Triangles: 2 marks Questions (CBSE Class 10)
56 different 2 marks questions on Triangles from CBSE Class 10 Maths board exams 2022–2026, newest first.
In the given figure, QR ∥ CB and RP ∥ AC. If BR = 10 cm, QA = 12 cm, BP = 12 cm and PC = 18 cm, then find the lengths of AR and QC.
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Answer: AR = 15 cm, QC = 8 cm
In △ A B C , RP ∥ AC, so by BPT R A B R = P C B P . A R 10 = 18 12 ⇒ A R = 15 cm.QR ∥ CB, so R B A R = QC A Q . 10 15 = QC 12 ⇒ QC = 8 cm.
In the given figure, AB ∥ DC. If OB = 3OD and CD = 1.8 cm, then find the length AB.
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Answer: AB = 5.4 cm
In △ A O B and △ C O D : ∠ O A B = ∠ O C D and ∠ O B A = ∠ O D C (alternate angles, AB ∥ DC). So △ A O B ∼ △ C O D (AA). C D A B = O D O B = 3 AB = 3 × 1.8 = 5.4 cm
In the given figure, △ O D C ∼ △ O B A . If ∠ B O C = 11 0 ∘ , ∠ O D C = 4 5 ∘ and AB = 2 CD, then find (i) m ∠ O A B (ii) OB : OD.
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Answer: (i) 6 5 ∘ (ii) 2 : 1
D, O, B are collinear, so ∠ D O C = 18 0 ∘ − 11 0 ∘ = 7 0 ∘ . In △ O D C : ∠ O C D = 18 0 ∘ − 7 0 ∘ − 4 5 ∘ = 6 5 ∘ . (i) △ O D C ∼ △ O B A gives ∠ O A B = ∠ O C D = 6 5 ∘ . (ii) O B O D = B A D C = 2 C D C D = 2 1 , so OB : OD = 2 : 1.
In the given figure, two triangles ABC and PQR are shown such that ∠ A = ∠ P and ∠ C = ∠ R . If AD ⊥ BC and PS ⊥ QR, then prove that (i) △ A D B ∼ △ P S Q (ii) AD × QS = BD × PS.
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Answer: (i) Proved. (ii) Proved.
In △ A B C and △ P QR , ∠ A = ∠ P and ∠ C = ∠ R , so ∠ B = ∠ Q (angle sum property). (i) In △ A D B and △ P S Q : ∠ A D B = ∠ P S Q = 9 0 ∘ and ∠ A B D = ∠ P QS . So △ A D B ∼ △ P S Q (AA). (ii) Hence P S A D = QS B D , i.e. AD × QS = BD × PS.
In △ A B C , D E ∥ B C . If A D = x , D B = x − 2 , A E = x + 2 and E C = x − 1 , then find the value of x .
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Answer: x = 4
By BPT, D B A D = E C A E . x − 2 x = x − 1 x + 2 x ( x − 1 ) = ( x + 2 ) ( x − 2 ) x 2 − x = x 2 − 4 x = 4 .
In the figure given above, △ A B C ∼ △ X Y Z , then find the values of x and y .
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Answer: x = 4.8 cm, y = 5 cm
Since △ A B C ∼ △ X Y Z , X Y A B = Y Z B C = X Z A C . Y Z B C = 7.2 6 = 6 5 .x 4 = 6 5 , so x = 5 24 = 4.8 cm.6 y = 6 5 , so y = 5 cm.
In the given figure, △ A H K ∼ △ A B C . If AK = 10 cm, BC = 3.5 cm and HK = 7 cm, find the length of AC.
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Answer: AC = 5 cm
Since △ A H K ∼ △ A B C , A C A K = B C H K . A C 10 = 3.5 7 = 2 AC = 5 cm
In the given figure, XY ∥ QR, X Q P Q = 3 7 and PR = 6.3 cm. Find the length of YR.
