Math, asked by srinjoydas396, 7 months ago

Find the area of the shaded region. Here ABC is a right-angled triangle where AB = 8 cm,
BC = 6 cm and ADC, BEC and AFB are semicircles.​

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Answers

Answered by bhuvanahemu
0

Answer:

which is shaded here say

Answered by bhagyashreechowdhury
9

Given:

ABC is a right-angled triangle where AB = 8 cm,  BC = 6 cm and ADC, BEC and AFB are semicircles.​

To find:

The area of the shaded region

Solution:

Finding the area of the ΔABC:

We know,

  • Area of a triangle = ½ × base × height

∴ Area of triangle ABC is,

= ½ × AB × BC

substituting the given values of AB & BC

= ½ × 8 × 6

= 4 × 6

= 24 cm²

Finding the area of semi-circle AFB:

We know,

  • Area of a semi-circle = ½ × \pi r^2

∴ Area of semi-circle AFB is,

= ½ × \pi r^2

substituting the value of r = \frac{diameter}{2} = \frac{AB}{2} = \frac{8}{2}  = 4

= ½ × \frac{22}{7} × 4²

= \frac{22\times 16}{14}

= 25.14 cm²

Finding the area of semi-circle BEC:

∴ Area of semi-circle BEC is,

= ½ × \pi r^2

substituting the value of r = \frac{diameter}{2} = \frac{BC}{2} = \frac{6}{2}  = 3

= ½ × \frac{22}{7} × 3²

= \frac{22\times 9}{14}

= 14.14 cm²

Finding the area of semi-circle ADC:

Using the Pythagoras theorem in Δ ABC, we get

AB² + BC² = AC²

⇒ 8² + 6² = AC²

⇒ 64 + 36 = AC²

⇒ 100 = AC²

⇒  AC = √100

⇒ AC = 10 cm

∴ Area of semi-circle ADC is,

= ½ × \pi r^2

substituting the value of r = \frac{diameter}{2} = \frac{AC}{2} = \frac{10}{2}  = 5

= ½ × \frac{22}{7} × 5²

= \frac{22\times 25}{14}

= 39.28 cm²

Finding the area of the shaded region:

∴ The area of the shaded region is,

= [Area of triangle ABC] + [Area of semi-circle AFB] + [Area of semi-circle BEC] + [Area of semi-circle ADC]

= [24 cm²] + [25.14 cm²] + [14.14 cm²] + [39.28 cm²]

= 102.56 cm²

Thus, \boxed{\boxed{\bold{Area\:of\:shaded\:region\:is\:\underline{102.56\:cm^2}}}}.

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