Triangles: 1 mark Questions (CBSE Class 10)
70 different 1 mark questions on Triangles from CBSE Class 10 Maths board exams 2022–2026, newest first.
In the given figure, PQ ∥ BC. If AP : AB = 3 : 7 then, AQ : QC equals :
(A) 3 : 7(B) 3 : 10(C) 7 : 3(D) 3 : 4
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Answer: (D) 3 : 4
AP : AB = 3 : 7, so AP : PB = 3 : 4. By BPT, QC A Q = P B A P = 4 3 .
Which of the following statements is not always true ?
(A) Two circles are similar.(B) Two isosceles right triangles are similar.(C) Two rectangles are similar.(D) Two equilateral triangles are similar.
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Answer: (C) Two rectangles are similar.
All circles, all isosceles right triangles (angles 45°, 45°, 90°) and all equilateral triangles have the same shape, so they are always similar. Two rectangles have equal angles, but their sides need not be in the same ratio (e.g. 1 × 2 and 1 × 3). So 'Two rectangles are similar' is not always true.
In the given figure, DE ∥ BC. If AD : AB = 1 : 3 and AE = 2.5 cm, then AC equals
(A) 7.5 cm(B) 5 cm(C) 10 cm(D) 2.5 cm
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Answer: (A) 7.5 cm
Since DE ∥ BC, A B A D = A C A E (BPT). 3 1 = A C 2.5 , so AC = 7.5 cm.
If △ A B C and △ D E F are similar such that 2 A B = D E and B C = 8 cm, then EF is equal to :
(A) 4 cm(B) 8 cm(C) 12 cm(D) 16 cm
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Answer: (D) 16 cm
Since △ A B C ∼ △ D E F , D E A B = E F B C . D E A B = 2 1 , so E F 8 = 2 1 .E F = 16 cm.
Shown below are three triangles. The measures of two adjacent sides and included angle are given for each triangle : Which of these triangles are similar ?
(A) △ R P Q and △ X Z Y (B) △ R P Q and △ M N L (C) △ X Z Y and △ M N L (D) △ R P Q , △ X Z Y and △ M N L are similar to one another
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Answer: (A) △ R P Q and △ X Z Y
Each triangle has an included angle of 6 0 ∘ , so compare the ratios of the sides containing it. △ R P Q : P R : P Q = 6 : 4 = 3 : 2 .△ X Z Y : Z X : Z Y = 9 : 6 = 3 : 2 .△ M N L : N M : N L = 4 : 3 .So Z X P R = Z Y P Q = 3 2 and ∠ P = ∠ Z ; by SAS, △ R P Q ∼ △ X Z Y . △ M N L is not similar to either.
In triangles ABC and PQR, ∠ A = ∠ Q and ∠ B = ∠ R , then AB : AC is equal to :
(A) PQ : PR(B) PQ : QR(C) QR : QP(D) PR : QR
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Answer: (C) QR : QP
By AA similarity, △ A B C ∼ △ QR P (A ↔ Q, B ↔ R, C ↔ P). So QR A B = QP A C . Hence AB : AC = QR : QP.
In the given figure △ A B C is shown, in which DE ∥ BC. If AD = 5 cm, DB = 2.5 cm and DE = 8 cm, then the length of BC is :
(A) 10 cm(B) 6 cm(C) 12 cm(D) 7.5 cm
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Answer: (C) 12 cm
Since DE ∥ BC, △ A D E ∼ △ A B C . B C D E = A B A D = 7.5 5 = 3 2 B C = 2 3 × 8 = 12 cm
In Δ DEF, AB ∥ EF. The value of x is :
(A) 0, 2(B) 2 only(C) – 2(D) 1
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Answer: (B) 2 only
From the figure: DA = 2 x , AE = 3 x + 1 , DB = x , BF = 2 x − 2 1 By BPT, A E D A = B F D B : 3 x + 1 2 x = 2 x − 2 1 x 2 x ( 2 x − 2 1 ) = x ( 3 x + 1 ) , so 4 x 2 − x = 3 x 2 + x x 2 − 2 x = 0 , so x = 0 or x = 2 x = 0 would make DA = 0 , so x = 2 only
It is given that Δ ABC ~ Δ QRP such that AB = 9 cm, BC = 5 cm and PR = 2 cm. Length of side QR is :
(A) 0.9 cm(B) 18 5 cm(C) 9 10 cm(D) 3.6 cm
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Answer: (D) 3.6 cm
△ A B C ∼ △ QR P gives QR A B = R P B C QR 9 = 2 5 , so QR = 5 18 = 3.6 cm
In the given figure, OA × OB = OC × OD. Which of the following option is correct ?
