Question

ABCDEFG এর পরিসীমা ও ক্ষেত্রফল কত?​

Answer 1

Given:

ABCDEFG is a polygon with measurements given in the figure

To find:

Perimeter and Area

Construction:

Draw DH ⊥ AB

Draw CI ⊥ AB

Solution:

From the given figure, we get

AB = 15 cm

AG = 12 cm

DE = GE = 5 cm

DC = HI = 7 cm

GE = AH = 5 cm

CI = AG - DE = 12 - 5 = 7 cm

BI = AB - [AH + HI] = 15 - [5 + 7] = 3 cm

Consider Δ BIC, using Pythagoras Theorem, we get

BC² = CI² + BI²

⇒ BC² = 7² + 3²

⇒ BC = $\sqrt{49 + 9}$

⇒ BC = $\sqrt{58}$

⇒ BC = 7.6 cm

Consider arc EFG, we will find the length of the arc EFG,

i.e., The circumference of the arc EFG is,

= $\frac{1}{2} \times 2\pi r$

= $\pi r$

= $\pi \times \frac{d}{2}$

= $\frac{22}{7} \times \frac{5}{2}$

= 7.8 cm

Finding the perimeter of ABCDEFG:

∴ The perimeter of ABCDEFG is,

= AB + BC + CD + DE + arc(EFG) + AG

= 15 + 7.6 + 7 + 5 + 7.8 + 12

= 54.4 cm

Finding the area of ABCDEFG:

∴ The area of ABCDEFG is,

= [Area of Δ BIC] + [Area of square DCIH] + [Area of rectangle AHEG] + [Area of semicircle EFG]

= [$\frac{1}{2} \times BI\times CI$] + [CD²] + [GE × AG] + [$\frac{1}{2} \pi r^2$]

substituting the values

= [$\frac{1}{2} \times 7\times 3$] + [7²] + [12 × 5] + [$\frac{1}{2} \times \frac{22}{7} \times (\frac{5}{2} )^2$]

= 10.5 + 49 + 60 + 9.8

= 129.3 cm²

Thus,

$\boxed{\bold{The \:perimeter\:is\:\underline{54.4\:cm}}}\\\\\boxed{\bold{The \:area\:is\:\underline{129.3\:cm^2}}}$

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