44/45 ### Geometry ProblemsBelow are some geometry problems designed to help students practice calculating areas and volumes of various shapes.#### Problem 44**Question:**The area formula for a regular polygon is \( \frac{1}{2}ap \). Find the area of a regular pentagon with side length 15 cm and apothem 10 cm.**Options:**- A. 750 cm²- B. 375 cm²- C. 900 cm²- D. 450 cm²**Explanation:**Here, \( a = \) apothem = 10 cm,\( p = \) perimeter = 5 × 15 cm = 75 cm.Using the formula for the area of a regular polygon:\[ \text{Area} = \frac{1}{2} \times 10 \times 75 = 375 \, \text{cm}² \]The correct answer is:- **B. 375 cm²**#### Problem 45**Question:**Sphere \( B \) has 9 times the surface area of sphere \( A \). How many times the volume of sphere \( A \) is the volume of sphere \( B \)?**Options:**- A. 9- B. \(27\pi\)- C. \(9\pi\)- D. 27**Explanation:**If Sphere \( B \) has 9 times the surface area of Sphere \( A \), then the ratio of their radii will be \( \sqrt{9} = 3 \). The volume ratio will be \(3^3 = 27 \).The correct answer is:- **D. 27**#### Problem 46**Question:**Find the area of this trapezoid using \( A = \frac{1}{2}h(b_1 + b_2) \). The given dimensions are:- Height \( h = 9 \, \text{ft} \)- Base 1 \( b_1 = 7 \, \text{ft} \)- Base 2 \( b_2 = 11 \, \text{ft} \)**Explanation:**Applying the formula to find the area:\[ \text{Area} = \frac{1}{2} \times 9 \times (7 + 11) \]\[ \text

To find area of pentagon & volume of sphere B

44. The area formula for a regular polygon is ap. Find the area of a regular 2 pentagon with side length 15 cm and apothem 10 cm. A. 750 cm² B. 375 cm² C. 900 cm² D. 450 cm² 45. Sphere B has 9 times the surface area of sphere A. How many times the volume of sphere A is the volume of sphere B? A. 9 B. 27 C. 9 D. 27 46. Find the area of this trapezoid using A=h(b₁ + b₂) 9 ft 11 ft A 7 ft FRE 9x B of 094 to 8 aut al ainen ni 0 euiban riw