Grade 8 Mensuration. Answer t he quest ions. Choose correct answer(s) f rom given choice. For more such worksheets visit

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1 Grade 8 Mensuration For more such worksheets visit ID : cn-8-mensuration [1] Answer t he quest ions (1) We draw a square inside a rectangle. The ratio of the rectangle's width the square's side is 5:1, and the ratio of the rectangle's length to its width is 2:1. What percentage of the rectangle's area is inside the square? (2) A square park has an area of 900 sq. m. It has to be f enced. The f encing will require use of a wire that must be able to enclose the park 4 times, and each circuit of the wire would be 4 % greater than the perimeter of the park. What is the length of wire needed f or this? (3) A cuboid has width 3, height 10 and length 7. If you had to construct a cube of integer sides that had a larger volume than the cuboid, what would the size of the edge of the cube? Choose correct answer(s) f rom given choice (4) Find the amount of water that can be f illed in this cuboid. (All measurements are in cm). a. 264 cm 3 b. 259 cm 3 c. 252 cm 3 d. 246 cm 3 (5) A wire f rame is bent into a circle of radius 56 is reshaped as a rhombus. What is the length of the side of the resulting rhombus? Assume π = 22/7 a. 89 b. 84 c. 86 d. 88

2 (6) In the rhombus shown below angle PRS = X + 10 and angle QSP = 6X - 12, then what is the value of X? ID : cn-8-mensuration [2] a. 2 b. 4 c. 6 d. 1 (7) If a rhombus is re-shaped such that one of its diagonal increases by 5%, while other diagonal decreases by 5%. Find the percentage change in the area of rhombus. a. 2.5% decrease b. 0.25% decrease c. 0.25% increase d. 2.5% increase (8) Some workers are painting a hall with length, breadth and height of 20 m, 45 m and 10 m respectively. If they can paint m 2 area f rom one liters of paint, f ind the amount of paint required to paint the walls and ceiling of hall. a. 28 liters b. 22 liters c. 16 liters d. 21 liters (9) A square and an equilateral triangle have the same perimeter. The diagonal of the square is 3 2 cm. What is the area of the triangle? a. 12 b. 4 3 c. 4 2 d. 4 3 Fill in the blanks (10) The area of f ollowing trapezoid = cm 2. (All measurements are in cm). (11) If the sides of a square are increased by 30%, then the area of the square increases by %.

3 (12) What is the area of a rectangle with length 4 m 8 dm 7 cm and width as 9 m 7 dm 4 cm is ID : cn-8-mensuration [3] sq. m. (13) If radius and height of a closed cylinder are 7 cm and 17 cm respectively, its surf ace area = cm 2 (Assume π = 22/7) (14) If diagonals of a rhombus shaped park are 10 and 24 meters, the cost of f encing the park boundary at the rate of 4 per meter = cm. (15) In a trapezoid of area 30 cm 2, the length of two parallel sides are in ratio 3:1. If the height of trapezoid is 3 cm, the length of larger of the two parallel sides is cm Edugain ( All Rights Reserved Many more such worksheets can be generated at

4 Answers ID : cn-8-mensuration [4] (1) 2 % (2) m Let's assume that a be the length of the park. Theref ore the area of a square park = a 2 = 900 a 2 = 30 2 a = 30 m Now the perimeter of a square = 4a = 4 X 30 = 120 m Since each circuit of the wire be 4% greater than the perimeter of the park. Theref ore the length of the each circuit of the wire = perimeter of the square park + 4% of the perimeter of the square park 4 = X = = = m The f encing will require use of a wire that must be able to enclose the park 4 times. Theref ore the length of wire needed f or this = 4 X = m. (3) 6

5 (4) c. 252 cm 3 ID : cn-8-mensuration [5], If we look at the f igure, we notice that the length, width and height of the cuboid are 6 cm, 7 cm and 6 cm respectively. The amount of water that can be f illed in the cuboid will be equal to the volume of the cuboid. The volume of the cuboid = Length Width Height = 6 cm 7 cm 6 cm = 252 cm 3 Thus, 252 cm 3 water can be f illed in this cuboid.

6 (5) d. 88 ID : cn-8-mensuration [6] A wire f rame of some length was f irst bent into a circle and then reshaped as a rhombus: Wire Circle Rhombus Let us f irst f ind the length of the wire f rame. We know that the total length of the boundary of a circle is called its circumf erence and is given by: Circumf erence = 2πr, where r is the radius of the circle. Since the circle is f ormed by the wire f rame, the length of the wire f rame = 2πr = [It is given that the radius of the circle is 56 and π = 22/7] 7 = 352 Now, we know that the same wire f rame with length 352 is reshaped as a rhombus. A rhombus has 4 sides and all sides are equal. This means, the length of a side of the rhombus will be 352 divided by 4. That is: 352/4 = 88 Thus the length of the side of the resulting rhombus is 88. (6) a. 2 (7) b. 0.25% decrease

7 (8) b. 22 liters ID : cn-8-mensuration [7] It is given that the length (l), breadth (b) and height (h) of the hall be 20 m, 45 m and 10 m respectively. Area of the walls and ceiling of the hall = Area of the walls of the hall + Area of the ceiling of the hall = 2(bh + hl) + lb = 2{(45 10) + (10 20)} + (20 45)} = 2( ) = 2(650) = = 2200 m 2 It is also given that the amount of paint required to paint m 2 area = 1 liters The amount of paint required to paint 1 m 2 area = 1 liters The amount of paint required to paint 2200 m 2 area = = 22 liters Thus, the amount of paint required to paint the walls and ceiling of hall is 22 liters.

