Question

The diagonal of a quadrikateral shaped field is 24m,and perpendicular dropped on it from the remaining opposite vertices are 10m and 14m. Find the area of the fields .write answer step by step.​

Answer 1

$\huge{ \mathfrak{ \pink{Correct \: Question:- }}}$

• The diagonal of a quadrilateral shaped field is 24m,and perpendicular dropped on it from the remaining opposite vertices are 10m and 14m. Find the area of the fields .

$\huge{ \mathfrak{ \green{Given:-}}}$

• Length of diagonal d = 24

• Length of perpendicular dropped on BD
• $h_1 = 10m \\ h_2 = 14m$

$\huge { \mathfrak { \orange{To \: find:- }}}$

• Area of the field

$\huge { \mathfrak{ \red{✍Solution:- }}}$

$\\ \: \: \: \: \: \: \: \: \: \: \: \mapsto \bf \: Area \: of \: field= \frac{d}{2} (h_1 + h_2)$

$\\ \: \: \: \: \: \: \: \: \: \: \mapsto \: \frac{24}{2} \times (10 + 14) {m}^{2}$

$\\ \: \: \: \: \: \: \: \: \: \: \mapsto \: 12 \times 24 {m}^{2}$

$\\ \: \: \: \: \: \: \: \: \: \: \mapsto \: 288 {m}^{2}$

$\boxed{ \sf{ \red{ \therefore \: the \: area \: of \: field \: is{ \boxed{ \mathfrak{ \pink{ 288m²}}}}}}}$

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Step-by-step explanation:

$⠀\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}$

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

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