Grade 8 Geometry Problems and Questions
with with Solutions and Explanations

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

  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.
    A = 2 radius π height + π radius2 + π radius2
    = 2 5 π 34 + π 52 + π 52
    = 340 π + 25 π + 25 π = 390 π square centimeters
    = 1224.6 square centimeters (using π = 3.14)
    The volume V of the given cylinder is equal to
    V = π radius2 height
    = π 25 34 = 850 π cubic centimeters
    = 2669 cubic centimeters (using π = 3.14)

  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.
    A = π radius slant height + π radius2
    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.
    S2 = 152 + 52 = 250 S = 15.8 cm
    A = π 5 15.8 + π 25
    = 104 π square centimeters
    = 327 square centimeters (using π = 3.14)
    The volume V of the cone is given by.
    V = (1/3) π r2 h
    = (1/3) π 52 15
    = 125 π cubic centimeters
    = 393 cubic centimeters

  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
    150 / 6 = 25 square feet
    Since the face of a cube is square, if x is the length of an edge of the cube, then
    x2 = 25 square feet
    and x = 5 feet
    The volume V of the cube is equal to
    V = 5 5 5 = 125 cubic feet

  4. Which two angles are complementary?

    1. 21 degrees and 78 degrees
    2. 58 degrees and 22 degrees
    3. 67 degrees and 23 degrees
    4. 140 degrees and 40 degrees

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

    1. 21 + 78 = 99
    2. 58 + 22 = 80
    3. 67 + 23 = 90
    4. 140 + 40 = 180

    Angles 67 and 23 are complementary

  5. Which two angles are not supplementary?

    1. 30 degrees and 150 degrees
    2. 5 degrees and 175 degrees
    3. 89 degrees and 91 degrees
    4. 23 degrees and 177 degrees

    Solution
    Two angles are supplementary if the sum of their measures is equal to 180

    1. 30 + 150 = 180
    2. 5 + 175 = 180
    3. 89 + 91 = 180
    4. 23 + 177 = 200

    Angles 23 and 177 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
    A = (1/2)(base1 + base2) height = (1/2)(27 + 13) h
    The area A of the trapezoid is equal to 400 square cm. Hence
    (1/2)(27 + 13) h = 400
    (1/2)(40) h = 400
    20 h = 400
    h = 20 cm

  7. Find the width w of the parallelogram so that its area is equal to 600 square feet.

    geometry problem 7

    .
    Solution
    The area A of a parallelogram is equal to
    A = W height = W 20 = 600
    W = 600 / 20 = 30 feet

  8. Find area of the shaded shape.

    geometry problem 8

    .
    Solution
    The shaded (blue) shape is made up of a large rectangle of length 40 and width 25 from which a smaller rectangle (left) of width 14 and width 12 and a triangle (right) of base 25 and height 10 have been cut. Hence the area of this shape is equal to
    Area of large rectangle - area of smaller rectangle - area of triangle
    = 40 25 - 14 12 - (1/2) 25 10 = 707 square cm

  9. Find the total surface area of the box open at the top.

    geometry problem 9

    .
    Solution
    Area of the open box is equal to
    The areas of the front and back faces + the areas of the left and right faces + the area ot bottom
    = 2(12 11) + 2(9 11) + 9 12 = 570 square cm

  10. If the quadrilateral ABCD is a parallelogram, what are the coordinates of point D?

    geometry problem 10

    .
    Solution
    Since ABCD is a parallelogram sides AB and DC are parallel. Because points A and B have the same y coordinate, AB is parallel to the x axis. Hence DC is also parallel to the x axis and therefore the y coordinate of point D is the same as the y coordinate of C and is equal to 2. Also distance AB must be equal to distance CD. Hence if a is the x coordinate of C, then
    7 - a = 8 - 2
    x = 1
    Point C has coordinates (1 , 2)

  11. Which of these are right triangles?

    geometry problem 11

    . Solution
    The unknown angle in triangle A) is equal to:
    180 - (62 +28) = 90
    Triangle A) is a right triangle.
    To be a right triangle, the sides of triangle B) must satisfy Pythagora's theorem.
    32 + 3.92 ? = 52
    left side: 32 + 3.92 = 24.21
    right side: 52 = 25
    Triangle B) is not a right triangle.
    To be a right triangle, the sides of triangle C) must satisfy Pythagora's theorem.
    92 + 122 ? = 152
    left side: 92 + 122 = 225
    right side: 152 = 225
    Triangle C) is a right triangle.

  12. Find x if triangle ABC is a right triangle with angle B = 90.

    geometry problem 12

    .
    Solution
    Angle BAC is equal to
    180 - (47 + 90) = 43
    ∠ABC + x + 35 = 180
    x = 180 - (43 + 35) = 102

  13. Find all the unknown sides x, y, z and w if all three triangles are similar.

    geometry problem 13

    .
    Solution
    When two triangles are similar, their corresponding angles are equal and their corresponding sides are proportional. The corresponding sides of the triangle on the left and the one in the middle gives:
    12 / 17 = 36 / y and 12 / 3 = 36 / x
    Solve the above equations to find
    y = 51 and x = 9
    The corresponding sides of the triangle on the right and the one in the middle gives:

    12 / 17 = z / 204 and 17 / 3 = 204 / w
    Solve the above equations to find
    z = 144 and w = 36

  14. A quadrilateral with vertices (-2,6) , (6,8) , (9,2) and (4,-1) is reflected on the x axis. What are the coordinates of the vertices of the quadrilateral after reflection?
    Solution
    A reflection on the x axis of a point with coordinates (a , b) on the x axis gives the point (a , -b). Hence the 4 points are reflected into the points:
    (-2,-(6)) , (6,-(8)) , (9,-(2)) and (4,-(-1))
    simplify to
    (-2,-6) , (6,-8) , (9,-2) and (4,1)

  15. The side of cube A is 3 times the side of cube B. The volume of cube A is 3,375 cubic feet. Find the volume of cube B.
    Solution
    Let x be the side of cube B. Hence the side of cube A is 3 x. The volume of cube A is given by
    Va = (3 x) (3 x) (3 x)
    Simplify the right side to obtain
    Va = 27 x 3
    The volume of cube B is given by Vb = x x x = x 3
    The ratio of Vb and Va is given by
    Vb / Va = x 3 / 27 x 3
    Simplify to obtain
    Vb / Va = 1 / 27
    Va is given, hence
    Vb / 3375 = 1 / 27
    Multiply both sides by 3375
    3375 Vb / 3375 = 3375 1 / 27
    Simplify to obtain
    Vb = 3375 / 27 = 125 cubic feet

  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 rctangles A and B respectively. Since the two rectangles are similar, we have the following proportionality
    Wa / 24 = Wb / 96
    The areas Aa and Ab of rectangles A and B respectively are given by
    Aa = 24 Wa and Wb = 96 Wb
    The ratio Aa / Ab is given by
    Aa / Ab = 24 Wa / 96 Wb = (24 / 96) (Wa / Wb)
    The equation Wa / 24 = Wb / 96 obtained above may be written as
    Wa / Wb = 24 / 96
    We now substitute to obtain
    Aa / Ab = (24 / 96) (24 / 96) = 1 / 16

More Middle School Math (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers
More High School Math (Grades 10, 11 and 12) - Free Questions and Problems With Answers
More Primary Math (Grades 4 and 5) with Free Questions and Problems With Answers
Author - e-mail
Home Page
Updated: 6 August 2018 (A Dendane)