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Mensuration Test - 8

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Mensuration Test - 8
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Weekly Quiz Competition
  • Question 1
    1 / -0
    If the side of a square park is 5 m then its perimeter is______
    Solution
    We know,
    Perimeter $$=4 \ast side $$ = $$ 4\ast 5$$ = $$20m$$
  • Question 2
    1 / -0
    Perimeter of square garden is 444 sq m. Then its side measures:
    Solution
    Perimeter of square garden $$\displaystyle =  444 \displaystyle =  4 \displaystyle \times  side$$

                                           $$\displaystyle side= \frac{444}{4}= 111m $$
  • Question 3
    1 / -0
    Area of square whose side measures 13 m is
    Solution
    We know,
    Area of a square = $$ Side^2$$
    = 13 $$\displaystyle \times $$ 13 $$\displaystyle =  $$ 169 sq m
  • Question 4
    1 / -0
    Area of a rectangle whose length is $$13$$ m and breadth is $$12$$ m is
    Solution
    Here, $$ABCD$$ is a rectangle.

    $$AB,CD$$ are length of a rectangle and $$AD,BC$$ breadth of a rectangle.

    $$\therefore$$  $$AB=CD=13m$$ and $$AD=BC=12m$$

    $$\Rightarrow$$  Length of a rectangle $$(l)=13m$$
    $$\Rightarrow$$  Breadth of a rectangle $$(b)=12m$$

    $$\Rightarrow$$  Area of rectangle $$ABCD=l\times b$$
                                                       $$=13m\times 12m$$
                                                       $$=156\,m^2$$

  • Question 5
    1 / -0
    A rectangular playground which is $$250\ m$$ long and $$20\ m$$ broad is to be fenced with wire.How much wire needed?
    Solution
    Wire needed would be perimeter of the playground:
    We know,
    Perimeter of rectangle:  $$ 2(l+b) $$
                                        $$=250 + 20 + 250 + 20 = 540\ m$$
  • Question 6
    1 / -0
    A postage stamp is $$12\ mm$$ in breadth and $$20\ mm$$ in length. Find the area of stamp in $$cm^{2}$$
    Solution
    As we know that,
    $$1\ cm=10\ mm$$
    So, $$1\ mm=0.1\ cm$$

    Given:-
    Length $$= 20\ mm$$
                 $$=20\times 0.1\ cm$$
                 $$=2\ cm$$
    Breadth $$= 12\ mm$$
                   $$=12\times 0.1\ cm$$
                   $$=1.2\ cm$$

    Area of stamp $$=$$ Length $$\times$$ Breadth
                             $$=2\times 1.2$$
                             $$=2.4\ cm^2$$

    Hence, option $$D$$ is correct.
  • Question 7
    1 / -0
    Fill in the blanks:
    Perimeter of the square of side $$4$$ cm = _____ cm
    Solution
    Given, Side of a square = $$4$$ cm

    Perimeter of a square $$= 4 \times side$$
    So, perimeter of the square $$= 4 \times 4 \ cm= 16 cm$$

    Hence, the perimeter of the square is $$16\ cm$$
  • Question 8
    1 / -0
    An athelete takes $$5$$ rounds of a rectangular park, $$50$$ m long and $$250$$ m wide. Then the total distance covered by him is ______ .
    Solution
    Given, Breadth of the rectangular park $$b=250$$ m 
                and Length $$l=50$$ m

    Perimeter of the park $$=2(b+l)$$

                                         $$=2\times (250+50)$$

                                         $$=2\times 300$$

                                         $$=600$$ m

    Given: Athlete take $$5$$ rounds around the rectangular park

    So, the total distance covered by the athlete $$=$$ $$5\times 600$$ m $$=3000$$ m
  • Question 9
    1 / -0
    How many paving stones each measuring $$2.5\text{ m}$$ $$\times 2\text{ m}$$ are required to pave a rectangular courtyard $$30\text{ m}$$ long and $$16.5\text{ m}$$ wide?
    Solution
    Area of paving stones $$=2.5 \times 2=5 \text{ m}^2$$
    Area of courtyard $$=30 \times 16.5=495 \text{ m}^2$$
    Now,
    Number of paving stones required $$=$$ area of courtyard $$÷$$ area of paving stone
    $$= 495 ÷ 5$$
    $$= 99 $$ paving stones
    So option C is the correct answer.
  • Question 10
    1 / -0
    The figure is made up of $$2$$ rectangles
    What is the perimeter of the big rectangle?

    Solution
    $$\Rightarrow$$  Length of a big rectangle $$(l)=(7r+3+3)cm=(7r+6)cm$$
    .
    $$\Rightarrow$$  Breadth of a big rectangle $$(b)=(4r+3+3)cm=(4r+6)cm$$

    $$\Rightarrow$$  Perimeter of big rectangle $$=2(l+b)cm$$
                                                       $$=2[(7r+6)+(4r+6)]cm$$
                                                       $$=2[11r+12]cm$$
                                                       $$=22r+24\,cm$$
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