(A) ∠ A = ∠ C (B) ∠ A = ∠ B (C) ∠ A = ∠ D (D) Δ O A D ∼ Δ O B C
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Answer: (A) ∠ A = ∠ C
OA × OB = OC × OD gives O C O A = O B O D ∠ A O D = ∠ C O B (vertically opposite angles)So △ A O D ∼ △ C O B (SAS similarity), with A ↔ C, D ↔ B Hence ∠ A = ∠ C (and ∠ D = ∠ B ) (D) would need O B O A = O C O D , which is not given
It is given that △ A B C ∼ △ E D F . Which of the following is not true ?
(A) Perimeter of △ E D F Perimeter of △ A B C = E D A B (B) E D A B = E F A C (C) ∠ A = ∠ D , ∠ C = ∠ F (D) A C A B + B C = E F D E + D F
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Answer: (C) ∠ A = ∠ D , ∠ C = ∠ F
△ A B C ∼ △ E D F gives the correspondence A ↔ E , B ↔ D , C ↔ F .So ∠ A = ∠ E , not ∠ D . (A), (B) and (D) follow from E D A B = D F B C = E F A C . Hence (C) is not true.
In the given figure, D E ∥ B C . If D B A D = 3 1 and AC = 6 cm, then length AE is
(A) 1.5 cm(B) 1 cm(C) 2 cm(D) 3 cm
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Answer: (A) 1.5 cm
By BPT, E C A E = D B A D = 3 1 . So A E = 4 1 × A C = 4 1 × 6 = 1.5 cm.
ABCD is a parallelogram such that AF = 7 cm, FB = 3 cm and EF = 4 cm, length FD = equals
(A) 4 21 cm(B) 3 28 cm(C) 7 12 cm(D) 5.5 cm
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Answer: (A) 4 21 cm
A D ∥ B C , so in △ A F D and △ E F B : ∠ D A F = ∠ B E F (alternate angles) and ∠ A F D = ∠ E F B (vertically opposite).△ A F D ∼ △ E F B (AA), so E F A F = F B F D 4 7 = 3 F D ⇒ F D = 4 21 cm
Devansh proved that △ A B C ∼ △ P QR using SAS similarity criteria. If he found ∠ C = ∠ R , then which of the following was proved true ?
(A) A B A C = P Q P R (B) A C B C = QR P R (C) B C A C = P Q P R (D) B C A C = QR P R
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Answer: (D) B C A C = QR P R
In SAS similarity, the sides including the equal angles must be proportional. The sides including ∠ C are AC and BC; the sides including ∠ R are PR and QR. So P R A C = QR B C , i.e. B C A C = QR P R .
In the given figure, P Q ∥ Y Z such that XP : PY = 2 : 3. If PQ = 5 cm, then YZ equals
(A) 12.5 cm(B) 10 cm(C) 15 cm(D) 7.5 cm
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Answer: (A) 12.5 cm
P Q ∥ Y Z , so △ X P Q ∼ △ X Y Z (AA).X Y X P = 2 + 3 2 = 5 2 Y Z P Q = 5 2 , so Y Z = 2 5 × 5 = 12.5 cm
In the given figure, A B ∥ E F . If AB = 24 cm, EF = 36 cm and DA = 7 cm, then AE equals
(A) 2.5 cm(B) 10.5 cm(C) 3.5 cm(D) 3 14 cm
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Answer: (C) 3.5 cm
A B ∥ E F , so △ D A B ∼ △ D E F (AA).D E D A = E F A B = 36 24 = 3 2 D E = 2 3 × 7 = 10.5 cmA E = D E − D A = 10.5 − 7 = 3.5 cm
Which of the following is not the criterion for similarity of triangles ?
(A) AAA(B) SSS(C) SAS(D) RHS
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Answer: (D) RHS
The criteria for similarity of triangles are AAA (AA), SSS and SAS. RHS is a criterion for congruence, not for similarity.
From the figures given below, which of the following is true about the measure of ∠ P ?
