# CAT 2020 Set-2 | Question: 70

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Let $\text{C}$ be a circle of radius $5 \; \text{meters}$ having center at $\text{O}.$ Let $\text{PQ}$ be a chord of $\text{C}$ that passes through points $\text{A}$ and $\text{B}$ where $\text{A}$ is located $4 \; \text{meters}$ north of $\text{O}$ and $\text{B}$ is located $3 \; \text{meters}$ east of $\text{O}.$ Then, the length of $\text{PQ}$, in meters, is nearest to

1. $7.2$
2. $7.8$
3. $6.6$
4. $8.8$

Given that, radius of circle $= 5 \; \text{m}.$

Now, using the given information we can draw the diagram.

In $\triangle \text{AOB} :$

$\frac{1}{2} \times 4 \times 3 = \frac{1}{2} \times 5 \times \text{OR}$

$\Rightarrow \boxed{ \text{OR} = 2 .4 \; \text{m}}$

$\text{OR}$ bisects chord $\text{PQ} \Rightarrow \boxed{ \text{PR = RQ}}$

Now, we can take out $\triangle \text{ORP},$ and observe.

Using the Pythagorean theorem $:$

$\text{(OP)}^{2} = \text{(PR)}^{2} + \text{(OR)}^{2}$

$\Rightarrow (5)^{2} = \text{(PR)}^{2} +(2 .4)^{2}$

$\Rightarrow 25 = \text{(PR)}^{2} + 5 .76$

$\Rightarrow \text{(PR)}^{2} = 25 – 5 .76$

$\Rightarrow \text{PR} = \sqrt{ 25 – 5 .76}$

$\Rightarrow \text{PR} = \sqrt{19 .24}$

$\Rightarrow \boxed{\text{PR} = 4 .39 \; \text{m}}$

Thus, $\text{PQ} = \text{PR} + \text{RQ}$

$\Rightarrow \text{PQ} = \text{PR} + \text{PR} \quad [\because \text{PR = RQ}]$

$\Rightarrow \text{PQ} = 2 \text{PR}$

$\Rightarrow \text{PQ} = 2 (4 .39)$

$\Rightarrow \text{PQ} = 8 .78 \; \text{m}$

$\Rightarrow \boxed{\text{PQ} \approx 8 .8 \; \text{m}}$

$\therefore$ The length of $\text{PQ} \; \text{in meters},$ is nearest to $8 .8.$

Correct Answer$: \text{D}$

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