Question

homework #1
show all calculations for each problem.
1 determine the areas of the figures.

Answer:

To solve the area problems, we'll analyze each figure one by one. Let's start with the first figure (a) which appears to be a parallelogram.

Figure (a): Parallelogram

A parallelogram’s area is given by \( \text{Area} = \text{base} \times \text{height} \).

• From the diagram, base \( b = 40 \, \text{m} \), height \( h = 20 \, \text{m} \) (assuming the height corresponding to the base of 40 m).
• Area \( = 40 \times 20 = 800 \, \text{m}^2 \).

Figure (b): Trapezoid

A trapezoid’s area is \( \text{Area} = \frac{1}{2} \times (a + b) \times h \), where \( a, b \) are the two parallel sides, and \( h \) is the height.

• Parallel sides: \( a = 10 \, \text{ft} \), \( b = 20 \, \text{ft} \); height \( h = 30 \, \text{ft} \).
• Area \( = \frac{1}{2} \times (10 + 20) \times 30 = \frac{1}{2} \times 30 \times 30 = 450 \, \text{ft}^2 \).

Figure (c): Triangle (or Parallelogram? Wait, the diagram shows a triangle with base 30 m and height 18 m? Wait, no—if it’s a triangle, area is \( \frac{1}{2} \times \text{base} \times \text{height} \). Wait, the sides are 30 m, 30 m, and height 18 m? Wait, maybe it’s a triangle:

• Base \( = 30 \, \text{m} \), height \( = 18 \, \text{m} \).
• Area \( = \frac{1}{2} \times 30 \times 18 = 270 \, \text{m}^2 \).

Figure (d): Quadrilateral (Two Right Triangles?)

The figure has two right angles. Let’s split it into two right triangles:

• First triangle: legs \( 25 \, \text{ft} \) and \( 65 \, \text{ft} \). Area \( = \frac{1}{2} \times 25 \times 65 = 812.5 \, \text{ft}^2 \).
• Second triangle: legs \( 50 \, \text{ft} \) and \( 60 \, \text{ft} \). Area \( = \frac{1}{2} \times 50 \times 60 = 1500 \, \text{ft}^2 \).
• Total area \( = 812.5 + 1500 = 2312.5 \, \text{ft}^2 \).

Figure (e): Trapezoid (or Composite Figure?)

The figure has a trapezoid with bases \( 20 \, \text{m} \) and \( 10 \, \text{m} \), and the average length of the top/bottom? Wait, the base is 15 m? Wait, maybe it’s a trapezoid with height 15 m? Wait, no—let’s re-examine. The figure has a height of 20 m, a lower base of 15 m, and a top base of 30 m? Wait, maybe it’s a trapezoid:

• Bases \( a = 30 \, \text{m} \), \( b = 15 \, \text{m} \); height \( h = 10 \, \text{m} \) (the middle height). Wait, no—maybe it’s a composite of a trapezoid and a triangle? Alternatively, use the formula for the area of a trapezoid: \( \text{Area} = \frac{1}{2} \times (a + b) \times h \). If \( a = 30 \, \text{m} \), \( b = 15 \, \text{m} \), \( h = 10 \, \text{m} \), then \( \text{Area} = \frac{1}{2} \times (30 + 15) \times 10 = 225 \, \text{m}^2 \). But this might be incorrect—need clearer dimensions.

Figure (f): Right Triangle

A right triangle’s area is \( \frac{1}{2} \times \text{leg}_1 \times \text{leg}_2 \).

• Legs: \( 150 \, \text{ft} \) and (using Pythagoras, but wait, hypotenuse is 170 ft? Wait, \( 150^2 + x^2 = 170^2 \)? \( 22500 + x^2 = 28900 \) → \( x^2 = 6400 \) → \( x = 80 \, \text{ft} \).
• Area \( = \frac{1}{2} \times 150 \times 80 = 6000 \, \text{ft}^2 \).

Figure (g): Composite Figure (Triangle and Trapezoid?)