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Answer: YR = 2.7 cm
Since XY ∥ QR, by BPT X Q P Q = Y R P R . 3 7 = Y R 6.3 Y R = 7 6.3 × 3 = 2.7 cm
The diagonals of a quadrilateral ABCD intersect each other at the point O such that O C A O = O D B O . Show that quadrilateral ABCD is a trapezium.
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Answer: Proved.
Through O draw OE ∥ DC meeting AD at E In △ A D C , OE ∥ DC, so by BPT E D A E = O C A O Given O C A O = O D B O , so E D A E = O D B O In △ D A B this means E and O divide DA and DB in the same ratio, so by the converse of BPT, EO ∥ AB Hence AB ∥ EO ∥ DC, so AB ∥ DC and ABCD is a trapezium
Two right triangles PRQ and PSQ are drawn on the same hypotenuse PQ. If PR and QS intersect at T, prove that ST × TQ = PT × TR.
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Answer: Proved.
In △ P S T and △ QR T : ∠ P S T = ∠ QR T = 9 0 ∘ ∠ P T S = ∠ QT R (vertically opposite angles)So △ P S T ∼ △ QR T (AA similarity) R T S T = QT P T , so ST × TQ = PT × TR
In the given figure, Δ A B E ≅ Δ A C D . Prove that Δ A D E ∼ Δ A B C .
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Answer: Proved.
△ A B E ≅ △ A C D , so AB = AC and AE = AD (CPCT)Hence A B A D = A C A E In △ A D E and △ A B C , ∠ D A E = ∠ B A C (common) So △ A D E ∼ △ A B C (SAS similarity)
D is a point on the side BC of △ A B C such that ∠ C A B = ∠ C D A . Show that C A 2 = C B × C D .
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Answer: Proved.
In △ C A B and △ C D A : ∠ C A B = ∠ C D A (given) and ∠ C is common.So △ C A B ∼ △ C D A (AA similarity). Hence C D C A = C A C B , i.e. C A 2 = C B × C D .
In the given figure, DEFG is a square. △ A B C is right angle triangle with ∠ A = 9 0 ∘ . Prove that A G × D G = A F × D B .
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Answer: Proved.
In △ A GF and △ D B G : ∠ G A F = 9 0 ∘ = ∠ B D G (DEFG is a square, so G D ⊥ B C ). GF ∥ B C , so ∠ A GF = ∠ GB D (corresponding angles).Hence △ A GF ∼ △ D B G (AA). So D B A G = D G A F , i.e. A G × D G = A F × D B .
In the given figure, A B ∥ D E and A C ∥ D F . Show that △ A B C ∼ △ D E F . If BC = 10 cm, EB = CF = 5 cm and AB = 7 cm, then find the length DE.
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Answer: DE = 14 cm
A B ∥ D E , so ∠ A B C = ∠ D E F (corresponding angles, with transversal EF).A C ∥ D F , so ∠ A C B = ∠ D F E (corresponding angles).Hence △ A B C ∼ △ D E F (AA similarity). E F = E B + B C + C F = 5 + 10 + 5 = 20 cmA B D E = B C E F = 10 20 = 2 , so D E = 2 × 7 = 14 cm
In the given figure, if PQ ∥ RS, then prove that △ P O Q ∼ △ S O R .
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Answer: Proved.
In △ P O Q and △ S O R : ∠ QP O = ∠ R S O (alternate angles, PQ ∥ RS, PS a transversal).∠ P QO = ∠ S R O (alternate angles, PQ ∥ RS, QR a transversal).Hence △ P O Q ∼ △ S O R (AA similarity).
In the given figure, △ O S R ∼ △ O QP , ∠ R O Q = 12 5 ∘ and ∠ O R S = 7 0 ∘ . Find the measures of ∠ O S R and ∠ O QP .
