If the perimeter of a rectangle is p and its diagonal is d, the d…

MCQ

If the perimeter of a rectangle is p and its diagonal is d, the difference between the length and width of the rectangle is:

✓
$\frac{\sqrt{8\text{d}^{2}-\text{p}^{2}}}{2}$
B
$\frac{\sqrt{8\text{d}^{2}+\text{p}^{2}}}{2}$
C
$\frac{\sqrt{6\text{d}^{2}-\text{p}^{2}}}{2}$
D
$\frac{\sqrt{6\text{d}^{2}+\text{p}^{2}}}{2}$

✓

Answer

Correct option: A.

$\frac{\sqrt{8\text{d}^{2}-\text{p}^{2}}}{2}$

Perimeter of the rectangle $= P2(l + b) = P \Rightarrow $
$1+\text{b}=\frac{\text{P}}{2}\rightarrow$
$(1)$ diagonal of the rectangle $ = \text{d}\sqrt{1^2+\text{b}^2}=\text{d}$
$\Rightarrow{1}^{2}+\text{b}^{2}=\text{d}^{2}$
$(1)^2 ⟹ d^2 + 2lb =$
$\frac{\text{p}^{2}}{4}$ $\Rightarrow 2lb =\frac{\text{p}^2 - 4\text{d}^2}{4}$
$⟹ l^2 + b^2− 2lb = d^2$
$=\frac{\text{p}^2 - 4\text{d}^2}{4}$
$\Rightarrow(1-\text{b})^{2}=$$=\frac{\text{8d}^2 - \text{pd}^2}{4}$
$\Rightarrow (1 - b)$
$\frac{\sqrt{8\text{d}^{2}-\text{p}^{2}}}{2}$
$\therefore$ Difference between length and breadth $ = \frac{\sqrt{8\text{d}^{2}-\text{p}^{2}}}{2}$

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