Perimeter and Area Test

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Perimeter and Area Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

A wire is bent to form a square of side $$22\, cm$$ If the wire is rebent to form a circle, its radius is.

A
$$22\, cm$$

B
$$14\, cm$$

C
$$11\, cm$$

D
$$7\, cm$$

Solution

Option (b) is correct.
Given, side of a square is $$22\, cm$$ The wire has same length.
Hence, perimeter of square arid circumference of circle are equal.
ACQ,
Circumference of circle $$=$$ Perimeter of square
$$= 2 \times \pi \times r = 4\times (side)$$
$$ \Rightarrow 2\times \dfrac {22}{7}\times r = 4\times 22$$

$$\Rightarrow r =\dfrac {4\times 22 \times 7}{2\times 22}$$
$$ \Rightarrow r= 14\, cm$$
Therefore, $$14\, cm$$ is the radius of the circle.
Question 2
1 / -0

In Fig., EFGH is a parallelogram, altitudes FK and Fl are 8 cm and ,10 cm respectively. If $$EF =10\, cm$$, then area of EFGH is

A
$$ 20\, cm^2$$

B
$$ 32\, cm^2$$

C
$$ 40\, cm^2$$

D
$$ 80\, cm^2$$

Solution

Option (c) is correct
Area of parallelogram (EFGH) $$=$$ Base $$\times$$ corresponding height
$$ = 10\times 4$$
$$ = 40\, cm^2$$
Question 3
1 / -0

Area of triangle PQR is $$100\, cm^2$$ (Fig ). If altitude QT is $$10\, cm$$, then its base PR is.

A
$$ 20 \,cm$$

B
$$ 15 \,cm$$

C
$$ 10 \,cm$$

D
$$5 \,cm$$

Solution

Option (a) is correct.
Give, area of triangle $$PQR =100\, cm^2$$.
We know drat Area of triangle $$PQR \times \dfrac {1}{2} PR \times QT$$

$$ \Rightarrow 100 = \dfrac {1}{2} \times PR \times 10$$

$$ \Rightarrow PR =\dfrac {100\times 2}{10}$$

$$ \Rightarrow PR = 20\, cm$$
Question 4
1 / -0

In Fig, if $$PR = 12\, cm, QR = 6\, cm$$ and $$PL= 8\, cm$$, then QM is

A
$$ 6\, cm$$

B
$$ 9\, cm$$

C
$$4\, cm$$

D
$$ 5\, cm$$

Solution

Given that $$PL =8\, cm$$
$$PR =12\,cm$$
$$ QR = 6\, cm$$
Now, in right-angled triangle PLR
Using Pythagoras theorem,
$$ \Rightarrow PR^2 = PL^2 + LR^2$$
$$ \Rightarrow LR^2 = 144 - 64$$
$$ =\sqrt {80} = 4\sqrt {5}\ cm$$
$$LR = LQ +QR$$
$$ \Rightarrow LQ = LR - QR = (4\sqrt{5} - 6)\ cm$$

Now area of triangle PLR
$$A =\dfrac {1}{2}\times \left ( 4\sqrt {5} \right )\times 8$$
$$= 16\sqrt{5}\ cm^2$$

Now, area of triangle PLQ
$$A =\dfrac {1}{2}\times \left ( 4\sqrt {5} - 6\right )\times 8$$
$$= (16\sqrt{5}-24) \,cm^2$$

Hence, area of triangle PLR $$=$$ Area of triangle PLQ $$ +$$ Area of triangle PQR

So, area of triangle PLR $$-$$ Area of triangle PLQ $$ =$$ Area of triangle PQR
$$ \Rightarrow 16 \sqrt {5} - (16 \sqrt {5} -24)$$,
$$ \Rightarrow $$ Area of triangle PQR $$= 24 \, cm^2$$

$$ \Rightarrow \dfrac {1}{2}\times PR\times QM = 24$$

$$ \Rightarrow \dfrac {1}{2} \times 12 \times QM =24$$

$$ \therefore QM = 4\,cm$$
Question 5
1 / -0

Area of circular garden with diameter $$8\, m$$ is.

