LIWA

Triangles

CBSE Class 10 · Maths · Chapter 6

Past board exam questions

CBSE 2027 sample paper · 1 mark

The perimeters of two similar triangles are 56 cm and 70 cm respectively. If one side of the first triangle is 14 cm, then the corresponding side of the second triangle (in cm) is:

  1. 5
  2. 7.5
  3. 10
  4. 17.5
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17.5

CBSE 2027 sample paper · 1 mark

A ladder 15 m long reaches a window 12 m above the ground. The distance of the foot of the ladder from the base of the wall is

  1. 8 m
  2. 9 m
  3. 10 m
  4. 13 m
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9 m

CBSE 2027 sample paper · 1 mark

The perimeter of two similar triangles is 28 cm and 35 cm respectively. If one side of the first triangle is 8cm, then the corresponding side of the second triangle is

  1. 10 cm
  2. 12cm
  3. 14cm
  4. 16cm
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10 cm

CBSE 2027 sample paper · 1 mark

If in ∆ABC and ∆PQR, ∠B = ∠Q, ∠R = ∠C and AB = 2PQ, then the two triangles are

  1. Congruent but not similar
  2. Similar but not congruent
  3. Neither congruent nor similar
  4. Congruent as well as similar
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Similar but not congruent

CBSE 2027 sample paper · 1 mark

ASSERTION (A): If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. REASON (R): Line drawn from midpoint of one side of triangle parallel to another will bisect the third side.

  1. Both Assertion (A) and Reason (R) are true and (R) is the correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true but (R) is not the correct explanation of Assertion (A).
  3. Assertion (A) is true but Reason (R) is false.
  4. Assertion (A) is false but Reason (R) is true.
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Both Assertion (A) and Reason (R) are true but (R) is not the correct explanation of Assertion (A).

CBSE 2027 sample paper · 2 marks

ABCD is a trapezium in which AB is parallel to DC and its diagonals intersect each other at the point O. Show that (AO)/(BO) = (CO)/(DO).

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In ∆AOB and ∆COD: ∠OAB = ∠OCD (alternate interior angles, AB ∥ DC) ∠OBA = ∠ODC (alternate interior angles) ∴ ∆AOB ∼ ∆COD (AA criterion) ⇒ (AO)/(CO) = (BO)/(DO), i.e. (AO)/(BO) = (CO)/(DO).

CBSE 2027 sample paper · 5 marks

Prove that if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.

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Given: In ∆ABC, a line DE ∥ BC intersects AB at D and AC at E. To prove: (AD)/(DB) = (AE)/(EC). Construction: Join BE and CD. Draw DM ⊥ AC and EN ⊥ AB. Proof: ar(∆ADE) = (1)/(2) × AD × EN and ar(∆BDE) = (1)/(2) × DB × EN, so (ar(ADE))/(ar(BDE)) = (AD)/(DB) ...(1) ar(∆ADE) = (1)/(2) × AE × DM and ar(∆DEC) = (1)/(2) × EC × DM, so (ar(ADE))/(ar(DEC)) = (AE)/(EC) ...(2) ∆BDE and ∆DEC are on the same base DE and between the same parallels BC and DE, so ar(∆BDE) = ar(∆DEC) ...(3) From (1), (2) and (3): (AD)/(DB) = (AE)/(EC). Hence the other two sides are divided in the same ratio.

CBSE 2027 sample paper · 5 marks

Prove that a line drawn parallel to one side of a triangle to intersect the other two sides in distinct points, divides the other two sides in the same ratio. Using the above theorem solve the following: PQRS is a trapezium with PQ || SR. X and Y are points on non-parallel sides PS and QR respectively such that XY || PQ. Show that (PX)/(XS) = (QY)/(YR).

Show answer
Theorem. Given: In ∆ABC, a line DE ∥ BC intersects AB at D and AC at E. To prove: (AD)/(DB) = (AE)/(EC). Construction: Join BE and CD. Draw EN ⊥ AB and DM ⊥ AC. Proof: ar(∆ADE) = (1)/(2) × AD × EN and ar(∆BDE) = (1)/(2) × DB × EN, so (ar(ADE))/(ar(BDE)) = (AD)/(DB) … (1) ar(∆ADE) = (1)/(2) × AE × DM and ar(∆DEC) = (1)/(2) × EC × DM, so (ar(ADE))/(ar(DEC)) = (AE)/(EC) … (2) ∆BDE and ∆DEC are on the same base DE and between the same parallels BC and DE, so ar(∆BDE) = ar(∆DEC) … (3) From (1), (2) and (3): (AD)/(DB) = (AE)/(EC). Application. Join PR, intersecting XY at M. Since XY ∥ PQ and …