Ex 6.1, 5 (vii)

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Ex 6.1, 5 (vii) Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) Here, we have 3 semicircles of different Diameters Let’s find the Hypotenuse of the Right angled triangle By Pythagoras Theorem π‘―π’šπ’‘π’π’•π’†π’π’–π’”π’†^𝟐=πŸ–^𝟐+πŸ”^𝟐 π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’^2=64+36 π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’^2=100 π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’^2=10^2 π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’=√(10^2 " " ) π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’=10 So, our figure looks like Now, Perimeter of shape = Circumference of semicircle with Diameter 10 cm + Circumference of semicircle with Diameter 8 cm + Circumference of semicircle with Diameter 6 cm = Circumference of semicircle with Radius 5 cm + Circumference of semicircle with Radius 4 cm + Circumference of semicircle with Radius 3 cm = 𝟏/𝟐 Γ— 𝟐 Γ— 𝝅 Γ— πŸ“+𝟏/𝟐 Γ— 𝟐 Γ— 𝝅 Γ— πŸ’+𝟏/𝟐 Γ— 𝟐 Γ— 𝝅 Γ— πŸ‘ = πœ‹ Γ— 5+πœ‹ Γ— 4+ πœ‹ Γ— 3 = πœ‹ Γ— (5+4+3) = 𝝅 Γ— 𝟏𝟐 = 𝟐𝟐/πŸ• Γ— 𝟏𝟐 = πœ‹ Γ— 12 = 22/7 Γ— 12 = 264/7 = 37.71 cm

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