(A) ∠ P = 6 0 ∘ (B) ∠ P = 8 0 ∘ (C) ∠ P = 4 0 ∘ (D) The measure of ∠ P cannot be determined
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Answer: (C) ∠ P = 4 0 ∘
A B QR = 3.8 7.6 = 2 , B C P Q = 6 12 = 2 , A C P R = 3 3 6 3 = 2 .So △ A B C ∼ △ QR P (SSS), with C ↔ P . ∠ C = 18 0 ∘ − 8 0 ∘ − 6 0 ∘ = 4 0 ∘ .Hence ∠ P = ∠ C = 4 0 ∘ .
In the given figure, if DE ∥ BC, AD = 1.5 cm, DB = 3 cm and EC = 2 cm, the length of AC is :
(A) 1.5 cm(B) 3 cm(C) 3.5 cm(D) 4.5 cm
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Answer: (B) 3 cm
Since DE ∥ BC, by BPT D B A D = E C A E . 3 1.5 = 2 A E ⇒ A E = 1 cm.AC = AE + EC = 1 + 2 = 3 cm.
Assertion (A) : All congruent triangles are similar. Reason (R) : In congruent triangles, the ratio of corresponding sides is 1 : 1.
(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 Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Congruent triangles have equal corresponding angles and corresponding sides in the ratio 1 : 1. So their corresponding sides are proportional, which makes them similar. Both A and R are true and R explains A.
Which types of triangles are always similar ?
(A) Right-angled triangles(B) Acute-angled triangles(C) Isosceles triangles(D) Equilateral triangles
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Answer: (D) Equilateral triangles
All angles of every equilateral triangle are 6 0 ∘ . So any two equilateral triangles are similar (AAA). The other types need not have equal angles.
What values of x and y will make △ A B C similar to △ QR P in the figures given below ?
(A) x = 6 , y = 5 (B) x = 5 , y = 6 (C) x = 6 , y = 6 (D) x = 12 , y = 3.2
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Answer: (B) x = 5 , y = 6
△ A B C ∼ △ QR P means A ↔ Q, B ↔ R, C ↔ P.So QR A B = R P B C = P Q C A , i.e. 4 2 = x 2.5 = y 3 . x = 5 cm and y = 6 cm.
In △ ABC, PQ ∥ BC. It is given that AP = 2.4 cm, PB = 3.6 cm and BC = 5.4 cm. PQ is equal to :
(A) 2.7 cm(B) 1.8 cm(C) 3.6 cm(D) 2.16 cm
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Answer: (D) 2.16 cm
PQ ∥ BC, so △ A P Q ∼ △ A B C (AA). A B = 2.4 + 3.6 = 6 cm.B C P Q = A B A P , so P Q = 6 2.4 × 5.4 = 2.16 cm.
In △ A B C , DE ∥ BC. If AE = ( 2 x + 1 ) cm, EC = 4 cm, AD = ( x + 1 ) cm and DB = 3 cm, then value of x is
(A) 1(B) 2 1 (C) − 1 (D) 3 1
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Answer: (B) 2 1
By BPT, D B A D = E C A E , so 3 x + 1 = 4 2 x + 1 . 4 x + 4 = 6 x + 3 , so 2 x = 1 and x = 2 1 .
Assertion (A) : △ A B C ∼ △ P QR such that ∠ A = 6 5 ∘ , ∠ C = 6 0 ∘ . Hence ∠ Q = 5 5 ∘ . Reason (R) : Sum of all angles of a triangle is 18 0 ∘ .
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not correct explanation for Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
R is true (angle sum property). In △ A B C , ∠ B = 18 0 ∘ − 6 5 ∘ − 6 0 ∘ = 5 5 ∘ by R. Corresponding angles of similar triangles are equal, so ∠ Q = ∠ B = 5 5 ∘ . A is true. R is used to find the angle, so R explains A.
In triangles A B C and D E F , ∠ B = ∠ E , ∠ F = ∠ C and A B = 3 D E . Then, the two triangles are :
(A) congruent but not similar(B) congruent as well as similar(C) neither congruent nor similar(D) similar but not congruent
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Answer: (D) similar but not congruent
∠ B = ∠ E and ∠ C = ∠ F , so △ A B C ∼ △ D E F (AA similarity).A B = 3 D E , so the corresponding sides are not equal and the triangles are not congruent.
