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[2026 Boards Sample Paper] Prove that if a line is drawn parallel to - CBSE Class 10 Sample Paper for 2026 Boards - Maths Standard

part 2 - Question 33 - CBSE Class 10 Sample Paper for 2026 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10
part 3 - Question 33 - CBSE Class 10 Sample Paper for 2026 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10
part 4 - Question 33 - CBSE Class 10 Sample Paper for 2026 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10
part 5 - Question 33 - CBSE Class 10 Sample Paper for 2026 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10 part 6 - Question 33 - CBSE Class 10 Sample Paper for 2026 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10

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Question 33 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. Using the above theorem prove that a line through the point of intersection of the diagonals and parallel to the base of the trapezium divides the non parallel sides in the same ratio.Let’s prove the theorem first Theorem If a line is drawn parallel to one side of a triangle to intersect the other two side in distinct points, the other two sides are divided in the same ratio. Given: Δ ABC where DE ∥ BC To Prove: 𝐴𝐷/𝐷𝐵 = 𝐴𝐸/𝐸𝐶 Construction: Join BE and CD Draw DM ⊥ AC and EN ⊥ AB. Proof: Now, ar (ADE) = 1/2 × Base × Height = 1/2 × AD × EN ar (BDE) = 1/2 × Base × Height = 1/2 × DB × EN Divide (1) and (2) "ar (ADE)" /"ar (BDE)" = (1/2 " × AD × EN" )/(1/2 " × DB × EN " ) "ar (ADE)" /"ar (BDE)" = "AD" /"DB" ar (ADE) = 1/2 × Base × Height = 1/2 × AE × DM ar (DEC) = 1/2 × Base × Height = 1/2 × EC × DM Divide (3) and (4) "ar (ADE)" /"ar (DEC)" = (1/2 " × AE × DM" )/(1/2 " × EC × DM " ) "ar (ADE)" /"ar (DEC)" = "AE" /"EC" Now, ∆BDE and ∆DEC are on the same base DE and between the same parallel lines BC and DE. ∴ ar (BDE) = ar (DEC) Hence, "ar (ADE)" /"ar (BDE)" = "ar (ADE)" /"ar (DEC)" "AD" /"DB" = "AE" /"EC" Hence Proved Now, let’s look at our question Hence in ΔPQR, prove that a line ℓ intersects the sides PQ and PR of a ∆PQR at L and M respectively such that LM II QR. If PL = 5.7 cm, PQ = 15.2 cm and MR = 5.5 cm, then find the length of PM (in cm) Using theorem If a line is drawn parallel to one side of a triangle to intersect the other two side in distinct points, the other two sides are divided in the same ratio 𝑷𝑳/𝑳𝑸=𝑷𝑴/𝑴𝑹 (𝟓.𝟕)/((𝟏𝟓.𝟐 − 𝟓.𝟕))=𝑷𝑴/(𝟓.𝟓) 5.7/9.5=𝑃𝑀/5.5 (57/10)/(95/10)=𝑃𝑀/5.5 57/95=𝑃𝑀/5.5 57/95 × 5.5=𝑃𝑀 57/95 ×55/10=𝑃𝑀 𝑃𝑀=57/19 ×11/10 𝑃𝑀=3 ×11/10 𝑃𝑀=33/10 𝑷𝑴=𝟑.𝟑 𝒄𝒎

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 15 years. He provides courses for Maths, Science and Computer Science at Teachoo