8 (9) d. 4 3 ID : cn-8-mensuration [8] Lets assume s and t are the sides of square and triangle respectively as shown in the f ollowing f igures. Square Equilateral triangle According to question the diagonal AC of the square is 3 2 cm. Now in right angle triangle ABC AB 2 + BC 2 = AC 2 s 2 + s 2 = (3 2) 2 2(s) 2 = (3 2) 2 s 2 = (3 2)2 2 s 2 = 9 cm Perimeter of the given square = 4s Perimeter of the given equilateral triangle = 3t According to question the perimeter of a square is equal to the perimeter of an equilateral triangle. Theref ore 3t = 4s Squaring both sides (3t) 2 = (4s) 2 9t 2 = 16s 2 t 2 = 16 9 [Since s 2 = 9] 9 t 2 = 16 cm Now the area of an equilateral traingle = t2 3 4 = 16 3 [Since t 2 = 16 ] 4 = 4 3 cm 2 Theref ore the area of the triangle is 4 3 cm 2.

9 (10) ID : cn-8-mensuration [9] It is given that, Length of parallel sides are 16 cm and 19 cm. Distance between parallel sides is 9 cm. Area of trapezoid = 1 2 (Sum of parallel sides) (Distance between parallel sides) = 1 2 ( ) 9 = = Theref ore, the area of the trapezoid is cm 2. (11) 69 Lets assume s and A are the side of the square and area of the square respectively. According to question the sides of the square are increased by 30% 30 theref ore the side of the square = s + s s + 30s = = 1.3s Now the area of the square = (1.3s) 2 = 1.69s 2 = 1.69A [Since A = s 2 ] The area of the square increase = 1.69A - A = 0.69A Theref ore you can say that the area of the square increase = 0.69 = 69% %

10 (12) ID : cn-8-mensuration [10] To calculate the area of a rectangle f irst of all we have to change the length and width of a rectangle into same unit. Since 1cm = 1 m and 1dm = 1 10 m theref ore the length of a rectangle = 4 m 8 dm 7 cm = and the width of a rectangle = 9 m 7 dm 4 cm = Now the area of a rectangle = length width = = sq. m = 4.87 m = 9.74 m (13) 1056 It is given that the radius (r) and height (h) of cylinder is 7 cm and 17 cm respectively. The surf ace area of the closed cylinder = 2πr(r + h) = (7 + 17) 7 = 44(24) = 1056 cm 2 Thus, the surf ace area of the closed cylinder is 1056 cm 2.

11 (14) 208 ID : cn-8-mensuration [11] The f ollowing f igure shows the rhombus shaped park, It is given that, the diagonal of the rhombus shaped park are 10 and 24 meters. Theref ore, AC = 10 meters, BD = 24 meters We know that the diagonals of a rhombus bisect each other at right angles. Theref ore, the OC = AC/2 = 10/2 = 5 meters, OD = BD/2 = 24/2 = 12 meters, and COD = 90 Now, in right angled triangle COD, CD = (OC 2 + OD 2 ) = ( ) = = 169 = 13 meters We know that the all sides of a rhombus are congruent. The perimeter of the rhombus ABCD = 4(CD) = 4 13 = 52 meters Step 5 The cost of f encing the one meter park boundary = 4 The cost of f encing the 52 meters park boundary = 52 4 = 208 Step 6 Thus, the cost of f encing the park boundary is 208.

12 (15) 15 ID : cn-8-mensuration [12] Let's assume that 'a' and 'b' are the lengths of the parallel sides of the trapezoid and 'h' is the height of the trapezoid. Then, the area of the trapezoid = a + b h. 2 It is given that the length of the two parallel sides of the trapezoid are in ratio 3:1. Let's assume 'x' is the ratio f actor of the parallel sides. Then, the parallel sides 'a' and 'b' of the trapezoid can be considered to be 3x and 1x respectively. It is also given that the area of the trapezoid = 30 cm 2, and the height of the trapezoid(h) = 3 cm. 3x + 1x Thus, the area of the trapezoid = = 4x = 4x = 4x 3 60 = x 3 4 x = 5 Step 5 Now that we know the value of 'x', a = 3x = 3 5 = 15 cm, b = 1x = 1 5 = 5 cm. Step 6 Thus, the length of the larger of the two parallel sides is 15 cm.

(1) The perimeter of a trapezoid of 10 cm height is 35 cm. If the sum of non-parallel sides is 25 cm,

Grade 8 Mensuration For more such worksheets visit www.edugain.com ID : ww-8-mensuration [1] Answer t he quest ions (1) The perimeter of a trapezoid of 10 cm height is 35 cm. If the sum of non-parallel