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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. $8.8$
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Given that, radius of circle $ = 5 \; \text{m}.$

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

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 \cdot 4 \; \text{m}} $

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

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

DIAGRAM

Using the pythgoras’ theorem$:$

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

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

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

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

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

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

$ \Rightarrow \boxed{\text{PR} = 4 \cdot 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 \cdot 39) $

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

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

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

Correct Answer$: \text{D}$
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