Answer:
Part A
The perimeter of the of the shaded region is (3·π + 6 + 6·(√2)) cm
Part B
[tex]The \ area \ of \ the \ figure = 4\dfrac{1}{2} \cdot \pi + 18 \ cm^2[/tex]
Step-by-step explanation:
Part A
The perimeter of the the shaded region is given as follows;
The perimeter of the the shaded region = The perimeter of the semicircle + The perimeter of the right triangle with legs 6 cm each - The length of the dotted line
The perimeter of the semicircle = π × r
We note that the diameter of the semicircle, D = [tex]\overline{AB}[/tex] = 6 cm
The radius of the semicircle, r = D/2 = 6 cm/2 = 3 cm
∴ The perimeter of the semicircle = π × 3 cm = 3·π cm
The perimeter of the right triangle with legs 6 cm each = The length of the 2 6 cm legs + The length of the hypotenuse side
By Pythagoras' theorem, the length of the hypotenuse side = √(6² + 6²) = 6·(√2) cm
∴ The perimeter of the right triangle = 6 cm + 6 cm + 6·(√2) cm = (12 + 6·(√2)) cm
The length of the dotted line = The the length of side [tex]\overline{AB}[/tex] of the square from which the right triangle ABC is based
∴ The length of the dotted line = The length of [tex]\overline{AB}[/tex] = 6 cm
∴ The perimeter of the figure = 3·π cm + (12 + 6·(√2)) cm - 6 cm = (3·π + 6 + 6·(√2)) cm
The perimeter of the of the shaded region = (3·π + 6 + 6·(√2)) cm
Part B
Given that the figure is made up of portion of a square and a semicircle, we have;
BC ≅ AB = 6 cm
The area of semicircle BC with radius BC/2 = 3 is 1/2×π×r² = 1/2×π×3² = 4.5·π cm²
Triangle ABC = 1/2 × Area of square from which ABC is cut
The area of triangle ABC = 1/2×Base ×Height = 1/2×AB×BC = 1/2×6×6 = 18 cm²
The area of the figure = The area of semicircle BC + The area of triangle ABC
The area of the figure = 4.5·π cm² + 18 cm² = [tex]\dfrac{9 \cdot \pi +36}{2} \ cm^2[/tex]
