ID : in-8-mensuration [1] Class 8 Mensuration For more such worksheets visit www.edugain.com Answer the questions (1) The diagonals of a rhombus are 14 cm and 10 cm. Find the area of the rhombus. () A solid cylinder of radius 4 cm and height 1 cm is melted and re-shaped as a cuboid. If length and breadth of the cuboid are 16 cm and 11 cm respectively, find the height of the cuboid. (Assume π = /7) (3) Radius of two circular wheels is 18 cm and 1 cm, respectively. The wheels are connected as gear, so they rotate with each other. If the second wheel makes 900 revolutions, find the number of revolutions made by the first wheel. (4) Ajoy is painting the following box. If he painted all surfaces except bottom side of the box, find the area of surface painted by him. (All measurements are in cm). (5) The diagonal of following quadrilateral ABCD is 30 cm long, and length of perpendiculars dropped on diagonal are 15 cm and 15 cm. Find the area of quadrilateral. Copyright 017 www.edugain.com
ID : in-8-mensuration [] (6) The one third part of a cylindrical vessel of radius 1 cm and height 4 cm is filled with water. Find the volume of water in the vessel. (Assume π = /7) (7) If a rhombus is re-shaped such that one of its diagonal increases by 4%, while other diagonal decreases by 4%. Find the percentage change in the area of rhombus. (8) Area of parallelogram ABCD is x cm. If E, F, G and H are mid-points of the sides, find the area of EFGH. (9) The perimeter of a trapezoid of 5 cm height is 9 cm. If the sum of non-parallel sides is 19 cm, find the area of trapezoid. (10) Some workers are painting a hall with length, breadth and height of 50 m, 5 m and 8 m respectively. If they can paint 50 m area from one liters of paint, find the amount of paint required to paint the walls and ceiling of hall. (11) One of the diagonal of following polygon is 0 cm long, and length of perpendiculars dropped on diagonal are 8, 8 and 1 cm. Find the area of polygon. (1) A wire frame is bent into a circle of diameter 49 is reshaped as a rhombus. What is the length of the side of the resulting rhombus? Assume π = /7 (13) If diagonal of a rhombus are in ratio :3, and its area is 108 cm. Find the larger diagonal of the rhombus. (14) Find area of following parallelogram (All measurements are in cm). (15) Aditya folds a rectangular paper of size 88 cm 4 cm in cylindrical shape such that height of the cylinder is 4 cm. Find the volume of cylinder. (Assume π = /7) Copyright 017 www.edugain.com
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Answers ID : in-8-mensuration [4] (1) 70 cm We know that the area of a rhombus is calculated by multiplying the length of the diagonals and then dividing by. Step We have been told that the diagonals of the rhombus are 14 cm and 10 cm. 14 10 Therefore, the area of the rhombus = = 140 = 70 cm Therefore, the area of the rhombus is 70 cm. Copyright 017 www.edugain.com
() 6 cm ID : in-8-mensuration [5] Since the cuboid is made by melting the cylinder, the volume of the cuboid is equal to the volume of the cylinder. Step It is given that the radius, and height of the cylinder are 4 cm, 1 cm respectively. Therefore, the volume of the cylinder = πr h = 7 4 4 1 cm 3 = 1056 cm 3 Let s assume the length, breadth and height of the cuboid are l, b and h respectively. It is given that, l = 16 cm b = 11 cm The volume of the cuboid = l b h = 16 11 h = 176h Since volume of cylinder is equal to that of cuboid, 176h = 1056 h = 1056 176 h = 6 cm Step 5 Therefore, the height of the cuboid is 6 cm. Copyright 017 www.edugain.com
(3) 600 revolutions ID : in-8-mensuration [6] Distance covered by the wheel in a revolution is equal to the perimeter of the wheel. Step According to the question, the radius of the first and the second wheel is 18 cm and 1 cm, respectively. Perimeter of the second wheel = π 1 = 4π cm Distance covered by the second wheel in one revolution = 4π cm Distance covered by the second wheel in 900 revolutions = 900π 4 = 1600π cm Perimeter of the first wheel = π 18 = 36π cm Or, we can say that the number of revolutions covered by the first wheel in 36π cm = 1 revolution 1 The number of revolutions covered by the first wheel in 1 cm = revolutions 36π The number of revolutions covered by the first wheel in 1600π cm = = 600 revolutions Thus, the number of revolutions made by the first wheel are 600. 1 36π 1600π Copyright 017 www.edugain.com
