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CAT 2021 Slot 3 — QA Question 6

Basics of QuadrilateralsEasy

A park is shaped like a rhombus and has area 96 sq m. If 40 m of fencing is needed to enclose the park, the cost, in INR, of laying electric wires along its two diagonals, at the rate of ₹125 per m, is 

Answer & solution

Answer: 3500

Solution

Easy

A rhombus has four equal sides, so the perimeter gives the side. Its area is 12d1d2\tfrac12 d_1 d_2, and its diagonals bisect each other at right angles, so half-diagonals and a side form a right triangle. From the product d1d2d_1 d_2 and the sum of squares d12+d22d_1^2+d_2^2, get d1+d2d_1+d_2 — exactly the wire length to be costed.

1

Side and diagonal product. Perimeter =40=40 m gives the side; area =96=96 gives d1d2d_1 d_2.

side=404=10 m12d1d2=96(area of rhombus)⇒ d1d2=192\begin{aligned} &\text{side}=\frac{40}{4}=10\text{ m}\\ &\frac12 d_1 d_2=96 \quad\text{(area of rhombus)}\\ &\Rightarrow\ d_1 d_2=192 \end{aligned}
2

Sum of squares of diagonals. The diagonals bisect each other perpendicularly, so a half of each diagonal and a side form a right triangle.

(d12)2+(d22)2=102(Pythagoras)⇒ d12+d22=400\begin{aligned} &\left(\frac{d_1}{2}\right)^2+\left(\frac{d_2}{2}\right)^2=10^2 \quad\text{(Pythagoras)}\\ &\Rightarrow\ d_1^2+d_2^2=400 \end{aligned}
3

Get d1+d2d_1+d_2. Combine the product and the sum of squares using the identity (d1+d2)2=d12+d22+2d1d2(d_1+d_2)^2=d_1^2+d_2^2+2d_1 d_2.

(d1+d2)2=400+2(192)=784(from steps 1, 2)⇒ d1+d2=28 m\begin{aligned} &(d_1+d_2)^2=400+2(192)=784 \quad\text{(from steps 1, 2)}\\ &\Rightarrow\ d_1+d_2=28\text{ m} \end{aligned}
4

Cost of the wire. Total wire length along both diagonals is d1+d2d_1+d_2, charged at ₹125\textsf{₹}125/m.

Cost=125×28=3500\begin{aligned} &\text{Cost}=125\times 28=3500 \end{aligned}
Cost=₹ 3500\text{Cost}=\textsf{₹}\,3500
CAT 2021 Slot 3 QA Q6: A park is shaped like a rhombus and has area 96 sq m. If 40 m of fencing is needed to enclose the park, the co — Solution | TheCATExam