Question

Find the ratio of the shaded area to the area of the quadrilateral ABCD.

• A. $2+\sqrt{6}:\sqrt{6}$
• B. $3:2+\sqrt{6}$
• C. $\sqrt{6}:2+\sqrt{6}$
• D. $\sqrt{6}:4+\sqrt{6}$

Answer 1

The correct option is C $\sqrt{6}:2+\sqrt{6}$

$Hero{n}^{\prime }sFormula:$

Area of a triangle $=\Delta =\sqrt{s(s-a)(s-b)(s-c)}$

where $a,b,c$ are sides and $s$ is the semi perimeter. $[s=\frac{a+b+c}{2}]$

For $△ABC$,

$a=28m,b=24m,c=20ms=\frac{28+24+20}{2}=36m$

Area of $△ABC=\Delta ABC=\sqrt{36(36-28)(36-24)(36-20)}=\sqrt{36\times 8\times 12\times 16}=96\sqrt{6}{m}^2$

For $△ACG$,

$a=40m,b=32m,c=24ms=\frac{40+32+24}{2}=48m$

Area of $△ACG=\Delta ACG=\sqrt{48(48-40)(48-32)(48-24)}=\sqrt{48\times 8\times 16\times 24}=16\times 24=384m^2$

But $\Delta ACD=\frac{1}{2}\Delta ACG=\frac{1}{2}\times 384=192m^2$ ($\because CD$ is the median of $△ACG$ )

$\Rightarrow$ Area of $qlABCD=\Delta ACD+\Delta ABC=192+96\sqrt{6}{m}^2$

So, the required ratio $=\frac{Areaof△ABC}{AreaofqlABCD}=\frac{96\sqrt{6}}{192+96\sqrt{6}}=\frac{\sqrt{6}}{2+\sqrt{6}}.\left[C\right]$