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Question
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The most accurate way to determine the age of a tree is to count the annual rings of wood growth. One such trunk has been shown.
To make an identification mark, the forest department has painted segment ACBA, as shown in the given figure. If the chord AB makes an angle of 90° at the centre O and the radius of the trunk is 21 cm, find the: |
- length of the chord AB.
- area of the \(\triangle OAB\).
- area of the segment ACBA.
- perimeter of the sector OACBO.
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Solution
i. Length of the chord AB
In right \(\triangle AOB\), \(OA = OB = 21\) cm and \(\angle AOB = 90^\circ\).
\[AB^2 = OA^2 + OB^2\]
\[= 21^2 + 21^2\]
\[= 441 + 441\]
\[= 882\]
\[AB = \sqrt{882}\]
\[= 21\sqrt{2}\ \text{cm} \]
ii. Area of \(\triangle OAB\)
\[\text{Area} = \frac{1}{2} \times OA \times OB\]
\[= \frac{1}{2} \times 21 \times 21\]
\[= 220.5\ \text{cm}^2\]
iii. Area of the segment ACBA
Area of sector OACBO:
\[\frac{90}{360} \times \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 21 \times 21\]
\[= 346.5\ \text{cm}^2 \]
\[\text{Area of segment} = 346.5 - 220.5\]
\[= 126\ \text{cm}^2 \]
iv. Perimeter of the sector OACBO \[ = OA + OB + \text{arc } ACB \]
\[\text{Arc length} = \frac{1}{4} \times 2\pi r\]
\[= \frac{1}{4} \times 2 \times \frac{22}{7} \times 21\]
\[= 33\ \text{cm} \]
\[\text{Perimeter} = 21 + 21 + 33\]
\[= 75\ \text{cm} \]
