1) cm, find the value of x.

2. In the given figure, BADC. Show that DOAB ~ DODC. If AB = 4 cm, CD = 3 cm, OC = 5·7 cm and OD = 3·6 cm, find OA and OB.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

3. In the given figure, ÐABC = 90° and BD ^ AC. If AB = 5·7 cm, BD = 3·8 cm and CD = 5·4 cm, find BC.
4. In the given figure, DABC ~ DPQR and AM, PN are altitudes, whereas AX and PY are medians. Prove that NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

5. In the given figure, BCDE, area (DABC) = 25 cm2, area (trap. BCED) = 24 cm2 and DE = 14 cm. Calculate the length of BC.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

6. In DABC, ÐC = 90°. If BC = a, AC = b and AB = c, find :

(i) c when a = 8 cm and b = 6 cm.

(ii) a when c = 25 cm and b = 7 cm

(iii) b when c = 13 cm and a = 5 cm

7. The sides of a right triangle containing the right angle are (5x) cm and (3x _ 1) cm. If the area of triangle be 60 cm2, calculate the length of the sides of the triangle.

8. Find the altitude of an equilateral triangle of side NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev cm.

9. In the adjoining figure (not drawn to scale), PS = 4 cm, SR = 2 cm, PT = 3 cm and QT = 5cm.

(i) Show that DPQR ~ DPST. (ii) Calculate ST, if QR = 5·8 cm.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

10. In the given figure, ABPQ and ACPR. Prove that BCQR.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

11. In the given figure, AB and DE are perpendicular to BC. If AB = 9 cm, DE = 3 cm and AC = 24 cm, calculate AD.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

12. In the given figure, DEBC. If DE = 4 cm, BC = 6 cm and area (DADE) = 20 cm2, find the area of DABC.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

13. A ladder 15 m long reaches a window which is 9 m above the ground on one side of the street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 12 m high. Find the width of the street.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

14. In the given figure, ABCD is a quadrilateral in which BC = 3 cm, AD = 13 cm, DC = 12 cm and ÐABD = ÐBCD = 90°. Calculate the length of AB.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

15. In the given figure, ÐPSR = 90°, PQ = 10 cm, QS = 6 cm and RQ = 9 cm, calculate the length of PR.

16. In a rhombus PQRS, side PQ = 17 cm and diagonal PR = 16 cm. Calculate the area of the rhombus.

17. From the given figure, find the area of trapezium ABCD.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

18. In a rhombus ABCD, prove that AC2 + BD2 = 4AB2.

19. A ladder 13 m long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the wall, find the distance of the other end of the ladder from the ground.

Long Answer Type Questions 

1. In the given figure, it is given that ÐABD = ÐCDB = ÐPQB = 90°. If AB = x units, CD = y units and PQ = z units, prove that NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

2. In the adjoining figure, ABCD is a parallelogram, P is a point on side BC and DP when produced meets AB produced at L. Prove that: (i) DP : PL = DC : BL (ii) DL : DP = AL : DC.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

3. In the given figure, ABCD is a parallelogram, E is a point on BC and the diagonal BD intersects AE at F. Prove that: DF × FE = FB × FA.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

4. In the adjoining figure, ABCD is a parallelogram in which AB = 16 cm, BC = 10 cm and L is a point on AC such that CL : LA = 2 : 3. If BL produced meets CD at M and AD produced at N, prove that:

(i) DCLB ~ DALN (ii) DCLM ~ DALB

5. In the given figure, medians AD and BE of DABC meet at G and DFBE. Prove that

(i) EF = FC (ii) AG : GD = 2 : 1.

6. In the given figure, the medians BE and CF of DABC meet at G. Prove that:

(i) DGEF ~ DGBC and therefore, BG = 2GE. (ii) AB × AF = AE × AC.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

7. In the given figure, DEBC and BD = DC.

(i) Prove that DE bisects ÐADC.

(ii) If AD = 4·5 cm, AE = 3·9 cm and DC = 7-5 cm, find CE.

(iii) Find the ratio AD : DB.

8. O is any point inside a DABC. The bisectors of ÐAOB, ÐBOC and ÐCOA meet the sides AB, BC and CA in points D, E and F respectively. Prove that AD·BE·CF = DB·EC·FA.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

9. In the figure, DEBC.

(i) Prove that DADE and DABC are similar.

