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MathsCBSE Class 10 · Standard · Board exam 2024, set 30/4/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2024: question 33(a)

Set in the CBSE Class 10 board exam of 2024, paper 30/4/1, section D, question 33(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(a) · 5 marks · Long answerTriangles

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ABE CFB.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 30/4/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Two pairs of equal angles are enough for similarity.

Hint 2

Use a property of opposite angles of a parallelogram, and AD ∥ BC with transversal BE.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

  1. Figure: parallelogram ABCD with AD produced beyond D to E; BE cuts CD at F. Triangles ABE and CFB are marked.

    2 marks
  2. In and : [opposite angles of parallelogram ABCD are equal, ]

    1 mark
  3. [alternate angles, since AE ∥ BC (AD ∥ BC) and BE is a transversal]

    1 mark
  4. Therefore [AA similarity criterion].

    1 mark

Answer: △ABE ∼ △CFB by AA similarity

Where the marks are

  • Correct figure (2 marks)
  • ∠EAB = ∠BCF, opposite angles of parallelogram (1 mark)
  • ∠AEB = ∠CBF, alternate angles as AE ∥ BC (1 mark)
  • Concludes similarity by AA (1 mark)

Revise the chapter

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.