Question

10. The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropper
on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field

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Answer 1

Answer:

132

Step-by-step explanation:

1/2×d(h1+h2)

1/2×24×(8+3)

1/2×24×11

12×11

132

Answer 2

✪Question:-

• The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropper on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.

✹Given:-

• Length of diagonal=d=24
• Length of perpendicular dropped on BD.
• h1= 13 m
• h2= 8m

✱To find:-

• The area of field

❖Solution:-

$\sf \: \: Area \: of \: ABCD = \frac{d}{2} (h1 + h2)$

$= \frac{24}{2} (13 + 8)$

$= 12 \times 21$

$= 252 \: {m}^{2}$

$\boxed{ \underline{ \mathrm{ \orange{Area \: of \: quadrilateral \: is \: 252 m²}}}}$

Step-by-step explanation:

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

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