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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 $cm^{2}$ . Find the cost of painting.

Answers (1)

[Rs. 960]

Let ABCD be a rhombus thus AB = BC = CD = DA = x (Let)

$We \; have\; perimeter \; of \; rhombus = 40 cm$

$\Rightarrow AB + BC + CD + DA = 40 cm$

$\Rightarrow x + x + x + x = 40 cm$

$\Rightarrow 4x = 40 cm$

$\Rightarrow x = \frac{40}{4} cm$

$\Rightarrow x = 10 cm$

$\therefore sides \; of \; rhombus \; AB = BC = CD = DA = 10 cm$

Area of rhombus = 2 $\times$ Ar( $\Delta$ ABC) [diagonal of rhombus divides it into two triangles of equal area]

Now, we find area of triangle using Heron’s formula

$In\; \Delta ABC, S=\frac{a+b+c}{2}= \frac{10+10+12}{2}= \frac{32}{2}= 16cm$

$Area \; of\; \Delta ABC =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$

$=\sqrt{16\left ( 16-10 \right )\left ( 16-10 \right )\left ( 16-12 \right )}$

$= \sqrt{16\times 6\times 6\times 4}$

$= \sqrt{8\times 8\times 6\times 6\times 4}$

$= \sqrt{4\times 4\times 6\times 6\times 4}$

$= 4\times 6\sqrt{2\times 2}$

$= 4\times 6\times 2= 48cm^{2}$

Now, Area of rhombus = 2 $\times$ Ar( $\Delta$ ABC)

$= 2\times 40cm^{2}= 96cm^{2}$

We find the cost of painting

Thus,

$\because cost\; of \; painting\; the\; sheet \; of\; 1 cm^{2} = Rs.\; 5$

$\therefore cost \; of\; painting\; the \; sheet\; of \; 96\; cm^{2}= Rs.\; 96 \times 5 = Rs.\; 480$

Hence, the cost of the painting both sides of the sheet = 2 $\times$ 480 = Rs. 960.

Posted by

infoexpert21

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