Question

Question 8
The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order) respectively, and the angle between the first two sides is a right angle. Find its area.

Answer 1

Given ABCD is a quadrilateral having sides AB = 6cm, BC = 8cm, CD = 12cm, DA = 14cm.
Now join AC.

We have,$\Delta ABC$ right-angled at B.
Now, $A{C}^2=A{B}^2+B{C}^2$ [by Pythagoras theorem]
$=6^2+8^2=36+64=100$
$\Rightarrow$ AC = 10cm [taking positive square]
$\therefore$ Area of quadrilateral ABCD = $Areaof\Delta ABC+Areaof\Delta ACD$
$Now,Areaof\Delta ABC=\frac{1}{2}\times AB\times BC$
$\because areaoftriangle=\frac{1}{2}(base\times height)$
$=\frac{1}{2}\times 6\times 8=24c{m}^2$
In $\Delta ACD$, AC = a = 10cm, CD = b = 12 cm
And DA = c = 14cm
Now, semi-perimeter of $\Delta ACD$,
$s=\frac{a+b+c}{2}=\frac{10+12+14}{2}=\frac{36}{2}=18cm$
$Areaof\Delta ACD=\sqrt{s(s-a)(s-b)(s-c)}\left[byHero{m}^{\prime }sformula\right]$
$=\sqrt{18(18-10)(18-12)(18-14)}$
$=\sqrt{18\times 8\times 6\times 4}=\sqrt{(3{)}^2\times 2\times 4\times 2\times 3\times 2\times 4}$
$=3\times 4\times 2\sqrt{3\times 2}=24\sqrt{6}c{m}^2$

Area of quadrilateral ABCD = Area of $\Delta ABC$ + Area of $\Delta ACD$
$=24+24\sqrt{6}=24(1+\sqrt{6})c{m}^2$
Hence, the area of quadrilateral is $24(1+\sqrt{6})c{m}^2$