ID : in-8-mensuration [7] (4) 488 cm If we look at the figure, we notice that the length, width and height of the box are 10 cm, 11 cm and 9 cm respectively. Step The surface area painted by him = The total surface area of the box - The area of the bottom side of the box = {(10 11) + (11 9) + (9 10)} - (10 11) = 598-110 = 488 cm Therefore, the area of surface painted by him is 488 cm. Copyright 017 www.edugain.com
ID : in-8-mensuration [8] (5) 450 cm Given, Length of the diagonal, AC = 30 cm Length of the perpendiculars, BE and DF are 15 cm and 15 cm Step We know that the area of a triangle = Base Height Area of the triangle ABC = AC BE = 30 15 = 5 cm Area of the triangle ACD = AC DF = 30 15 = 5 cm Thus, the area of the quadrilateral ABCD = Area of the triangle ABC + Area of the triangle ACD = 5 + 5 = 450 cm Copyright 017 www.edugain.com
ID : in-8-mensuration [9] (6) 44 cm 3 It is given that the radius and the height of the cylindrical vessel are 1 cm and 4 cm respectively. Step The volume of the cylindrical vessel = πr h = /7 1 1 4 = 13 cm 3 It is also given that the volume of water in the cylindrical vessel is one third of the volume of the cylinder vessel. The volume of water in the vessel = 1/3 (The volume of the cylindrical vessel) = 1/3 13 = 44 cm 3 Thus, the volume of water in the vessel is 44 cm 3. (7) 0.16% decrease Let s assume the length of the diagonals BD and AC of the rhombus ABCD are p and q respectively. Step The area of the rhombus = pq According to the question one of its diagonal increases by 4%, while other diagonal decreases by 4%. The new length of the diagonal BD = p + p 4 = p + 0.04p = (1 + 0.04)p 100 Copyright 017 www.edugain.com
The new length of the diagonal AC = q - q Step 5 4 100 = q - 0.04q = (1-0.04)q ID : in-8-mensuration [10] Now, the area of the rhombus = (1 + 0.04)p (1-0.04)q = (1-0.04 )pq...[since, (a + b)(a - b) = a - b ] = pq - 0.0016pq Step 6 Change in area = New area of the rhombus - The area of the rhombus pq - 0.0016pq = - pq = pq - 0.0016pq - pq = -0.0016pq Step 7 % Change in area = -0.0016pq Change in area The area of the rhombus 100 = pq 100 = -0.16% Step 8 Thus, the area of the rhombus is decreased by 0.16%. (8) x/ cm It is given that, E, F, G and H are respectively the mid-points of the sides of the parallelogram Copyright 017 www.edugain.com
ABCD. ID : in-8-mensuration [11] Step Let's join the midpoints E, F, G and H and join OF, OG, OH and OE. Also, join the diagonals AC and BD to intersect at O. In ΔBCD, F and G are the mid-points of BC and DC respectively. FG BD -------(1) [In a triangle, the line segment joining the mid-points of any two sides is parallel to the third side.] In Δ BAD, E and H are the mid-points of AB and AD respectively. EH BD -------() [In a triangle, the line segment joining the mid-points of any two sides is parallel to the third side] Step 5 From (1) and (), we get, EH BD FG Hence, EH FG --------(3) Step 6 Similarly, we can prove that, EF HG ---------(4) Step 7 From (3) and (4), we can say that, the quadrilateral EFGH is a parallelogram [Since opposite sides are parallel] Step 8 A quadrilateral is a parallelogram if its opposite sides are equal. F is the mid-point of CB and O is the mid-point of CA, FO BA FO CG ---------(5) [BA CD, Opposite sides of a parallelogram are parallel] BA CG and FO = 1 1 BA = CD [Opposite sides of a parallelogram are equal] FO = CG ---------(6) [G is the mid-point of CD] Step 9 From (5) and (6), we can sat that, the quadrilateral OFCG is a parallelogram. OP = PC [Diagonals of a parallelogram bisect each other] Δ OPF and Δ CPF have equal bases and have a common vertex F. Their altitudes are also the same. Copyright 017 www.edugain.com
