Question

Find the perimeter and area of the quadrilateral ABCD in which AB = 17 cm, AD = 9 cm, CD = 12 cm, $\angle$ ACB = ${90}^o$ and AC = 15 cm.

Answer 1

Given that: AB = 17cm

AD = 9cm

CD = 12cm

AC = 15cm

and ∠ACB = 90°

In right angled triangle ACB,

$(BC{)}^2=(17{)}^2-(15{)}^2=289-225=64BC=8cm$

We have to find the area of quadrilateral ABCD.

Area of Quadrilateral ABCD = Area of ΔACB + Area of ΔACD ....... (1)

Now, Area of right-angled triangle ACB

$=\frac{1}{2}\times base\times height=\frac{1}{2}\times 15\times 8=60c{m}^2$ ------(2)

We will calculate the area of triangle ACD using Heron's Formula as :

Area of $△ACD=\sqrt{s(s-a)(s-b)(s-c)}$

Where a, b, c are the sides of triangle and S is the semiperimeter of the triangle and is given by

$s=\frac{a+b+c}{2}$

Here, a = 15cm, b = 12cm, c = 9cm.

$s=\frac{15+12+19}{2}=\frac{36}{2}=18cm$

$Areaof△ACD=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{18(18-15)(18-12)(18-9)}=\sqrt{18\times 3\times 6\times 9}=\sqrt{18\times 18\times 9}=18\times 3=54c{m}^2-----\left(3\right)$

∴ From (1),

Area of quadrilateral ABCD = (60 + 54) $c{m}^2$

=114 $c{m}^2$