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.

• A. 114 cm${}^2$
• B. 60 cm${}^2$
• C. 54 cm${}^2$
• D. 114 cm${}^2$

Answer 1

The correct option is A 114 cm${}^2$

Given that: AB = 17cm

AD = 9cm

CD = 12cm

AC = 15cm

and ∠ACB = 90°

In right angled triangle ACB,

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

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

Now, Area of right-angled triangle ACB:

$=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times 15\times 8=60{\text{cm}}^2$

The area of triangle ACD using Heron's Formula as :

$\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$

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.

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

$\therefore \text{Area of}△ACD=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{18(18-15)(18-12)(18-9)}=\sqrt{18\times 3\times 6\times 9}=\sqrt{18\times 18\times 9}=18\times 3=54{\text{cm}}^2$

$\therefore$ Area of quadrilateral ABCD = (60 + 54) cm${}^2$

= 114 cm${}^2$