What is the area of quadrilateral ABCD, (shown below) in which sides AB and BC are equal, sides AD and CD are 5 cm and 13 cm, respectively, and side AD is perpendicular to the diagonal AC?
Show answer & explanation
Correct answer: B
Explanation
The correct answer is option 2 i.e. 78 cm2.
AB = BC, AD = 5 cm, AB = 10 cm, and CD = 13 cm
∠DAC = 90°
In ΔADC, AC2 = CD2 - AD2 = 132 - 52 = 169 - 25 = 144
⇒ AC = √144 = 12 cm
Area (ΔADC) = (1/2) × AD × AC = (1/2) × 5 × 12 = 30 cm2
Since AB = BC = 10 cm, triangle ABC is an isosceles triangle.
AE = EC = AC/2 = 12/2 = 6 cm
So, BE2 = BC2 - EC2 = 102 - 62 = 100 - 36 = 64
⇒ BE = √64 = 8 cm
Area (ΔABC) = (1/2) × AC × BE = (1/2) × 12 × 8 = 48 cm2
Area of quadrilateral ABCD = Area (ΔADC) + Area (ΔABC) = 30 + 48 = 78 cm2
Therefore, the area of quadrilateral ABCD is 78 cm2.