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 Pi height + Pi radius2 + Pi radius2

    = 2 5 Pi 34 + 3.14 52 + Pi 52

    = 340 Pi + 25 Pi + 25 Pi = 390 Pi square centimeters

    = 1224.6 square centimeters

    The volume V of the given cylinder is equal to

    V = Pi radius2 height

    = Pi 25 34 = 850 Pi cubic centimeters

    = 2669 cubic centimeters (using Pi = 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 = Pi radius slant height + Pi 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 = Pi 5 15.8 + Pi 25

    = 104 Pi square centimeters

    = 327 square centimeters (using Pi = 3.14)

    The volume V of the cone is given by.

    V = (1/3) Pi r2 h

    = (1/3) Pi 52 15

    = 125 Pi 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. br> 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 .


  12. Find x if triangle ABC is a right triangle.

    geometry problem 12 .


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

    geometry problem 13 .


  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?

  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.

  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.


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Updated: 6 April 2009 (A Dendane)