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Flower beds look beautiful growing in gardens. One such circular garden of radius \(r \mathrm{~m}\), has two segments with flowers. One segment which subtends an angle of 90° at the centre is full

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Question

Flower beds look beautiful growing in gardens. One such circular garden of radius \(r \mathrm{~m}\), has two segments with flowers. One segment which subtends an angle of 90° at the centre is full of red roses, while the other segment with central angle 60° is full of yellow coloured flowers. It is given that the combined area of the two segments of flowers is \(256 \frac{2}{3} \mathrm{~m}^{2}\).

Based on the above information, answer the following questions.

  1. Write an equation representing the total area of the two segments in terms of r.
  2. Find the value of r.
  3. Find the area of the segment with red roses.
  4. Find the area of the segment with yellow flowers.
Case Study
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Solution

i. Equation for total area of the two segments

Area of segment with red roses 90°: \[\left(\frac{\pi r^2}{4} - \frac{1}{2}r^2 \right)\ \text{m}^2\]

Area of segment with yellow flowers 60°: \[\left( \frac{\pi r^2}{6} - \frac{\sqrt{3}}{4}r^2 \right)\ \text{m}^2\]

Total area: \[\left[ \left( \frac{\pi r^2}{4} - \frac{1}{2}r^2 \right) + \left( \frac{\pi r^2}{6} - \frac{\sqrt{3}}{4}r^2 \right) \right] = 256\frac{2}{3}\ \text{m}^2\]

ii. Value of r

Combine terms: \[\left( \frac{5\pi}{12} - \frac{2+\sqrt{3}}{4} \right) r^2 = \frac{770}{3}\]

Using \(\pi = \frac{22}{7}\) and \(\sqrt{3} \approx 1.732\): 

\[ \frac{5 \times \frac{22}{7} - 6 - 3\sqrt{3}}{12}, r^2 = \frac{770}{3} \] 

\[ \frac{\frac{110}{7} - 11.196}{12}, r^2 = \frac{770}{3} \] 

\[ \frac{110 - 78.372}{84}, r^2 = \frac{770}{3} \] 

\[ \frac{31.628}{84}, r^2 = \frac{770}{3} \] 

\[ r^2 = \frac{770}{3} \times \frac{84}{31.628} \approx 681.674\] 

\[r \approx 26.1\ \text{m}\]

iii. Area of the segment with red roses \[ = \frac{\pi r^2}{4} - \frac{1}{2}r^2 \]

\[= \frac{1}{4} \times \frac{22}{7} \times (26.1)^2 - \frac{1}{2} \times (26.1)^2\] 

\[= 535.236 - 340.605\]

\[= 194.631\ \text{m}^2\]

iv. Area of the segment with yellow flowers \[= \frac{\pi r^2}{6} - \frac{\sqrt{3}}{4}r^2\] 

\[= \frac{1}{6} \times \frac{22}{7} \times (26.1)^2 - \frac{1.732}{4} \times (26.1)^2\] 

\[= 356.824 - 294.964\]

\[= 61.86\ \text{m}^2\]

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Chapter 20: Additional Questions - Area of Circle, Sector and Segment [Page 1042]

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R.S. Aggarwal Mathematics [English] Class 10
Chapter 20 Additional Questions
Area of Circle, Sector and Segment | Q 5. | Page 1042
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