Question

The perimeter of a rhombus is 100 cm and one of its diagonals measures 40 cm, find its area.

Answer 1

Answer:

Area of rhombus will be 600 cm^2.

Step-by-step explanation:

Perimeter of a rhomus = 4 * side (which same as perimeter of a square)

100 = 4* side

Hence, each side of rhombus = 25cm

Now draw the rhombus. Let it be ABCD. Join the diagonals AC and BD. And label the intersection point as O. Take any one diagonal to be equal to 40cm (say BD).

Now rhombus has one property that its diagonals bisect each other at right angles.

Therefore, AO = OC and OB = OD = 20cm

Hence angle AOB = angle BOC = angle COD = angle AOD = 90 (making each triangle a right triangle)

Apply pythagorus theorem in triangle AOB where OB = 20cm and AB = 25 cm and angle O is 90

AB ^2 = OB^2 + AO^2

This will give you AO = 15cm = OC

Now you have both the diagonals AC = 30 cm (AO + OC) and BD = 40cm

Area of rhombus = (1/2) * (product of diagonals) = (1/2) * (AC * BD)

Solve this and you will get Area = 600 cm ^2