MCQ
Mark $(\checkmark)$ against the correct answer in the following: The area of a rectangular carpet is $120m^2$ and its perimeter is $46m$. The length of its diagonal is:
Hint: $l + b = 23$ and $lb = 120$ Diagonal $=\sqrt{\text{l}^2+\text{b}^2}=\sqrt{289}$ $=\sqrt{17\times17}=17$
  • A
    $15m$
  • B
    $16m$
  • $17m$
  • D
    $20m$

Answer

Correct option: C.
$17m$
Area of rectangular carpet $= 120cm^2$
Perimeter $= 46m$
Now $2(l + b)$
$= 46m$
$\Rightarrow \text{l}+\text{b}=\frac{46}{2}=23$
and $lb = 120$
$\therefore (\text{l}-\text{b})^2=(\text{l}+\text{b})^2-4\text{lb}$
$=(23)^2-4\times120$
$=529-480$
$=49=(7)^2$
$\therefore \text{l}-\text{b}=7$
and $l + b = 23$
Adding we get, $2l = 30$
$\Rightarrow \text{l}=\frac{30}{2}$
$=15$
$\therefore b = 23 - 15 = 8$
Now diagonal $=\sqrt{\text{l}^2+\text{b}^2}$
$=\sqrt{(15)^2+(8)^2}$
$=\sqrt{225+64}$
$=\sqrt{289}$
$=17\text{m}$

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