(ii) Given that AD = BD, calculate DE, if BC = 4.5 cm.

10. In the adjoining figure, ABCD is a trapezium in which ABDC and AB = 2 DC. Determine the ratio of the areas of DAOB and DCOD.

 

11. In the adjoining figure, LM is parallel to BC. AB = 6 cm, AL = 2cm and AC = 9 cm. Calculate :

(i) the length of CM.

(ii) the value of

 

12. In the given figure, DEBC and DE : BC = 3 : 5. Calculate the ratio of the areas of DADE and the trapezium BCED.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

13. In DABC, D and E are mid-points of AB and AC respectively. Find the ratio of the areas of DADE and DABC.

14. In a DPQR, L and M are two points on the base QR, such that ÐLPQ = ÐQRP and ÐRPM = ÐRQP. Prove that (i) DPQL ~ DRPM (ii) QL·RM = PL·PM (iii) PQ2 = QL·QR

15. In the adjoining figure, the medians BD and CE of a DABC meet at G.

Prove that:

(i) DEGD ~ DCGB

(ii) BG = 2 GD from (i) above.

16. In the adjoining figure, PQRS is a parallelogram with PQ = 15 cm and RQ = 10 cm. L is a point on RP such that RL : LP = 2 : 3. QL produced meets RS at M and PS produced at N. Find the lengths of PN and RM.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

17. In DPQR, LMQR and PM : MR = 3 : 4. Calculate:

(i) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev and then NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev;

(ii)

(iii) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

18. In DABC, ÐB = 90° and D is the mid point of BC.

Prove that :

(i) AC2 = AD2 + 3CD

(ii) BC2 = 4(AD2 _ AB2)

19. In DABC, if AB = AC and D is a point on BC. Prove that AB2 _ AD2 = BD × CD.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Similar Triangle Exercise (X)-CBSE

ANSWER KEY

Very Short Answer Type Questions 

1. (i) (ii) 4 cm 2. (i) Yes, (ii) No, (iii) No, (iv) Yes 3. 9 cm 4. 13 m 5. PQ = 4 cm  6. AC = 5.6 cm 7. 50 cm2 8. 4.5 cm 9. 50 m 10. 17 m 11. 8 m

Short Answer Type Questions 

1. (i) 3.6 cm, (ii) 2.1 cm, (iii) x = 4 2. OA = 4.8 cm, OB = 7.6 cm 3. 8.1 cm 5. 10 cm 6. (i) 10 cm, (ii) 24 cm, (iii) 12 cm 7. 15cm, 8 cm, 17cm 8. 7.5 cm  9. 2.9 cm 11. 16 cm 12. 45 cm2 13. 21 m 14. 4 cm 15. 17 cm 16. 240 cm2 17. 14 cm2 19. 12 m

Long Answer Type Questions 

7. (ii) 6.5 cm, (ii) 3 : 5 9. DE = 1.5 cm 10. 4 : 1 11. (i) 6 cm, (ii)  12. 9 : 16 13. 1 : 4

16. PN = 15 cm, RM = 10 cm 17. (i) (ii) 3 : 7 (iii) 10 : 7

Exercise-3 (for school/board exams)

Previous years board questions

Very Short Answer Type Questions 

1. DABC and DDEF are similar, BC = 3 cm, EF = 4 cm and area of DABC = 54 cm2. Determine the area of DDEF. Delhi-1996 

2. In DABC, CE ^ AB, BD ^ AC and CE & BD intersect at P, considering triangles BEP and CPD, prove that BP × PD = EP × PC. Delhi-1996C 

3. A right triangle has hypotenuse of length q cm and one side of length p cm. If (q _ p ) = 2, express the length of third side of the right triangle in terms of q.  AI-1996C 

4. In the given figure, ABC is a triangle in which AB = AC, D and E are points on the sides AB and AC respectively, such that AD = AE. Show that the points B, C, E and D are concyclic.  AI-1996C 

5. In a DABC, AB = AC and D is a point on side AC, such that BC2 = AC × CD. Prove that BD = BC.Your Target is to  AI-1997 

6. DABC is right angled at B. On the side AC, a point D is taken such that AD = DC and AB = BD. Find the measure of ÐCAB.  Delhi -1998 

