Previous Year Questions: Triangles | Mathematics (Maths) Class 10 PDF Download

Ans: Let ABC be an isosceles triangle with length of base BC = a 

⇒ 16a = 32 × 6 ⇒ a = 12 cm 
∴ BC = 12 cm, AB = 10 cm, AC= 10 cm 

Draw AD ⊥ BC so that BD = DC =  12/2 = 6 cm
ln Δ ABD, 
AD² = AB² - BD² (By Pythagoras theorem) 
⇒ AD² = 10² - 6² = 100 - 36 
⇒ AD² = 64 ⇒ AD = 8 cm 
So, height of ΔABD, h = 8 cm 

Ans: 
Let ABC be a right angled triangle. 
Since, the perimeter the right triangle is 60 cm, 
AB + BC+ CA = 60 
⇒ AB + BC+ 25 = 60    {∵ AC = 25 cm}
⇒ AB + BC = 35    ... (i) 
ln ΔABC, 
AC² = AB² + BC² {By using Pythagoras theorem} 
⇒ (25)² = (AB + BC)² - 2(AB)(BC) 
⇒ 2(AB)(BC) = (35)² - (25)²   {∵ Using(i)} 
⇒ (AB)(BC) = 300    ... (ii) 
Now, 
(AB - BC)² = (AB + BC)² - 4(AB)(BC) 
⇒ (AB - BC)² = (35)² - 4 × 300 {∵ Using(i)} 
⇒  AB - BC = √25 = 5    ...(iii)

Adding eqn (i) & (iii), we get 
2AB = 40  ⇒ AB = 20 cm 
From (i), AB+ BC= 35 
⇒ BC = 35 - 20 = 15 cm 
Hence, lengths of other two sides of the triangle are 20 cm and 15cm. 

Ans: (c) 
Given, PQ ||XY || BC, AP = 2 cm, PX = 1.5 cm, BX = 4 cm and QY = 0.75 cm. 

Similarly, AQ = AP/2 = 1cm and CY BX/2 = 4/2 cm 
So, AQ + CY= 1 + 2 = 3 cm

Ans: (c)
Given, ΔABC ∼ ΔPQR, ∠A = 30°, ∠Q = 90° 
Since , ΔABC ∼ ΔPQR 
∠A = ∠P = 30°, ∠B = ∠Q = 90° 
So, ∠C = ∠R = 180° - 90° - 30° = 60° 
⇒ (LR + LB) = (60° + 90°) = 150° 

Ans: (c) 
In ΔABC, DE||BC
By Basic Proportionality Theorem, 

Ans: Statement: If a line is drawn parallel to one side of a triangle, then it divides the other two sides in the same ratio. 
Given: In ΔABC, ∵ IIBC, which intersects AB and AC at 'D' and 'E' respectively.
To prove:  AD/DB = AE/EC 
Construction: Draw OM ⊥ AC and EN ⊥ AB. Also, join B to E and C to D. 

Ans: (d)
Given, in triangles 
ΔDEF and ΔPQR 
∠D = ∠Q 
∠R =∠E 
ΔDEF ∼ ΔPQR

(If triangles are similar then corresponding sides are proportional) 
Hence, option (d) is not true. 

Ans: (c) 
In ΔLMN, ∠L + ∠M + ∠N = 180°
∠L + 130° + 28° = 180° ⇒ ∠L + 158° = 180°
⇒ ∠L = 22°
In ΔLMN and ΔABC, ∠L = ∠A, ∠M = ∠B
∴ ΔLMN ∼ ΔABC [By AA similarity criterion]

Ans: (c)
In ΔABC and ΔDAC
∠BAC = ∠ADC = 90°
∠C = ∠C (Common)
So, ΔABC ∼ ΔADC (By AA similarity)

⇒ x² = 16 × 4 = 64 
⇒ x = 8 cm

Ans: (d) 
In ΔABC and ΔDEF
∠B = ∠E (Given)
∠C = ∠F (Given)
∴ ΔABC ~ ΔDEF [By AA similarity criterion]

Ans: 
Since, ΔABC ~ ΔPQR 

Ans:
Given,  QR/QS = QT/PQ
and ∠1 = ∠2 ... (i)
Since, ∠1 = ∠2 ... (ii)
∴ PQ = PR ... (iii)
∴ QR/QS = QT/PQ [Using (i) and (iii)]
∴ PS || RT    [By converse of Basic Proportionality Theorem]
∴ ∠QPS = ∠QTR    [Corresponding angles]
and ∠QSP = ∠QRT    [Corresponding angles]
∴ △PQS ~ △TQR    [By AA similarity criterion]
Hence, proved.

