Question

ABCD is a rhombus AC= 12cm BD = 20 cm, then the area of rhombus is
(A) 240
(B) 120
(C) 60
(D) 180

Answer 1

Hint: Assume that the diagonals AC and BD are intersecting at point O. Use the property that the diagonals of a rhombus bisect each other at right angles. The length of OA = The length of OC = 6 cm. Similarly, the length of OD = The length of OB = 10 cm. Use the formula, Area = \[\dfrac{1}{2}\times Base\times Perpendicular\] and calculate the area of \[\Delta BOA\] . Similarly, calculate the area of \[\Delta COB\] , \[\Delta COD\] , and \[\Delta DOA\] . At last, calculate the area of rhombus ABCD.

Complete step by step answer:
According to the question, we are given that ABCD is a rhombus where AC and BD are its diagonals.
The length of the diagonal AC = 12 cm …………………………………(1)
The length of the diagonal BD = 20 cm ………………………………….(2)
We know the property that the diagonals of a rhombus bisect each other at right angles ………………………………………(3)
Let us assume that the diagonals AC and BD of rhombus ABCD is intersecting at point O.

Using the property shown in equation (3), we have
The length of OA = The length of OC = 6 cm ……………………………………….(4)
The length of OB = The length of OD = 10 cm ………………………………………….(5)
Now, in \[\Delta BOA\] , we have
The length of base OA = 6 cm ………………………………………..(6)
The length of perpendicular OB = 10 cm ……………………………………….(7)
We know the formula for the area of triangle, Area = \[\dfrac{1}{2}\times Base\times Perpendicular\] ………………………………….(8)
Now, from equation (6), equation (7), and equation (8), we get
The area of \[\Delta BOA\] = \[\dfrac{1}{2}\times 6cm\times 10cm=30c{{m}^{2}}\] ………………………………………….(9)
Similarly, in \[\Delta COB\] , we have
Base = OC = 6 cm …………………………..(10)
Perpendicular = OB = 10 cm ……………………………………..(11)
Now, from equation (8), equation (10), and equation (11), we get
The area of \[\Delta COB\] = \[\dfrac{1}{2}\times 6cm\times 10cm=30c{{m}^{2}}\] ……………………………………….(12)
Similarly, in \[\Delta COD\] , we have
Base = OC = 6 cm …………………………..(13)
Perpendicular = OD = 10 cm ……………………………………..(14)
Now, from equation (8), equation (13), and equation (14), we get
The area of \[\Delta COD\] = \[\dfrac{1}{2}\times 6cm\times 10cm=30c{{m}^{2}}\] ……………………………………….(15)
Similarly, in \[\Delta DOA\] , we have
Base = OA = 6 cm …………………………..(16)
Perpendicular = OD = 10 cm ……………………………………..(17)
Now, from equation (8), equation (16), and equation (17), we get
The area of \[\Delta DOA\] = \[\dfrac{1}{2}\times 6cm\times 10cm=30c{{m}^{2}}\] ……………………………………….(18)
Area of rhombus ABCD = Area of \[\Delta BOA\] + Area of \[\Delta COB\] + Area of \[\Delta COD\] + Area of \[\Delta DOA\] ………………………………………………..(19)
Now, from equation (9), equation (12), equation (15), equation (18), and equation (19), we get
Area of rhombus ABCD = \[30c{{m}^{2}}+30c{{m}^{2}}+30c{{m}^{2}}+30c{{m}^{2}}=120c{{m}^{2}}\] .
Therefore, the area of the rhombus ABCD is \[120c{{m}^{2}}\] .
Hence, the correct option is (B).

Note:
The best way to approach this type of question where we are asked to calculate the area of rhombus and the lengths of the diagonals are given is to use the formula of area of rhombus. The area of rhombus = \[\dfrac{1}{2}\times \] Product of its diagonals.