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A parallelogram $ABCD$ has area $48$ sqcm. If the length of $CD$ is $8$ cm and that of $AD$ is $s$ cm, then which one of the following is necessarily true?

1. $s\geq6$
2. $s\neq6$
3. $s\leq6$
4. $5\leq s\leq7$

### 1 comment

$\textrm{Option 1, the value of S couldn't be less then 6.}$

Given that, area of parallelogram $\text{ABCD}$ is $48 \; \text{sq cm.}$

And, $\text{CD} = 8 \; \text{cm}, \; \text{AD} = s \; \text{cm}$

We can draw the parallelogram :

The area of parallelogram $\text{ABCD} = 2 \times \text {The area of triangle ACD}$

$\Rightarrow 2 \times \text{The area of triangle ACD = 48}$

$\Rightarrow \text{The area of triangle ACD} = 24$

$\Rightarrow \frac{1}{2} \times \text{AD} \times \text{CD} \times \sin \theta = 24$

$\Rightarrow s \times 8 \times \sin \theta = 48$

$\Rightarrow s \times \sin \theta = 6$

$\Rightarrow \sin \theta = \frac{6}{s}$

As,  $– 1 \leq \sin \theta \leq 1$

But length can’t be negative.

So, $0 < \sin \theta \leq 1$

$\Rightarrow 0 < \frac{6}{s} \leq 1$

$\Rightarrow 6 \leq s$

$\Rightarrow \boxed {s \geq 6\; \text{cm}}$

Correct Answer $: \text{A}$

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