7. In a DABC, P and Q are points on the sides AB and AC respectively such that PQ is parallel to BC. Prove that median AD, drawn from A to BC, bisects PQ.  AI-1998 

8. Two poles of height 7 m and 12 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tips.  AI-1998C 

9. In a DABC, D and E are points on AB & AC respectively such that DE is parallel to BC and AD : DB = 2 : 3. Determine Area (DADE) : Area (DABC). Foreign-1999 

10. In the given figure, ÐA = ÐB and D & E are points on AC and BC respectively such that AD = BE, show that DE || AB.  Delhi-1999 

11. In figure, Ð1 = Ð2 and Ð3 = Ð4. Show that PT . QR = PR . ST. Foreign-2000

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

12. In figure, LM || NQ and LN || PQ. If MP = MN, find the ratio of the areas of DLMN and DQNP.

Foreign-2000 

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

13. ABC is an isosceles triangle right angled at B. Two equilateral triangles BDC and AEC are constructed with side BC and AC. Prove that area of DBCD = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev area of DACE. Delhi-2001 

14. The areas of two similar triangles are 81 cm2 and 49 cm2 respectively. If the altitude of the first triangle is 6.3 cm, find the corresponding altitude of the other. AI-2001

15. L and M are the mid-points of AB and BC respectively of DABC, right-angled at B. Prove that

4LC2 = AB2 + 4BC2.  AI-2001;Foreign-2001 

16. The areas of two similar triangles are 121 cm2 and 64 cm2 respectively. If the median of the first triangle is 12.1 cm, find the corresponding median of the other.  AI-2001 

17. In an equilateral triangle ABC, AD is the altitude drawn from A on side BC. Prove that 3AB= 4AD2.

18. (i) Prove that the equilateral triangles described on the two sides of a right angled triangle are together equal to the equilateral triangle on the hypotenuse in terms of their areas. AI-2002

(ii) P is a point in the interior of DABC. X, Y and Z are points on lines PA, PB and PC respectively such that XY || AB and XZ || AC. Prove that YZ || BC. AI-2002 ; Delhi-2003 [NCERT]

(iii) D and E are points on the sides AB and AC respectively of DABC such that DE is parallel to BC and AD : DB = 4 : 5. CD and BE intersect each other at F. Find the ratio of the areas of DDEF and DBCF.  AI-2000 ; AI-2003

(iv) P, Q are respective points on sides AB and AC of triangle ABC. If AP = 2 cm, PB = 4 cm, AQ = 3 cm and QC = 6 cm, prove that BC = 3PQ.  Foreign-2003 

19. D is a point on the side BC of DABC such that ÐADC = ÐBAC. Prove that NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev.

20. ABCD is a trapezium in which AB || DC. The diagonals AC and BD intersect at O. Prove that NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

21. In a DABC, AD ^ BC and . Prove that ABC is a right triangle, right angled at A. Foreign -2004 

22. In a right angled triangle ABC, ÐA = 90° and AD ^ BC. Prove that AD2 = BD × CD. Delhi -2004C, 2006 

23. In fig., AB || DE and BD || EF. Prove that DC= CF × AC. AI-2004C; Delhi-2007

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

24. If one diagonal of a trapezium divides the other diagonal in the ratio of 1 : 2, prove that one of the parallel sides is double the other.  Foreign-2005 

25. In DABC, AD ^ BC, prove that AB2 + CD2 = AC2 + DB2. Delhi-2005C, AI-2006 [NCERT] 

26. Prove that the sum of the squares of the sides of a rhombus is equal to sum of the squares of its diagonals.  AI-2005C [NCERT] 

27. In figure, S and T trisect the side QR of a right triangle PQR. Prove that 8PT2 = 3PR2 + 5PS2.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

OR

If BL and CM are medians of a triangle ABC right-angled at A, then prove that 4(BL2 + CM2) = 5 BC2. AI-2006 C; Foreign 2009

28. In the fig, P and Q are points on the sides AB and AC respectively of DABC such that AP = 3.5 cm, PB = 7 cm, AQ = 3 cm and QC = 6 cm. If PQ = 4.5 cm, find BC. Delhi-2008 

