Find the area of a rhombus having each side equal to 13cm and one of the diagonals equal to 24cm.
Answers
Use the Pythagoras theorem by dividing the rhombus into two triangles. After using it find the value of the other diagonal and then put the formula area of the rhombus =1/2*product of diagonals.
Hope this helps you buddy!!
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
