find the area of a rombuz having each side is 13cm and one of whose diagonals is 24cm
Answers
Answer: 120 cm²
Step-by-step explanation:
Let us name the rhombus ABCD. (Refer to the image.)
All sides of a rhombus are equal.
Side of the rhombus = AD = 13 cm
Diagonals of a rhombus perpendicularly bisect each other. Let the point of intersection of the diagonals be = O. (Refer to the image.)
Let us assume that the diagonal AC = 24 cm. (Given in the question)
AO = OC (∵ Diagonals bisect each other.)
AO = OC = 1/2 of AC
∴ AO = 12 cm
ΔAOD is a right-angled triangle. (∵ Diagonals are perpendicular to each other.)
According to pythagoras theorem,
AO² + OD² = AD²
(12cm)² + OD² = (13cm)²
144 cm² + OD² = 169 cm²
OD² = 169 cm² - 144 cm²
OD² = 25 cm²
OD = √25 cm²
∴ OD = 5 cm
Diagonal BD = 2 × OD (∵ Diagonals bisect each other.)
∴ BD = 2 × 5 cm = 10 cm
Area of rhobmus = 1/2 × Product of the diagonals
Area of ABCD = 1/2 × 24 cm × 10 cm
∴ Area of ABCD = 120 cm²
