CBSE Class 10 Mathematics Basic 2023: question 33(a)

Hints

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

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. Area of sector AOB $=\frac{\theta }{360^{\circ }}\times \pi r^2=\frac{90}{360}\times \frac{22}{7}\times 7\times 7$

   1 mark

2. $=\frac{77}{2}$ cm² $=38.5$ cm²

   ½ mark

3. Area of right $△AOB=\frac{1}{2}\times OA\times OB=\frac{1}{2}\times 7\times 7=\frac{49}{2}$ cm²

   1 mark

4. Area of shaded segment = sector − triangle $=\frac{77}{2}-\frac{49}{2}=\frac{28}{2}=14$ cm²

   1 mark

5. Length of minor arc AB $=\frac{\theta }{360^{\circ }}\times 2\pi r=\frac{90}{360}\times 2\times \frac{22}{7}\times 7=11$ cm

   1½ marks

Answer: Shaded area = 14 cm²; minor arc AB = 11 cm

Revise the chapter

• Chapter: Areas Related to Circles (notes, formulas, where marks are lost and every past question)
• Topic: Area of a Sector of a Circle