The figure has a height of 4 m, 6 m, and 8 m, with a base of 40 m and a middle segment of 15 m. Let’s split it into two triangles and a trapezoid? Alternatively, use the formula for the area of a triangle: Wait, maybe it’s a triangle with base 40 m and height 8 m? No, the middle height is 6 m. Alternatively, use the formula for the area of a polygon: \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \). Wait, this is unclear—need more precise dimensions.

Figure (h): Right Triangle

Legs: Let’s assume the legs are 325 ft and (wait, the right angle is at the bottom, so legs are, say, 300 ft and 125 ft? Wait, no—if hypotenuse is 325 ft, and one leg is 300 ft, then the other leg is \( \sqrt{325^2 - 300^2} = \sqrt{105625 - 90000} = \sqrt{15625} = 125 \, \text{ft} \).

• Area \( = \frac{1}{2} \times 300 \times 125 = 18750 \, \text{ft}^2 \).

Figure (i): Sector Minus a Triangle

The figure is a sector with radius \( r = 28 \, \text{cm} \) and a right angle (90°) removed.

• Area of full circle: \( \pi r^2 = \pi (28)^2 = 784\pi \, \text{cm}^2 \).
• Sector area (3/4 of circle, since 360° - 90° = 270°): \( \frac{270}{360} \times 784\pi = \frac{3}{4} \times 784\pi = 588\pi \approx 1847.26 \, \text{cm}^2 \).

Figure (j): Semicircle? Wait, a circle with radius 100 ft and a base of 100 ft? No, it’s a circle with diameter 100 ft? Wait, radius \( r = 50 \, \text{ft} \), and a base of 100 ft (a straight line). Wait, maybe it’s a semicircle? No, the base is 100 ft (diameter), so area of the circle: \( \pi (50)^2 = 2500\pi \, \text{ft}^2 \), but it’s a semicircle? Wait, the diagram shows a “flat” base, so maybe it’s a semicircle with diameter 100 ft: \( \text{Area} = \frac{1}{2} \pi (50)^2 = 1250\pi \approx 3927 \, \text{ft}^2 \).

Figure (k): Sector with Radius 20 cm, Angle 240° (360° - 120°)

• Area of sector: \( \frac{\theta}{360} \times \pi r^2 \), where \( \theta = 240^\circ \), \( r = 20 \, \text{cm} \).
• Area \( = \frac{240}{360} \times \pi (20)^2 = \frac{2}{3} \times 400\pi = \frac{800\pi}{3} \approx 837.76 \, \text{cm}^2 \).

Figure (l): Composite Figure (Trapezoid + Semicircle)

• Trapezoid: Bases \( 50 \, \text{ft} \) and \( 100 \, \text{ft} \), height \( 50 \, \text{ft} \) (assuming). Area \( = \frac{1}{2} \times (50 + 100) \times 50 = 3750 \, \text{ft}^2 \).
• Semicircle: Diameter \( 100 \, \text{ft} \), radius \( 50 \, \text{ft} \). Area \( = \frac{1}{2} \pi (50)^2 = 1250\pi \approx 3927 \, \text{ft}^2 \).
• Total area \( = 3750 + 3927 = 7677 \, \text{ft}^2 \).

Final Answers (for each figure):

(a) \( \boldsymbol{800 \, \text{m}^2} \)
(b) \( \boldsymbol{450 \, \text{ft}^2} \)
(c) \( \boldsymbol{270 \, \text{m}^2} \) (if triangle)
(d) \( \boldsymbol{2312.5 \, \text{ft}^2} \)
(f) \( \boldsymbol{6000 \, \text{ft}^2} \)
(h) \( \boldsymbol{18750 \, \text{ft}^2} \)
(i) \( \boldsymbol{588\pi \, \text{cm}^2 \approx 1847 \, \text{cm}^2} \)
(k) \( \boldsymbol{\frac{800\pi}{3} \, \text{cm}^2 \approx 838 \, \text{cm}^2} \)

(Note: Some figures lack clear dimensions, so these are approximate based on visible labels. For precise answers, confirm all measurements.)