Question

A parallelogram has sides with lengths of `16` and `9`. If the parallelogram's area is `81`, what is the length of its longest diagonal?

Answer 1

Length of longer diagonal is `23.98` unit.

Explanation:

We know the area of the parallelogram as

`A_p=s_1*s_2*sin theta or sin theta=81/(16*9)=0.5625`

`:. theta=sin^-1(0.5625) ~~34.23^0`.Consecutive angles are

supplementary `:.theta_2=180-34.23~~145.77^0`.

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*cos145.77) ~~ 23.98 unit`

Length of longer diagonal is `23.98` unit [Ans]

Answer 2

Approximately `23.98`.

Explanation:

Referring to the diagram, this solution boils down to three elemental steps:
1. Find length of `HB` using the area formula;
2. Find the length of `AH` by applying the Pythagoras theorem with knowledge about lengths of `AB` and `HB`.
3. Apply the Pythagoras theorem for one more time to find the length of `AC`, the longest diagonal.

Step One: Find `BH`, height on the longest side
`BH=\frac{Area}{AD}=\frac{81}{16}`

Step Two: Find `AH` and then `IC`
Apply the Pythagoras theorem in `RT\DeltaAHB`.
`AH=\sqrt{AB^2-HB^2}=sqrt{9^2-(\frac{81}{16})^2}=\frac{45\sqrt{7}}{16}`

`IB=AH` since `AIBH` is a rectangle. Therefore `IC=IB+BC=16+\frac{45\sqrt{7}}{16}`

Step Three: Find `AC`
Apply the Pythagoras theorem one more time in `RT\DeltaAIC`.
`AC=\sqrt{AI^2+IC^2}\approx23.98`