question_answer

The area of an isosceles triangle ABC with AB = AC and altitude AD = 3 cm is 12 square cm. What is its perimeter?

A) 18 cm                          

B) 16 cm

C) 14 cm                          

D) 12 cm

Correct Answer: A

Solution :

Area of the \[\Delta ABC=\frac{1}{2}\times b\times h\] \[\Rightarrow 12=\frac{1}{2}\times b\times 3\therefore b=\frac{12\times 2}{3}=8cm\] Here, BD = CD =\[\frac{b}{2}=\frac{8}{2}\] 4 cm In right angled AABD, by Pythagoras theorem, \[AB=\sqrt{B{{D}^{2}}+A{{D}^{2}}}\,\,\Rightarrow \,\,a=\sqrt{{{4}^{2}}+{{3}^{2}}}\] \[\sqrt{16+9}=\sqrt{25}\]= 5 cm Now, perimeter of isosceles triangle \[= 2a+b= 2 \times  5 + 8 =10 + 8 = 18 cm\]