Rhombus ABCD has perimeter 148, and one of its diagonals has length 24. How long is the other diagonal?
Rhombus ABCD has perimeter 148, and one of its diagonals has length 24.
How long is the other diagonal?
Let e one of its diagonal = 24
Let f the other diagonal = ?
Let a be the side of the rhombus. All sides are equal.
\(\mathbf{a= \ ? }\\ \begin{array}{|rcll|} \hline 4a &=& \text{perimeter} \\ a &=& \frac{ \text{perimeter} } {4} \\ a &=& \frac{ 148 } {4} \\ \mathbf{a} & \mathbf{=} & \mathbf{37} \\ \hline \end{array}\)
\(\mathbf{f= \ ? }\\ \begin{array}{|rcll|} \hline \mathbf{4a^2} & \mathbf{=} & \mathbf{e^2+f^2 } \\\\ 4\cdot 37^2 &=& 24^2+f^2 \\ 4\cdot 1369 &=& 576 + f^2 \\ 5476 &=& 576 + f^2 \quad & | \quad -576 \\ 5476-576 &=& f^2 \\ 4900 &=& f^2 \\ 70 &=& f \\\\ \mathbf{f} & \mathbf{=} & \mathbf{70} \\ \hline \end{array}\)
The other diagonal is 70
Here's another method....
The sides are all equal, so one side = 148/ 4 = 37
And the diagonals meet at right angles......so 1/2 the length of the known diagonal = 1/2 * 24 = 12
And this half diagonal and one of the rhombus sides will form the leg and hypotenuse of a right triangle.....and the other half diagonal will form the other leg.....and its length = sqrt [ 37^2 - 12^2] = sqrt [ 1225 ] = 35
So......the length of the other diagonal = 2 * 35 = 70