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Mensuration Test - 10

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Mensuration Test - 10
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  • Question 1
    1 / -0
    The length of a rectangle which is $$25\ cm$$ is equal to the side of a square, and the area of the rectangle is $$125\ cm^2$$ less than the area of the square. What is the breadth of the rectangle?
    Solution
    Area of the square $$= (25 \times 25)\ cm^2= 625\ cm^2$$
    $$\therefore$$ Area of the rectangle $$= (625 - 125)\  cm^2$$$$=500 \ cm^2$$
    $$\therefore$$ Breadth of the rectangle $$=\dfrac{Area}{Length}=\displaystyle \frac{500}{25} cm= 20\, cm$$
  • Question 2
    1 / -0
    The perimeter of a rectangle is $$170$$ m and its length is $$50$$ m. Then the breadth is:
    Solution
    Let the breadth of the rectangle is $$b$$. 
    Perimeter of the rectangle $$=2(l+b)=170 m$$      {Given}
    $$\Rightarrow2(50+b)=170$$
    $$\Rightarrow 50+b=85$$
    $$ \Rightarrow b=35m$$
  • Question 3
    1 / -0
    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 verandah is:

    Solution
    Area of outer rectangle $$= 19\, \times\, 16\, =\, 304\, m^2$$
    Area of inner rectangle $$= 15\, \times\, 12\, =\, 180\, m^2$$
    $$\therefore$$ Reqd. area $$= (304 -180)\, m^2\, =\, 124\, m^2.$$
  • Question 4
    1 / -0
    The perimeter of a rectangular plot whose length is $$75\ m$$ and breadth is $$50\ m$$ is ...........
    Solution
    The perimeter of the rectangular plot = $$2\times(length + breadth)$$
                                                                      $$=2\times(75+70)$$
                                                                      $$=2\times 145$$
                                                                      $$=250\ m$$
  • Question 5
    1 / -0
    The cost of levelling a rectangular ground at Rs $$1.25$$ per sq meter is Rs $$900$$. If the length of the ground is $$30$$ meters, then the width is
    Solution
    Surface area of floor=$$\dfrac{total cost}{cost\quad per\quad m^2}$$
    =$$\dfrac{900}{1.25}=720 m^2$$
    $$30 \times$$ (width)=$$720 m^2$$
    $$width=24 m$$
  • Question 6
    1 / -0
    Find the perimeter of the following figure?

    Solution
    Perimeter of figure = sum of all sides $$=100+100+100+100+100+100$$
                                                                  $$=600m$$
  • Question 7
    1 / -0
    Raghu walks around a park everyday. How far does he walk in one round?

    Solution
    In $$1$$ round Raghu walks $$(66+75+102+85)m=328m$$
  • Question 8
    1 / -0
    The length and breadth of a rectangular plot are 900 m and 700 m respectively If three rands of fence is fixed around the field at the cost of Rs.8 per meter the total amount spend is
    Solution
    l $$\displaystyle = $$ 900 b $$\displaystyle = $$ 700
    Perimeter $$\displaystyle = $$ 2(900 + 700)
    $$\displaystyle = $$ 2(1600)
    $$\displaystyle = $$ 3200
    3 rands fence:
    $$\displaystyle = $$ 3(3200)
    $$\displaystyle = $$ 9600m
    $$\displaystyle = $$ 9600 $$\displaystyle \times $$ 8 $$\displaystyle = $$ Rs 76,800
  • Question 9
    1 / -0
    The perimeter of a square is ____ times the length of the side.
    Solution
    As we know that all the sides of a squqre are equal. 
    So its perimeter will be $$4$$ times the length of the side.
    Hence, the answer is $$4$$.
  • Question 10
    1 / -0
    Find the sum of the perimeters of the figures given above.

    Solution
    Perimeter is simply the sum of sides of figure.
    Perimeter for first figure(triangle) $$=(60+50+60)=170\,cm$$
    Perimeter for first figure(rectangle) $$=(60+40+60+40)=200\,cm$$
    So, the sum of perimeter $$=170+200=370$$ cm 
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