the length of one side of rhombus is 41 CM and it's area is 720 cm2. find the sum of the length of diagonals of rhombus
Answers
Answer:
We have been that length of one side of a rhombus is 41 cm and its area is 720 cm square and we are asked to find the sum of the length of its diagonal.
Let P and Q be diagonals of rhombus.
In order to solve this problem we will use rhombus area formula and Pythagorean theorem.
\text{Area of rhombus}=\frac{\text{Product of both diagonals}}{2}Area of rhombus=
2
Product of both diagonals
\begin{gathered}\text{Area of rhombus}=\frac{P\cdot Q}{2}=720\\{P\cdot Q}=720\cdot 2=1440\end{gathered}
Area of rhombus=
2
P⋅Q
=720
P⋅Q=720⋅2=1440
Using Pythagorean theorem we get,
\begin{gathered}(\frac{P}{2})^{2}+(\frac{Q}{2})^{2}=41^{2}\\\frac{P^{2}}{4}+\frac{Q^{2}}{4}=1681\\P^{2}+Q^{2}=1681\cdot 4=6724\end{gathered}
(
2
P
)
2
+(
2
Q
)
2
=41
2
4
P
2
+
4
Q
2
=1681
P
2
+Q
2
=1681⋅4=6724
Now we will use perfect square formula to find the sum of both diagonals.
(P+Q)^{2}=P^{2}+2PQ+Q^{2}(P+Q)
2
=P
2
+2PQ+Q
2
Upon substituting our findings in the formula we will get,
\begin{gathered}(P+Q)^{2}=6724+2\cdot 1440\\(P+Q)^{2}=6724+2880\\(P+Q)^{2}=9604\\(P+Q)=\sqrt9604\end{gathered}
(P+Q)
2
=6724+2⋅1440
(P+Q)
2
=6724+2880
(P+Q)
2
=9604
(P+Q)=
9
604
(P+Q)=98(P+Q)=98
Therefore, the sum of diagonals is 98.