Question

A field in the shape of a rhombus has the distances between pairs of opposite vertices as 14 m and 48 m.
What is the cost of fencing the field at Rs 20 per metre
(A) Rs. 1800
(B) Rs. 1900
(C) Rs. 2000
(D) None of these​

Answer 1

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$\huge\bold{\underline{\underline{{Answer:-}}}}$

Length of one side=√{(48÷2)^2+(14×2)^{2} M

➢√{576+49} M

➢√625 M

➢ 25 M

Total perimeter of the field=25×4 M =100 M

Therefore total cost of fencing will be=(100×20)/- =2000/-

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Answer 2

$\large\red{\tt{Question:-}}$

A field in the shape of a rhombus has the distances between pairs of opposite vertices as 14 m and 48 m.

What is the cost of fencing the field at Rs 20 per metre?

(A) Rs. 1800

(B) Rs. 1900

(C) Rs. 2000

(D) None of these

$\large\red{\bf{Answer:-}}$

Length of one side:-

$\sqrt{ { (\frac{48}{2}) }^{2} + {(14 \times 2)}^{2} } m$

$= > \sqrt{576 + 49} \: m$

$= > \sqrt{625} \: m$

$= > 25 \: m$

Total perimeter of the field = 25 × 4 M = 100 M

Rate of fencing = Rs 20 per meter.

Therefore, total cost of fencing will be

=> (100×20)

=> Rs 2000.

Thanks.