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)
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. (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
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, 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 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
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.
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, GC||BD and GE||BF. 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 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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3. What is the Pythagorean theorem and how is it related to right triangles? |
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