Question

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Answer 1

Answer:

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Answer 2

Answer:

Area of ABCDEF = 65/2 cm² = 32.5 cm²

Step-by-step explanation:

Given: AD=8cm , AJ=6cm , AI=5cm , AH=3cm , AG=2.5cm , EI=4cm , JC=2cm , FG=HB=3cm

⇒ ID = AD-AI = 8-5

or ID=3cm

⇒ DJ = AD-AJ = 8-6

or DJ=2cm

⇒ HJ = AD-AH-JD = 8-3-2

or HJ=3cm

And, GI = AD-AG-ID = 8-2.5-3

or GI=2.5cm

According to the figure,

Area of ABCDEF = area of ΔABH + area of trapezium BCJH + area of ΔCDJ + area of ΔDEI + area of trapezium EFGI + area of ΔAGF

w.k.t., area of Δ = $\frac{1}{2}*Base*height$

And area of trapezium = $\frac{1}{2}*(Sum of parallel sides)*(perpendicular distance between two parallel sides)$

⇒ Area of ABCDEF = $\frac{1}{2}*(AH)*(BH)$ + $\frac{1}{2}*(BH+CJ)*(HJ)$ + $\frac{1}{2}*(DJ)*(CJ)$ + $\frac{1}{2}*(ID)*(EI)$ + $\frac{1}{2}*(FG+EI)*(GI)$ + $\frac{1}{2}*(AG)*(FG)$

⇒ Area of ABCDEF = $\frac{1}{2}*3*3$ + $\frac{1}{2}*(3+2)*(3)$ + $\frac{1}{2}*(2)*(2)$ + $\frac{1}{2}*(3)*(4)$ + $\frac{1}{2}*(3+4)*(2.5)$ + $\frac{1}{2}*(2.5)*(3)$

⇒ Area of ABCDEF = $\frac{9}{2}$ + $\frac{15}{2}$ + $\frac{4}{2}$ + $\frac{12}{2}$ + $\frac{17.5}{2}$ + $\frac{7.5}{2}$ = $\frac{9+15+4+12+17.5+7.5}{2}$

⇒ Area of ABCDEF = 65/2 cm² = 32.5 cm²

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