A parallelogram has sides with lengths of #15 # and #8 #. If the parallelogram's area is #48 #, 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 #48  #, what is the length of its longest diagonal?

Wyatt Anderson · Accepted Answer

The length of longer diagonal is #22.56 unit # We know the area of the parallelogram as #A_p=s_1*s_2*sin theta or sin theta=48/(15*8)=0.4 :. theta=sin^-1(0.4)=23.58^0:.#consecutive angles are supplementary #:.theta_2=180-23.58=156.42^0#. 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*cos156.42) =22.56 unit #[Ans]

Caroline Aucoin · Answer

undefined To find the length of the longest diagonal of the parallelogram, we can use the formula for the area of a parallelogram, which is given by $ \text{Area} = \text{base} \times \text{height} $.

Given that the area is 48 and one side (base) is 15, we can rearrange the formula to solve for the height: $ \text{height} = \frac{\text{Area}}{\text{base}} = \frac{48}{15} = 3.2 $.

Now, to find the length of the longest diagonal, we can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (longest side) is equal to the sum of the squares of the other two sides.

Let the other side of the parallelogram be $ a $. Then, the length of the longest diagonal $ d $ can be found using $ d^2 = 15^2 + a^2 $, where $ a = \sqrt{d^2 - 15^2} $.

Substitute the known values: $ a = \sqrt{d^2 - 225} $.

Since the height of the parallelogram is perpendicular to the base, it forms a right triangle with one of the diagonals. Thus, we have another right triangle with legs $ 3.2 $ and $ a $, and hypotenuse $ d $.

Using the Pythagorean theorem again, we have $ d^2 = 3.2^2 + a^2 = 10.24 + a^2 $.

Substitute $ a = \sqrt{d^2 - 225} $ into the equation: $ d^2 = 10.24 + (d^2 - 225) $.

Solve for $ d $: $ d^2 = 10.24 + d^2 - 225 $, $ 225 = 10.24 $, $ d^2 = 235.24 $, $ d = \sqrt{235.24} $, $ d \approx 15.35 $.

Therefore, the length of the longest diagonal of the parallelogram is approximately 15.35 units.

Carson Arnold · Answer

undefined To find the length of the longest diagonal of the parallelogram, use the formula for the area of a parallelogram: $ \text{Area} = \text{base} \times \text{height} $. Since the parallelogram has sides of length 15 and 8, let's assume 15 is the base. Rearrange the formula to solve for the height: $ \text{height} = \frac{\text{Area}}{\text{base}} $. Plug in the given values: $ \text{height} = \frac{48}{15} = 3.2 $. Now, use the Pythagorean theorem to find the length of the longest diagonal. The longest diagonal $d$ satisfies $d^2 = 8^2 + 3.2^2$, so $d = \sqrt{8^2 + 3.2^2}$. Calculate this to find the length of the longest diagonal.