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Mensuration Exercise 1

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  1. Question 1 of 27
    1. Question

    In an isoscele right angled triangle, the perimeter is 20 metre. Find its area.

    Hint

    In an isoscele right angled triangle,

    Area = 1/23.3 × perimeter²

    = 1/23.3 × 20² = 17.167 m²

  2. Question 2 of 27
    2. Question

    The area of a triangle is 615 m². If one of its sides is 123 metre, find the length of the perpendicular dropped on that side from opposite vertex.

    Hint

    In a triangle,

    Area mensuration-q-41382.png

    ⇒ mensuration-q-41376.png

    ∴ Length of perpendicular

    mensuration-q-41370.png = 10 m.

  3. Question 3 of 27
    3. Question

    The altitude drawn to the base of an isosceles triangle is 8 cm and the perimeter is 32 cm. The area of the triangle is

    Hint

    Let ABC be the isosceles triangle and AD be the altitude.

    Let AB = AC = x.

    Then, BC = (32 – 2x).

    mensuration-q-42057.png

    Since, in an isosceles triangle, the altitude bisects the base. So,

    BD = DC = (16 – x).

    In ∆ADC, AC² = AD² + DC²

    ⇒ x² = (8)² + (16 – x)²

    ⇒ 32x = 320

    ⇒ x = 10.

    ∴ BC = (32 – 2x)

    = (32 – 20) cm

    = 12 cm.

    Hence, required area

    = mensuration-q-42051.png

    mensuration-q-42045.png

  4. Question 4 of 27
    4. Question

    In a triangle ABC, points P, Q and R are the mid-points of the sides AB, BC and CA respectively. If the area of the triangle ABC is 20 sq. units, find the area of the triangle PQR

    Hint

    Consider for an equilateral triangle. Hence ∆ABC consists of 4 such triangles with end points on mid points AB, BC and CA

    mensuration-q-44287.png

    ⇒ ¼ ar (∆ABC) = ar (∆PQR)

    ⇒ ar (∆PQR) = 5 sq. units

  5. Question 5 of 27
    5. Question

    How many squares are there in a 5 inch by 5 inch square grid, if the grid is made up one inch by one inch squares?

    Hint

    Required number of squares 5²/1² = 25

  6. Question 6 of 27
    6. Question

    If the ratio of areas of two squares is 9 : 1, the ratio of their perimeter is:

    Hint

    Let the area of two squares be 9x and x respectively.

    So, sides of both squares will be

    mensuration-q-41436.png and mensuration-q-41430.png respectively. [since, side = mensuration-q-41424.png]

    Now, perimeters of both squares will be

    mensuration-q-41418.png and mensuration-q-41412.png respectively. [since, perimeter = 4 × side]

    Thus, ratio of their perimeters

    mensuration-q-41406.png = 3 : 1

  7. Question 7 of 27
    7. Question

    The area of a square field is 576 km². How long will it take for a horse to run around at the speed of 12 km/h?

    Hint

    Area of field = 576 km². Then,

    each side of field

    = mensuration-q-41400.png

    Distance covered by the horse Perimeter of square field

    = 24 × 4 = 96 km

    ∴ mensuration-q-41394.png

    = 96/12= 8 h

  8. Question 8 of 27
    8. Question

    A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, then what is the area of the parking space in square feet?

    Hint

    Clearly, we have:

    ℓ = 9 and ℓ + 2b

    = 37 or b = 14.

    ∴ Area = (ℓ × b)

    = (9 × 14) sq. ft.

    = 126 sq. ft.

  9. Question 9 of 27
    9. Question

    A farmer wishes to start a 100 square metres rectangular vegetable garden. Since he has only 30 m barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. The dimension of the garden is:

    Hint

    We have : 2b + ℓ = 30

    ⇒ ℓ = 30 – 2b.

    Area = 100 m²

    ⇒ ℓ × b = 100

    ⇒ b(30 – 2b) = 100

    ⇒ b² – 15b + 50 = 0

    ⇒ (b – 10) (b – 5) = 0

    ⇒ b = 10 or b = 5.

    When b = 10, ℓ = 10 and when b = 5, ℓ = 20.

    Since the garden is rectangular,

    so its dimension is 20 m × 5 m.

  10. Question 10 of 27
    10. Question

    A rectangular plot 15 m ×10 m, has a path of grass outside it. If the area of grassy pathway is 54 m², find the width of the path.

    Hint

    mensuration-q-42104.png

    Let the width of the path = W m

    then, length of plot with path

    = (15 + 2w) m

    and breadth of plot with path

    = (10 + 2 W) m

    Therefore, Area of rectangular plot (without path)

    = 15 × 10 = 150 m²

    and Area of rectangular plot (with path)

    = 150 + 54 = 204 m²

    Hence, (15 + 2w) × (10 + 2w) = 204

    ⇒ 4w² + 50 w – 54 = 0

    ⇒ 2w² + 25 w – 27 = 0

    ⇒ (w – 2) (w + 27) = 0

    Thus w = 2 or –27

    ∴ with of the path = 2 m

  11. Question 11 of 27
    11. Question

    The area of a rectangular field is 144 m². If the length had been 6 metres more, the area would have been 54 m² more. The original length of the field is

    Hint

    Let the length and breadth of the original rectangular field be x m and y m respectively.