In the given figure, P Q ∥ B C . If P B A P = 13 4 and A C = 20.4 cm, then the length of AQ is :
(A) 2.8 cm(B) 5.8 cm(C) 3.8 cm(D) 4.8 cm
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Answer: (D) 4.8 cm
By BPT, QC A Q = P B A P = 13 4 So A C A Q = 17 4 A Q = 17 4 × 20.4 = 4.8 cm
Which of the following statements is incorrect ?
(A) Two congruent figures are always similar.(B) A square and a rhombus of the same area are always similar.(C) Two equilateral triangles are always similar.(D) Two similar triangles need not be congruent.
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Answer: (B) A square and a rhombus of the same area are always similar.
Congruent figures are always similar, all equilateral triangles are similar, and similar triangles need not be congruent: (A), (C), (D) are correct. A rhombus need not have right angles, so its angles need not equal those of a square; equal area does not make them similar. So (B) is incorrect
If in two triangles △ D E F and △ P QR , ∠ D = ∠ Q and ∠ R = ∠ E , then which of the following is not true ?
(A) QR D E = P Q D F (B) P R E F = P Q D F (C) R P E F = QR D E (D) P Q D E = R P E F
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Answer: (D) P Q D E = R P E F
∠ D = ∠ Q , ∠ E = ∠ R , so △ D E F ∼ △ QR P (AA).Hence QR D E = R P E F = QP D F . Options (A), (B), (C) follow from this; (D) P Q D E = R P E F does not.
The measurements of △ L M N and △ A B C are shown in the figure given below. The length of side AC is :
(A) 16 cm(B) 7 cm(C) 8 cm(D) 4 cm
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Answer: (C) 8 cm
In △ L M N , ∠ L = 18 0 ∘ − 13 0 ∘ − 2 8 ∘ = 2 2 ∘ . ∠ L = ∠ A = 2 2 ∘ and ∠ M = ∠ B = 13 0 ∘ , so △ L M N ∼ △ A B C (AA).A B L M = A C L N gives 5 45 = A C 72 .A C = 45 72 × 5 = 8 cm.
In the given figure, in △ A B C , A D ⊥ B C and ∠ B A C = 9 0 ∘ . If BC = 16 cm and DC = 4 cm, then the value of x is :
(A) 4 cm(B) 5 cm(C) 8 cm(D) 3 cm
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Answer: (C) 8 cm
△ A D C ∼ △ B A C (AA: ∠ C common, ∠ A D C = ∠ B A C = 9 0 ∘ ).So B C A C = A C D C , i.e. A C 2 = B C × D C = 16 × 4 = 64 . x = A C = 8 cm.
△ A B C and △ P QR are shown in the adjoining figures. The measure of ∠ C is :
(A) 14 0 ∘ (B) 8 0 ∘ (C) 6 0 ∘ (D) 4 0 ∘
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Answer: (D) 4 0 ∘
R Q A B = 7.6 3.8 = 2 1 , QP B C = 12 6 = 2 1 , P R C A = 6 3 3 3 = 2 1 So △ A B C ∼ △ R QP (SSS), and ∠ C = ∠ P ∠ P = 18 0 ∘ − 8 0 ∘ − 6 0 ∘ = 4 0 ∘ , so ∠ C = 4 0 ∘
E and F are points on the sides AB and AC respectively of a △ A B C such that E B A E = F C A F = 2 1 . Which of the following relation is true ?
(A) E F = 2 B C (B) B C = 2 E F (C) E F = 3 B C (D) B C = 3 E F
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Answer: (D) B C = 3 E F
By the converse of BPT, E F ∥ B C , so △ A E F ∼ △ A B C . A B A E = 1 + 2 1 = 3 1 B C E F = 3 1 , so B C = 3 E F
Which of the following statements is false ?
(A) Two right triangles are always similar.(B) Two squares are always similar.(C) Two equilateral triangles are always similar.(D) Two circles are always similar.
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Answer: (A) Two right triangles are always similar.
Squares, equilateral triangles and circles each have a fixed shape, so any two are similar. Two right triangles need not have equal acute angles (e.g. 3 0 ∘ -6 0 ∘ -9 0 ∘ and 4 5 ∘ -4 5 ∘ -9 0 ∘ ), so (A) is false.
