A parallelogram has sides with lengths of #16 # and #15 #. If the parallelogram's area is #75 #, what is the length of its longest diagonal?

Question

A parallelogram has sides with lengths of #16  # and #15  #.  If the parallelogram's area is #75  #, what is the length of its longest diagonal?

Theodore Abad · Accepted Answer

Length of longer diagonal is #30.61(2dp) unit # We know the area of the parallelogram as #A_p=s_1*s_2*sin theta or sin theta=75/(16*15)=0.3125 :. theta=sin^-1(0.3125)=18.21^0#. Consecutive angles are supplementary #:.theta_2=180-18.21=161.79^0#. Longer diagonal can be found by applying cosine law:#d_l= sqrt(s_1^2+s_2^2-2*s_1*s_2*costheta_2)=sqrt(16^2+15^2-2*16*15*cos161.79) =30.61(2dp) unit #[Ans]

Wyatt Anderson · Answer

undefined To find the length of the longest diagonal of a parallelogram with sides of lengths 16 and 15 and an area of 75, we can use the formula for the area of a parallelogram, which is base times height. Let's denote the length of the longest diagonal as $ d $.

We know that the area of the parallelogram is 75, and one of the sides (base) is 16. So we can find the height of the parallelogram using the formula for the area.

$$ \text{Area} = \text{Base} \times \text{Height} $$

$$ 75 = 16 \times \text{Height} $$

$$ \text{Height} = \frac{75}{16} $$

Now, we have the height of the parallelogram. Using the Pythagorean theorem, we can find the length of the longest diagonal $ d $. In a parallelogram, the diagonal forms a right triangle with the two adjacent sides.

$$ d^2 = \text{Side}_1^2 + \text{Side}_2^2 $$

$$ d^2 = 16^2 + \left(\frac{75}{16}\right)^2 $$

$$ d^2 = 256 + \frac{5625}{256} $$

$$ d^2 = \frac{4096 + 5625}{256} $$

$$ d^2 = \frac{9721}{256} $$

$$ d \approx \sqrt{\frac{9721}{256}} $$

$$ d \approx \sqrt{37.99609375} $$

$$ d \approx 6.164414 $$

So, the length of the longest diagonal of the parallelogram is approximately $ 6.16 $ (rounded to two decimal places).