    Area of the original field

    = x × y = 144 m²

    ∴ mensuration-q-42092.png … (i)

    If the length had been 6 m more, then area will be

    (x + 6) y = 144 + 54

    ⇒ (x + 6) y = 198 … (ii)

    Putting the value of x from eq (i) in eq (ii), we get

    mensuration-q-42086.png

    ⇒ 144 + 6y = 198

    ⇒ 6y = 54 ⇒ y = 9 m

    Putting the value of y in eq (i) we get x = 16 m

  12. Question 12 of 27
    12. Question

    The length and breadth of a playground are 36 m and 21 m respectively. Poles are required to be fixed all along the boundary at a distance 3 m apart. The number of poles required will be

    Hint

    Given, playground is rectangular.

    Length = 36 m, Breadth = 21 m

    Now, perimeter of playground

    = 2( 21 + 36) = 114

    Now, poles are fixed along the boundary at a distance 3m.

    ∴ Required no. of poles = ¹¹⁴⁄₃ = 38

  13. Question 13 of 27
    13. Question

    The cost of carpeting a room 18 m long with a carpet 75 cm wide at Rs 4.50 per metre is Rs 810. The breadth of the room is:

    Hint

    Length of the carpet

    = mensuration-q-42453.png

    mensuration-q-42446.png

    Area of the room = Area of the carpet

    = mensuration-q-42439.png

    ∴ Breadth of the room

    = mensuration-q-42433.png

    = 7.5 m.

  14. Question 14 of 27
    14. Question

    If the perimeter and diagonal of a rectangle are 14 and 5 cms respectively, find its area.

    Hint

    In a rectangle,

    mensuration-q-42427.png

    ⇒ mensuration-q-42421.png

    49 = 25 + 2 × area

    mensuration-q-42415.png

  15. Question 15 of 27
    15. Question

    A square carpet with an area 169 m² must have 2 metres cut-off one of its edges in order to be a perfect fit for a rectangular room. What is the area of rectangular room?

    Hint

    Side of square carpet

    mensuration-q-42478.png

    mensuration-q-42472.png

    After cutting of one side,

    Measure of one side

    = 13 – 2 = 11 m

    and other side = 13 m (remain same)

    ∴ Area of rectangular room

    = 13 × 11 = 143 m²

  16. Question 16 of 27
    16. Question

    A picture 30 inch × 20 inch has a frame 2½ inch wide. The area of the picture is approximately how many times the area of the frame?

    Hint

    mensuration-q-42466.png

    Length of frame

    = 30 + 2.5 × 2 = 35 inch

    Breadth of frame

    = 20 + 2.5 × 2 = 25 inch

    Now, area of picture

    = 30 × 20 = 600 sq. inch

    Area of frame

    = (35 × 2.5) + (25 × 2.5) = 150

  17. Question 17 of 27
    17. Question

    The floor of a rectangular room is 15 m long and 12 m wide. The room is surrounded by a verandah of width 2 m on all its sides. The area of the verandah is:

    Hint

    Area of the outer rectangle = 19 × 16 = 304 m²

    mensuration-q-42523.png

    Area of the inner rectangle

    = 15 × 12 = 180 m²

    Required area

    = (304 – 180) = 124 m²

  18. Question 18 of 27
    18. Question

    A typist uses a paper 12 inch by 5 inch length wise and leaves a margin of 1 inch at the top and the bottom and a margin of ½ inch on either side. What fractional part of the paper is available to him for typing?

    Hint

    Area of paper

    = 12 × 5 = 60 sq. inch

    Area of typing part

    mensuration-q-42517.png

    = (12 – 2) × (5 – 1)

    = (10 × 4) sq. inch

    ∴ Required fraction

    mensuration-q-42510.png

  19. Question 19 of 27
    19. Question

    A rectangular lawn 70 m × 30 m has two roads each 5 metres wide, running in the middle of it, one parallel to the length and the other parallel to the breadth. Find the cost of gravelling the road at the rate of Rs 4 per square metre.

    Hint

    mensuration-q-42504.png

    Total area of road

    = Area of road which parallel to length + Area of road which parallel to breadth – overlapped road

    = 70 × 5 + 30 × 5 – 5 × 5

    = 350 + 150 – 25

    = 500 – 25 = 475 m²

    ∴ Cost of gravelling the road

    = 475 × 4 = Rs 1900

  20. Question 20 of 27
    20. Question

    The length and breadth of the floor of the room are 20 feet and 10 feet respectively. Square tiles of 2 feet length of different colours are to be laid on the floor. Black tiles are laid in the first row on all sides. If white tiles are laid in the one-third of the remaining and blue tiles in the rest, how many blue tiles will be there?

