A parallelogram has sides with lengths of #16 # and #9 #. If the parallelogram's area is #64 #, what is the length of its longest diagonal?

Question

A parallelogram has sides with lengths of #16  # and #9  #.  If the parallelogram's area is #64  #, what is the length of its longest diagonal?

Claire Akin · Accepted Answer

Longer diagonal is #24.39# unit.

We know the area of the parallelogram as #A_p=s_1*s_2*sin theta or sin theta=64/(16*9)=0.444 :. theta=sin^-1(0.444)=26.39^0:.#consecutive angles are supplementary #:.theta_2=180-26.39=153.61^0#. 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+9^2-2*16*9*cos153.61) =24.39 # unit [Ans]

Evelyn Arboleda · Answer

undefined The length of the longest diagonal of a parallelogram can be found using the formula:

$$ \text{Longest diagonal} = \sqrt{a^2 + b^2 + 2ab\cos(\theta)} $$

where $ a $ and $ b $ are the lengths of the sides of the parallelogram, and $ \theta $ is the angle between them.

Given that the sides of the parallelogram are 16 and 9, and the area is 64, we can find the value of $ \theta $ using the area formula:

$$ \text{Area} = ab\sin(\theta) $$

Substituting the given values:

$$ 64 = 16 \times 9 \times \sin(\theta) $$

$$ \sin(\theta) = \frac{64}{144} $$

$$ \sin(\theta) = \frac{4}{9} $$

Using the inverse sine function to find $ \theta $:

$$ \theta = \sin^{-1}\left(\frac{4}{9}\right) $$

$$ \theta \approx 26.57^\circ $$

Now, we can use the law of cosines to find the length of the longest diagonal:

$$ \text{Longest diagonal} = \sqrt{16^2 + 9^2 + 2(16)(9)\cos(26.57^\circ)} $$

$$ \text{Longest diagonal} \approx \sqrt{256 + 81 + 288\cos(26.57^\circ)} $$

$$ \text{Longest diagonal} \approx \sqrt{337 + 288\left(\frac{4}{9}\right)} $$

$$ \text{Longest diagonal} \approx \sqrt{337 + 128} $$

$$ \text{Longest diagonal} \approx \sqrt{465} $$

$$ \text{Longest diagonal} \approx 21.56 $$

So, the length of the longest diagonal is approximately 21.56 units.

Parker Allen · Answer

undefined To find the length of the longest diagonal of the parallelogram, you can use the formula for the area of a parallelogram: Area = base * height. In this case, the base is the length of one of the sides, which is 16 units, and the height is the length of the other side, which is 9 units. So, you can rearrange the formula to solve for the height: height = Area / base.

Substitute the given values: height = 64 / 16 = 4 units.

Now, you can use the Pythagorean theorem to find the length of the longest diagonal. The diagonal, the height, and one side of the parallelogram form a right triangle.

Let d be the length of the longest diagonal.

Using the Pythagorean theorem: $d^2 = 16^2 + 4^2$

$d^2 = 256 + 16$

$d^2 = 272$

$d = \sqrt{272}$

$d ≈ 16.49$

So, the length of the longest diagonal of the parallelogram is approximately 16.49 units.