Question

Find the area of the quadrilateral ABCD , in which AB = 7 cm, BC = 6 cm, CD = 12 cm, DA = 15 cm and AC = 9 cm.

• A. 54.98 sq. cm
• B. 64.98 sq. cm
• C. 74.98 sq. cm
• D. 84.98 sq. cm

Answer 1

The correct option is C 74.98 sq. cm

The diagonal AC divides the quadrilateral ABCD into two triangles ABC and ACD.

Area of quad ABCD = Area of $△ABC$ + Area of $△ACD$
The area of a triangle with sides a, b and c can be calculated using Heron's formula, given by:
$Area=\sqrt{s(s-a)(s-b)(s-c)}$
where, s is the semi-perimeter.

For $△ABC$, we have,
Semi-perimeter = $\frac{6+7+9}{2}=11$ cm
So, $ar\left(△ABC\right)=\sqrt{11(11-6)(11-7)(11-9)}$
$=\sqrt{11\times 5\times 4\times 2}$
$=\sqrt{440}$ sq. cm
$=20.98$ sq. cm

For $△ACD$, we have,
Semi-perimeter = $\frac{9+12+15}{2}=18$ cm
So, $ar\left(△ACD\right)=\sqrt{18(18-9)(18-12)(18-15)}$
$=\sqrt{18\times 9\times 6\times 3}$
$=54$ sq. cm

Hence,
Area of quad ABCD = 20.98 sq. cm + 54 sq. cm = 74.98 sq. cm