Given
the perimeter of a rhombus is 560 meters
one of its diagonals has a length of 76 meters.
Procedure
The perimeter of a rhombus is:
P=4L
[tex]\begin{gathered} P=4L \\ 560=4L \\ L=\frac{560}{4} \\ L=140 \end{gathered}[/tex]if d1=76 and L=140, we can find d2
[tex]\begin{gathered} (\frac{d1}{2})^2+(\frac{d2}{2})^2=L^2 \\ (\frac{76}{2})^2+(\frac{d2}{2})^2=140^2 \\ (\frac{d2}{2})^2=140^2-(\frac{76}{2})^2 \\ \frac{d2}{2}^{}=\sqrt{19600-1444} \\ d2=269.48 \end{gathered}[/tex]the area is:
[tex]\begin{gathered} A=\frac{d2\cdot d1}{2} \\ A=\frac{269.48\cdot76}{2} \\ A=10240 \end{gathered}[/tex]The answer is:
Part A: the length of the other diagonal is 269.48
Part B: The area is 10240 square meters.
