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Mensuration Test 6
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Mensuration Test 6
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  • Question 1/15
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    The perimeter of a right angle triangle is 60 cm. If the first side is 2 cm greater than the second side and the third side is 2 cm less than half of the second side, then find the area of the right angle triangle.

    Solutions

    Let the sides of the triangle be x cm, x + 2 cm and x/2 – 2 cm

    Perimeter of the triangle = x + x + 2 + x/2 – 2 cm = 5x/2 cm = 60 cm (given)

    ⇒ x = 120/5 = 24

    Sides of the triangle are 24 cm, 26 cm and 10 cm

    Since it is a right angled triangle,

    ∴ Area of Triangle = 1/2 × 24 × 10 cm2 = 120 cm2

  • Question 2/15
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    The perimeter of a rhombus is 104 m and its area is 480 m2. Find the ratio of the lengths of its diagonals.
    Solutions

    Let the length of side of rhombus be a m

    Perimeter of the rhombus = 4a m = 104 m

    ⇒ a = 26 m

    Let the length of diagonals be x m and y m

    Since diagonals of a rhombus bisect each other at 90ᵒ

    Using Pythagoras theorem, we get

    ⇒ (x/2)2 + (y/2)2 = 262

    ⇒ x2 + y2 = 2704      ----(1)

    Area of rhombus = 1/2 × x × y m2 = 480 m2

    ⇒ xy = 960

    ⇒ y = 960/x

    Putting value of y in equation (1), we get

    ⇒ x2 + 921600/x2 = 2704

    ⇒ x4 - 2704x2 + 921600 = 0

    ⇒ x4 - 400x2 - 2304x2 + 921600 = 0

    ⇒ (x2 - 2304)(x2 - 400) = 0

    ⇒ x2 = 400       or         x2 = 2304

    ⇒ x = 20          or         x = 48

    When x = 20, y = 48 and when y = 20, x = 48

    ∴ Required ratio = 20 : 48 = 5 : 12

  • Question 3/15
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    The perimeter of a trapezium shaped ground is 5.2 km. The non - parallel sides and one of the parallel sides are of length 1 km each. Find the area of the ground.

    Solutions

    Let the length of other parallel side be a km.

    Perimeter of the ground = 1 + 1 + 1 + a km = 3 + a km = 5.2 km

    ⇒ a = 2.2 km

    Let the length of altitude be x km.

    Using Pythagoras theorem, we get

    x2 + 0.62 = 12

    ⇒ x2 = 1 - 0.36 = 0.64

    ⇒ x = 0.8

    ∴ Area of the trapezium = 1/2 × (1 + 2.2) × 0.8 km2 = 1.28 km2

  • Question 4/15
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    The perimeter of two parallelograms A and B is 26 cm and 44 cm respectively and length of altitudes are 4 cm and 6 cm respectively. The smaller side of B is twice of the smaller side of A and the base of B is 4 cm greater than the base of A. Then the difference in their areas is

    Solutions

    Let the length of each side of A be x cm and y cm.

    Then the length of each side of B will be 2x cm and y + 4 cm.

    Perimeter of A = 2(x + y) cm = 26 cm

    ⇒ x + y = 13       ____(1)

    Perimeter of B = 2(2x + y + 4) cm = 44 cm

    ⇒ 2x + y + 4 = 22

    ⇒ 2x + y = 18       ____(2)

    Subtracting equation (1) from (2), we get

    x = 5

    y = 13 - 5 = 8

    Length of each side of parallelogram A is 5 cm and 8 cm

    Length of each side of parallelogram B is 10 cm and 12 cm.

    Area of A = 8 × 4 cm2 = 32 cm2

    Area of B = 12 × 6 cm2 = 72 cm2

    ∴ Difference between areas = 72 - 32 cm2 = 40 cm2

  • Question 5/15
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    A square with diagonal of 12 cm is cut from all sides by 1 cm width. Find the diagonal of the new square.

