Solutions to Grade 8 Geometry Problems

Detailed solutions and full explanations to Grade 8 Geometry Problems and Questions are presented below.

1. Find the total surface area and the volume of a closed cylindrical container with radius 5 cm and a height of 34 cm.

Solution

The total surface area A of a cylinder is given by the sum of the area of the lateral surface plus the areas of the base and bottom of the container.

\begin{align} A &= 2 \times \text{radius} \times \pi \times \text{height} + \pi \times \text{radius}^2 + \pi \times \text{radius}^2 \\ &= 2 \times 5 \times \pi \times 34 + \pi \times 5^2 + \pi \times 5^2 \\ &= 340\pi + 25\pi + 25\pi = 390\pi \text{ square centimeters} \\ &= 1224.6 \text{ square centimeters (using } \pi = 3.14\text{)} \end{align}

The volume V of the given cylinder is equal to

\begin{align} V &= \pi \times \text{radius}^2 \times \text{height} \\ &= \pi \times 25 \times 34 = 850\pi \text{ cubic centimeters} \\ &= 2669 \text{ cubic centimeters (using } \pi = 3.14\text{)} \end{align}

2. Find the total surface area and the volume of a closed conical container with radius 5 cm and a height of 15 cm. (Round your answer to the nearest unit.)

Solution

The total surface area A of a cone is given by the sum of the area of the lateral surface plus the area of the base.

\begin{align} A &= \pi \times \text{radius} \times \text{slant height} + \pi \times \text{radius}^2 \end{align}

The slant height S, the height of the cone (h = 15 cm) and the radius (r = 5 cm) of its base form a right triangle where S may be found using Pythagor's theorem as follows.

\begin{align} S^2 &= 15^2 + 5^2 = 250 \\ S &= 15.8 \text{ cm} \end{align}
\begin{align} A &= \pi \times 5 \times 15.8 + \pi \times 25 \\ &= 104\pi \text{ square centimeters} \\ &= 327 \text{ square centimeters (using } \pi = 3.14\text{)} \end{align}

The volume V of the cone is given by

\begin{align} V &= \frac{1}{3} \times \pi \times r^2 \times h \\ &= \frac{1}{3} \times \pi \times 5^2 \times 15 \\ &= 125\pi \text{ cubic centimeters} \\ &= 393 \text{ cubic centimeters} \end{align}

3. A cube has a total surface area of the six faces equal to 150 square feet. What is the volume of the cube?

Solution

The surface area of one square face is equal to

\begin{align} \frac{150}{6} = 25 \text{ square feet} \end{align}

Since the face of a cube is square, if x is the length of an edge of the cube, then

\begin{align} x^2 &= 25 \text{ square feet} \\ x &= 5 \text{ feet} \end{align}

The volume V of the cube is equal to

\begin{align} V = 5 \times 5 \times 5 = 125 \text{ cubic feet} \end{align}

4. Which two angles are complementary?

  1. \(21^{\circ} \) and \(78^{\circ} \)
  2. \(58^{\circ} \) and \(22^{\circ} \)
  3. \(67^{\circ} and \(23^{\circ} \)
  4. \(140^{\circ} \) and \(40^{\circ} \)

Solution

Two angles are complementary if the sum of their measures is equal to 90°

  1. \(21^{\circ} + 78^{\circ} = 99^{\circ}\)
  2. \(58^{\circ} + 22^{\circ} = 80^{\circ}\)
  3. \(67^{\circ} + 23^{\circ} = 90^{\circ}\)
  4. \(140^{\circ} + 40^{\circ} = 180^{\circ}\)

Angles \( 67^{\circ} \) and \( 23^{\circ} \) are complementary

5. Which two angles are not supplementary?

  1. \(30^{\circ}\) and \( 150^{\circ} \)
  2. \( 5^{\circ} \) and \( 175^{\circ} \)
  3. \( 89 ^{\circ} \) and \( 91^{\circ} \)
  4. \(23^{\circ}\) and \( 177^{\circ} \)

Solution

Two angles are supplementary if the sum of their measures is equal to \( 180^{\circ} \)

  1. \(30^{\circ} + 150^{\circ} = 180^{\circ}\)
  2. \(5^{\circ} + 175^{\circ} = 180^{\circ}\)
  3. \(89^{\circ} + 91^{\circ} = 180^{\circ}\)
  4. \(23^{\circ} + 177^{\circ} = 200^{\circ}\)

Angles \(23^{\circ}\) and \( 177^{\circ} \) are not supplementary.

6. Find the height h of the trapezoid so that its area is equal to 400 square cm.

geometry problem 6

Solution

The area A of a trapezoid is given by the formula

\begin{align} A &= \frac{1}{2}(\text{base}_1 + \text{base}_2) \times \text{height} \\ &= \frac{1}{2}(27 + 13) \times h \end{align}

The area A of the trapezoid is equal to 400 square cm. Hence

\begin{align} \frac{1}{2}(27 + 13) \times h &= 400 \\ \frac{1}{2}(40) \times h &= 400 \\ 20 \times h &= 400 \\ h &= 20 \text{ cm} \end{align}

16. The length of rectangle A is 24 cm and the length of rectangle B is 96 cm. The two rectangles are similar. Find the ratio of the area of A to the area of B.

Solution

Let Wa and Wb be the width of rectangles A and B respectively. Since the two rectangles are similar, we have the following proportionality

\begin{align} \frac{W_a}{24} = \frac{W_b}{96} \end{align}

The areas Aa and Ab of rectangles A and B respectively are given by

\begin{align} A_a = 24 \times W_a \text{ and } A_b = 96 \times W_b \end{align}

The ratio Aa / Ab is given by

\begin{align} \frac{A_a}{A_b} = \frac{24 \times W_a}{96 \times W_b} = \frac{24}{96} \times \frac{W_a}{W_b} \end{align}

The equation Wa / 24 = Wb / 96 obtained above may be written as

\begin{align} \frac{W_a}{W_b} = \frac{24}{96} \end{align}

We now substitute to obtain

\begin{align} \frac{A_a}{A_b} = \frac{24}{96} \times \frac{24}{96} = \frac{1}{16} \end{align}

Answers to the Above Questions

  1. area = 390p square cm, volume = 850p cubic cm
  2. area = 327 square cm, volume = 393 cubic cm
  3. volume = 125 cubic feet
  4. C) \( 67^{\circ} \) and \( 23^{\circ} \) degrees are complementary
  5. D) \( 23^{\circ} \) and \( 177^{\circ} \) are not supplementary
  6. h = 20 cm
  7. w = 30 feet
  8. area = 707 square cm
  9. surface area = 570 square cm
  10. (1, 2)
  11. triangles A) and C) are right triangles
  12. \( x = 102^{\circ}\)
  13. x = 9, y = 51, z = 144, w = 36
  14. vertices after reflection (-2,-6), (6,-8), (9,-2) and (4,1)
  15. volume of cube B = 125 cubic feet
  16. ratio of area of A to area of B = 1:16

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