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Answer: ∠ O S R = 5 5 ∘ , ∠ O QP = 5 5 ∘
∠ R O Q is an exterior angle of △ O S R (S, O, Q are collinear).So ∠ O S R + ∠ O R S = 12 5 ∘ , giving ∠ O S R = 12 5 ∘ − 7 0 ∘ = 5 5 ∘ . Since △ O S R ∼ △ O QP , corresponding angles are equal: ∠ O QP = ∠ O S R = 5 5 ∘ .
D is a point on side BC of △ A B C such that ∠ A D C = ∠ B A C . Prove that ( C A ) 2 = C B ⋅ C D .
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Answer: Proved.
In △ A D C and △ B A C : ∠ A D C = ∠ B A C (given) and ∠ A C D = ∠ B C A (common angle C).So △ A D C ∼ △ B A C (AA similarity). Corresponding sides are proportional: C B C A = C A C D . Hence ( C A ) 2 = C B ⋅ C D .
A vertical pole of height 10 m casts a shadow of 15 m on the ground and at the same time, a tower casts a shadow of 45 m on the ground. Find the height of the tower.
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Answer: 30 m
At the same time the sun's rays make the same angle with the ground, so the pole–shadow triangle and the tower–shadow triangle are similar (AA). 45 h = 15 10 .h = 30 m.
In the given figure ∠ A D E = ∠ A C B and D B A D = E C A E . Prove that △ A B C is an isosceles triangle.
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Answer: Proved.
Since D B A D = E C A E , by the converse of the basic proportionality theorem, DE ∥ BC. So ∠ A D E = ∠ A B C (corresponding angles). Given ∠ A D E = ∠ A C B , hence ∠ A B C = ∠ A C B . Sides opposite equal angles are equal, so AC = AB. Hence △ A B C is isosceles.
Two chords BA and CD intersect at point P outside the circle. Prove that △ P D A ∼ △ P B C
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Answer: Proved.
A, D, C, B lie on the circle, so ADCB is a cyclic quadrilateral and ∠ A D C + ∠ A B C = 18 0 ∘ . P, D, C are collinear, so ∠ P D A + ∠ A D C = 18 0 ∘ (linear pair). Hence ∠ P D A = ∠ A B C = ∠ P B C . In △ P D A and △ P B C : ∠ P is common and ∠ P D A = ∠ P B C . So △ P D A ∼ △ P B C (AA similarity).
The diagonal BD of parallelogram ABCD is divided by segment AE in the ratio 1 : 2. If BE = 1.8 cm, find the length of AD.
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Answer: AD = 3.6 cm
AE meets BD at O with BO : OD = 1 : 2 (as in the figure). In △ B O E and △ D O A : ∠ B O E = ∠ D O A (vertically opposite) and ∠ O B E = ∠ O D A (alternate angles, BC ∥ AD). So △ B O E ∼ △ D O A (AA) and D A B E = D O B O = 2 1 . AD = 2 × 1.8 = 3.6 cm.
In a trapezium ABCD, AB ∥ DC and its diagonals intersect at O. Prove that O C O A = O D O B .
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Answer: Proved.
In △ A O B and △ C O D : ∠ O A B = ∠ O C D (alternate angles, AB ∥ DC)∠ O B A = ∠ O D C (alternate angles)So △ A O B ∼ △ C O D (AA similarity). Hence O C O A = O D O B (corresponding sides of similar triangles are proportional).
In the given figure, AB ∥ DE and BD ∥ EF. Prove that DC2 = CF × AC.
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Answer: Proved.
In △ C A B , DE ∥ AB, so by BPT C A C D = C B C E ... (1) In △ C D B , FE ∥ DB, so by BPT C D C F = C B C E ... (2) From (1) and (2), C A C D = C D C F . Hence DC2 = CF × AC.
The perimeters of two similar triangles are 22 cm and 33 cm respectively. If one side of first triangle is 9 cm, then find the length of corresponding side of the second triangle.
