Question

Find the area of a rhombus whose perimeter is 40 cm and one of the diagonal is 16 cm

Answer 1

one side can be calculated by dividing 40 by 4 i.e 10 . and after that we know that diagonals bisects each other at 90° , so there froms a right angled triangle in hypotenuse will be the side and one side be 16/2 i.e 8 and after that third side will be calculated by Pythagoras theorem and after that area can be calculated by the formula 1/2 × base ×height

Answer 2

We have
Perimeter of Rhombus = 40cm
One diagonal = 16 cm
Therefore
4* side = 40
Side = 10 cm
So by heron's formula
S = 16 + 10 + 10 /2 = 18
√ 18 ( 18 - 16 ) ( 18 - 10 ) ( 18 - 10 )
= √ 18 * 2 * 8 * 8
= 8 √ 2 * 3 * 3 * 2
= 8 * 2 * 3
= 48 sq cm
Area of one triangle = 48
Area of two triangles = 48 * 2 = 96 sq cm
Area of rhombus = 96 sq cm