Question

A string of length $12$cm is bent first into a square $PQRS$ and then into an isosceles triangle $PQT$ by keeping the side $PQ$ of the square as base then what is area of the square $PQRS$: area of the triangle $PQT=$

• A. $3:2$
• B. $\sqrt{2}:1$
• C. $2:3$
• D. $1:\sqrt{2}$

Answer 1

The correct option is B $\sqrt{2}:1$

Solution:

Length of string = 12

First bent into square, then perimeter $4\times side=12$

$\Rightarrow$ side = 3

Then bent into isosceles triangle, with base as square side

perimeter = $2x+3=12$

$\Rightarrow x=4.5$

Where x is the equal side of triangle

Now height of $A{h}^2=\sqrt{(4.5{)}^2-(3/2{)}^2}=18$

$\Rightarrow h=3\sqrt{2}$

$\frac{Areaofsquare}{Areaoftriangle}=\frac{3\times 3}{\frac{1}{2}\times 9\times \sqrt{2}}=\frac{2}{\sqrt{2}}$

$=\frac{\sqrt{2}}{1}=\sqrt{2}:1$