Perimeter and Area Test

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Perimeter and Area Test - 31

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Weekly Quiz Competition

Question 1
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In a parallelogram the base and height are is in the ratio of $$5 : 2$$. If the area of the parallelogram is $$360\ m^{2}$$, find its base and height.

A
$$30\ m, 12\ m$$.

B
$$20\ m, 12\ m$$.

C
$$30\ m, 10\ m$$.

D
None of these

Solution

The ration of the base and height of the parallelogram is 5 : 2
$$ \therefore $$ let, the base and height be 5x & 2x respectively.
We know, area of a parallelogram is given by, $$ \dfrac{1}{2} \textrm{base} \times \textrm{height} $$.
$$ \therefore $$ according to the question,
$$ \dfrac{1}{2} \times {5x} \times {2} = 360 $$ [ The area is given by 360 $$m^2$$ ]
$$ \Rightarrow x^2 = 36 $$
$$ \Rightarrow x = 6 [ \therefore $$ base & height cannot be negative ]
$$ \therefore $$ The base is ( 5 X 6 ) = 30 m . and the height is ( 2 X 6 ) = 12 m .
$$ \therefore $$ option Ais correct.
Question 2
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In Figure $$ABCD$$ is a square then find the area of the shaded region.

A
$$800\ cm^{2}$$.

B
$$600\ cm^{2}$$.

C
$$200\ cm^{2}$$.

D
None of these

Solution

to find out the area of shaded region let us follow the method;

Area of shaded region $$=$$ Area of asquare $$ABCD-$$ Area of triangle $$ABE$$

Area of square of side $$a$$ is equal to $$a^2$$

for $$a=40cm$$

$$\therefore $$ Area of square $$ABCD=40 \times 40=1600cm^2$$

Area of triangle $$ABE=\dfrac{1}{2}\times$$ base$$\times$$ height

For base $$=40cm;height=40cm$$

Area of triangle $$ABE=\dfrac{1}{2}\times 20\times 40\Rightarrow 20\times 40 sqcm$$

$$\therefore$$ Area of triangle $$\Rightarrow 800cn^2$$

Now, Area of shaded region $$=$$ area of square $$ABCD -$$area of triangle $$ABE$$

$$\Rightarrow 1600cm^2-800cm^2=800cm^2$$

$$\therefore$$ Area of shaded region $$=800cm^2$$
Question 3
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Find the circumference of circle whose diameter is- $$17.5\ cm$$
Take $$\pi = \dfrac {22}{7}$$ in the above two questions.

A
$$55\ m$$

B
$$25\ m$$

C
$$59\ m$$

D
None of these

Solution

Given: Diameter of circle $$(d)=17.5\space\mathrm{m}$$

$$\Rightarrow$$ Radius of circle $$(r)=\dfrac{17.5}{2}=8.75\space\mathrm{m}$$

Let $$C$$ be the circumference of circle,

$$C=2\pi r\\=2\times\dfrac{22}{7}\times8.75\\=55\space\mathrm{m}$$

So, $$\text{A}$$ is the correct option.
Question 4
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If the circumference of a circle is $$264\ cm$$, find its radius. Take $$\pi = \dfrac {22}{7}$$.

A
$$42\ cm$$.

B
$$22\ cm$$.

C
$$32\ cm$$.

D
None of these

Solution

Let the radius be $$r\ cm$$
Given Circumference of a circle is $$264\ cm$$
$$\implies 2\pi r=264$$
$$\implies 2\times \dfrac{22}{7}\times r=264$$
$$\implies r=\dfrac{264\times 7}{44}=42$$
So the radius is $$42\ cm$$
Question 5
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Rohit's family is planning to build a concrete basketball court in their backyard. How many square metres will be left in the backyard for grass?

A
$$60$$ sq. cm

B
$$35$$ sq. cm

C
$$31$$ sq. cm

D
$$24$$ sq. cm

Solution

$$1$$sq. m$$=1\square$$
Area of the concrete basketball court$$=7\times 5=35$$sq. m
Area of the backyard$$=11\times 6=66$$sq. m
Area of the backyard left for grass $$=(66-35)$$sq. m$$=31$$sq. m.
Question 6
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The area of the shaded region in the figure is ______ .

