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Mensuration Test 1
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Mensuration Test 1
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  • Question 1/10
    1 / -0

    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.

    Solutions
    Hint

    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

  • Question 2/10
    1 / -0

    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:

    Solutions

    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.

  • Question 3/10
    1 / -0

    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?

    Solutions

    Clearly, we have:

    ℓ = 9 and ℓ + 2b

    = 37 or b = 14.

    ∴ Area = (ℓ × b)

    = (9 × 14) sq. ft.

    = 126 sq. ft.

  • Question 4/10
    1 / -0

    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?

    Solutions

    Area of field = 576 km². Then,

    each side of field

    = 

    Distance covered by the horse Perimeter of square field

    = 24 × 4 = 96 km

    ∴ 

    = 96/12= 8 h

  • Question 5/10
    1 / -0

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

    Solutions

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

    So, sides of both squares will be

     and  respectively. [since, side = ]

    Now, perimeters of both squares will be

     and  respectively. [since, perimeter = 4 × side]

    Thus, ratio of their perimeters

     = 3 : 1

  • Question 6/10
    1 / -0

    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?

    Solutions

    Required number of squares 5²/1² = 25

  • Question 7/10
    1 / -0

    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

    Solutions

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

    ⇒ ¼ ar (∆ABC) = ar (∆PQR)

    ⇒ ar (∆PQR) = 5 sq. units

  • Question 8/10
    1 / -0

    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

    Solutions

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

    Let AB = AC = x.

    Then, BC = (32 – 2x).

    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

    = 

  • Question 9/10
    1 / -0

    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.

    Solutions

    In a triangle,

    Area 

    ⇒ 

    ∴ Length of perpendicular

     = 10 m.

  • Question 10/10
    1 / -0

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

    Solutions

    In an isoscele right angled triangle,

    Area = 1/23.3 × perimeter²

    = 1/23.3 × 20² = 17.167 m²

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