Question

area of rhombus is 840 cm². if the perimeter of the rhombus is 148 cm, then find the sum of the length of its two diagonals​

Answer 1

Solution

Refer to attachment for figure

One of diagonal AC= $d_1$

Another diagonal DB = $d_2$

Side of the rhombus = a cm

Area of the rhombus = 840 cm²

$\implies \dfrac{d_1d_2 }{2} = 840$

$\implies d_1d_2 = 840 \times 2$

$\implies d_1d_2 = 1680$

Perimeter of the rhombus = 148 cm

i.e 4a = 148

⇒ a = 148/4

⇒ a = 37 cm

From figure,

Consider ΔBOC

In rhombus diagonals bisect each other perpendicularly

Therefore,

• ∠BOC = 90°
• OC = AC/2 = d1/2
• OB = DB/2 = d2/2
• BC = a

So, ΔBOC is a Right angled triangle.

By pythagoras theorem

⇒ OC² + OB² = BC²

Substituting the value

$\implies \bigg( \dfrac{d_1}{2} \bigg)^2 + \bigg( \dfrac{d_2}{2} \bigg)^2 = a^2$

$\implies \dfrac{(d_1)^{2} }{4}+\dfrac{(d_2)^2}{4} = a^2$

$\implies \dfrac{(d_1)^{2} + (d_2)^2}{4} = a^2$

$\implies (d_1)^{2} + (d_2)^2 = 4a^2$

$\implies (d_1 + d_2 )^{2} - 2d _2d_2= 4a^2$

[ Because (a + b)² = a² + b² + 2ab

⇒ (a + b)² - 2ab = a² + b²

⇒ a² + b² = (a + b)² - 2ab ]

Substituting a = 37 and d1d2 = 1680 in the equation

$\implies (d_1 + d_2 )^{2} - 2(1680)= 4(37)^2$

$\implies (d_1 + d_2 )^{2} - 3360= 4(1369)$

$\implies (d_1 + d_2 )^{2} - 3360= 5476$

$\implies (d_1 + d_2 )^{2} = 5476 + 3360$

$\implies (d_1 + d_2 )^{2} = 8836$

$\implies d_1 + d_2 = \sqrt{ 8836}$

$\implies d_1 + d_2 = 94$

i.e Sum of the length of the 2 diagonals = 94 cm

Hence, sum of the length of the 2 diagonals is 94 cm.