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If a rhombus has area $12 \; \text{sq cm}$ and side length $5 \; \text{cm},$ then the length, $\text{in cm},$ of its longer diagonal is

  1. $\sqrt{13} + \sqrt{12}$
  2. $\sqrt{37} + \sqrt{13}$
  3. $\frac{\sqrt{37} + \sqrt{13}}{2}$
  4. $\frac{\sqrt{13} + \sqrt{12}}{2}$
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Given that, the area of rhombus $= 12\; \text{cm}^{2},$ and side length $= 5\;\text{cm.}$

Let the diagonal length of the rhombus be $2a\;\text{cm},\;2b\;\text{cm}, (a>b)$

Let’s draw the diagram.



$\triangle \text{AOB}$ is a right-angle triangle. We can apply the Pythagorean theorem.

$\text{(Hypotenuse)}^{2} = \text{(Perpendicular)}^{2} + \text{(Base)}^{2}$

$\Rightarrow 5^{2} = a^{2} + b^{2}$

$\Rightarrow \boxed{a^{2} + b^{2} = 25}\quad \longrightarrow (1)$

The area of rhombus $= \dfrac{\text{Product of diagonal}}{2}$

$\Rightarrow 12 = \frac{2a \times 2b}{2}$

$\Rightarrow \boxed{ab = 6}$

From equation $(1),$

$a^{2} + b^{2} = 25$

$\Rightarrow (a+b)^{2} – 2ab = 25$

$\Rightarrow (a+b)^{2} – 2(6) = 25\quad [\because ab=6]$

$\Rightarrow (a+b)^{2} = 25 + 12$

$\Rightarrow (a+b)^{2} = 37$

$\Rightarrow \boxed{a+b = \sqrt{37}}\quad \longrightarrow (2)$

Again, from equation $(1),$

$a^{2} + b^{2} = 25$

$\Rightarrow (a-b)^{2} + 2ab = 25$

$\Rightarrow (a-b)^{2} + 2(6) = 25\quad [\because ab=6]$

$\Rightarrow (a-b)^{2} = 25 - 12$

$\Rightarrow (a-b)^{2} = 13$

$\Rightarrow \boxed{a-b = \sqrt{13}}\quad \longrightarrow (3)$

Adding the equation $(2)$ and $(3).$

$\begin{array}{} a+b = \sqrt{37} \\ a-b = \sqrt{13} \\\hline  \end{array}$

$\boxed{2a = \sqrt{37} + \sqrt{13}}$

Subtract the equation $(3),$ from equation $(2).$

$\begin{array}{} a+b = \sqrt{37} \\ a-b = \sqrt{13} \\  – \;\;\; + \qquad – \\\hline  \end{array}$

$\boxed{2b = \sqrt{37} - \sqrt{13}}$

$\therefore$ The length of the longer diagonal $= 2a = (\sqrt{37} + \sqrt{13})\;\text{cm.}$

Correct Answer $:\text{B}$

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