• The area of an isosceles $$\Large \triangle ABC$$ with AB = AC and altitude AD = 3 cm is 12 sq cm. What is its perimeter?
 A) 18 cm B) 16 cm C) 14 cm D) 12 cm

 A) 18 cm

Let AB = CA = a cm and base = b cm

Now, area of the $$\Large \triangle ABC$$= $$\Large \frac{1}{2}\times b\times h$$

=> 12= $$\Large \frac{1}{2}\times b\times 3$$

b= $$\Large \frac{12\times 2}{3}$$=8cm

Here, BD=CD=$$\Large \frac{b}{2}=\frac{8}{2}$$=4cm

In right angled $$\Large \triangle ABD$$ , by Pythagoras theorem,

AB=$$\Large \sqrt{BD^{2}+AD^{2}}$$ => a=$$\Large \sqrt{4^{2}+3^{2}}$$

=$$\Large \sqrt{16+9}=\sqrt{25}$$=5cm

Now, perimeter of an isosceles triangle

= 2a + b = $$\Large 2\times 5+8$$

= 10 + 8 = 18 cm

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