Results on Area of Similar Triangles, Pythagoras Theorem - Triangles, Class 10, Mathematics | Extra Documents, Videos & Tests for Class 10 PDF Download

RESULTS ON AREA OF SIMILAR TRIANGLES

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

Given : ΔABC ~ ΔDEF
To prove : 

 Construction : Draw AL ⊥ BC and DM ⊥ EF.

Proof :

STATEMENT		REASON
1.		Area of Δ = (1/2) × Base × Height
2.	In ΔALB and ΔDME, we have (i) ∠ALB = ∠DME (ii) ∠ABL = ∠DEM ∴   ΔALB ~ ΔDME ⇒	Each equal to 90° ΔABC ~ ΔDEF ⇒ ∠B = ∠E AA-axiom Corresponding sides of similar Ds are proportional.
3.	ΔABC ~ ΔDEF	Given.
4.	Substituting = in 1, we get :	Corresponding sides of similar Δs are proportional.
5.	Combining 3 and 5, we get :	From 2 and 3.

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

Given : ΔABC ~ ΔDEF, AL ⊥ BC and DM ⊥ EF.

To prove :

Proof :

 Hence, proved.

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

Given : ΔABC ~ ΔDEF and AP, DQ are their medians.

To prove : 

Proof : 

Hence, proved.

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

Given : ΔABC ~ ΔDEF and AX, DY are their

bisectors of ∠A and ∠D respectively.

To prove :

Proof : 

Ex.11 It is given that ΔABC ~ ΔPQR, area (ΔABC) = 36 cm² and area (ΔPQR) = 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.

Let BC = x cm. Then,

Hence BC = 7.2 cm

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

Sol. Area (ΔAPQ)
= Area (trap. PBCQ) [Given]
⇒ Area (ΔAPQ) = [Area (ΔABC) – Area (ΔAPQ)]
⇒ 2 Area (ΔAPQ) = Area (ΔABC)

Now, in ΔAPQ and ΔABC, we have
∠PAQ = ∠BAC [Common ∠A]
∠APQ = ∠ABC [PQPBC, corresponding ∠s are equal]  
∴ ΔAPQ ~ ΔABC.
We known that the areas of similar Δs are proportional to the squares of their corresponding sides.

                    Using (i)]

⇒  i.e., AB = √2· AP

⇒ AB = √2 (AB – PB) ⇒ √2PB = ( √2 – 1) AB
⇒ 

 ∴ PB : AB = (√2 – 1) :√2

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 ΔABC and ΔDEF be the given triangles in which AB = AC, DE = DF, ∠A = ∠D and  

Draw AL ⊥ BC and DM ⊥ EF

Now

 = 1 and  = 1        [∵ AB = AC and DE = DF]

⇒

 ∴ In ΔABC and ΔDEF, we have

and  ∠A = ∠D
⇒ ΔABC ~ ΔDEF       [By SAS similarity axiom]

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

∴ 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 ΔABC ~ ΔDEF and area (ΔABC) = area (ΔDEF).
Since the ratio of the areas of two similar Ds is equal to the ratio of the squares on their corresponding sides, we have

   = 1              [∵ Area (ΔABC) = Area (ΔDEF)]


⇒ AB² = DE², AC² = DF² and BC² = EF²
⇒ AB = DE, AC = DF and BC = EF  

∴ ΔABC @ ΔDEF [By SSS congruence]

Ex.15 In fig, the line segment XY is parallel to side AC of ΔABC and it divides the triangle into two parts of equal areas. Find the ratio

Sol. We are given that XY ║ AC.

⇒ ∠1 = ∠3 and  ∠2 = ∠4 [Corresponding angles]
⇒ ΔBXY ~ ΔBAC [AA similarity]

              [By theorem] ...(i)
Also, we are given that ar (ΔBXY) = (1/2)
× ar (ΔBAC) ⇒              ...(ii) 

From (i) and (ii), we have

           ...(iii)

Now,

Hence,

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 ΔABC in which ∠B = 90°.
To prove : AC² = AB² + BC².                                                           

Construction : From B, Draw BD ⊥ AC. D

Proof : 

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 ΔABC in which AB2 + BC2 = AC2

To prove : ∠B = 90°                                                       

Construction : Draw a ΔDEF in which   DE = AB, EF = BC and ∠E = 90°
Proof : 

Hence, ∠B = 90°

Ex.16 If ABC is an equilateral triangle of side a, prove that its altitude =  

Sol. ΔABC 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]
⇒ ΔABD @ ΔACD          [By RHS congruence]
⇒ BD = CD ⇒ BD = DC = (1/2) BC = (a/2)


From right triangle ABD, AB² = AD² + BD² ⇒ a² = AD² +
⇒ AD² = a² –
⇒ AD = 

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

AC² = AB² + BC² + 2BC × BD

Sol. In ΔADB, ∠D = 90°.  

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

In ΔADC, ∠D = 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, ∠B = 90°. D and E are any points on AB and BC respectively.

Prove that : AE² + CD² = AC² + DE². 

Sol. In ΔABE, ∠B = 90°
 ∴ AE² = AB² + BE²                 ...(i)
In ΔDBC, ∠B = 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 EOF ║ AB. 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)

[∵ BF = AE & DE = CF]

From (i) and (ii),
we get OA² + OC² = OB² + OD².

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

Let BC = a, CA = b, AB = c and CD = p, where CD ⊥ AB.
Prove that: (i) cp = ab (ii)

Sol. (i) Area of ΔABC = (1/2)
 AB × CD = (1/2) cp

Also, area of ΔABC = (1/2) BC × AC = (1/2) ab.
 ∴ (1/2) cp = (1/2) ab. ⇒ cp = ab
(ii) cp = ab ⇒ p = (ab/c)
⇒
⇒                    [∵ c² = a² + b²]
⇒

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 ΔABC in which AD is a median.
To prove : 
 or AB² + AC² = 2(AD² + BD²)
Construction : Draw AE ⊥ BC.
Proof : ∵ AD is median    ∴ BD = DC
Now,      AB² + AC² = (AE² + BE²) + (AE² + CE²) = 2AE² + BE² + CE²   

= 2[AD² – DE²] + BE² + CE²


Hence, Proved.