A parallelogram has sides with lengths of $18$ $18$ and $4$ $4$ . If the parallelogram's area is $36$ $36$ , what is the length of its longest diagonal?

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A parallelogram has sides with lengths of $18$ $18$ and $4$ $4$ . If the parallelogram's area is $36$ $36$ , what is the length of its longest diagonal?

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Answer 1 · 2016-04-03T10:56:36

Length of longest diagonal is $21.56$ $21.56$ units.

Explanation:

Area of a parallelogram is given by $a \times b \times sin θ$ $axxbxxsintheta$ ,

where $a$ $a$ and $b$ $b$ are two sides of a parallelogram and $θ$ $theta$ is the angle included between them.

As sides are $18$ $18$ and $4$ $4$ and area is $36$ $36$ we have

$18 \times 4 \times sin θ = 36$ $18xx4xxsintheta=36$ or $sin θ = \frac{36}{18 \times 4} = \frac{1}{2}$ $sintheta=36/(18xx4)=1/2$

Hence $θ = 30^{\circ}$ $theta=30^@$ and two angles of parallelogram are $30^{\circ}$ $30^@$ and $150^{\circ}$ $150^@$ .

Then smaller diagonal of parallelogram would be given by

$\sqrt{a^{2} + b^{2} - 2 a b {cos 30}^{\circ}} = \sqrt{18^{2} + 4^{2} - 2 \times 18 \times 4 \times (\frac{\sqrt{3}}{2})}$ $sqrt(a^2+b^2-2abcos30^@)=sqrt(18^2+4^2-2xx18xx4xx(sqrt3/2)$

= $\sqrt{324 + 16 - 72 \times \sqrt{3}} = \sqrt{215.2923} = 14.67$ $sqrt(324+16-72xxsqrt3)=sqrt215.2923=14.67$

and larger diagonal of parallelogram would be given by

$\sqrt{a^{2} + b^{2} - 2 a b {cos 150}^{\circ}} = \sqrt{18^{2} + 4^{2} + 2 \times 18 \times 4 \times (\frac{\sqrt{3}}{2})}$ $sqrt(a^2+b^2-2abcos150^@)=sqrt(18^2+4^2+2xx18xx4xx(sqrt3/2)$

= $\sqrt{324 + 16 + 72 \times \sqrt{3}} = \sqrt{464.7077} = 21.56$ $sqrt(324+16+72xxsqrt3)=sqrt464.7077=21.56$

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A parallelogram has sides with lengths of $18$ $18$ and $4$ $4$ . If the parallelogram's area is $36$ $36$ , what is the length of its longest diagonal?

1 Answer

Explanation:

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A parallelogram has sides with lengths of $18$ $18$ and $4$ $4$ . If the parallelogram's area is $36$ $36$ , what is the length of its longest diagonal?