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

29. In fig, ÐM = ÐN = 46° Express x in terms of a, b and c where a, b and c are lengths of LM, MN and NK respectively. Delhi-2009 

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

30. In figure, DABD is a right triangle, right-angled at A and AC ^ BD. Prove that AB2 = BC. BD.AI-2009 

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

31. In a DABC, DE||BC. If DE = BC and area of DABC = 81 cm2, find the area of DADE.Foregin-2009

Short Answer Type Questions 

1. P and Q are points on the sides CA and CB respectively of a DABC right-angled at C. Prove that

AQ2 + BP= AB2 + PQ2.  Delhi-1996, 2007 

2. ABC is a right triangle, right angled at B. AD and CE are the two medians drawn from A and C respectively. If AC = 5 cm and AD = cm, find the length of CE.  AI -1997 

3. In DABC, if AD is the median, show that AB2 + AC2 = 2 [AD2 + BD2]. Delhi-1997, 98 

4. In the given figure, M is the mid-point of the side CD of parallelogram ABCD. BM, when joined meets AC in L and AD produced in E. Prove that EL = 2BL. AI -1998; Delhi-1999, AI-2009 

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

5. ABC is a right triangle, right-angled at C. If p is the length of the perpendicular from C to AB and a, b, c have the usual meaning, then prove that (i) pc = ab (ii) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev Delhi-1998, 98C 

6. In an equilateral triangle PQR, the side QR is trisected at S. Prove that 9PS2 = 7PQ2.AI-1998, 98C [NCERT]

7. If the diagonals of a quadrilateral divide each other proportionally, prove that it is trapezium.Foreign-1999

8. In an isosceles triangle ABC with AB = AC, BD is a perpendicular from B to the side AC. Prove that BD2 _ CD2 = 2CD . AD. Foreign-1999 

9. ABC and DBC are two triangles on the same base BC. If AD intersect BC at O. Prove that NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

AI-19999C; Delhi-2005 

10. In DABC, ÐA is acute. BD and CE are perpendiculars on AC and AB respectively. Prove that AB × AE = AC × AD.  AI-2003 

11. Points P and Q are on sides AB and AC of a triangle ABC in such a way that PQ is parallel to side BC. Prove that the median AD drawn from vertex A to side BC bisects the segment PQ.

12. If the diagonals of a quadrilateral divide each other proportionally, prove that it is a trapezium.

OR

Two Ds' ABC and DBC are on the same base BC and on the same side of BC in which ÐA = ÐD = 90°. If CA and BD meet each other at E, show that AE.EC = BE.ED.  Delhi-2008 

13. D and E are points on the sides CA and CB respectively of DABC right-angled at C. Prove that AE2 + BD2 = AB2 + DE2.

OR

In fig. DB ^ BC, DE ^ AB and AC ^ BC. Prove that NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev.  AI-2008 

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

14. E is a point on the side AD produced of a ||gm ABCD and BE intersects CD at F. Show that DABE ~ DCFB. Foreign-2008 

15. In fig, DABC is right angled at C and DE ^ AB. Prove that DABC ~ DADE and hence find the lengths of AE and DE.

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

OR

In fig, DEFG is a square and ÐBAC = 90°. Show that DE2 = BD × EC Delhi-2009 

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

16. In fig, AD ^ BC and BD = CD. Prove that 2CA2 = 2AB2 + BC2. 

AI-2009

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

17. In fig, two triangles ABC and DBC lie on the same side of base BC. P is a point on BC such that PQ||BA and PR||BD. Prove that QR||AD. Foreign-2009 

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Long Answer Type Questions 

1. In a right triangle ABC, right-angled at C, P and Q are points on the sides CA and CB respectively which divide these sides in the ratio 1 : 2. Prove that  AI-1996C

(i) 9AQ2 = 9AC2 + 4BC2 (ii) 9BP= 9BC2 + 4AC2 (iii) 9(AQ2 + BP2) = 13AB2.

2. The ratio of the areas of similar triangles is equal to the ratio of the squares on the corresponding sides, prove. Using the above theorem, prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal.Delhi-1997C; 2005C; Foreign-2003 

3. Perpendiculars OD, OE and OF are drawn to sides BC, CA and AB respectively from a point O in the interior of a DABC. Prove that :

(i) AF2 + BD2 + CE2 = OA2 + OB2 + OC2 _ OD2 _ OE2 _ OF2.

