CBSE Class 10 Mathematics Standard 2023: question 35

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. The horse is tied at a corner of the square and the rope (5 m) is shorter than the side (15 m). The corner angle is $90^{\circ }$, so the grazing region is a quadrant (sector of angle $90^{\circ }$) of radius $r=5$ m.

   Grazing area $=\frac{90^{\circ }}{360^{\circ }}\pi r^2=\frac{1}{4}\times 3.14\times (5{)}^2$

   1 mark

2. $=\frac{1}{4}\times 3.14\times 25=19.625{\text{m}}^2$

   1 mark

3. With a 10 m rope (still less than 15 m), the region is again a quadrant, now of radius 10 m.

   New grazing area $=\frac{1}{4}\times 3.14\times (10{)}^2$

   1 mark

4. $=\frac{1}{4}\times 314=78.5{\text{m}}^2$

   1 mark

5. Increase in grazing area $=78.5-19.625=58.875{\text{m}}^2$

   1 mark

Answer: Grazing area = 19.625 m²; increase in grazing area = 58.875 m²

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