Perimeter and Area Test

Result

Perimeter and Area Test - 32

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If the circumference of a circle is $$60$$ cm more than its diameter, then the circumference is?

A
$$14\pi$$cm

B
$$28\pi$$cm

C
$$35\pi$$cm

D
$$42\pi$$cm

Solution

Given: Circumference of circle $$=60+$$ Diameter

$$\Rightarrow 2\pi r=60+2r$$

$$\Rightarrow$$ $$2\times \displaystyle\frac{22}{7}\times r=60+2r$$

$$\Rightarrow$$ $$\displaystyle\frac{22}{7}\times r=30+r$$

$$\Rightarrow$$ $$30=\displaystyle\frac{22}{7}r-r$$

$$\Rightarrow$$ $$30=\displaystyle\frac{15r}{7}$$

$$\Rightarrow r=14$$

Hence circumference $$=2\pi r$$

$$=2\times \dfrac{22}{7} \times 14$$

$$=88$$
Question 2
1 / -0

The length of a rectangle is $$\left( \cfrac { 6 }{ 5 } \right) $$th of its breadth. It its perimeter is $$132m$$, its area will be ______ .

A
$$1080{m}^{2}$$

B
$$640{m}^{2}$$

C
$$1620{m}^{2}$$

D
$$2160{m}^{2}$$

Solution

$$\Rightarrow$$ Let breadth of a rectangle be $$x$$.
$$\Rightarrow$$ Then length of a rectangle be $$\dfrac{6}{5}x$$.
$$\Rightarrow$$ Perimeter of a rectangle = $$2(length+breadth)$$
$$\therefore$$ $$132=2(\dfrac{6}{5}x+x)$$

$$\therefore$$ $$66=\dfrac{6+5}{5}x$$

$$\therefore$$ $$x=66\times \dfrac{5}{11}$$
$$\therefore$$ $$x=30\,m$$
$$\Rightarrow$$ So, breadth = $$30\,m$$ and length = $$\dfrac{6}{5}\times 30=36\,m$$
$$\Rightarrow$$ Area of rectangle = $$length\times breadth$$
$$\therefore$$ Area of rectangle = $$36\times 30=1080\,m^2$$
Question 3
1 / -0

Ishika has designed a small oval race track for her remote control car. Her design is shown in the figure. What is the total distance around the track? Round your answer to the nearest whole cm.

A
$$205\ cm$$

B
$$305\ cm$$

C
$$100\ cm$$

D
$$245\ cm$$

Solution

Total distance around the track $$=$$ Perimeter of the given track

The track is a combination of a rectangular path and two semi-circular paths.

Hence perimeter of the given track $$= 52 + 52 + 2\pi r$$

$$= 52 + 52 + 2\times \dfrac {22}{7}\times 16$$

$$= 104 + 100.57$$

$$ = 204.57 \approx 205\ cm$$.
Question 4
1 / -0

The number of envelopes that can be made out of sheet of paper $$384cm$$ by $$172cm$$ if each envelope requires a piece of paper of size $$16cm$$ by $$12cm$$, is _________ .

A
$$340$$

B
$$344$$

C
$$338$$

D
$$342$$

Solution

$$\Rightarrow$$ Length and breadth of a sheet is $$384\,cm$$ and $$172\,cm$$.
$$\Rightarrow$$ Area of a sheet = $$length\times breadth$$
$$\therefore$$ Area of a sheet = $$384\,cm\times 172\,cm=66048\,cm^2$$
$$\Rightarrow$$ Length of a envelope is $$16\,cm$$ and breadth is $$12\,cm$$
$$\Rightarrow$$ Area of envelope = $$length\times breadth$$
$$\therefore$$ Area of a envelope = $$16\,cm\times 12\,cm=192\,cm^2$$
$$\Rightarrow$$ $$Number\, of\, envelopes = \dfrac{Area\,of\,sheet}{Area\,of\,envelope}$$

$$\therefore$$ $$Number\, of\, envelopes =\dfrac{66048}{192}=344$$
Question 5
1 / -0

The area of a square is numerically equal to the perimeter of the square, then the side of square is ______ .

A
$$2$$ units

B
$$3$$ units

C
$$4$$ units

D
$$5$$ units

Solution

$$\Rightarrow$$ Let $$'s'$$ be the side of the square.
$$\rightarrow$$ According to the given condition,
$$\Rightarrow$$ $$Area\,of\,square=Perimeter\,of\,square$$
$$\Rightarrow$$ $$s^2=4s$$
$$\therefore$$ $$s=4.$$
$$\therefore$$ Side of square is $$4\,units.$$
Question 6
1 / -0

All the figures given below consist of the four squares of equal size. Which figure has the smallest perimeter?

A

B

C

D

Solution

$$\Rightarrow$$ We have given that Four squares in diagram has equal size.
$$\Rightarrow$$ Let the each square be $$1\times 1$$ unit square.
$$\Rightarrow$$ Perimeter of Option A diagram = $$1+1+1+1+1+1+1+1=8$$
$$\Rightarrow$$  Perimeter of Option B diagram = $$1+1+1+1+1+1+1+1+1+1=10$$
$$\Rightarrow$$  Perimeter of Option C diagram = $$1+1+1+1+1+1+1+1+1+1=10$$
$$\Rightarrow$$  Perimeter of Option D diagram = $$1+1+1+1+1+1+1+1+1+1=10$$
$$\therefore$$ Option $$A$$ diagram has the smallest perimeter.
Question 7
1 / -0

What is the perimeter of the shaded area in the given figure?

A
$$18$$ units

B
$$19$$ units

C
$$16$$ units

D
$$9$$ units
Question 8
1 / -0

The length of a plot is 45 m and breadth is 35 m . Find the cost of fencing it at Rs. 20 per m

A
Rs. 3200/-

B
Rs. 3500/-

C
Rs. 3300/-

D
Rs. 1652/-

Solution

L=$$45 $$m , B=$$35$$ m
Perimeter of rectangle =$$ 2(L+B)$$
= $$2\times(45+35)$$=$$2\times(80)$$=$$160$$
Cost of fencing is $$Rs.20$$ per m
Cost of fencing the plot=$$20\times160$$=$$Rs.3200$$
Question 9
1 / -0

Of the following statements, the one that is incorrect is:

A
Doubling the length of a given rectangle doubles the area.

B
Doubling the altitude of a triangle doubles the area.

C
Doubling the radius of a given circle doubles the area .

D
Doubling the divisor of a fraction and dividing its numerator by 2 changes the quotient.

Solution

We know, Area of circle $$A=\pi r^2$$

When the radius is doubled, $$R=2r$$

Then, New area $$=\pi R^2$$

$$=\pi (2r)^2$$

$$=4A$$

Area of a circle is directly proportional to square of its radius, so doubling radius will quadruple its area.
Question 10
1 / -0

If a bicycle wheel makes $$5000$$ revolution in moving $$11$$ km, then diameter of wheel is

A
$$110$$ cm

B
$$120$$ cm

C
$$70 $$ cm

D
$$100$$ cm

Solution

Let radius of wheel be $$r㎝$$. Then, distance travel in $$1$$ revolution$$=2\pi r$$
Total distance$$=11㎞=1100m=1100000㎝$$
Total revolutions$$=5000$$
$$\Rightarrow 5000\times 2\pi r=1100000\Rightarrow r=\cfrac { 1100000\times 7 }{ 2\times 22\times 5000 } =35㎝$$
$$\therefore $$Diameter of wheel $$=2\times 35=70㎝$$