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.
- 2 marks
Figure: parallelogram ABCD with AD produced beyond D to E; BE cuts CD at F. Triangles ABE and CFB are marked.
- 1 mark
In and : [opposite angles of parallelogram ABCD are equal, ]
- 1 mark
[alternate angles, since AE ∥ BC (AD ∥ BC) and BE is a transversal]
- 1 mark
Therefore [AA similarity criterion].
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
- Chapter: Triangles (notes, formulas, where marks are lost and every past question)
- Topic: Problems on Similar Triangles
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