Question

In the given isosceles right angled triangle UVW, a square PQRS is inscribed as shown in the figure. If PV:VS=2:1, what is the ratio of areas of the square to the outer triangle UVW?

• A. 3:5
• B. 4:7
• C. 2:5
• D. 3:7

Answer 1

The correct option is C 2:5

Using variables

In the given figure, we draw QT || VW.

$\Delta$PTQ and $\Delta$PVS are congruent (A,A,A and side)

Hence PT=Y, QT=X=UT (also since $\Delta$UTQ and $\Delta$UVW are similar).

Thus UV=UW=2X+Y. Area of square: X${}^2$+Y${}^2$; Area of ΔUVW= $\frac{1}{2}$ $\times$ (2X+Y)${}^2$.

Given X=2Y, (X${}^2$+Y${}^2$): ( $\frac{1}{2}$ $\times$ (2X+Y)${}^2$) = 5Y${}^2$: ( 25Y${}^2$$\times$$\frac{1}{2}$ ); = $\frac{2}{5}$.

Using numbers,You can solve the problem faster as follows:

PV=2, SV=1 => PS= √5 => Area of square =5

PTQ congruent to PVS

PT=1 and QT=2

UT=TQ (45-45-90) $\Rightarrow$ UT= 2

Now UV= 5 $\Rightarrow$ WV=5 (45-45-90)

UW=5$\sqrt{2}$

$\left(\frac{Areaofsquare}{Areaoftriangle}\right)$ = $\left(\frac{5}{\left(\frac{25}{2}\right)}\right)$