Question 138

PQRS is a rectangle. T is a point on PQ such that RTQ is an isosceles triangle and PT = 5 QT.If the area of triangle RTQ is $$12 \sqrt{3}$$ sq.cm, then the area of the rectangle PQRS is:

Solution

PQRS is the rectangle, having length $$6x$$ cm and breadth = $$x$$ cm

Area of $$\triangle$$ RQT = $$\frac{1}{2}\times x\times x=12\sqrt3$$

=> $$x^2=24\sqrt3$$ --------------(i)

Now, area of rectangle PQRS = $$6x\times x=6x^2$$

= $$6\times24\sqrt3=144\sqrt3$$ $$cm^2$$

=> Ans - (C)

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