Perimeter and Area Test

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Perimeter and Area Test - 22

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Weekly Quiz Competition

Question 1
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A wire is bent into the shape as shown. It is made up of $$5$$ semicircles. What is the length of the wire? (Take $$\pi =3.14$$)

A
$$27\ cm$$

B
$$42.39\ cm$$

C
$$45\ cm$$

D
$$27.92\ cm$$

Solution

Let the length of the wire be $$l.$$

The length of the wire is equal to the arc length of the semi-circles drawn on the diameters $$AB,\ BC,\ CD,\ EF$$ and $$ED.$$

$$l= \pi \times 1.5 +\pi \times 3 +\pi \times 4.5 +\pi \times 3 +\pi \times 1.5 $$
$$= \pi \times \left ( 1.5+3+4.5+3+1.5 \right )$$
$$ = 3.14 \times 13.5 $$
$$= 42.39\ cm$$
Question 2
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The number of revolutions a wheel of diameter $$40 cm$$ makes in travelling a distance of $$176 m$$ is: $$\displaystyle \left ( \pi =\frac{22}{7} \right )$$

A
$$140$$

B
$$150$$

C
$$160$$

D
$$166$$

Solution

Diameter of wheel$$=40cm,$$ Radius $$=20$$cm
Distance covered in $$1$$ revolution$$=2\pi r=2\times \cfrac { 22 }{ 7 } \times 20\\ =\cfrac { 880 }{ 7 } cm$$
total distance$$ =176m=17600cm$$
$$ \therefore $$ No. of revolutions$$=\cfrac { 17600 }{ 880/7 } =140$$
Question 3
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Find the area of the figure shown above.

A
$$9$$ sq cm

B
$$10$$ sq cm

C
$$11$$ sq cm

D
$$15$$ sq cm

Solution

Area $$=$$ (Area of rectangle $$ABDC$$) $$\times 2+$$ Area of rectangle $$ DGEF$$
$$=2\times \left( 3\times 1 \right) +5\times 1$$
$$=6+5$$
$$=11$$ sq cm
Question 4
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Samuel wanted to implant some vertical stones along the boundary of his plot at a distance of $$10$$ m each. If length of the plot is $$30$$ m and the breadth is $$15$$ m, then the number of stones used is:

A
$$450$$

B
$$45$$

C
$$9$$

D
$$10$$

Solution

Perimeter of plot$$=2(l+b)=2(30+15)=90m$$
Distance between two stones$$=10m$$
Therefore, number of stones$$=\cfrac { 90 }{ 10 } =9$$
Question 5
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Area of the shaded figure is:

A
$$2400\ sq. m$$

B
$$48\ sq. m$$

C
$$50\ sq. m$$

D
$$98\ sq. m$$

Solution

We know that area of rectangle of sides $$l$$ & $$b$$ = $$ l \times b $$
Area of the shaded region
$$=$$ Area of $$A+$$ Area of $$B$$
$$=8\times 6+10\times 5$$
$$=48+50=98\ sq.\ m$$
Question 6
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The adjacent sides of a parallelogram are $$8\ m$$ and $$5\ m$$. The distance between the longer sides is $$4\ m$$. What is the distance between the shorter sides?

A
$$6.4\ m$$

B
$$6\ m$$

C
$$8.6\ m$$

D
$$None\ of\ these$$

Solution

Area of the parallelogram $$=$$ Longer side $$\times$$ Distance between them
$$= 8 m \times 4 m = 32 \displaystyle\ m^{2}$$
Also, area of the parallelogram $$=$$ Shorter side $$\times$$ Distance between shorter sides
$$ \displaystyle \Rightarrow 32\ m^{2}=5\ m \times$$ Distance between shorter sides
$$ \displaystyle \Rightarrow $$ Distance between shorter sides $$ \displaystyle \frac{32\ m^{2}}{5\ m}$$$$= \displaystyle 6.4\ m$$
Question 7
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Expenditure incurred in cultivating a square field at the rate of Rs. $$170$$ per hectare is Rs. $$680$$. What would be the cost of fencing the field at the rate of Rs. $$3$$ per meter?

