Question

$ABCD$ is a rectangle and lines $DX,DY$ and $XY$ are drawn as shown. Area of $\Delta AXD$ is $5$, Area of $\Delta BXY$ is $4$ and area of $\Delta CYD$ is $3$. If the area of $\Delta DXY$ can be expressed as $\sqrt{x}$, where $x\in$ N then $x$ is equal to:

• A. $72$
• B. $75$
• C. $84$
• D. $96$

Answer 1

The correct option is C $84$

Let $AB=CD=l$ and $AD=b$

Let $AX=m$ and $CY=n$

$\therefore BX=l-m$ and $BY=b-n$

Area of rectangle $ABCD$ $=l\times b$ ...(1)

Area of triangle = $\frac{1}{2}\times$ base $\times$ height

$A\left(\Delta AXD\right)=5$

$\Rightarrow \frac{1}{2}AX\times AD$

$\therefore m\cdot b=10$ ...(2)

Similarly,

$A\left(\Delta CYD\right)=\frac{1}{2}l\cdot n=3\Rightarrow l\cdot n=6$ and

$A\left(\Delta BXY\right)=\frac{1}{2}(l-m)(b-n)=4\Rightarrow (l-m)(b-n)=8$ ...(3)

Now,

$A\left(\square ABCD\right)$ = $A\left(\Delta AXD\right)+A\left(\Delta BXY\right)+A\left(\Delta CYD\right)+A\left(\Delta DXY\right)$

$A\left(\Delta DXY\right)=A\left(\square ABCD\right)-5-4-3=l\cdot b-12$ ...(4)

$A\left(\Delta DXY\right)=l\cdot b-\frac{1}{2}[m\cdot b+l\cdot n+(l-m\left)\right(b-n\left)\right]$ ...(5)

$\therefore l\cdot b-12$ = $l\cdot b-\frac{1}{2}[m\cdot b+l\cdot n+(l-m\left)\right(b-n\left)\right]$ ...from (4) & (5)

$\therefore 24=l\cdot b+m\cdot n$

$\therefore 24=l\cdot b+(\frac{10}{b}\cdot \frac{6}{l})$

On solving, we get

$l\cdot b=12+\sqrt{84}$

Hence, $A\left(\Delta DXY\right)=\sqrt{84}$