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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 $: \text{A}$