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(a) The given figure shows a circle with centre O and radius $4 \text{ cm}$ circumscribed by $\triangle ABC$. BC touches the circle at D such that $BD = 6 \text{ cm}, DC = 10 \text{ cm}$. Find the length of AE.
OR
(b) PA and PB are tangents drawn to a circle with centre O. If $\angle AOB = 120^\circ$ and $OA = 10 \text{ cm}$, then
(i) Find $\angle OPA$.
(ii) Find the perimeter of $\triangle OAP$.
(iii) Find the length of chord AB.
OR
(b) PA and PB are tangents drawn to a circle with centre O. If $\angle AOB = 120^\circ$ and $OA = 10 \text{ cm}$, then
(i) Find $\angle OPA$.
(ii) Find the perimeter of $\triangle OAP$.
(iii) Find the length of chord AB.
Show SolutionHide Solution↓
(a) Let $AE = x \Rightarrow AF = x$ and $CE = 10 \text{ cm}, BF = 6 \text{ cm}$ (1 mark)
$s = \frac{16 + 10 + x + 6 + x}{2} = 16 + x$ ($\frac{1}{2}$ mark)
$\therefore \text{Area of } \triangle ABC = \sqrt{(16 + x)(x)(6)(10)}$ (1 mark)
Also, area of $\triangle ABC = \frac{1}{2} [16 \times 4 + (10 + x)4 + (6 + x)4]$ (1 mark)
Equating (i) & (ii), we get $x = \frac{64}{11}$ or $5.8$ ($1\frac{1}{2}$ marks)
Length of $AE = \frac{64}{11} \text{ cm or } 5.8 \text{ cm}$
OR
(b) (i) $\angle OPA = 30^\circ$ (1 mark)
(ii) In $\triangle OAP, \sin 30^\circ = \frac{10}{OP} \Rightarrow OP = 20 \text{ cm}$ (1 mark)
$\tan 30^\circ = \frac{10}{PA} \Rightarrow PA = 10\sqrt{3} \text{ cm}$ (1 mark)
$\therefore \text{Perimeter of } \triangle OPA = (30 + 10\sqrt{3}) \text{ cm}$ (1 mark)
(iii) $PA = PB$ and $\angle APB = 60^\circ \Rightarrow \triangle APB$ is an equilateral triangle ($\frac{1}{2}$ mark)
$\therefore PA = AB = 10\sqrt{3} \text{ cm}$ ($\frac{1}{2}$ mark)
$s = \frac{16 + 10 + x + 6 + x}{2} = 16 + x$ ($\frac{1}{2}$ mark)
$\therefore \text{Area of } \triangle ABC = \sqrt{(16 + x)(x)(6)(10)}$ (1 mark)
Also, area of $\triangle ABC = \frac{1}{2} [16 \times 4 + (10 + x)4 + (6 + x)4]$ (1 mark)
Equating (i) & (ii), we get $x = \frac{64}{11}$ or $5.8$ ($1\frac{1}{2}$ marks)
Length of $AE = \frac{64}{11} \text{ cm or } 5.8 \text{ cm}$
OR
(b) (i) $\angle OPA = 30^\circ$ (1 mark)
(ii) In $\triangle OAP, \sin 30^\circ = \frac{10}{OP} \Rightarrow OP = 20 \text{ cm}$ (1 mark)
$\tan 30^\circ = \frac{10}{PA} \Rightarrow PA = 10\sqrt{3} \text{ cm}$ (1 mark)
$\therefore \text{Perimeter of } \triangle OPA = (30 + 10\sqrt{3}) \text{ cm}$ (1 mark)
(iii) $PA = PB$ and $\angle APB = 60^\circ \Rightarrow \triangle APB$ is an equilateral triangle ($\frac{1}{2}$ mark)
$\therefore PA = AB = 10\sqrt{3} \text{ cm}$ ($\frac{1}{2}$ mark)