Question

The sides of a quadrilateral ABCD are $6cm,8cm,12cm$ and $14cm$ (taken in order), respectively. The angle between the first two sides is a right angle. Its area is $24(\sqrt{m}+1)c{m}^2$. The value of $m$ is:

Answer 1

The sides of a quadrilateral ABCD are $6cm,8cm,12cm$ and $14cm$ (taken in order).

i.e. $\angle ABC={90}^o$

Now, let us join the points A & C.

So, we get two triangles namely $△ABC$ (a right angled triangle) and $△ACD.$

Applying Pythagoras theorem in $△ABC$, we get

$AC=\sqrt{{AB}^2+{BC}^2}=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10cm$

So, the area of $△ABC=\frac{1}{2}\times base\times height$

$=\frac{1}{2}\times AB\times BC$

$=\frac{1}{2}\times 6\times 8=24$

Now, in $△ACD$, we have

$AC=10cm,CD=12cm,AD=14cm.$

According to Heron's formula the area of triangle $\left(A\right)=\sqrt{\left[s\right(s-a\left)\right(s-b\left)\right(s-c\left)\right]}$

where, $2s=(a+b+c).$

Here, $a=10cm,b=12cm,c=14cm$

$s=\frac{(10+12+14)}{2}=\frac{36}{2}=18$

Area of $△ACD=\sqrt{[18\times (18-10)(18-12)(18-14)]}$

$=\sqrt{(18\times 8\times 6\times 4)}$

$=\sqrt{(2\times 3\times 3\times 2\times 2\times 2\times 2\times 3\times 2\times 2)}$

$=\sqrt{\left[\right(2\times 2\times 2\times 2\times 2\times 2\times 3\times 3)\times 2\times 3]}$

$=2\times 2\times 2\times 3\times \sqrt{6}$

$=24\sqrt{6}$

So, total area of quadrilateral ABCD $=△ABC+△ACD$

$=24+24\sqrt{6}$

$=24(\sqrt{6}+1)$