Ans: 
Given: ∠APC = ∠BAC
In ΔABC and ΔPAC,
∠BAC = ∠APC    (Given)
∠ACB = ∠ACP    (Common)
∴ By AA criterion, ΔABC ~ ΔPAC.
Thus, corresponding sides are proportional.
∴ AC/CP = BC/AC
∴ AC² = BC·CP
Hence, proved.

Ans: 
In ΔAPC and ΔBQC, we have
∠PAC = ∠QBC     (Each 90°)
ΔACP = ΔBCQ     (Common)
∴ ΔAPC ~ ΔBQC     (By AA similarity)
⇒ AP/BQ = AC/BC
⇒ x/y = AC/BC     ...(i)
Again, in ΔACR and ΔABQ, we have
∠ACR = ∠ABQ     (Each 90°)
∠CAR = ∠BAQ (Common)
∴ ΔACR ~ ΔABQ     (By AA similarity)
⇒ AC/AB = CR/BQ
⇒ AC/AB = Z/Y   ...(ii)
From (i) and (ii), we have

Hence, proved.

Ans: Given, ΔABC and ΔPQR in which AD and PM are the medians such that

∴ ΔABD ~ ΔPQM     [By SSS similarity]
⇒ ∠B = ∠Q
Now, in ΔABC and ΔPQR,
AB/PQ = BC/QR    [Given]
∠B = ∠Q (Proved above)
∴ ΔABC ~ ΔPQR     (By SAS similarity)

Ans: 
In ΔDAF and ΔBEF,
∠DAF = ∠BEF     [Alternate interior angles]
∠AFD = ∠EFB     [Vertically opposite angles] ∴ ΔDAF ~ ΔBEF     [By AAA similarity criterion]
∴ DF/BF = AF/EF [Corresponding sides of similar triangles] 
⇒ DF ×  EF = AF × BF 

Ans: 

Given, In ΔABC, AD ⊥ BC
and AD² = BD × DC
⇒ AD × AD = BD × DC
⇒ AD/BD = DC/AD
In ΔABD and ΔCAD,
AD/BD = DC/AD and ∠ADB = ∠ADC = 90° 
⇒ ΔABD ~ ΔCAD
⇒ ∠BAD = ∠ACD and ∠ABD = ∠CAD
Now, in ΔABD, ∠BAD + ∠CAD = 90°
⇒ ∠BAD + ∠BAD = 90°    [∵ ∠ABD = ∠CAD]
⇒ ∠BAC = 90°

Ans:

In ΔADB and ΔCDA,
∠ADB = ∠CDA     [Each 90°]
∠DAB = ∠DCA     [Each (90° - B)]
∴ ΔADB ~ ΔCDA     [By AA similarity]
Also, AD/CD = BD/AD
⇒ AD² = BD × CD
⇒ AD² = 8 × 2        [∵ BO = BC - CD] 
⇒ AD = 4 cm

Ans: (b) 

Consider the triangles ΔADE and ΔABC.

Since DE || BC, the corresponding angles are equal.

Hence, 
∠ADE = ∠ABC

∠AED = ∠ACB

Thus, by AA similarity, we find that ΔADE and ΔABC are similar.

So, ΔADE ~ ΔABC.

When two triangles are similar, their corresponding sides are proportional.

This gives the equation:
AE/AC = DE/BC = AD/AB

From this, we write:
AD/AB = DE/BC

We know that:
AB = AD + BD = 2 + 3 = 5 cm

Substituting values:
2/5 = DE/7.5

Using cross multiplication:
2 × 7.5 = 5 × DE
DE = (15/5)
DE = 3 cm

Ans:

Ans:
Let the measures of the two angles be x° and y° (x > y).
Given:
x + y = 180° ........ (i)     (Sum of Supplementary angles are 180°)
Also, 
x - y = 18°   ........ (ii)      (Given)

Adding both the equations, we get:
2x = 198°
x = 99°
By putting value of x in equation (i)
we get,
 99° + y = 180°
y = 180° - 99° = 81°
Hence, y = 81° and x = 99°.

Ans:

Given: In ΔABC, DE || BC
To Prove: DB/AD =EC/AE

Proof:
In ΔABC and ΔADE

∠AED = ∠ACB (Corresponding angles)
∠ADE = ∠ABC (Corresponding angles)
∠EAD is common to both the triangles 

∴ ΔAED ~ ΔACB by AAA similarity

⇒ AC / AE = AB / AD   ( ∴ corresponding sides of similar triangles are proportional) ...............(i)
⇒ Also, AC = AE + EC and AB = AD + BD
Putting these values in (i), we get

⇒ (AE + EC) / AE = (AD + BD) / AD

⇒ EC / AE = BD / AD

Hence proved.