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Answer: 13.5 cm
For similar triangles, the ratio of perimeters equals the ratio of corresponding sides. 33 22 = x 9 , so x = 22 9 × 33 = 13.5 cm.
If △ A B C ∼ △ P QR in which A B = 6 cm, B C = 4 cm, A C = 8 cm and P R = 6 cm, then find the length of ( P Q + QR ) .
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Answer: P Q + QR = 7.5 cm
Since △ A B C ∼ △ P QR : P Q A B = QR B C = P R A C = 6 8 = 3 4 . P Q = 4 3 × 6 = 4.5 cm.QR = 4 3 × 4 = 3 cm.P Q + QR = 4.5 + 3 = 7.5 cm.
In the given figure, QS QR = P R QT and ∠1 = ∠2 , show that △ P QS ∼ △ T QR .
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Answer: Proved.
In △ P QR , ∠1 = ∠2 , i.e. ∠ P QR = ∠ P R Q , so P Q = P R (sides opposite equal angles). Given QS QR = P R QT . Replace P R by P Q : QS QR = P Q QT , so QR QS = QT P Q . In △ P QS and △ T QR : T Q P Q = QR QS and ∠ P QS = ∠ T QR (= ∠1 , common). Hence △ P QS ∼ △ T QR (SAS similarity).
In the given figure, D is a point on the side BC of △ A B C such that ∠ A D C = ∠ B A C . Show that C A 2 = C D ⋅ C B .
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Answer: Proved.
In △ A D C and △ B A C : ∠ A D C = ∠ B A C (given) and ∠ A C D = ∠ B C A (common) So △ A D C ∼ △ B A C (AA similarity) Corresponding sides are proportional: C B C A = C A C D Hence C A 2 = C D ⋅ C B
In the given figure, O A ⋅ O B = O C ⋅ O D . Show that ∠ A = ∠ C and ∠ B = ∠ D .
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Answer: Proved.
Given O A ⋅ O B = O C ⋅ O D , so O C O A = O B O D In △ A O D and △ C O B : ∠ A O D = ∠ C O B (vertically opposite angles) and O C O A = O B O D So △ A O D ∼ △ C O B (SAS similarity) Corresponding angles are equal: ∠ A = ∠ C and ∠ D = ∠ B
In the given figure, S Q P S = T R P T and ∠ P S T = ∠ P R Q . Prove that △ P QR is an isosceles triangle.
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Answer: Proved.
Since S Q P S = T R P T , by the converse of the Basic Proportionality Theorem, S T ∥ QR So ∠ P S T = ∠ P QR (corresponding angles) Given ∠ P S T = ∠ P R Q , hence ∠ P QR = ∠ P R Q Sides opposite equal angles are equal: P R = P Q So △ P QR is isosceles
In the given figure, △ A B E ≅ △ A C D . Prove that △ A D E ∼ △ A B C .
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Answer: Proved.
△ A B E ≅ △ A C D , so by CPCT A B = A C and A E = A D Hence A B A D = A C A E In △ A D E and △ A B C : ∠ D A E = ∠ B A C (common) and A B A D = A C A E So △ A D E ∼ △ A B C (SAS similarity)
A 1.5 m tall boy is walking away from the base of a lamp post which is 12 m high, at the speed of 2.5 m/sec. Find the length of his shadow after 3 seconds.
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Answer: 14 15 m (about 1.07 m)
Distance walked in 3 s = 2.5 × 3 = 7.5 m Let the shadow be s m. The boy and the lamp post are both vertical, so the two right triangles formed with the tip of the shadow are similar (AA). 12 1.5 = 7.5 + s s 12 s = 1.5 s + 11.25 ⇒ 10.5 s = 11.25 ⇒ s = 14 15 m ≈ 1.07 m
In parallelogram ABCD, side AD is produced to a point E and BE intersects CD at F. Prove that △ A B E ∼ △ C F B
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Answer: Proved.