In the adjoining figure, ABCD is a trapezium in which X Y ∥ A B ∥ C D . If A X = 3 2 A D , then C Y : Y B =
(A) 2 : 3(B) 3 : 2(C) 1 : 3(D) 1 : 2
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Answer: (D) 1 : 2
A X = 3 2 A D ⇒ X D = 3 1 A D , so D X : X A = 1 : 2 .Since X Y ∥ A B ∥ C D , the parallel lines cut the two transversals proportionally. C Y : Y B = D X : X A = 1 : 2
In the adjoining figure, P Q ∥ X Y ∥ B C , A P = 2 cm, P X = 1.5 cm and B X = 4 cm. If Q Y = 0.75 cm, then A Q + C Y =
(A) 6 cm(B) 4.5 cm(C) 3 cm(D) 5.25 cm
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Answer: (C) 3 cm
Parallel lines cut the two sides proportionally, so Q Y A Q = P X A P and Q Y C Y = P X B X . A Q = 1.5 2 × 0.75 = 1 cmC Y = 1.5 4 × 0.75 = 2 cmA Q + C Y = 1 + 2 = 3 cm
Given △ A B C ∼ △ P QR , ∠ A = 3 0 ∘ and ∠ Q = 9 0 ∘ . The value of ( ∠ R + ∠ B ) is
(A) 9 0 ∘ (B) 12 0 ∘ (C) 15 0 ∘ (D) 18 0 ∘
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Answer: (C) 15 0 ∘
Corresponding angles are equal: ∠ B = ∠ Q = 9 0 ∘ and ∠ R = ∠ C . ∠ C = 18 0 ∘ − 3 0 ∘ − 9 0 ∘ = 6 0 ∘ , so ∠ R = 6 0 ∘ .∠ R + ∠ B = 6 0 ∘ + 9 0 ∘ = 15 0 ∘
Assertion (A) : A line drawn parallel to any one side of a triangle intersects the other two sides in the same ratio. Reason (R) : Parallel lines cannot be drawn to any side of a triangle.
(A) Both Assertion (A) and Reason (R) are correct and Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are correct but Reason (R) is not the correct explanation of Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (C) Assertion (A) is true, but Reason (R) is false.
By the Basic Proportionality Theorem, a line parallel to one side of a triangle meeting the other two sides in distinct points divides them in the same ratio. So A is true. A line parallel to any side of a triangle can always be drawn, so R is false. Answer: (C)
If in the given figure, D E ∥ B C . If A D = 2.8 cm, D B = 2.1 cm and E C = 4.8 cm, then the value of x is :
(A) 3.6 cm(B) 2.4 cm(C) 6.4 cm(D) 4.8 cm
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Answer: (C) 6.4 cm
By the basic proportionality theorem, D B A D = E C A E . 2.1 2.8 = 4.8 x x = 2.1 2.8 × 4.8 = 6.4 cm.
In the given figure, if △ A B C ∼ △ QP R , then the value of x is :
(A) 5.3 cm(B) 4.6 cm(C) 2.3 cm(D) 4 cm
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Answer: (C) 2.3 cm
△ A B C ∼ △ QP R gives QP A B = P R B C = QR A C .QR A C = 3 6 = 2 , so P R B C = 2 .x = P R = 2 4.6 = 2.3 cm.
It is given that △ A B C ∼ △ D E F . If ∠ A = 5 5 ∘ , ∠ E = 4 5 ∘ , then ∠ C is :
(A) 8 0 ∘ (B) 9 0 ∘ (C) 5 5 ∘ (D) 4 5 ∘
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Answer: (A) 8 0 ∘
Corresponding angles are equal, so ∠ B = ∠ E = 4 5 ∘ . ∠ C = 18 0 ∘ − 5 5 ∘ − 4 5 ∘ = 8 0 ∘ .
In the given figure, in △ A B C , DE ∥ BC. If AD = 2x cm, AE = (x + 2) cm, DB = 4 cm, EC = 3 cm, then the value of x is :
(A) 3(B) 2(C) 6(D) 4
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Answer: (D) 4
By BPT, D B A D = E C A E . 4 2 x = 3 x + 2 , so 6 x = 4 x + 8 .x = 4 .
