A parallelogram has sides with lengths of #15 # and #8 #. If the parallelogram's area is #15 #, what is the length of its longest diagonal?

Question

A parallelogram has sides with lengths of #15  # and #8  #.  If the parallelogram's area is #15  #, what is the length of its longest diagonal?

Sadie Adams · Accepted Answer

Longer diagonal is #22.96# unit.

We know the area of the parallelogram as #A_p=s_1*s_2*sin theta or sin theta=15/(15*8)=0.125 :. theta=sin^-1(0.125)=7.18^0:.#.Consecutive angles are supplementary #:.theta_2=180-7.18=172.82^0#. 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(15^2+8^2-2*15*8*cos172.82) =22.96# unit[Ans]

Sadie Allen · Answer

undefined To find the length of the longest diagonal of a parallelogram, we can use the formula:

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

In a parallelogram, the base and the height are perpendicular to each other. So, the height of the parallelogram is the length of the line segment perpendicular to the base and passing through one of its vertices.

Given that the area of the parallelogram is 15 and one of its sides is 15, we can rearrange the formula to solve for the height:

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

$$ \text{Height} = \frac{15}{15} = 1 $$

Now, we can use the Pythagorean theorem to find the length of the longest diagonal. In a parallelogram, the diagonals bisect each other and form two congruent right triangles. The length of the longest diagonal is the hypotenuse of one of these right triangles.

Using the lengths of the sides given (15 and 8), we can find the length of the other side of the right triangle:

$$ \text{Other side} = \sqrt{15^2 - 8^2} = \sqrt{225 - 64} = \sqrt{161} $$

Now, we have two sides of the right triangle formed by the diagonal and the sides of the parallelogram. We can find the length of the diagonal (the hypotenuse) using the Pythagorean theorem:

$$ \text{Diagonal} = \sqrt{(\text{Base})^2 + (\text{Other side})^2} $$

$$ \text{Diagonal} = \sqrt{15^2 + (\sqrt{161})^2} $$

$$ \text{Diagonal} = \sqrt{225 + 161} $$

$$ \text{Diagonal} = \sqrt{386} $$

Therefore, the length of the longest diagonal of the parallelogram is $ \sqrt{386} $.