In △ A B E and △ C F B : ∠ B A E = ∠ F C B (opposite angles of parallelogram ABCD, ∠ A = ∠ C )A E ∥ B C (AD ∥ BC and E lies on AD produced), with BE as transversal, so ∠ A E B = ∠ C B F (alternate angles)Hence △ A B E ∼ △ C F B (AA similarity).
AD and PS are medians of triangles ABC and PQR respectively such that △ A B D ∼ △ P QS . Prove that △ A B C ∼ △ P QR .
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Answer: Proved.
△ A B D ∼ △ P QS ⇒ P Q A B = QS B D and ∠ B = ∠ Q .AD and PS are medians, so B D = 2 1 B C and QS = 2 1 QR . Hence P Q A B = QR /2 B C /2 = QR B C . In △ A B C and △ P QR : P Q A B = QR B C and included ∠ B = ∠ Q . By SAS similarity, △ A B C ∼ △ P QR .
AD and PS are angle bisectors of ∠ A and ∠ P of triangles ABC and PQR. If △ A B C ∼ △ P QR , prove that △ A C D ∼ △ P R S .
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Answer: Proved.
△ A B C ∼ △ P QR ⇒ ∠ A = ∠ P and ∠ C = ∠ R .AD and PS bisect ∠ A and ∠ P , so ∠ C A D = 2 1 ∠ A = 2 1 ∠ P = ∠ R P S . In △ A C D and △ P R S : ∠ C A D = ∠ R P S and ∠ A C D = ∠ P R S . By AA similarity, △ A C D ∼ △ P R S .
In △ A B C and △ P QR , AD and PS are altitudes such that △ A B D ∼ △ P QS and △ A C D ∼ △ P R S . Prove that △ A B C ∼ △ P QR .
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Answer: Proved.
△ A B D ∼ △ P QS ⇒ ∠ B = ∠ Q (corresponding angles).△ A C D ∼ △ P R S ⇒ ∠ C = ∠ R (corresponding angles).In △ A B C and △ P QR : ∠ B = ∠ Q and ∠ C = ∠ R . By AA similarity, △ A B C ∼ △ P QR .
P is a point on the side BC of △ A B C such that ∠ A P C = ∠ B A C . Prove that A C 2 = B C ⋅ C P .
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Answer: Proved.
In △ B A C and △ A P C : ∠ B A C = ∠ A P C (given) and ∠ A C B = ∠ P C A (common angle C)So △ B A C ∼ △ A P C (AA similarity). Corresponding sides are proportional: A C B C = P C A C Hence A C 2 = B C ⋅ C P .
In the adjoining figure, B D A D = E C A E and ∠ B D E = ∠ C E D , prove that △ A B C is an isosceles triangle.
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Answer: Proved.
B D A D = E C A E , so by the converse of BPT, D E ∥ B C .Hence ∠ A D E = ∠ A B C and ∠ A E D = ∠ A C B (corresponding angles). ∠ A D E = 18 0 ∘ − ∠ B D E and ∠ A E D = 18 0 ∘ − ∠ C E D ; since ∠ B D E = ∠ C E D , ∠ A D E = ∠ A E D .So ∠ A B C = ∠ A C B , giving A B = A C . Hence △ A B C is isosceles.
In the adjoining figure, A P = 1 cm, B P = 2 cm, A Q = 1.5 cm and A C = 4.5 cm. Prove that △ A P Q ∼ △ A B C . Hence find the length of PQ, if B C = 3.6 cm.
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Answer: Proved; PQ = 1.2 cm
A B = A P + P B = 1 + 2 = 3 cmA B A P = 3 1 and A C A Q = 4.5 1.5 = 3 1 ∠ P A Q = ∠ B A C (common)So △ A P Q ∼ △ A B C (SAS similarity). B C P Q = 3 1 ⇒ P Q = 3 3.6 = 1.2 cm
In the given figure, A P ⊥ A B and B Q ⊥ A B . If OA = 15 cm, BO = 12 cm and AP = 10 cm, then find the length of BQ.