Ans:

Since AD and PM are medians of ΔABC and ΔPQR,
∴ BD = 1/2 BC and QM = 1/2 QR ……….(1)

Given that,
AB / PQ = BC / QR = AD / PM ……….(2)

From (1) and (2),
AB / PQ = BD / QM = AD / PM ……….(3)

In ΔABD and ΔPQM,
AB / PQ = BD / QM
By SSS criterion of proportionality,
ΔABD ~ ΔPQM
∴ ∠B = ∠Q (Corresponding Sides of Similar Triangles) ……….(4)

In ΔABC and ΔPQR,
AB / PQ = BC / QR (From 2)
∠B = ∠Q (From 4)
By SAS criterion of proportionality,
ΔABC ~ ΔPQR

Ans:

i. Given: diagonal BD bisects ∠B and ∠D

To prove: ΔABD ~ ΔCBD

Proof: In ΔABD and ΔCBD

∠ABD = ∠CBD … (BD bisects ∠B)

∠ADB = ∠CDB … (BD bisects ∠D)

Therefore, ΔABD ~ ΔCBD … (by AA rule)

Hence proved

ii. Since, ΔABD ~ ΔCBD

Therefore, ABBD = BCBD … (by cpst)

Hence, AB = BC

Hence proved.

Ans:

In ΔCAP and ΔCBQ

∠CAP = ∠CBQ = 90°

∠PCA = ∠QCB (common angle)

So, ΔCAP ~ ΔCBQ ... (By AA similarity Rule)

Hence, BQAP = BCAC ⇒ yx = BCAC  ... (i)  (Corresponding part of similar triangle)

Now, in ΔACR and ΔABQ

∠ACR = ∠ABQ = 90°

∠QAB = ∠RAC (common angle)

So, ΔACR ~ ΔABQ (By AA similarity Rule)

Hence, BQCR = ABAC ⇒ yz = ABAC ... (ii)  (Corresponding part of similar triangle)

On adding eq. (i) and (ii), we get

yx + yz = BCAC + ABAC

⇒ y(1/x + 1/z) = BC + ABAC

⇒ y(1/x + 1/z) = 1

1/x + 1/z = 1/y

Hence proved.

Ans: (b)
Sol: Since, PQ || BC

∴ APPB = AQQC .......... (By Proportionality Theorem)
⇒ 46 = 8QC

⇒ QC4 = 8 × 64

= 12 cm
Now length of AC = AQ+QC = 12+8 = 20 cm

Ans:

Given: AZ = 3 cm, ZC = 2 cm, BM = 3 cm, and MC = 5 cm.

Proof: 

We can calculate the lengths of AC and BC:

• AC = AZ + ZC = 3 + 2 = 5 cm,
• BC = BM + MC = 3 + 5 = 8 cm.

Next, in triangles AXY and ABM, we know that:

• ∠AXY = ∠ABM (corresponding angles since XZ is parallel to BC),
• ∠XAY = ∠BAM (common angle).

∴ By the AA similarity criterion (Angle-Angle similarity):
ΔAXY ~ ΔABM.

Since these triangles are similar, their corresponding sides are proportional:
AX / AB = XY / BM = AZ / AC.

From the proportionality, we substitute the known values:
AX / AB = AZ / AC, which simplifies to:
XY / BM = AZ / AC.

Substituting the known values of AZ = 3 cm and AC = 5 cm, we get:
XY / 3 = 3 / 5.

Now, solving for XY, we multiply both sides by 3:
XY = (3 × 3) / 5 = 9 / 5 = 1.8 cm.

Thus, the length of XY is 1.8 cm.

Ans: (d)

In ΔABC, AC² = AB² + BC²  (Using Pythagoras Theorem)
⇒ AC² = 2² + 3²
⇒ AC² = 4 + 9
⇒ AC= √13 cm
So, perimeter is (2 + 3 + √13)cm = (5 + √13), which is irrational.
Hence, Assertion in false but Reason is true.

Ans:

We have, PN . NR = QN²
∴ PN / QN = QN / NR …(i)

In ΔQNP and ΔRNQ,
PN / QN = QN / NR  (from (i))

And ∠PNQ = ∠RNQ (Each equal to 90°)

∴ ΔQNP ~ ΔRNQ [By SAS similarity criterion]
Then, ΔQNP and ΔRNQ are equiangular,
i.e., ∠PQN = ∠QRN
⇒ ∠QRN = ∠QPN

On adding both sides, we get:
∠PQN + ∠QRN = ∠QRN + ∠QPN
⇒ ∠PQR = ∠QRN + ∠QPN …(ii)

We know that the sum of angles of a triangle is 180°.