A
$$207\ \text{m}^{2}$$

B
$$158\ \text{m}^{2}$$

C
$$256\ \text{m}^{2}$$

D
$$60\ \text{m}^{2}$$

Solution

$$\Rightarrow$$ We can see, given figure is a square. There are $$4$$ equal small squares inside of large square.
$$\Rightarrow$$ Side of large square is $$16\,m$$
$$\therefore$$ Area of large square = $$16^2=256\,m^2$$
$$\Rightarrow$$ Side of small square is $$7\,m$$, there are $$4$$ small equal squares.
$$\therefore$$ Area of small $$4$$ squares = $$4\times 7^2=196\,m^2$$
$$\therefore$$ The area of the shaded region = Area of large square - Area of $$4$$ small squares.
$$\therefore$$ The area of the shaded region = $$256\,m^2-196m^2=60\,m^2$$
Question 7
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A square and a rectangular plot of land have the same perimeter. If the square is of side $$40m$$ and the rectangle is of length $$5$$ decameter. Then area of rectangle is _______ .

A
$$1500{m}^{2}$$

B
$$1600{m}^{2}$$

C
$$200{m}^{2}$$

D
$$150{m}^{2}$$

Solution

$$\Rightarrow$$ Length of a rectangle $$=5$$ decameter $$=50\,m$$
$$\Rightarrow$$ Side of square = $$40\,m$$
$$\therefore$$ Perimeter of squares = $$4\times 40=160\,m$$
$$\Rightarrow$$ It is given that, perimeter of square = perimeter of a rectangle.
$$\Rightarrow$$ Perimeter of rectangle = $$2(l+b)$$
$$\therefore$$ $$160=2(50+b)$$
$$\therefore$$ $$80=50+b$$
$$\therefore$$ $$b=30\,m$$
$$\Rightarrow$$ Area of a rectangle $$=l\times b=50\times 30=1500\,m^2$$
Question 8
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If the area of a rectangle is equal to the area of a square and if one side ($$l$$) of the rectangle is equal to the perimeter of the square, then the other side ($$b$$) of rectangle is _______ .

A
$$side\div 2$$

B
$${side}^{2}\div 2$$

C
$$side \div 4$$

D
$$side\div 3$$

Solution

Perimeter of square$$=4\times $$ side
Then, $$l=4\times $$ side
Area of rectangle=Area of square$$\Rightarrow l\times b=$$ side $$\times $$side
$$\Rightarrow \left( 4\times side \right) \times b=$$ side $$\times $$side $$\Rightarrow b=\cfrac { side\times side }{ 4\times side } =\cfrac { side }{ 4 } $$
Question 9
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If the circumference of a circle is $$33\ cm$$, find its diameter.

A
$$10.5\ cm$$.

B
$$12.5\ cm$$.

C
$$16.5\ cm$$.

D
None of these

Solution

Given: Circumference of circle $$=33\space\mathrm{cm}$$
$$\Rightarrow 2\pi r=33$$
$$\Rightarrow 2\times\dfrac{22}{7}\times r=33$$
$$\Rightarrow r=33\times\dfrac{1}{2}\times\dfrac{7}{22}$$
$$\Rightarrow r=5.25\space\mathrm{cm}$$
$$\Rightarrow d=2r=2\times5.25=10.5\space\mathrm{cm}$$
$$\therefore$$ Diameter of circle $$d=10.5\space\mathrm{cm}$$
Hence. $$\text{A}$$ is the correct option.
Question 10
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The shaded part in the given figure is covered with cement. If its costs Rs. $$84$$ to cement an area of $$3$$ $$cm^2$$, find the total cost of cementing.

A
Rs. $$8820$$

B
Rs. $$8125$$

C
Rs. $$8610$$

D
Rs. $$8804$$

Solution

Area of larger rectangle $$=(32\times 25)cm^2=800cm^2$$
Also, area of smaller rectangle$$=485cm^2$$
$$\therefore$$ Area of shaded part $$=(800-485)cm^2=315cm^2$$
Now,
Cost of cementing of $$3cm^2=$$Rs. $$84$$
$$\therefore$$ Cost of cementing of $$1cm^2=(84\div 3)=$$Rs. $$28$$
$$\therefore$$ Cost of cementing of $$315cm^2$$
$$=$$Rs. $$(28\times 315)=$$Rs. $$8820$$.