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Answer: BQ = 8 cm
In △ O A P and △ O B Q : ∠ O A P = ∠ O B Q = 9 0 ∘ and ∠ A O P = ∠ B O Q (vertically opposite). So △ O A P ∼ △ O B Q (AA). O B O A = B Q A P , so 12 15 = B Q 10 .B Q = 15 10 × 12 = 8 cm.
D is a point on the side BC of △ A B C such that ∠ A D C = ∠ B A C . Show that A C 2 = B C × D C .
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Answer: Proved.
In △ A D C and △ B A C : ∠ A D C = ∠ B A C (given) and ∠ A C D = ∠ B C A (common). So △ A D C ∼ △ B A C (AA). Corresponding sides are proportional: B C A C = A C D C . Hence A C 2 = B C × D C .
In the given figure, O A ⋅ O B = O C ⋅ O D . Prove that △ A O D ∼ △ C O B .
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Answer: Proved.
Given O A ⋅ O B = O C ⋅ O D , so O C O A = O B O D . ∠ A O D = ∠ C O B (vertically opposite angles).In △ A O D and △ C O B , two sides are proportional and the included angles are equal. Hence △ A O D ∼ △ C O B (SAS similarity).
In the given figure, ∠ D = ∠ E and D B A D = E C A E . Prove that △ A B C is isosceles.
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Answer: Proved.
In △ A D E , ∠ A D E = ∠ A E D , so A D = A E (sides opposite equal angles). Given D B A D = E C A E and A D = A E , so D B = E C . A B = A D + D B = A E + E C = A C .Hence △ A B C is isosceles.
In the given figure, P Q ∥ R S . Prove that O P × O R = O Q × O S .
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Answer: Proved.
In △ O P Q and △ O S R : ∠ O P Q = ∠ O S R (alternate angles, P Q ∥ R S , transversal PS).∠ P O Q = ∠ S O R (vertically opposite angles).So △ O P Q ∼ △ O S R (AA similarity). Hence O S O P = O R O Q , i.e. O P × O R = O Q × O S .
In the given figure, ABCD is a quadrilateral. Diagonal BD bisects ∠ B and ∠ D both. Prove that : (i) △ A B D ∼ △ C B D (ii) AB = BC
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Answer: Proved.
(i) In △ A B D and △ C B D : ∠ A B D = ∠ C B D (BD bisects ∠ B )∠ A D B = ∠ C D B (BD bisects ∠ D )So △ A B D ∼ △ C B D (AA similarity). (ii) Corresponding sides are proportional: C B A B = B D B D = 1 . Hence AB = BC.
Diagonals AC and BD of a trapezium ABCD intersect at O, where AB ∥ DC. If O B D O = 2 1 , then show that AB = 2CD
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Answer: Proved.
In △ A O B and △ C O D : ∠ O A B = ∠ O C D and ∠ O B A = ∠ O D C (alternate angles, AB ∥ DC).So △ A O B ∼ △ C O D (AA similarity). Hence C D A B = O D O B = 2 , i.e. AB = 2CD.
In △ A B C , altitudes AD and BE are drawn. If AD = 7 cm, BE = 9 cm and EC = 12 cm then, find the length of CD.
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Answer: CD = 3 28 cm
In △ A D C and △ B E C : ∠ A D C = ∠ B E C = 9 0 ∘ and ∠ C is common. So △ A D C ∼ △ B E C (AA similarity). B E A D = C E C D , so 9 7 = 12 C D .CD = 9 84 = 3 28 cm.
In the given figure, E C E A = E D E B , prove that △ E A B ∼ △ E C D
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Answer: Proved.
In △ E A B and △ E C D : E C E A = E D E B (given).∠ A E B = ∠ C E D (vertically opposite angles).So by SAS similarity, △ E A B ∼ △ E C D .