Area(Δ OPF) = Area(Δ CPF) --------(7) ID : in-8-mensuration [1] 0 Similarly, Area(Δ OQF) = Area(Δ BQF) -------(8). 1 By adding (7) and (8), we get Area(Δ OPF) + Area(Δ OQF) = Area(Δ CPF) + Area(Δ BQF) Area of the parallelogram OQFP = Area(Δ CPF) + Area(Δ BQF) --------(9) Similarly, the area of the parallelogram OPGS = Area(Δ GPC) + Area(Δ DSG) -------(10) The area of the parallelogram OSHR = Area(Δ DSH) + Area(Δ HAR) -------(11) The area of the parallelogram OREQ = Area(Δ ARE) + Area(Δ EQB) -------(1) 3 By adding the corresponding sides of (9), (10), (11) and (1), we get, Area(parallelogram EFGH) = {Area(Δ CPF) + Area(Δ GPC)} + {Area(Δ DSG) + AreaΔ DSH)} + {Area(Δ HAR) + Area(Δ ARE)} + {Area(Δ BQF) + Area(Δ EQB)} = Area(Δ FCG) + Area(Δ GDH) + Area(Δ HAE) + Area(Δ EBF) = Area(parallelogram ABCD) - Area(parallelogram EFGH) Area(parallelogram EFGH) + Area(parallelogram EFGH) = Area(parallelogram ABCD) Area(parallelogram EFGH) = Area(parallelogram ABCD) Area of parallelogram EFGH = 1 Area of parallelogram ABCD 4 It is given that, the area of the parallelogram ABCD is x cm, The area of EFGH = 1 area(abcd) = x/ cm Copyright 017 www.edugain.com
ID : in-8-mensuration [13] (9) 5 cm The following figure shows the trapezoid ABCD. It is given that the height of the trapezoid ABCD = 5 cm, The perimeter of the trapezoid ABCD = 9 cm. The sum of the non-parallel sides of the trapezoid ABCD = BC + DA = 19 cm. Step The perimeter of the trapezoid ABCD = AB + BC + CD + DA 9 = AB + CD + 19 9-19 = AB + CD 10 = AB + CD AB + CD = 10 cm The area of the trapezoid ABCD = AB + CD h = 10 5 = 5 cm Thus, the area of the trapezoid is 5 cm. Copyright 017 www.edugain.com
(10) 49 liters ID : in-8-mensuration [14] It is given that the length (l), breadth (b) and height (h) of the hall be 50 m, 5 m and 8 m respectively. Step Area of the walls and ceiling of the hall = Area of the walls of the hall + Area of the ceiling of the hall = (bh + hl) + lb = {(5 8) + (8 50)} + (50 5)} = (00 + 400) + 150 = (600) + 150 = 100 + 150 = 450 m It is also given that the amount of paint required to paint 50 m area = 1 liters The amount of paint required to paint 1 m area = 1 50 liters The amount of paint required to paint 450 m area = 450 1 50 = 49 liters Thus, the amount of paint required to paint the walls and ceiling of hall is 49 liters. Copyright 017 www.edugain.com
(11) 4 cm ID : in-8-mensuration [15] We know that, the area of a triangle = 'The base of the triangle' 'The height of the triangle' The area of a trapezium = trapezium' The sum of the lengths of the parallel sides 'The height of the Step The following figure shows the required polygon, The area of the polygon = The area of the ΔABC + The area of the ΔAGD + The area of the ΔEFC + The area of the trapezium DEFG = 0 8 + 4 8 + 8 1 + (1+8) 8 160 + 3 + 96 + 160 = = 4 cm Thus, the area of the polygon is 4 cm. Copyright 017 www.edugain.com
(1) 38.5 ID : in-8-mensuration [16] A wire frame of some length was first bent into a circle and then reshaped as a rhombus: Wire Circle Rhombu Step Let us first find the length of the wire frame. We know that the total length of the boundary of a circle is called circumference and is given by: Circumference = πr, where r is the radius of the circle. Since the circle is formed by the wire frame, the length of the wire frame = πr = 7 4.5 [It is given that the radius of the circle is 49/ = 4.5 and π = /7] = 154 Now, we know that the same wire frame with length 154 is reshaped as a rhombus. A rhombus has 4 sides a sides are equal. This means, the length of a side of the rhombus will be 154 divided by 4. That is: 154/4 = 38.5 Thus the length of the side of the resulting rhombus is 38.5. Copyright 017 www.edugain.com
(13) 18 cm ID : in-8-mensuration [17] It is given that diagonal of a rhombus are in ratio :3. Therefore, let us assume that the diagonals of rhombus are x cm and 3x cm. Step Area of the rhombus = (Multiplication of diagonals) / 108 = (x)(3y) 108 = 6x x = 108 / 6 x = 36 x = 6... (Since x cannot be negative) Therefore, two diagonals are, x = 6 = 1 cm 3 x = 3 6 = 18 cm Hence larger diagonal of the rhombus is 18 cm. (14) 77 cm If we look at the figure, we notice that, the base of the parallelogram = 11 cm, and the height of the parallelogram = 7 cm. Step The area of the parallelogram = Base Height = 11 7 = 77 cm Copyright 017 www.edugain.com
ID : in-8-mensuration [18] (15) 464 cm 3 The paper has to be folded along the length in order to get the cylinder of height 4 cm and the perimeter of the base of the cylinder is equal to the length of the paper. Step Let the radius of cylinder be r cm. Since, the base of the cylinder is circular, its perimeter is πr. Therefore, πr = 88 /7 r = 88 44/7 r = 88 r = 88 7/44 r = 14 cm The radius of the cylinder = 14 cm Now, the volume of the cylinder = πr h = 7 14 14 4 = 464 cm 3 Copyright 017 www.edugain.com