A
$$12.56\, m^2$$

B
$$ 25.12 \, m^2$$

C
$$50.28 \, m^2$$

D
$$200.96 \, m^2$$

Solution

Given, diameter is $$8\;\text{m}$$
Hence, radius $$ =\dfrac{8}{2}\;\text{m}= 4\;\text{m}$$
Therefore, area of the circular garden
$$=\pi r^2$$
$$=\dfrac {22}{7}\times 4\times 4$$
$$= 50.28\;\text{m}^2$$
Question 6
1 / -0

The area of a parallelogram is 60 $$cm^2$$ and one of its altitude is 5 cm. The length of its corresponding side is

A
12 cm

B
6 cm

C
4 cm

D
2 cm

Solution

Area of a parallelogram =
$$ side \times altitude$$
$$ \Rightarrow a \times h $$ = 60
$$ \Rightarrow a \times 5 $$ = 60
$$ \Rightarrow a= \dfrac{60}{5} $$
$$ \Rightarrow a=12 cm $$
Question 7
1 / -0

The radius of a circle is $$5\ cm$$ then its area is

A
$$154\ cm^2$$

B
$$1386\ cm^2$$

C
$$550\ cm^2$$

D
$$308\ cm^2$$

E
None

Solution

Area of circle $$=$$ $${\pi r}^2$$
$$=3.14\times5\times5$$
$$=78.5\ cm^2$$

Hence, option $$E$$ is correct.
Question 8
1 / -0

Find the area of a ring-shaped region enclosed between two concentric circles of diameter $$8cm$$ and $$6cm$$

A
$$28\ cm^2$$

B
$$44\ cm^2$$

C
$$66\ cm^2$$

D
$$88\ cm^2$$

Solution

We know that area of a circle is $$πr^2$$, where $$r$$ be a radius of the circle.
Let us assume $$R = 8$$ $$cm$$ & $$r = 6$$ $$cm$$
$$\therefore$$ Required area $$= \pi (R^2 - r^2 ) = \pi (8^{2} - 6^{2}) = 28π \ cm^{2}$$.
Question 9
1 / -0

The area of circle whose circumference is $$44cm$$ is

A
$$77\ cm^2$$

B
$$308\ cm^2$$

C
$$168\ cm^2$$

D
$$154\ cm^2$$

Solution

Let $$C$$ is circumference of circle
Given $$C=44\quad cm$$

$$\therefore 2\pi r=44$$

$$\therefore 2\times \frac { 22 }{ 7 } \times r=44$$

$$\therefore r=\frac { 44\times 7 }{ 44 } $$

$$\therefore r=7\quad cm$$
Thus, radius of circle is 7 cm.

Now, Area of circle is given by,
$$A=\pi { r }^{ 2 }$$

$$\therefore A=\frac { 22 }{ 7 } \times { 7 }^{ 2 }$$

$$\therefore A=22\times 7$$

$$\therefore A=154\quad { cm }^{ 2 }$$

Thus, area of circle is $$154\quad { cm }^{ 2 }$$
Question 10
1 / -0

The external and internal radius of a circle path area $$5 m$$ and $$3m$$ respectively. The are of the circular path is

A
$$87 \pi m^2$$

B
$$16 \pi m^2$$

C
$$157 \pi m^2$$

D
$$237 \pi m^2$$

Solution

We know that area of a circle is $$πr^2$$, where $$r$$ be a radius of the circle.
Let us assume $$R = 5$$ $$m$$ & $$r = 3$$ $$m$$
$$\therefore$$ Required area $$= \pi (R^2 - r^2 ) = \pi (5^{2} - 3^{2}) = 16π \ m^{2}$$.

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Perimeter and Area Test - 36

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Perimeter and Area Test - 36

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SHARING IS CARING

Question 1

$$22\, cm$$

$$14\, cm$$

$$11\, cm$$

$$7\, cm$$

Solution

Question 2

$$ 20\, cm^2$$

$$ 32\, cm^2$$

$$ 40\, cm^2$$

$$ 80\, cm^2$$

Solution

Question 3

$$ 20 \,cm$$

$$ 15 \,cm$$

$$ 10 \,cm$$

$$5 \,cm$$

Solution

Question 4

$$ 6\, cm$$

$$ 9\, cm$$

$$4\, cm$$

$$ 5\, cm$$

Solution

Question 5

$$12.56\, m^2$$

$$ 25.12 \, m^2$$

$$50.28 \, m^2$$

$$200.96 \, m^2$$

Solution

Question 6

12 cm

6 cm

4 cm

2 cm

Solution

Question 7

$$154\ cm^2$$

$$1386\ cm^2$$

$$550\ cm^2$$

$$308\ cm^2$$

None

Solution

Question 8

$$28\ cm^2$$

$$44\ cm^2$$

$$66\ cm^2$$

$$88\ cm^2$$

Solution

Question 9

$$77\ cm^2$$

$$308\ cm^2$$

$$168\ cm^2$$

$$154\ cm^2$$

Solution

Question 10

$$87 \pi m^2$$

$$16 \pi m^2$$

$$157 \pi m^2$$

$$237 \pi m^2$$

Solution

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