(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2. Delhi-1997C, [NCERT] 

4. In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares on the other two sides, prove. Using the above theorem, determine the length of AD in terms of b and C. AI-1997 C 

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

5. If a line is drawn parallel to one side of a triangle, the other two sides are divided in the same ratio, prove. Use this result to prove the following : In the given figure, if ABCD is a trapezium in which AB || DC || EF, then NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev.  Foreign-1998 

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

6. State and prove pythagoras theorem. Use the theorem and calculate area (DPMR) from the given figure.

Delhi-1998C, 2006 

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

7. In a right-angled triangle, the square of hypotenuse is equal to the sum of the squares of the two sides. Given that ÐB of DABC is an acute angle and AD ^ BC. Prove that AC2 = AB2 + BC2 _ 2BC . BD.  Delhi-1999 

8. In a right triangle, prove that the square on the hypotenuse is equal to the sum of the squares on the other two sides. Using above, solve the following : In quadrilateral ABCD, find the length of CA, if CD ^ DB, AB ^ DB, CD = 6 m, DB = 12 m and AB = 11 m.  Delhi-2000 

9. Prove that the ratio of the areas of two similar triangles is equal to the squares of their corresponding sides. Using the above, do the following:

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

In fig. DABC and DPQR are isosceles triangles in which ÐA = ÐP. If NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev find NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev.

10. In a right-angled triangle, prove that the square on the hypotenuse is equal to the sum of the squares on the other two sides. Using the above result, find the length of the second diagonal of a rhombus whose side is 5 cm and one of the diagonals is 6 cm.  AI-2001 

11. In a triangle, if the square on one side is equal to the sum of the squares on the other two sides prove that the angle opposite the first side is a right angle.

Use the above theorem and prove thet following : In triangle ABC, AD ^ BC and BD = 3CD. Prove that 2AB= 2AC+ BC2. AI-2003 

13. In a right triangle, prove that the square on hypotenuse is equal to sum of the squares on the other two sides. Using the above result, prove the following : PQR is a right triangle, right angled at Q. If S bisects QR, show that PR2 = 4 PS2 _ 3 PQ2.  Delhi-2004C 

14. If a line is drawn parallel to one side of a trial prove that the other two sides are divided in the same ratio. Using the above result, prove from fig. that AD = BE if ÐA = ÐB and DE || AB. AI-2004C 

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

15. Prove that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides. Apply the above theorem on the following : ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm, prove that area of DAPQ is one-sixteenth of the area of DABC. Delhi-2005 

16. If a line is drawn parallel to one side of a triangle, prove that the other two sides are divided in the same ratio. Use the above to prove the following : In the given figure DE || AC and DC || AP. Prove that NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev.  AI-2005 

 

17. In a triangle if the square on one side is equal to the sum of squares on the other two sides, prove that the angle opposite to the first side is a right angle. Use the above theorem to prove the following :

In a quadrilateral ABCD, ÐB = 90°. If AD2 = AB2 + BC2 + CD2, prove that ÐACD = 90°. AI-2005 

18. If a line is drawn parallel to one side of a triangle, to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio. Using the above, prove the following : In figure, DE || BC and BD = CE. Prove that ABC is an isosceles triangle.  Delhi-2007, 2009 

NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

19. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. Use the above for the following : If the areas of two similar triangles are equal, prove that they are congruent. AI-2007 

20. Prove that the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides. Using the above result, prove the following :

In a DABC, XY is parallel to BC and it divides DABC into two parts of equal area. Prove that NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

ANSWER KEY

Similar Triangle Exercise (X)-CBSE

Very Short Answer Type Questions 

2. 96 cm2  3.NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev6. 60° 8. 13 m 9. 4.25 12. 9 : 4 14. 4.9 cm 16. 8.8 cm 18. (iii) 16 : 81

28. 13.5 cm 29.NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev31. 36 cm

Short Answer Type Questions 

2.NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev15. AE = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev, DE = NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

Long Answer Type Questions 

4.NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev6. 24 cm8. 13 cm 9. 3 : 4 10. 8 cm

Exercise-4 (for Olympiads)

Choose The Correct One 

1. In a triangle ABC, if AB, BC and AC are the three sides of the triangle, then which of the statements is necessarily true?