In the given figure △ A B C is shown. DE is parallel to BC. If AD = 5 cm, DB = 2.5 cm and BC = 12 cm, then DE is equal to
(A) 10 cm(B) 6 cm(C) 8 cm(D) 7.5 cm
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Answer: (C) 8 cm
DE ∥ BC, so △ A D E ∼ △ A B C (AA). B C D E = A B A D = 5 + 2.5 5 = 3 2 DE = 3 2 × 12 = 8 cm
The perimeters of two similar triangles ABC and PQR are 56 cm and 48 cm respectively. PQ/AB is equal to
(A) 8 7 (B) 7 6 (C) 6 7 (D) 7 8
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Answer: (B) 7 6
For similar triangles, the ratio of perimeters equals the ratio of corresponding sides. A B P Q = 56 48 = 7 6
AB and CD are two chords of a circle intersecting at P. Choose the correct statement from the following :
(A) △ A D P ∼ △ C B A (B) △ A D P ∼ △ B P C (C) △ A D P ∼ △ B C P (D) △ A D P ∼ △ C B P
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Answer: (D) △ A D P ∼ △ C B P
∠ D A P = ∠ D A B = ∠ D C B = ∠ B C P (angles in the same segment, on chord DB).∠ A D P = ∠ A D C = ∠ A B C = ∠ C B P (angles in the same segment, on chord AC).So A ↔ C, D ↔ B, P ↔ P and △ A D P ∼ △ C B P (AA).
If the diagonals of a quadrilateral divide each other proportionally, then it is a :
(A) parallelogram(B) rectangle(C) square(D) trapezium
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Answer: (D) trapezium
Let diagonals AC and BD of quadrilateral ABCD meet at O with O C A O = O D B O . Draw OE ∥ DC through O meeting AD at E. By BPT in △ A D C , E D A E = O C A O = O D B O . By the converse of BPT in △ A B D , EO ∥ AB, so AB ∥ DC. Hence the quadrilateral is a trapezium.
In △ A B C , 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 :
(A) 2.5(B) 3(C) 5(D) 6
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Answer: (B) 3
Since DE ∥ BC, △ A D E ∼ △ A B C (AA). B C D E = A B A D = 2 + 3 2 = 5 2 .D E = 5 2 × 7.5 = 3 cm.
In △ A B C , DE ∥ BC (as shown in the figure). If AD = 4 cm, AB = 9 cm and AC = 13.5 cm, then the length of EC is :
(A) 6 cm(B) 7.5 cm(C) 9 cm(D) 5.7 cm
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Answer: (B) 7.5 cm
Since DE ∥ BC, by BPT A B A D = A C A E . 9 4 = 13.5 A E , so AE = 6 cm.EC = AC − AE = 13.5 − 6 = 7.5 cm.
In the given figure, in △ A B C , DE ∥ BC. If AD = 2.4 cm, DB = 4 cm and AE = 2 cm, then the length of AC is :
(A) 3 10 cm(B) 10 3 cm(C) 3 16 cm(D) 1.2 cm
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Answer: (C) 3 16 cm
By BPT, D B A D = E C A E ⇒ 4 2.4 = E C 2 ⇒ E C = 3 10 cm. A C = A E + E C = 2 + 3 10 = 3 16 cm.
If a vertical pole of length 7.5 m casts a shadow 5 m long on the ground and at the same time, a tower casts a shadow 24 m long, then the height of the tower is :
(A) 20 m(B) 40 m(C) 60 m(D) 36 m
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Answer: (D) 36 m
The pole and its shadow and the tower and its shadow form similar right triangles (same sun angle). 24 h = 5 7.5 ⇒ h = 36 m.
Assertion (A) : ABCD is a trapezium with DC ∥ AB. E and F are points on AD and BC respectively, such that EF ∥ AB. Then E D A E = F C B F . Reason (R) : Any line parallel to parallel sides of a trapezium divides the non-parallel sides proportionally.
(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 Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Join AC meeting EF at G. In △ A D C , EG ∥ DC, so E D A E = GC A G (BPT). In △ C A B , GF ∥ AB, so GC A G = F C B F . Hence E D A E = F C B F : R is true and A is exactly R applied to this trapezium, so R explains A.
The sides of two similar triangles are in the ratio 4 : 7. The ratio of their perimeters is
(A) 4 : 7(B) 12 : 21(C) 16 : 49(D) 7 : 4
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Answer: (A) 4 : 7
In similar triangles, the corresponding sides are proportional. So the ratio of the perimeters equals the ratio of the corresponding sides = 4 : 7.
