Question

The area of a quadrilateral is 116.25 m. find the length of the diagonal, if the perpendiculars from two vertices to the diagnol are 8.7 m and 6.3 m.​

Answer 1

Answer:

Given :

• ➤ The area of a quadrilateral is 116.25 m.
• ➤ The perpendiculars from two vertices to the diagnol are 8.7 m and 6.3 m.

$\begin{gathered}\end{gathered}$

To Find :

• ➤ The lenght of diagonal.

$\begin{gathered}\end{gathered}$

Using Formula :

${\longrightarrow{\small{\underline{\boxed{\pmb{\sf{Area_{(Quadrilateral)}= \dfrac{1}{2} \times diagonal \times \big( Some \: of \: perpendiculars \: \big)}}}}}}}$

$\begin{gathered}\end{gathered}$

Solution :

♦️ Finding the diognal of quadrilateral by substituting the values in the formula :-

${\dashrightarrow{\small{\sf{Area_{(Quadrilateral)}= \dfrac{1}{2} \times diagonal \times \bigg( Some \: of \: perpendiculars \: \bigg)}}}}$

${\dashrightarrow{\small{\sf{116.25 \: {m}^{2} = \dfrac{1}{2} \times diagonal \times \bigg(8.7 + 6.3\bigg)}}}}$

${\dashrightarrow{\small{\sf{116.25 \: {m}^{2} = \dfrac{1}{2} \times diagonal \times \bigg(15\bigg)}}}}$

${\dashrightarrow{\small{\sf{\dfrac{116.25}{15} = \dfrac{1}{2} \times diagonal}}}}$

${\dashrightarrow{\small{\sf{\cancel{\dfrac{116.25}{15}} = \dfrac{1}{2} \times diagonal}}}}$

${\dashrightarrow{\small{\sf{7.75 \: m = \dfrac{1}{2} \times diagonal}}}}$

${\dashrightarrow{\small{\sf{Diagonal = 7.75 \times 2}}}}$

${\dashrightarrow{\small{\sf{Diagonal = 15.5 \: m}}}}$

${\dashrightarrow{\small{\underline{\boxed{\sf{Diognal = 15.5 \: m}}}}}}$

∴ The diagonal of quadrilateral is 15.5 m.

$\begin{gathered}\end{gathered}$

Verification :

♦️ Checking our answer :-

${\dashrightarrow{\small{\sf{Area_{(Quadrilateral)}= \dfrac{1}{2} \times diagonal \times \bigg( Some \: of \: perpendiculars \: \bigg)}}}}$

${\dashrightarrow{\small{\sf{116.25 \: {m}^{2} = \dfrac{1}{2} \times 15.5 \times \bigg(8.7 + 6.3\bigg)}}}}$

${\dashrightarrow{\small{\sf{116.25 \: {m}^{2} = \dfrac{1}{2} \times 15.5 \times \bigg(15 \bigg)}}}}$

${\dashrightarrow{\small{\sf{116.25 \: {m}^{2} = \dfrac{1}{2} \times 15.5 \times 15 }}}}$

${\dashrightarrow{\small{\sf{116.25 \: {m}^{2} = \dfrac{1}{2} \times 232.5 \: {m}^{2} }}}}$

${\dashrightarrow{\small{\sf{116.25 \: {m}^{2} = \dfrac{1 \times 232.5}{2} \: {m}^{2} }}}}$

${\dashrightarrow{\small{\sf{116.25 \: {m}^{2} = \dfrac{232.5}{2} \: {m}^{2} }}}}$

${\dashrightarrow{\small{\sf{116.25 \: {m}^{2} = \cancel{\dfrac{232.5}{2}}}}}}$

${\dashrightarrow{\small{\sf{116.25 \: {m}^{2} = 116.25 \: {m}^{2} }}}}$

$\dashrightarrow{\underline{\boxed{\sf{LHS = RHS}}}}$

∴ Hence Verified!

$\begin{gathered}\end{gathered}$

Learn More :

$\boxed{\begin {minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {minipage}}$

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