In the given figure, △ A H K ∼ △ A B C . If AK = 8 cm, BC = 3.2 cm and HK = 6.4 cm, then find the length of AC.
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Answer: AC = 4 cm
Since △ A H K ∼ △ A B C , A C A K = B C H K . A C 8 = 3.2 6.4 = 2 .AC = 4 cm.
In the given figure, ABC and AMP are two right triangles, right angled at B and M, respectively. Prove that △ A B C ∼ △ A M P .
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Answer: Proved.
In △ A B C and △ A M P : ∠ A B C = ∠ A M P = 9 0 ∘ (given)∠ B A C = ∠ M A P (common angle A)So △ A B C ∼ △ A M P (AA similarity).
In the given figure, D E ∥ A C and E C B E = C P B C . Prove that D C ∥ A P .
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Answer: Proved.
In △ B A C , D E ∥ A C , so by BPT, D A B D = E C B E ...(1) Given E C B E = C P B C ...(2) From (1) and (2), D A B D = C P B C In △ B A P , D lies on BA and C lies on BP and they divide these sides in the same ratio. By the converse of BPT, D C ∥ A P .
In the given figure, L M ∥ C B and L N ∥ C D . Prove that A N A M = A D A B .
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Answer: Proved.
In △ A B C , L M ∥ C B , so by BPT, A B A M = A C A L ...(1) In △ A C D , L N ∥ C D , so by BPT, A D A N = A C A L ...(2) From (1) and (2), A B A M = A D A N Hence A N A M = A D A B .
In the adjoining figure, A, B and C are points on OP, OQ and OR respectively such that A B ∥ P Q and A C ∥ P R . Show that B C ∥ QR .
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Answer: Proved.
In △ O P Q , A B ∥ P Q , so by BPT A P O A = B Q O B ... (1) In △ O P R , A C ∥ P R , so by BPT A P O A = C R O C ... (2) From (1) and (2), B Q O B = C R O C . So in △ O QR , B and C divide OQ and OR in the same ratio; by the converse of BPT, B C ∥ QR .
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.
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Answer: XY = 1.8 cm
In △ A B C , X Z ∥ B C , so A B A X = A C A Z = 5 3 . In △ A B M , X Y ∥ B M , so △ A X Y ∼ △ A B M (AA). B M X Y = A B A X = 5 3 X Y = 5 3 × 3 = 5 9 = 1.8 cm
In the given figure, ABCD is a parallelogram. AE divides the line segment BD in the ratio 1 : 2. If BE = 1.5 cm, then find the length of BC.
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Answer: BC = 3 cm
AE meets BD at O with BO : OD = 1 : 2. B E ∥ A D (E lies on BC), so in △ B O E and △ D O A : ∠ B O E = ∠ D O A (vertically opposite) and ∠ O B E = ∠ O D A (alternate angles).△ B O E ∼ △ D O A (AA), so D A B E = D O B O = 2 1 .A D = 2 × 1.5 = 3 cm, and BC = AD = 3 cm.
In the given figure, ABC is a triangle in which DE∥ BC. If AD = x , DB = x − 2 , AE = x + 2 and EC = x − 1 , then find the value of x .
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Answer: x = 4
By BPT, D B A D = E C A E . x − 2 x = x − 1 x + 2 x ( x − 1 ) = ( x + 2 ) ( x − 2 ) x 2 − x = x 2 − 4 x = 4
Diagonals AC and BD of trapezium ABCD with AB∥ DC intersect each other at point O. Show that O C O A = O D O B .
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Answer: Proved.
In △ A O B and △ C O D : ∠ O A B = ∠ O C D (alternate angles, AB ∥ DC, AC transversal)∠ O B A = ∠ O D C (alternate angles, BD transversal)So △ A O B ∼ △ C O D (AA similarity). Hence O C O A = O D O B (corresponding sides of similar triangles are proportional).
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