$△ A B C$ is right-angled at $A . A D$ is perpendicular to $B C, A B = 5 c m, B C = 13 c m$ and $A C = 12 c m$ . Find the area of $△ A B C$ . Also, find the length of $A D$ .

30 c m^{2}; 4.62 c m

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20 c m^{2}; 4.62 c m

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40 c m^{2}; 4.62 c m

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None of these

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Solution

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A

$30 c m^{2}; 4.62 c m$ .

Area of $A B C = \frac{1}{2} \times b a s e \times h e i g h t$ where AB is the base & height = AC

$∴ A r e a = \frac{1}{2} \times 5 \times 12 = 30 s q c m .$

Consider $△ A B D, sin B = \frac{A D}{A B}$

Consider $△ A B C, sin B = \frac{A C}{B C}$

$∴ \frac{A D}{A B} = \frac{A C}{B C} \Rightarrow A D = \frac{5 \times 12}{13} = 4.62 c m$

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