(A) AB + BC < AC (B) AB + BC > AC (C) AB + BC = AC (D) AB2 + BC2 = AC2 .

2. The sides of a triangle are 12 cm, 8 cm and 6 cm respectively, the triangle is :

(A) acute (B) obtuse (C) right (D) can't be determined

3. In an equilateral triangle, the incentre, circumcentre, orthocentre and centroid are:

(A) concylic (B) coincident (C) collinear (D) none of these

4. In the adjoining figure D is the midpoint of BC of a DABC. DM and DN are the perpendiculars on AB and AC respectively and DM = DN, then the DABC is :

(A) right angled

(B) isosceles

(C) equilateral

(D) scalene

5. Triangle ABC is such that AB = 9 cm, BC = 6 cm, AC = 7.5 cm. Triangle DEF is similar to DABC, If EF = 12 cm then DE is :

(A) 6 cm (B) 16 cm (C) 18 cm (D) 15 cm

6. In DABC, AB = 5 cm, AC= 7 cm. If AD is the angle bisector of ÐA. Then BD : CD is:

(A) 25 : 49 (B) 49 : 25 (C) 6 : 1 (D) 5 : 7

7. In a DABC, D is the mid-point of BC and E is mid-point of AD, BF passes through E. What is the ratio of AF : FC?

(A) 1 : 1

(B) 1 : 2

(C) 1 : 3

(D) 2 : 3

8. In a DABC, AB = AC and AD ^ BC, then :

(A) AB < AD (B) AB > AD (C) AB = AD (D) AB £ AD

9. The difference between altitude and base of a right angled triangle is 17 cm and its hypotenuse is 25 cm. What is the sum of the base and altitude of the triangle is ?

(A) 24 cm (B) 31 cm (C) 34 cm (D) can't be determined

10. If AB, BC and AC be the three sides of a triangle ABC, which one of the following is true?

(A) AB _ BC = AC (B) (AB _ BC) > AC (C) (AB _ BA) < AC (D) AB2 _ BC2 = AC2

11. In the adjoining figure D, E and F are the mid-points of the sides BC, AC and AB respectively. DDEF is congruent to triangle :

(A) ABC

(B) AEF

(C) CDE , BFD

(D) AFE, BFD and CDE

12. In the adjoining figure ÐBAC = 60° and BC = a, AC = b and AB = c, then :

(A) a2 = b2 + c2

(B) a2 = b2 + c2 _ bc

(C) a2 = b2 + c2 + bc

(D) a2 = b2 + 2bc

13. If the medians of a triangle are equal, then the triangle is:

(A) right angled (B) isosceles (C) equilateral (D) scalene

14. The incentre of a triangle is determined by the:

(A) medians (B) angle bisectors

(C) perpendicular bisectors (D) altitudes

15. The point of intersection of the angle bisectors of a triangle is :

(A) orthocentre (B) centroid (C) incentre (D) circumcentre

16. A triangle PQR is formed by joining the mid-points of the sides of a triangle ABC. 'O' is the circumcentre of DABC, then for DPQR, the point 'O' is :

(A) incentre (B) circumcentre (C) orthocentre (D) centroid

17. If AD, BE, CF are the altitudes of DABC whose orthocentre is H, then C is the orthocentre of :

(A) DABH (B) DBDH (C) DABD (D) DBEA

18. In an equilateral DABC, if a, b and c denote the lengths of perpendiculars from A, B and C respectively on the opposite sides, then:

(A) a > b > c (B) a > b < c (C) a = b = c (D) a = c ¹ b

19. Any two of the four triangles formed by joining the midpoints of the sides of a given triangle are:

(A) congruent (B) equal in area but not congruent

(C) unequal in area and not congruent (D) none of these

20. The internal bisectors of ÐB and ÐC of DABC meet at O. If ÐA = 80° then ÐBOC is :

(A) 50° (B) 160° (C) 100° (D) 130°

21. The point in the plane of a triangle which is at equal perpendicular distance from the sides of the triangle is :

(A) centroid (B) incentre (C) circumcentre (D) orthocentre

22. Incentre of a triangle lies in the interior of :

(A) an isosceles triangle only (B) a right angled triangle only

(C) any equilateral triangle only (D) any triangle

23. In a triangle PQR, PQ = 20 cm and PR = 6 cm, the side QR is :

(A) equal to 14 cm (B) less than 14 cm (C) greater than 14 cm (D) none of these

24. If ABC is a right angled triangle at B and M, N are the mid-points of AB and BC, then 4 (AN2 + CM2) is equal to_

(A) 4AC2 (B) 6AC2 (C) 5AC2 (D) AC2

25. ABC is a right angle triangle at A and AD is perpendicular to the hypotenuse. Then is equal to :

(A) (B) (C) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev (D) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

26. Let ABC be an equilateral triangle. Let BE ^ CA meeting CA at E, then (AB2 + BC2 + CA2) is equal to :

(A) 2BE2 (B) 3BE2 (C) 4BE2 (D) 6BE2

27. If D, E and F are respectively the mid-points of sides of BC, CA and AB of a DABC. If EF = 3 cm, FD = 4 cm, and AB = 10 cm, then DE, BC and CA respectively will be equal to :

(A) 6, 8 and 20 cm (B) 4, 6 and 8 cm (C) 5, 6 and 8 cm (D) , 9 and 12 cm

28. In the right angle triangle ÐC = 90°. AE and BD are two medians of a triangle ABC meeting at F. The ratio of the area of DABF and the quadrilateral FDCE is :

(A) 1 : 1 (B) 1 : 2 (C) 2 : 1 (D) 2 : 3

29. The bisector of the exterior ÐA of DABC intersects the side BC produced to D. Here CF is parallel to AD.

(A) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev (B) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

(C) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev (D) None of these

30. The diagonal BD of a quadrilateral ABCD bisects ÐB and ÐD, then:

(A) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev (B) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

(C) AB = AD × BC (D) None of these

31. Two right triangles ABC and DBC are drawn on the same hypotenuse BC on the same side of BC. If AC and DB intersects at P, then

(A) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

(B) AP × DP = PC × BP

(C) AP × PC = BP × DP

(D) AP × BP = PC × PD

32. In figure, ABC is a right triangle, right angled at B. AD and CE are the two medians drawn from A and C respectively. If AC = 5 cm and AD = cm, find the length of CE:

(A) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev cm (B) 2.5 cm

(C) 5 cm (D) cm

 

33. In a DABC, AB = 10 cm, BC = 12 cm and AC = 14 cm. Find the length of median AD. If G is the centroid, find length of GA :

(A) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev (B) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev (C) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev (D) NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRev

34. The three sides of a triangles are given. Which one of the following is not a right triangle?

(A) 20, 21, 29 (B) 16, 63, 65

(C) 56, 90, 106 (D) 36, 35, 74

35. In the figure AD is the external bisector of ÐEAC, intersects BC produced to D. If AB = 12 cm, AC = 8 cm and BC = 4 cm, find CD.

(A) 10 cm

(B) 6 cm

(C) 8 cm

(D) 9 cm

36. In DABC, AB2 + AC2 = 2500 cm2 and median AD = 25 cm, find BC.

(A) 25 cm (B) 40 cm (C) 50 cm (D) 48 cm

37. In the given figure, AB = BC and ÐBAC = 15°, AB = 10 cm. Find the area of DABC.

(A) 50 cm2

(B) 40 cm2

(C) 25 cm2

(D) 32 cm2 

38. In the given figure, if NCERT Solutions - Similar Triangles, Class 10, Mathematics Class 10 Notes | EduRevand if AE = 10 cm. Find AB.

(A) 16 cm

(B) 12 cm

(C) 15 cm

(D) 18 cm

39. In the figure AD = 12 cm, AB = 20 cm and AE = 10 cm. Find EC.

(A) 14 cm

(B) 10 cm

(C) 8 cm

(D) 15 cm

40. In the given fig, BC = AC = AD, ÐEAD = 81°. Find the value of x.

(A) 45°

(B) 54°

(C) 63°

(D) 36°

41. What is the ratio of inradius to the circumradius of a right angled triangle?

(A) 1 : 2 (B) 1 : (C) 2 : 5 (D) Can't be determined






 

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