In the given figure, AB ∥ CD. If AB = 5 cm, CD = 2 cm and OB = 3 cm, then the length of OC is
(A) 2 15 cm(B) 3 10 cm(C) 5 6 cm(D) 5 3 cm
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Answer: (C) 5 6 cm
In △ A O B and △ D O C : ∠ A O B = ∠ D O C (vertically opposite angles). ∠ A B O = ∠ D C O (alternate angles, as AB ∥ CD).So △ A O B ∼ △ D O C (AA). D C A B = O C O B , so 2 5 = O C 3 .O C = 5 6 cm
In the above figure, the criterion of similarity by which △ A B C ∼ △ P QR is :
(A) SSA (Side – Side – Angle) Similarity(B) ASA (Angle – Side – Angle) Similarity(C) SAS (Side – Angle – Side) Similarity(D) AA (Angle – Angle) Similarity
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Answer: (C) SAS (Side – Angle – Side) Similarity
P Q A B = 4.4 2.2 = 2 1 and QR B C = 7 3.5 = 2 1 .The included angles are equal: ∠ B = ∠ Q = 5 0 ∘ . So the triangles are similar by the SAS criterion.
If △ A B C ∼ △ D E F and ∠ A = 4 7 ∘ , ∠ E = 8 3 ∘ , then ∠ C is equal :
(A) 4 7 ∘ (B) 5 0 ∘ (C) 8 3 ∘ (D) 13 0 ∘
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Answer: (B) 5 0 ∘
Corresponding angles of similar triangles are equal, so ∠ B = ∠ E = 8 3 ∘ . ∠ C = 18 0 ∘ − 4 7 ∘ − 8 3 ∘ = 5 0 ∘ .
△ A B C ∼ △ D E F and their perimeters are 32 cm and 24 cm respectively. If AB = 10 cm, then DE equals :
(A) 8 cm(B) 7.5 cm(C) 15 cm(D) 5 3 cm
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Answer: (B) 7.5 cm
In similar triangles, the ratio of perimeters equals the ratio of corresponding sides. D E A B = 24 32 , so DE = 32 10 × 24 = 7.5 cm.
In the given figure, AD = 2 cm, DB = 3 cm, DE = 2.5 cm and DE ∥ BC. The value of x is :
(A) 6 cm(B) 3.75 cm(C) 6.25 cm(D) 7.5 cm
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Answer: (C) 6.25 cm
DE ∥ BC, so △ A D E ∼ △ A B C (AA). A B A D = B C D E , i.e. 5 2 = x 2.5 .x = 2 2.5 × 5 = 6.25 cm.
In two triangles △ P QR and △ A B C , it is given that B C A B = P R P Q . For these two triangles to be similar, which of the following should be true ?
(A) ∠ A = ∠ P (B) ∠ B = ∠ Q (C) ∠ B = ∠ P (D) CA = QR
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Answer: (C) ∠ B = ∠ P
By SAS similarity, the angles included between the proportional sides must be equal. The angle between AB and BC is ∠ B ; the angle between PQ and PR is ∠ P . So we need ∠ B = ∠ P .
In the given figure, ABC is a triangle in which AD = 1.6 cm, BD = 4.8 cm, AE = 1.1 cm and EC = 2.2 cm. Then :
(A) DE ∥ BC(B) DE = 2 1 BC(C) DE = BC(D) DE is not parallel to BC (DE ∦ BC)
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Answer: (D) DE is not parallel to BC (DE ∦ BC)
D B A D = 4.8 1.6 = 3 1 .E C A E = 2.2 1.1 = 2 1 .Since D B A D = E C A E , by the converse of BPT DE is not parallel to BC. (Options B and C would need DE parallel to BC, so they are also ruled out.)
