Question

PLEASE SOLVE THIS OR MY LIFE WILL GO ​

Answer 1

$\large\underline{\sf{Solution-}}$

In quadrilateral ABCD,

• AC = 2.5 cm

• BE is perpendicular to AC

• DF is perpendicular to AC

• BE = 7.5 cm

• DF = 15 cm

We know, Area of quadrilateral is given by

$\boxed{ \rm{ \:Area_{(quadrilateral)} = \frac{1}{2} \times (Sum \: of \: perpendiculars) \times diagonal \: }}$

So, using this result, we have

$\rm \: Area_{(quadrilateral)} = \dfrac{1}{2} \times (BE + DF) \times AC$

So, on substituting the values, we get

$\rm \: Area_{(quadrilateral)} = \dfrac{1}{2} \times (7.5 + 15) \times 25$

$\rm \: Area_{(quadrilateral)} = \dfrac{1}{2} \times 22.5 \times 25$

$\rm \: Area_{(quadrilateral)} = 22.5 \times 12.5$

$\rm\implies \:\boxed{ \rm{ \:Area_{(quadrilateral)} = 281.25 \: {cm}^{2} \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

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