A
Rs. $$2400$$

B
Rs. $$3600$$

C
Rs. $$3000$$

D
Rs. $$2000$$

Solution

Total cost of cultivation$$=$$Rs.$$680$$
Cost per hectare$$=$$Rs. $$170$$
$$\therefore $$Area of cultivation$$=\cfrac { 680 }{ 170 } =4$$ hectare$$=4\times { 10 }^{ 4 }m^2$$
Now, field's shape is square, therefore area $$=side \times side$$
$$\Rightarrow 4\times { 10 }^{ 4 }={ (side) }^{ 2 }$$
$$\Rightarrow $$ side$$=\sqrt { 4\times { 10 }^{ 4 } } =200m$$
Perimeter of field$$=4\times 200=800m$$
$$\therefore $$ Cost of fencing $$800m=$$ Rs.$$\left( 3\times 800 \right) =$$ Rs.$$2400$$
Question 8
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If a wire is bent into the shape of a square, the area of the square is $$81$$ sq cm. When the wire is bent into a semicircular shape, what is the area of the semicircle? $$ \displaystyle \left ( \pi =\frac{22}{7} \right )$$

A
$$ \displaystyle 77 \ cm^{2}$$

B
$$80 \ cm^2$$

C
$$75 \ cm^2$$

D
$$70 \ cm^2$$

Solution

Let $$a$$ be the length of each side of the square.
Then, $$ \displaystyle a^{2}=81\Rightarrow a=9cm$$
$$ \displaystyle \therefore $$ Length of wire $$=$$ Perimeter of square
$$= 4\times 9 = 36 cm $$
When the wire is bent into a semi-circular shape then, perimeter of semi circle $$= \displaystyle \pi r+2r=36 $$

$$ \Rightarrow r=\cfrac{36}{\pi +2}=\cfrac{36}{\dfrac{22}{7}+2}$$
$$=\cfrac{36\times\:7}{\left ( 22+14 \right )}$$
$$=\cfrac{36\times\:7}{36}cm$$
$$=7cm $$

Now,
Area of the semicircle $$ \displaystyle =\frac{1}{2}\pi r^{2}=\frac{1}{2}\times\frac{22}{7}\times\:7\times\:7cm^{2}$$
$$ \displaystyle =77cm^{2}$$
Question 9
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Front side wall of a house consists of a rectangle of $$6$$ m $$\times$$ $$4$$ m surrounded by a semicircle with base $$4$$ m It has two isosceles triangles make with vertical sides of the rectangle The net area of the wall in sq m is :

A
$$\displaystyle 4\left ( 15+{\pi } \right )$$

B
$$\displaystyle 4\left ( 18+\pi \right )$$

C
$$\displaystyle 4\left ( 24+\pi \right )$$

D
$$\displaystyle 64+4\pi$$

Solution

Net area of wall$$=2\times $$ area of triangle + area of rectangle+area of semi circle

$$= 2 \times \dfrac{1}{2}\times base \times height + length \times breadth + \dfrac {1}{2} πr^2 $$

$$=2\times \cfrac { 1 }{ 2 } \times 6\times 6+6\times 4+\pi \times { (2) }^{ 2 }$$

$$=36+24+4\pi $$

$$=60+4\pi $$

$$=4\left( 15+\pi \right) m^2$$
Question 10
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Area of the shaded region is

A
$$96\ m^2$$

B
$$15\ m^2$$

C
$$81\ m^2$$

D
$$111\ m^2$$

Solution

We know that area of rectangle of sides $$l$$ & $$b$$ = $$ l \times b $$.
Area of the shaded region
$$=$$ Area of bigger rectangle $$-$$ area of smaller rectangle
$$=\left( 12\times 8 \right) -\left( 5\times 3 \right) $$
$$=96-15=81\ m^2$$