If △ A B C ∼ △ P QR with ∠ A = 3 2 ∘ and ∠ R = 6 5 ∘ , then the measure of ∠ B is :
(A) 3 2 ∘ (B) 6 5 ∘ (C) 8 3 ∘ (D) 9 7 ∘
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Answer: (C) 8 3 ∘
Corresponding angles of similar triangles are equal, so ∠ C = ∠ R = 6 5 ∘ . ∠ B = 18 0 ∘ − 3 2 ∘ − 6 5 ∘ = 8 3 ∘
In the given figure, D E ∥ B C . If AD = 2 units, DB = AE = 3 units and EC = x units, then the value of x is :
(A) 2(B) 3(C) 5(D) 2 9
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Answer: (D) 2 9
By BPT, D B A D = E C A E 3 2 = x 3 x = 2 9
In the given figure, A B ∥ P Q . If AB = 6 cm, PQ = 2 cm and OB = 3 cm, then the length of OP is :
(A) 9 cm(B) 3 cm(C) 4 cm(D) 1 cm
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Answer: (D) 1 cm
∠ O A B = ∠ O QP and ∠ O B A = ∠ O P Q (alternate angles), so △ O A B ∼ △ O QP .O P O B = QP A B O P 3 = 2 6 , so OP = 1 cm
In the given figure, ∠ A = ∠ C , AB = 6 cm, AP = 12 cm, CP = 4 cm. Then length of CD is :
(A) 2 cm(B) 6 cm(C) 8 cm(D) 18 cm
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Answer: (A) 2 cm
In △ A B P and △ C D P : ∠ A = ∠ C and ∠ A P B = ∠ C P D (vertically opposite). So △ A B P ∼ △ C D P (AA). C D A B = C P A P ⇒ C D 6 = 4 12 CD = 2 cm
In the given figure, △ A B C ∼ △ QP R . If AC = 6 cm, BC = 5 cm, QR = 3 cm and PR = x; then the value of x is :
(A) 3.6 cm(B) 2.5 cm(C) 10 cm(D) 3.2 cm
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Answer: (B) 2.5 cm
△ A B C ∼ △ QP R gives A ↔ Q, B ↔ P, C ↔ R.So QR A C = P R B C , i.e. 3 6 = x 5 . x = 2.5 cm
In △ A B C , PQ ∥ BC. If PB = 6 cm, AP = 4 cm, AQ = 8 cm, find the length of AC.
(A) 12 cm(B) 20 cm(C) 6 cm(D) 14 cm
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Answer: (B) 20 cm
By BPT, P B A P = QC A Q 6 4 = QC 8 ⇒ QC = 12 cmA C = A Q + QC = 8 + 12 = 20 cm
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) 3 25 cm
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Answer: (A) 8 cm
PQ ∥ AC, so B A B P = B C B Q (BPT / similar triangles BPQ and BAC). BA = BP + PA = 4 + 2.4 = 6.4 cm 6.4 4 = B C 5 ⇒ B C = 4 5 × 6.4 = 8 cm
In the given figure, DE ∥ BC. The value of x is :
(A) 6(B) 12.5(C) 8(D) 10
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Answer: (D) 10
Since DE ∥ BC, △ A D E ∼ △ A B C (AA). AB = AD + DB = 2 + 3 = 5 cm. A B A D = B C D E , so 5 2 = x 4 .x = 10.
In △ A B C and △ D E F , D E A B = F D B C . Which of the following makes the two triangles similar ?
(A) ∠ A = ∠ D (B) ∠ B = ∠ D (C) ∠ B = ∠ E (D) ∠ A = ∠ F
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Answer: (B) ∠ B = ∠ D
For SAS similarity, the included angles of the proportional sides must be equal. Angle included between AB and BC is ∠ B ; angle included between DE and FD is ∠ D . So ∠ B = ∠ D makes the triangles similar.
If △ P QR ∼ △ A B C ; PQ = 6 cm, AB = 8 cm and the perimeter of △ A B C is 36 cm, then the perimeter of △ P QR is
(A) 20.25 cm(B) 27 cm(C) 48 cm(D) 64 cm
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Answer: (B) 27 cm
Perimeters of similar triangles are in the ratio of corresponding sides. 36 Perimeter of △ P QR = 8 6 Perimeter of △ P QR = 36 × 4 3 = 27 cm
In the given figure, DE∥ BC. If AD = 3 cm, AB = 7 cm and EC = 3 cm, then the length of AE is
(A) 2 cm(B) 2.25 cm(C) 3.5 cm(D) 4 cm
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Answer: (B) 2.25 cm
DB = AB – AD = 7 – 3 = 4 cm. By BPT, D B A D = E C A E . 4 3 = 3 A E , so AE = 4 9 = 2.25 cm.
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