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Rhombus  ABCD has perimeter 148, and one of its diagonals has length 24. How long is the other diagonal?

 Aug 18, 2017
edited by Guest  Aug 18, 2017
 #1
avatar+26367 
+2

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

 

laugh

 Aug 18, 2017
 #2
avatar+128408 
+1

 

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

 

 

cool cool cool

 Aug 18, 2017

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