    Solutions

    As we know, diagonal of square = √2 × side of square

    ⇒ Side of initial square = 12/√2 = 6√2 cm

    When the square is cut 1 cm from all sides, the side is reduced by 2 cm,

    ⇒ Side of new square = (6√2 – 2) cm

    ∴ Diagonal of new square = (6√2 – 2) × √2 = (12 - 2√2) cm

  • Question 6/15
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    Manapa decided to construct 15 m wide road around a circular skating track which is having a circumference of 242 m. Find the area of that road.
    Solutions

    Let ‘r’ be the radius of the circular skating track

    According to the condition given in the problem,

    2πr = 242

    ⇒ r = 38.5 m

    Width of the road = 15 m

    Radius of the skating track with road = 38.5 + 15 = 53.5 m

    Area of the road = π × (53.52 – 38.52) = 22/7 × 1380 = 4337.14 m

    ∴ Area of the road is 4337.14m
  • Question 7/15
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    Calculate the area of an isosceles triangle whose length of equal sides is 15 cm and the base is 24 cm.
    Solutions

    Let the length of equal sides = ‘a’ and length of base = ‘b’

    Then, the area of the isosceles triangle \(= \;\frac{b}{4}\sqrt {4{a^2} - {b^2}}\)

    ⇒ Area of isosceles triangle\(= \frac{{24}}{4}\sqrt {\left( {4{{\left( {15} \right)}^2} - {{24}^2}} \right)} \)

    ∴ Area of isosceles triangle = 6 × 18 = 108 cm2
  • Question 8/15
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    Directions: Given below are two quantities named A & B. Based on the given informationyou have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose between the possible answers.

    Quantity A: ABC is right angled triangle with angle B as right angle and AB = 25 cm, BC = 60 cm. If G is the centroid of the triangle then find the value of BG.

    Quantity B : BC is a chord to the circle and OD is the radius of the circle in such a way that they meet at point A outside the circle. If BC = 20 cm, AB = 16 cm & OA = 40 cm then find the radius of the circle.
    Solutions

    Quantity A:

    Given that AB = 25 cm, BC = 60 cm

    ∴ AC = √(252+ 602) = 65 cm

    Since G is the centroid of the triangle, O will divide AC in 2 equal parts

    ∴ AO = OC = OB

    ∴ OB = 65/2 cm

    Since centroid will divide median BO in the ratio 2 ∶ 1

    ∴ BG = 2/3 × 65/2 = 65/3 cm

    Quantity B:

    Suppose r is the radius of the circle

    Given BC = 20 cm, AB = 16 cm & OA = 40 cm

    ∴ AC = 20 + 16 = 36 cm

    And AD = (40 – r) & AE = (40 + r) cm

    We know the theorem : AB × AC = AD × AE

    ∴ 16 × 36 = (40 – r) × (40 + r)

    ⇒ 576 = 1600 – r2

    ⇒ r2 = 1024

    ∴ r = 32 cm

    ∴ Quantity A < Quantity B
  • Question 9/15
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    In a parallelogram, the length of one diagonal and the perpendicular dropped on that diagonal are 30 and 20 metres respectively. Find its area.
    Solutions

    Area of a parallelogram = (Length of a diagonal) × (Length of perpendicular dropped onto that diagonal)

    ∴ Area of given parallelogram = 30 × 20 = 600 sq.m

    [A parallelogram can be viewed as two same triangles when cut along diagonal. So area of each triangle is half multiplied by diagonal multiplied by the perpendicular to that diagonal. Therefore, area of parallelogram is diagonal multiplied by the perpendicular to that diagonal.]
  • Question 10/15
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    The length, breadth and height of an open box is 60 cm, 50 cm and 30 cm respectively. Its thickness is 5 cm. What would be the cost of the box if 1 cm3 of wood costs Rs. 0.08?
    Solutions

    The outer dimensions of the box are as follows:

    Length = l1 = 60 cm; Breadth = b1 = 50 cm; Height = h1 = 30 cm

    Now, since the box is wooden and has a thickness of 5 cm, only the edges would be made of wood and there’ll be cavity inside the box. The dimensions of the cavity would be as shown in the figure.

    The dimensions of cavity are as follows:

    Length = l2 = 50 cm; Breadth = B2 = 40 cm; Height = H2 = 25 cm

    Now, Volume of wood used for box = volume of box – volume of cavity

    ⇒volume of wood used for the box = (l1 × b1 × h1) – (l2 × b2 × h2)

    ⇒volume of wood used for the box = (60 × 50 × 30) – (50 × 40 × 25)

    ⇒volume of wood used for the box = 90000 – 50000 = 40000 cm3

    ∴ Cost of wood = rate of wood × volume of wood

    ⇒ Cost of wood = 0.08 × 40000 = Rs. 3200
  • Question 11/15
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    The perimeter of a rectangle and equilateral triangle are same. Also, one of the sides of the rectangle is equal to the side of the triangle. The ratio of the area of the rectangle and the triangle is:
    Solutions

    Let length of rectangle = x units and breadth = y units

    ∴ side of triangle = y units

    Since their perimeters are equal,

    ⇒ 2x + 2y = 3y

    ⇒ 2x = y …………(i)

