Perimeter and Area Test

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Perimeter and Area Test - 15

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Weekly Quiz Competition

Question 1
1 / -0

$$80$$ students of the same height stand with both hands stretched all along the sides of a rectangular garden, each student covering a length of $$1.75$$ m. Then the perimeter of the garden is

A
$$1400$$ m

B
$$140$$ m

C
$$14$$ m

D
$$1400$$ km

Solution

It is given that $$80$$ students of same height stand with both hands stretched all along the sides of a rectangular garden, each student covering a length of $$1.7$$ m.
Perimeter of the garden $$=$$ Length covered by $$80$$ students.
$$ \Rightarrow 80\times $$ Length covered by single student.
$$ \Rightarrow 80\times 1.75=140 $$
So, option B is correct answer.
Question 2
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Which figure has an area of $$20$$sq. units and a perimeter of $$18$$ units?

A

B

C

D

Solution

Side of one square $$=1$$ unit
Area of one square $$=1\times 1=1$$ sq. unit
(A) Area of $$18$$ squares$$=18$$ sq. units
Perimeter $$=2(6+3)=18$$ units
(B) Area of $$18$$ squares $$=18$$ sq. units
Perimeter $$=2(9+2)=22$$ units
(C) Area of $$16$$ squares $$=16$$ sq. units
Perimeter $$=2(8+2)=20$$ units
(D) Area of $$20$$ squares $$=20$$sq. units
Perimeter $$=2(5+4)=18$$ units.
Question 3
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The given figure is made up of two rectangles. Find its total area.

A
$$144cm^2$$

B
$$216cm^2$$

C
$$240cm^2$$

D
$$300cm^2$$

Solution

Area of the bigger rectangle $$=(18\times 8)cm^2$$
$$=144cm^2$$
Area of the smaller rectangle $$=((30-18)\times (8-6)) = (12\times 6)cm^2$$
$$=72cm^2$$
$$\therefore$$ Total area = area of smaller rectangle + area of bigger rectangle $$=(72+144)cm^2=216cm^2$$.
Question 4
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If the outer and inner radii of a ring are $$10$$ cm and $$8$$ cm, then its area is nearly

A
$$113.443$$ sq. cm

B
$$113.343$$ sq. cm

C
$$113.243$$ sq. cm

D
$$113.143$$ sq. cm

Solution

We know that area of a circle is $$πr^2$$, where $$r$$ be a radius of the circle.
Let us assume $$R = 100$$ $$cm$$ & $$r = 8$$ $$cm$$
$$\therefore$$ Required area $$= \pi (R^2 - r^2 ) = \pi (100^{2} - 8^{2}) = 36π = 113.143\ cm^{2}$$.
Question 5
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A circular park, 42 m in diameter, has a path 3.5 m wide running around it on the outside. Find the cost of gravelling the path at Rs. 4 per $$m^2$$.

A
$$Rs. 2002$$

B
$$Rs. 1652$$

C
$$Rs. 1672$$

D
$$Rs. 2048$$

Solution

Internal Radius = 21 m, External Radius $$= 21 + 3.5 = 24.5$$ m
Area of path = $$\pi(24.5)^2- \pi(21)^2$$
$$\Rightarrow\pi[(24.5)^2-(21)^2]$$
$$= 500.5$$
Cost of travelling $$= 500.5 \times 4 = Rs.2002$$
Question 6
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Circumference $$C$$ of the circular ring is:
where $$C_{outer}$$ and $$C_{inner}$$ denotes the circumference of outer and inner circles respectively.

A
$$C_{outer}+C_{inner}$$

B
$$C_{outer}-C_{inner}$$

C
$$C_{outer}C_{inner}$$

D
$$2(C_{outer}+C_{inner})$$

Solution

Total circumference of the edges of the ring is sum of both inner and outer circumferences.

$$C=C_{outer}+C_{inner}$$
Question 7
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Perimeter of a triangle is the sum of the lengths of all the _____ sides.

A
$$2$$

B
$$4$$

C
$$5$$

D
none of these

Solution

$$\Rightarrow$$ Perimeter of a triangle is the sum of the lengths of all the $$3$$ sides.
$$\Rightarrow$$ In figure, ABC is a triangle.
$$\Rightarrow$$ Perimeter of $$\triangle ABC=AB+BC+CA$$
$$\therefore$$ Perimeter of a triangle is the sum of the lengths of all the three sides.
Question 8
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The shaded area enclosed between the two rings is called _____.

A
Circle

B
Spherical shell

C
Circular ring

D
Sphere

Solution

There are two circles, outer circle and inner circle
The shaded area is enclosed between these two circles or rings and it is also circular.
So, the shaded region enclosed between both the rings is called circular rings.
Hence, the answer is circular ring.
Question 9
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Find the area of the circular park of whose radius is $$4.5 m$$.

A
$$33.85\,{{m}^{2}} $$

B
$$53.55\,{{m}^{2}} $$

C
$$43.85\,{{m}^{2}} $$

D
$$63.585\,{{m}^{2}} $$

Solution

Given that,

Radius $$r=4.5\,m.$$

Then, find area of circular park

We know that,

$$ Area\,of\,circle\,=\,\pi {{r}^{2}}=3.14\times {{\left( 4.5 \right)}^{2}} $$

$$ =3.14\times 4.5\times 4.5=63.585\,{{m}^{2}} $$
Hence, this is the answer.
Question 10
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A circular ring is 5 cm wide. The difference of outer and inner radius is

A
$$5\:cm$$

B
$$2.5\:cm$$

C
$$10\:cm$$

D
none of these

Solution

Since, we are given width of a circular ring which is nothing but a difference of radii the outer and inner circle.
Hence, $$R - r = 5 cm$$.