Q3. The answer is 7/2 and 49/7.
a) The ratio of the perimeters of the similar figures is the ratio of their similar sides:
28 / 8 = (28÷4) / (8÷4) = 7 / 2
b) The ratio of the areas of the similar figures is the ratio of their perimeters raised to the second power:
(7/2)² = 7² / 2² = 49 / 4
Q6. The answer is 30π in.
The circumference (C) of the circle with radius r is:
C = 2 * π * r
According to the image, the radius of the circle is 15 in.
r = 15 in.
Therefore, to calculate the circumference, we must substitute the value for radius in the formula for the circumference:
C = 2 * π * 15 in = 30π in
Q7. The answer is 10.89π m².
The area of the circle (A) with radius r is:
A = r²π
According to the image, the diameter of the circle is 6.6 m. We know that the radius is half of the diameter, therefore:
r = 6.6 m / 2 = 3.3 m.
Now, substitute r in the formula for the area of the circle:
A = 3.3²π
A = 10.89π m²
Q8. The answer is 9.7 m².
Step 1. Calculate the area of the circle: A = r²π
The radius is the half of the diameter, so: r = d/2 = 4.6 m / 2 = 2.3 m.
The area of the circle is: A = (2.3 m)²π = 5.29π m²
Since π = 3.14, then A = 5.29 * 3.14 m² = 16.6 m²
Step 2. We know that the whole circle is 360° and its area is 16.6 m². The area of the sector with a central angle of 210° is A₂₁₀. Make a proportion:
360° : 16.6 m² = 270° : A₂₁₀
A₂₁₀ = 210° × 16.6 m² : 360°
A₂₁₀ = 9.7 m²
Q9. The answer is (270π + 81√3) m².
Step 1. Calculate the area of the whole circle:
A1 = r²π
r = 18 m
A1 = 18²π = 324π m²
Step 2. Calculate the section of the circle excluding the sector with the triangle with the angle of 60°.
If the whole circle is 360°, this sector is with the angle of 360° - 60° = 300°.
To calculate the area of this sector (A₃₀₀), we will make a proportion:
A1 : 360° = A₃₀₀ : 300°
324π : 360° = A₃₀₀ : 300°
A₃₀₀ = 324π * 300° : 360°
A₃₀₀ = 270π m²
Step 3. Calculate the area of the equilateral triangle:
A2 = √3 a² / 4
a = 18 m
A2 = √3 * 18²/4 = √3 * 81
A2 = 81√3 m²
Step 4. Sum up the areas of the sector with the angle of 300° (A₃₀₀) and the area of the triangle (A2) to get the area (A) of the shaded region:
A = A₃₀₀ + A2 = 270π m² + 81√3 m²
A = (270π + 81√3) m²