Question

Given a triangle whose vertices are at $(0,0),(4,4)$ and $(10,0)$. A square is drawn in it such that its base is on the x - axis and its two corners are on the 2 sides of the triangle. The area of square is equal to

• A. $\frac{400}{49}$
• B. $\frac{400}{25}$
• C. $\frac{625}{16}$
• D. $\frac{625}{49}$

Answer 1

The correct option is A $\frac{400}{49}$

Let two of vertices be $C(a,0)$ and $D(b,0)$

Side of the square $CD=b-a$

$CF=CD=b-a$

Co-ordinate of $F=(a,b-a)$

$DE=CD=b-a$

Co-ordinate of $E=(b,b-a)$

Equation of $OB$ is

$y-4=\frac{4-0}{4-0}(x-4)$

$\Rightarrow y-4=x-4$ or $y=x$

Since $F$ lies on $OB$ $\therefore$ it must satisfy the line $y=x$

$\Rightarrow b-a=a$ or $b=2a$ ......$\left(1\right)$

Equation of $AB$ is $y-4=\frac{4-0}{4-10}(x-4)$

$y-4=\frac{-2}{3}(x-4)$

$\Rightarrow -3y+12=2x-8$

or $2x+3y=20$

Since $E$ lies on $AB$ $\therefore$ it must satisfy the equation $3(b-a)+2b=20$

From eqn$\left(1\right)$ we have $3(2a-a)+2\times 2a=20$

$\Rightarrow 3a+4a=20$ or $a=\frac{20}{7}$

Since $b=2a$ we have $b=2\times \frac{20}{7}=\frac{40}{7}$

Side of square$=b-a=\frac{40}{7}-\frac{20}{7}=\frac{20}{7}$

Area of square$=s^2=\frac{20}{7}\times \frac{20}{7}=\frac{400}{49}$