Table of contents  
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: (2024)
(a) 2.5
(b) 3
(c) 5
(d) 6
Ans: (b)
Q2: In ΔABC, if AD ⊥ BC and AD^{2} = BD × DC, then prove that ∠BAC = 90º. (2024)
Ans:
Here,
AD ⊥ BC
and
AD^{2} = 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
Q3: in ΔABC, PQ  BC If PB = 6 cm, AP = 4 cm, AQ = 8 cm. find the length of AC. (2023)
(a ) 12 cm
(b) 20 cm
(c) 6 cm
(d) 14 cm
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, AC^{2} = AB^{2} + BC^{2}
⇒ AC^{2} = 2^{2} + 3^{2}
⇒ AC^{2} = 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 the figure given below, what value of x will make PQ  AB? (2022)
(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
Q7: 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°
Q8: 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
Q9: Aman goes 5 metres due west and then 12 metres due North. How far is he from the starting point? (2021)
View AnswerAns: 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
AC^{2} = AB^{2} + BC^{2}^{ }[By Pythagoras theorem]
AC^{2} = 5^{2} + 12^{2}
AC^{2} = 25 + 144 = 169
AC = 13m
So, Aman is 13 m away from his starting point.
Q10: All concentric circles are ___________ to each other. (2020)
View AnswerAns: All concentric circles arc similar to each other.
Q11: 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)
Q12: In figure, GCBD and GEBF. If AC = 3cm and CD = 7 cm, then find the value of AE / AF. (2019)
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,
Q13: The perpendicular from A on the side BC of a ΔABC intersects BC at D, such that DB = 3CD. Prove that 2AB^{2} = 2AC^{2} + BC^{2}. (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
AB^{2} = AD^{2 }+ BD^{2} _(i)
Similarly in ΔACD, we have AC^{2} = AD^{2} +CD^{2} _(ii)
Subtracting (ii) from (i), we get
AB^{2 } AC^{2} = BD^{2}  CD^{2} _(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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