    ∴ Area of rectangle = xy

    Area of triangle = y2√3/4

    Required ratio \( = \frac{{xy}}{{\frac{{{y^2}\sqrt 3 }}{4}}}\; = \frac{{4x}}{{y\sqrt 3 }} = \frac{{4x}}{{2x\sqrt 3 }} = \frac{2}{{\sqrt 3 }}\)
  • Question 12/15
    1 / -0

    The volume of the largest cylinder formed, when a rectangular sheet of aluminum of size 22 cm × 14 cm is rolled along its larger side, is
    Solutions

    Let the radius be r

    The length of the sheet = circumference of the circle

    ⇒ 2π r = 22

    \( \Rightarrow \;2\; \times \;\frac{{22}}{7} \times r\; = 22\) 

    \( \Rightarrow \;r\; = \;\frac{{22 \times 7}}{{22 \times 2}} = \frac{7}{2}\) 

    Height of the cylinder = 14 cm

    Volume of the cylinder = π r2 h

    \( = \frac{{22}}{7} \times \frac{7}{2} \times \frac{7}{2} \times 14\) 

    = 539 cm3

    Hence, the volume of the cylinder is 539 cm3

  • Question 13/15
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    A right circular cone with base radius 10 cm has a volume of 770 cubic cm. Having the same area as the curved surface area of this cone, a square is drawn. What will be the side length of this square? (Take π = 22/7)
    Solutions

    A right circular cone with base radius 10 cm has a volume of 770 cubic cm.

    Volume of right circular cone = (1/3) × π × square of radius × height

    ⇒ 770 = (1/3) × (22/7) × 100 × height

    ⇒ Height = (770 × 3 × 7)/(22 × 100) = 7.35 cm

    Slant height of cone = √ (square of radius of base of cone + square of height of cone) = √ (10 × 10 + 7.35 × 7.35) = √ 154.0225 = 12.41 cm

    Curved surface area = π × radius × slant height = (22/7) × 10 × 12.41 = 390 square cm

    With the same area, a square is drawn.

    ⇒ Area of square = 390 square cm

    Side length of square = √ 390 cm = 19.75 cm
  • Question 14/15
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    A hemisphere, a cylinder and a cone stand on the same base and 45% of height of hemisphere is equal to one - fourth of 180% of height of cylinder and height of cone is one - fourth of sum of 224% of height of hemisphere and 176% of height of cylinder. The ratio of the areas of their curved surface is

    Solutions

    Let radius of all three solids is r and height of hemisphere, cylinder and cone is h1, h2 and h3 respectively

    According to question,

    ⇒ 45% of h1 = ¼(180% of h2)

    ⇒ 45% of h1 = 45% of h2

    ⇒ h1 = h2

    Also,

    ⇒ h3 = ¼(224% of h1 + 176% of h2)

    ⇒ h3 = ¼(400% of h1) (∵ h1 = h2)

    ⇒ h3 = h1

    ⇒ h1 = h2 = h3 = h (Let)

    Also,

    Height of hemisphere = Radius of hemisphere

    ⇒ h = r

    Now,

    ⇒ CSA of Hemisphere : CSA of Cylinder : CSA of Cone = 2πr2 : 2πr × r : πr√(r2 + r2)

    ⇒ 2πr2 : 2πr2 : √2 πr2

    ∴ required ratio = √2 : √2 : 1

  • Question 15/15
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    A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, find the uniform thickness of the cylinder.
    Solutions

    Volume of sphere = \(\frac{4}{3}\pi {r^3}\)

    Where, r is radius of sphere.

    Volume of the given sphere = (4/3) × π × 63

    This volume is converted to the hollow cylinder.

    Let, the thickness of cylinder be x.

    External radius = 5 cm

    Internal radius = (5 – x) cm

    Area of circular base = πr2, where r is radius of base.

    Area of hollow base = (π × 52) – π (5 – x)2

    = π [52 – (5 – x)2]

    = π (5 + 5 – x) (5 – 5 + x)

    = πx(10 – x)

    Volume of cylinder = Area of base × height

    ∴(4/3) × π × 63 = 32 × π (10 – x)x

    ⇒ 9 = 10x – x2

    ⇒ x2 – 10x + 9 = 0

    ⇒ x2 – 9x – x + 9 = 0

    ⇒ x(x – 9) – 1(x – 9) = 0

    ⇒ (x – 1) (x – 9) = 0

    ⇒ x = 1 or x = 9

    x = 9 not possible, since the thickness cannot be greater than 5, which is the external radius.

    So, x = 1
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