Question

A rhombus-shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs 5 per cm². Find the cost of painting.

Answer 1

Let the sides of rhombus be of length x cm.



Perimeter of rhombus = 4x
⇒ 40 = 4x
⇒ x = 10 cm

Now,
In ∆ABC,
The sides of the triangle are of length 10 cm, 10 cm and 12 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{10+10+12}{2}=\frac{32}{2}=16\mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area}\mathrm{of}∆ABC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{16\left(16-10\right)\left(16-10\right)\left(16-12\right)} \\ =\sqrt{16\left(6\right)\left(6\right)\left(4\right)} \\ =48{\mathrm{cm}}^2...\left(1\right) \end{gathered}$$

In ∆ADC,
The sides of the triangle are of length 10 cm, 10 cm and 12 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{10+10+12}{2}=\frac{32}{2}=16\mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area}\mathrm{of}∆ADC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{16\left(16-10\right)\left(16-10\right)\left(16-12\right)} \\ =\sqrt{16\left(6\right)\left(6\right)\left(4\right)} \\ =48{\mathrm{cm}}^2...\left(2\right) \end{gathered}$$


∴ Area of the rhombus = Area of ∆ABC + Area of ∆ADC
= 48 + 48
= 96 cm²

The cost to paint per cm² = Rs 5
The cost to paint 96 cm² = Rs 5 × 96
= Rs 480
The cost to paint both sides of the sheet = Rs 2 × 480
= Rs 960

Hence, the total cost of painting is Rs 960.