In ΔPQR,
∠PQR + ∠PQR + ∠QR = 180°
⇒ ∠PQR + ∠QRN + ∠QRN = 180° [∴ ∠PQR = ∠QPN and ∠QRN = ∠QRN]
⇒ 2∠PQR = 180° [Using equation (ii)]

⇒ ∠PQR = 180° / 2 = 90°
∴ ∠PQR = 90°

Hence proved.

Ans: Given: ΔABC and ΔDBC are on the same base BC. AD and BC intersect at O. 
To Prove:
ar(ΔABC)ar(ΔDBC)= AODO

Construction: Draw AL ⊥ BC and DM ⊥ BC

Proof:
Now, in ΔALO and ΔDMO, we have 
∠ALO = ∠DMO = 90° 
∠AOL = ∠DOM (Vertically opposite angles) 
Therefore, ΔALO ~ ΔDMO (By AA criterion)

∴ ALDM = AODO (Corresponding sides of similar triangles are proportional)

ar(ΔABC)ar(ΔDBC) = 1/2 × BC × AL1/2 × BC × DM

⇒ ALDM = AODO

Hence, proved.

Ans: (a)
As PQ || AC by using basic proportionality theorem

⇒ BPPA = BQQC

⇒ 42.4 = 5QC

⇒ QC = 5 × 2.44

⇒ QC = 5 × 0.6

QC = 3 cm
∴ BC = BQ + QC
= 5 + 3
= 8 cm

Ans: (a)
Sol: Suppose PQ || AB
By using Proportionality theorem we have

CPPA = CQQB

⇒ x + 33x + 19 = x3x + 4

⇒ 3x² + 19x = 3x² + 9x + 4x + 12

⇒ 6x = 12
⇒ x = 2

Ans: (d)
Sol: Given Δ ABC and Δ PQR are similar.
Hence, ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R 
We know that, 
∠P + ∠Q + ∠R = 180° 
31° + ∠Q + 69° = 180° 
100° + ∠Q = 180° 
∠Q = 180° - 100° 
∠Q = 80°

Ans: (c)
Sol: Let AB be the pole and PQ be the tower
Let height of tower be h m
Now, ΔABC ∼ ΔPQR

∴ ABPQ = BCQR

⇒ 19h = 5751

⇒ h =  19 x 5157

⇒ h = 17m

Ans: 13 m
Let Aman starts from A point and continues 5 m towards west and readied at B point, from which he goes 12 m towards North reached at C point finally.
In ΔABC, we have

AC² = AB² + BC²  ............ [By Pythagoras theorem]
AC² = 5² + 12²
AC² = 25 + 144 = 169
AC = 13m
So, Aman is 13 m away from his starting point.

Ans: All concentric circles are similar to each other.

Ans: Given, PQIIBC
PQ = 3 cm, BC = 9 cm and AC = 7.5 cm
Since. PQ || BC
∴ ∠APQ = ∠ABC (Corresponding angles are equal)
Now,  in ΔAPQ and ΔABC
∠APQ =∠ABC        (Corresponding angles are equal)
∠A = ∠A                  (Common)
ΔAPQ ∼ ΔABC    (AA similarity)

∴ APAB = AQAC = PQBC

⇒ AQAC = 39

⇒ AQ7.5 = 13

⇒ AQ = 7.53 = 2.5 cm

Ans: In ΔEAB and  ΔECD

Acc to Converse of basic proportionality theorem, It states that if any line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

It is given that

Since, EA/EC = EB/ED 
∠1 = ∠2 [Vertically opposite angles] 
So, by SAS similarity criterion ΔEAB ~ ΔECD 
Hence, proved.

Ans: 3/10
Here in the given figure.
GC || BD and GE || BF
AC = 3 cm and CD = 7 cm
By Basic Proportionality theorem.

We get, ACCD = AEEF

∴ AE / EF = 3 / 7
⇒ AF / AE = 7 / 3
⇒ AE + EF / AE = 3 + 7 / 3

⇒ AF / AE = 10 / 3

Thus, AE / AF = 3 / 10.

Ans: 

In ΔABC, AD ⟂ BC and BD = 3CD

BD + CD = BC
3CD + CD = BC
4CD = BC
CD = (1/4) BC …… (1)

and, BD = (3/4) BC …… (2)

In ΔADC, ∠ADC = 90°

AC² = AD² + CD² [Using Pythagoras theorem]
AD² = AC² - CD² ….. (3)

In ΔADB, ∠ADB = 90°

AB² = AD² + BD² [Using Pythagoras theorem]
AB² = AC² - CD² + BD² [from equation (3)]
AB² = AC² + (3/4 BC)² - (1/4 BC)² [from equations (1) and (2)]
AB² = AC² + (9BC² - BC²)/16
AB² = AC² + 8BC²/16
AB² = AC² + 1/2 BC²

Thus, 2AB² = 2AC² + BC²