A parallelogram has sides with lengths of $16$ $16$ and $15$ $15$ . If the parallelogram's area is $60$ $60$ , what is the length of its longest diagonal?

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A parallelogram has sides with lengths of $16$ $16$ and $15$ $15$ . If the parallelogram's area is $60$ $60$ , what is the length of its longest diagonal?

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Answer 1 · 2016-03-24T02:18:25

Length of the longer diagonal $d = 30.7532$ $d=30.7532" "$ units

Explanation:

The required in the problem is to find the longer diagonal $d$ $d$

Area of the parallelogram $A = b a s e \cdot h e i g h t = b \cdot h$ $A=base * height=b*h$
Let base $b = 16$ $b=16$
Let other side $a = 15$ $a=15$
Let the height $h = \frac{A}{b}$ $h=A/b$

Solve for height $h$ $h$
$h = \frac{A}{b} = \frac{60}{16}$ $h=A/b=60/16$

$h = \frac{15}{4}$ $h=15/4$

Let $θ$ $theta$ be the larger interior angle which is opposite the longer diagonal $d$ $d$ .

$θ = 180^{\circ} - {sin}^{- 1} (\frac{h}{a}) = 180^{\circ} - {14.4775}^{\circ}$ $theta=180^@-sin^-1 (h/a)=180^@-14.4775^@$
$θ = {165.522}^{\circ}$ $theta=165.522^@$

By the Cosine Law, we can solve now for $d$ $d$

$d = \sqrt{(a^{2} + b^{2} - 2 \cdot a \cdot b \cdot cos θ)}$ $d=sqrt((a^2+b^2-2*a*b*cos theta))$
$d = \sqrt{(15^{2} + 16^{2} - 2 \cdot 15 \cdot 16 \cdot {cos 165.522}^{\circ})}$ $d=sqrt((15^2+16^2-2*15*16*cos 165.522^@))$
$d = 30.7532$ $d=30.7532" "$ units

God bless....I hope the explanation is useful.

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A parallelogram has sides with lengths of $16$ $16$ and $15$ $15$ . If the parallelogram's area is $60$ $60$ , what is the length of its longest diagonal?

1 Answer

Explanation:

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A parallelogram has sides with lengths of 1616 and 1515 . If the parallelogram's area is 6060 , what is the length of its longest diagonal?

1 Answer

Explanation:

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A parallelogram has sides with lengths of $16$ $16$ and $15$ $15$ . If the parallelogram's area is $60$ $60$ , what is the length of its longest diagonal?