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Previous Year Questions 2024 | |
Previous Year Questions 2023 | |
Previous Year Questions 2022 | |
Previous Year Questions 2021 | |
Previous Year Questions 2020 | |
Previous Year Questions 2019 |
Q1: In ΔABC, DE || BC (as shown in the figure). If AD = 2 cm, BD = 3 cm, BC = 7.5 cm, then the length of DE (in cm) is: (CBSE 2024)
(a) 2.5
(b) 3
(c) 5
(d) 6
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
Q2: In ΔABC, if AD ⊥ BC and AD2 = BD × DC, then prove that ∠BAC = 90º. (CBSE 2024)
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
Q4: In the given figure, XZ is parallel to BC. AZ = 3 cm, ZC = 2 cm, BM = 3 cm and MC = 5 cm. Find the length of XY. (2023)
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
Q5: Assertion (A) : The perimeter of ΔABC is a rational number.
Reason (R) : The sum of the squares of two rational numbers is always rational. (2023)
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
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.
Q6: In a ΔPQR, N is a point on PR, such that QN ⊥ PR. If PN × NR = QN2, prove that ∠PQR = 90°. (CBSE 2023)
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
Q7: In the given figure, ΔABC and ΔDBC are on the same base BC. If AD intersects BC at O, prove that . (CBSE 2023)
View Answer
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.
Q8: In the given figure, PQ || AC. If BP = 4 cm, AP = 2.4 cm and BQ = 5 cm, then length of BC is ______.
(a) 8 cm
(b) 3 cm
(c) 0.3 cm
(d) 25/3 cm (CBSE 2023)
Ans: (a)
As PQ || AC by using basic proportionality theorem
QC = 3 cm
∴ BC = BQ + QC
= 5 + 3
= 8 cm
(a) 2
(b) 3
(c) 4
(d) 5
Ans: (a)
Sol: Suppose PQ || AB
By Basic Proportionality theorem we have
⇒ 6x = 12
⇒ x = 2
So, for x = 2, PQ IIAB
Q10: If Δ ABC and Δ PQR are similar triangles such that ∠A = 31° and ∠R = 69°, then ∠Q is : (2022)
(a) 70°
(b) 100°
(c) 90°
(d) 80°
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°
Q11: A vertical pole of length 19 m casts a shadow 57 m long on the ground and at the same time a tower casts a shadow 51m long. The height of the tower is (2022)
(a) 171m
(b) 13 m
(c) 17 m
(d) 117 m
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.
Q14: In figure, PQ || BC, PQ = 3 cm, BC = 9 cm and AC = 7.5 cm. Find the length of AQ. (2020)
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)
Q15: In the given figure, EE/AC = EE/DB , prove that ΔEAB ~ ΔECD. (CBSE 2020)
Ans: In ΔEAB and ΔECD
Since, EA / EC = EB / ED
∠1 = ∠2 [Vertically opposite angles]
So, by SAS similarity rule ΔEAB ~ ΔECD
Hence, proved.
View Answer
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,
Q17: The perpendicular from A on the side BC of a ΔABC intersects BC at D, such that DB = 3CD. Prove that 2AB2 = 2AC2 + BC2. (2019)
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.
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