The perimeter of an isosceles triangle is 64 cm and each of the equal sides is 5/6 times the base. What is the area (in cm2) of the triangle?

Let the length of base = $$6x$$ cm

=> Length of each equal side = $$\frac{5}{6}\times6x=5x$$ cm

=> Perimeter = $$6x+5x+5x=16x=64$$

=> $$x=\frac{64}{16}=4$$

=> Base = $$b=24$$ cm and side = $$a=20$$ cm

Now, height of an isosceles triangle = $$h=\sqrt{(a)^2-(\frac{b}{2})^2}$$

=> $$h=\sqrt{(20)^2-(12)^2}$$

=> $$h=\sqrt{400-144}=\sqrt{256}=16$$ cm

$$\therefore$$ Area of isosceles triangle = $$\frac{1}{2}\times(b)\times(h)$$

= $$\frac{1}{2}\times24\times16=192$$ $$cm^2$$

=> Ans - (B)