    Hint

    Area left after laying black tiles

    = [(20 – 4) × (10 – 4)] sq. ft.

    = 96 sq. ft.

    Area under white tiles

    mensuration-q-42882.png sq. ft

    = 32 sq. ft.

    Area under blue tiles

    = (96 – 32) sq. ft = 64 sq. ft.

    Number of blue tiles

    = mensuration-q-42876.png

  21. Question 21 of 27
    21. Question

    The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12km / hr completes one round in 8 minutes, then the area of the park (in sq. m) is:

    Hint

    Perimeter = Distance covered in 8 min.

    mensuration-q-42911.png

    Let length = 3x metres and breadth = 2x metres.

    Then, 2 (3x + 2x) = 1600 or x = 160.

    ∴ Length = 480 m and Breadth = 320 m.

    ∴ Area = (480 × 320) m² = 153600 m².

  22. Question 22 of 27
    22. Question

    In measuring the side of a square, an error of 5% in excess is made. The error % in the calculated area is:

    Hint

    ℓ × b = 100 m²

    ⇒ mensuration-q-42939.png

    mensuration-q-42933.png

    Therefore, mensuration-q-42927.png

    ⇒ b² – 15b + 50 = 0

    ⇒ b = 10, 5

    If we take b = 10, then garden becomes a square

    Therefore, b = 5 m.

  23. Question 23 of 27
    23. Question

    The length of a room is double its breadth. The cost of colouring the ceiling at Rs 25 per sq. m is Rs 5,000 and the cost of painting the four walls at Rs 240 per sq. m is Rs 64,800. Find the height of the room.

    Hint

    Let the length of the room be ℓ m

    Then its, breadth = ℓ/2

    Therefore, mensuration-q-42963.png

    ⇒ ℓ² = 400

    ⇒ ℓ = 20 m

    Also, mensuration-q-42957.png

    ⇒ 3ℓh = 270

    ⇒ mensuration-q-42951.png

  24. Question 24 of 27
    24. Question

    A rectangular plate is of 6 m breadth and 12 m length. Two apertures of 2 m diameter each and one apertures of 1 m diameter have been made with the help of a gas cutter. What is the area of the remaining portion of the plate?

    Hint

    Given, Length = 12 m and Breadth = 6 m

    ∴ Area of rectangular plate

    = 12 × 6 = 72 m²

    mensuration-q-44229.png

    Since, two apertures of 3 m diameter each have been made from this plate.

    ∴ Area of these two apertures

    = π(1)² + π(1)²

    = π + π = 2 π

    Area of 1 aperture of 1m diameter

    = mensuration-q-44223.png

    ∴ Total area of aperture

    = mensuration-q-44217.png = ⁹⁹⁄₁₄

    ∴ Area of the remaining portion of the plate

    = 72 – ⁹⁹⁄₁₄ sq. m

    = ⁹⁰⁹⁄₁₄ sq. m ≈ 64.5 sq.m

  25. Question 25 of 27
    25. Question

    Four sheets 50 cm × 5 cm are arranged without overlapping to form a square having side 55 cm. What is the area of inner square so formed?

    Hint

    mensuration-q-44211.png

    Side of the inner square

    = 55 – 10 = 45

    ∴ Area of inner square

    = 45 × 45 = 2025 sq. m.

  26. Question 26 of 27
    26. Question

    In the adjoining figure, AC + AB = 5 AD and AC – AD = 8. Then the area of the rectangle ABCD is
    43833.png

    Hint

    mensuration-q-44205.png

    AC + AB = 5AD

    ⇒ AC + a = 5b …(1)

    AC – AD = 8

    ⇒ AC = b + 8 …(2)

    Using (1) and (2) ,

    a + b + 8 = 5b

    ⇒ a + 8 = 4b …(3)

    Using Pythagorous theorem,

    a² + b² = (b + 8)² = b² + 64 + 16b

    ⇒ a² = 16b + 64 = (4b – 8)² = 16b² + 64 – 64b

    ⇒ 16b² – 80b = 0

    ⇒ b = 0 or 5

    Putting b = 5 in (3),

    a = 4b – 8 = 20 – 8 = 12

    Area of rectangle = 12 × 5 = 60

  27. Question 27 of 27
    27. Question

    ABCD is a square, F is the mid-point of AB and E is a point on BC such that BE is one-third of BC. If area of ∆FBE = 108 m², then the length of AC is:

    Hint

    Let the side of the square be x, then,

    BE = x/3 and BF = x/2

    mensuration-q-44187.png

    Area of ∆FEB mensuration-q-44181.png

    Now, mensuration-q-44175.png

    ⇒ x² = 108 × 12 = 1296

    In ∆ADC, we have

    AC² = AD² + DC²

    = x² + x² = 2x²

    = 2 × 1296 = 2592

    ⇒ AC mensuration-q-44169.png

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