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Perimeter and Area Test - 2

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Perimeter and Area Test - 2
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  • Question 1
    1 / -0

    A field is in the form of a parallelogram whose base is 420 m and altitude is 3.6 dam. Find the cost of watering the field at 10 paise per sq m.

    Solution

    Area of Parallelogram = b x h
    b = 420 m h =3.6 dam = 36 m ∴ 1 dam = 10 m
    Area of paralellogram = 420 x 36 = 15120
    cost of waterin in Rs. = 15120 x 0.10    ∴1 Rs. = 100 Paisa
                                      = 1512 Rs.

  • Question 2
    1 / -0

    If ABCD is a rectangle having length 30 cm and breadth 20 cm, E, F and G are midpoints of AB, CD and AD respectively, find the area of the unshaded part.

    Solution

    Area of rectangle =l×b=30×20 =600cm2
    Area of ΔDGF  =12×b×h =75cm2
    Similarly area of ΔAGE=75cm2
    Area of unshaded region =600−(75+75)cm2 =450cm2

  • Question 3
    1 / -0

    If ABCD is a parallelogram, what is the ratio of areas of parallelogram ABCD and ΔABC?

    Solution

    The diagonal of Parallelogram bisects it in two equal parts or two congruent triangles . So area of any trialgles is half of the area of parallelogram. Hence, the ratio of areas of parallelogram ABCD and ΔABC is 2:1

  • Question 4
    1 / -0

    A farmer had a rectangular plot measuring 500 m by 100 m. If he fences the plot 4 times with barbed wire, what length of wire was used?

    Solution

    Dimensons of field - L=500m b=100m
    Wire required for fencing field one time = 2 (l+b) = 2 (500+100) = 2 x 600 = 1200 m
    Four time fencing Length = 4 x 1200 = 4800 m

  • Question 5
    1 / -0

    If the area of a circle is 2464 m2, find its diameter,      

    Solution

    ( πd2)/4 =2464 ⇒  d=√3136 =56m

  • Question 6
    1 / -0

    Find the altitude of a triangle whose base   is 24cm, and area is 672 cm2

    Solution

    Area of Triangle = 1/2 x b x h
    672 =1/2 x 24 x h
    672 = 12h
    h= 672/12
    56 cm

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