Question

please solve this bro mathdude500​

Answer 1

Answer:

area of trapezoid EFDC = 1/2 ×(11+5)×4

= 32 cm2

area of 2nd trapezoid = 32 cm2 ( because both

are same)

area of rectangle = 11×5= 55 cm2

total area = 32 + 32+ 55 = 119 cm2

Answer 2

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

Given that, ABCDEFGH is a regular octagon with BC = 5 cm.

So, AB = BC = CD = DE = EF = FG = GH = HA = 5 cm

Now, further given that CF = BG = 11 cm

Now, Octagon ABCDEFGH consist of three geometrical figures namely trapezium CDEF, rectangle BCFG and trapezium ABGH.

Now, Consider

$\rm \: Area_{(CDEF)}$

$\rm \: = \: \dfrac{1}{2}(DE + CF) \times distance \: between \: parallel \: lines$

$\rm \: = \: \dfrac{1}{2} \times (5 + 11) \times 4$

$\rm \: = \: 16 \times 2$

$\rm \: = \: 32 \: {cm}^{2}$

Now, Consider

$\rm \: Area_{(ABGH)}$

$\rm \: = \: \dfrac{1}{2}(AH + BG) \times distance \: between \: parallel \: lines$

$\rm \: = \: \dfrac{1}{2} \times (5 + 11) \times 4$

$\rm \: = \: 16 \times 2$

$\rm \: = \: 32 \: {cm}^{2}$

Now, Consider

$\rm \: Area_{(BCFG)}$

$\rm \: = \: BC \times CF$

$\rm \: = \: 11 \times 5$

$\rm \: = \: 55 \: {cm}^{2}$

Now, Consider

$\rm \: Area_{(ABCDEFGH)}$

$\rm \: = \: Area_{(CDEF)} +Area_{(BCFG)} + Area_{(ABGH)}$

$\rm \: = \: 32 + 55 + 32$

$\rm \: = \: 119 \: {cm}^{2}$

Hence,

$\color{green}\rm\implies \:Area_{(ABCDEFGH)} = 119 \: {cm}^{2}$

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

Additional Information :-

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