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

Ans: (b) 
In ΔABC, DE || BC 
AD = 2 cm 
BD = 3 cm
∴ AB = AB + BD
= (2 + 3) cm 
AB = 5 cm 
Now, ∠ADE = ∠ABC, ∠AED = ∠ACB [Corresponding angles] 
So by AA prop. ΔADE ∼ ΔABC
⇒ AD/AB = DE/BC
⇒ 2/5 = DE/7.5
DE = 3 cm

Ans:

Here,
AD ⊥ BC

and
AD² = BD × DC
i.e., AD × AD = BD × DC
AD/DC = BD/AD
and ∠ADB = ∠CDA [Each 90°]
⇒ ∆ADB ~ ∆CDA
∠1 = ∠2 [By CPST]
∠3 = ∠4 (i)
In ∆AD C,
∠3 + ∠ADC + ∠1 = 180°
∠3 + 90° + ∠1 = 180°
∠1 = 180° – 90° – ∠3
∠1 = 90° – ∠3
∠BAC = ∠1 + ∠4
= 90° – ∠3 + ∠3
[∵∠4 = ∠3 From eqn. (i)]
i.e., ∠BAC = 90°
Hence, Proved

Ans: (b)
Sol: Since, PQ || BC
∴  [By Thales theorem]
⇒
= 12 cm

Ans: Given, AZ = 3 cm, ZC = 2 cm. BM = 3 cm and MC = 5 cm
In ΔABC, XZ || BC

Now, AC = AZ + ZC = 3 + 2 = 5cm
BC = BM + MC = 3 + 5 = 8 cm and
In ΔAXY and ΔABM
∠AXY = ∠ABM (Corresponding angles are equal as XZ || BC)
∠XAY = ∠BAM (Common)
∴ ΔAXY ∼ ΔABM (By AA similarity criterion)

(Corresponding sides of similar triangles)
From (i) and (ii), we get


= 1.8cm

Ans: (d)

In ΔABC, AC² = AB² + BC²
⇒ 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: In ΔPQR, QN ⊥ PR and PN × NR = QN²
In ΔQNP and ΔRNQ, 
∠1 = ∠2 = 90° 
QN² = NR × NP (Given) 
QN × QN = NR × NP 
QN / NR = NP / QN 
∴ DQNP ~ DRNQ (By SAS similarity criterion) 
∠P = ∠RQN = x …(i) 
∠PQN = ∠R = ∠y …(ii)
In ΔPQR, we have         
∠P + ∠PQR + ∠R = 180º         
∠x + ∠x + ∠y + ∠y = 180º 
2∠x + 2∠y = 180º 
2(∠x + ∠y) = 180° 
∠x + ∠y = 90º 
∠PQR = 90º, 
Hence, proved

Ans: Given: ΔABC and ΔDBC are on the same base BC. AD and BC intersect at O. 
To Prove:

Construction: Draw AL ⊥ BC and DM ⊥ BC

Now, in ΔALO and ΔDMO, we have 
∠ALO = ∠DMO = 90° 
∠AOL = ∠DOM (Vertically opposite angles) 
Therefore, ΔALO ~ ΔDMO
∴  (Corresponding sides of similar triangles are proportional)

Hence, proved.

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

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

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

⇒ 6x = 12
⇒ x = 2
So, for x = 2, PQ IIAB 

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

⇒ 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 arc 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)

Ans: In ΔEAB and  ΔECD

Since, EA / EC = EB /  ED 
∠1 = ∠2 [Vertically opposite angles] 
So, by SAS similarity rule 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,

Ans: We have, ΔABC such that AD⊥BC. ΔABC Intersect SC at D such that BD = 3CD.
In right ΔADB, by Pythagoras theorem, we have
AB² = AD²  +  BD²    _(i)
Similarly in ΔACD, we have AC² = AD² +CD²    _(ii)
Subtracting (ii) from (i), we get
AB² - AC² = BD² - CD²    _(iii)
Now, BC = DB + CD = 4CD    [∵ BD = 3CD]

Substituting the value of BD and CD in eqn.(iii) we get

Hence proved.