CBSE Class 10 Mathematics Standard sample paper 2023-24: question 22

Hints

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Step-by-step solution

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1. Let $AB=DC=a$ (opposite sides of a parallelogram). P divides AB in the ratio 2:3, so $AP=\frac{2}{5}a$.

   ½ mark

2. Q divides DC in the ratio 4:1, so $DQ=\frac{4}{5}a$ and $CQ=\frac{1}{5}a$.

   ½ mark

3. O is the point where the diagonal AC meets PQ. In $△APO$ and $△CQO$: $\angle PAO=\angle QCO$ [alternate angles, $AB\parallel DC$] and $\angle AOP=\angle COQ$ [vertically opposite]. So $△APO\sim △CQO$ [AA similarity].

   ½ mark

4. Hence $\frac{AO}{CO}=\frac{AP}{CQ}=\frac{\frac{2}{5}a}{\frac{1}{5}a}=2$. So $OC=\frac{1}{2}OA$. Proved.

   ½ mark

Answer: $△APO\sim △CQO$ gives $AO:OC=2:1$, so $OC=\frac{1}{2}OA$.

Revise the chapter

• Chapter: Triangles (notes, formulas, where marks are lost and every past question)
• Topic: Basic Proportionality Theorem (Thales Theorem)