Perimeter and Area Test

Result

Perimeter and Area Test - 23

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Question 1
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What is the area of a figure formed by a square of side 8 cm and an isosceles triangle with the base as one side of the square and the other equal sides as 5 cm?

A
$$ \displaystyle 70 \ cm^{2}$$

B
$$ \displaystyle 76 \ cm^{2}$$

C
$$ \displaystyle 82 \ cm^{2}$$

D
$$ \displaystyle 64 \ cm^{2}$$

Solution

Construction: Draw EF perpendicular to AB which will act as a median.
From figure:
Area of given figure$$=ar\triangle ABE+ar \ \square ABCD$$

Now, $$ar\triangle ABE=2\times ar\triangle AFE$$ {$$\because$$ its an isosceles triangle, therefore, EF will act as the median}

$$\triangle AFE$$ is a right-angled triangle, so by Pythagoras theorem

$$EF=\sqrt { A{ E }^{ 2 }+A{ F }^{ 2 } } $$

$$=\sqrt { { \left( 5 \right) }^{ 2 }+{ \left( 4 \right) }^{ 2 } } =\sqrt { 9 } =3cm$$

$$\therefore ar\triangle AFE=\cfrac { 1 }{ 2 } \times $$ base $$\times $$ height

$$=\cfrac { 1 }{ 2 } \times 4\times 3=6cm^2$$

$$\therefore ar\triangle ABE=2\times 6=12cm^2$$

Also, area of square $$ABCD = (8\times 8) cm^2$$

Now, area of figure$$=12+{ (8) }^{ 2 }=76cm^2$$
Question 2
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The perimeter of an equilateral triangle and a square are same then $$\displaystyle \frac{area\, of \Delta }{area\, of\ \square }=$$

A
$$\displaystyle \frac{4}{3}$$

B
$$1$$

C
$$1.5$$

D
$$<1$$

Solution

Let side of equilateral $$\triangle =a$$
Side of square$$=b$$
Since, perimeter of triangle = perimeter of square
$$\therefore 3a=4b\Rightarrow \cfrac { a }{ b } =\cfrac { 4 }{ 3 } $$

$$\therefore \cfrac { ar\triangle }{ ar\triangle } =\cfrac { \cfrac { \sqrt { 3 } }{ 4 } \times { a }^{ 2 } }{ { b }^{ 2 } } =\cfrac { \sqrt { 3 } }{ 4 } \times { \left( \cfrac { a }{ b } \right) }^{ 2 }=\cfrac { \sqrt { 3 } }{ 4 } \times \cfrac { 16 }{ 9 } =\cfrac { 4 }{ 3\sqrt { 3 } } <1$$
Question 3
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A square and a rectangular field with measurements as given in the following figure have the same perimeter. Which field has a larger area?

A
Square has larger area, square $$=3600\;m^2$$.

B
Rectangle has larger area, square $$=3600\;m^2$$.

C
Square has larger area, square $$=1600\;m^2$$.

D
Square has larger area, square $$=3800\;m^2$$.

Solution

Let $$x$$ be the breadth of the rectangle.
It is given that the perimeter of a rectangle = perimeter of the square.
$$\therefore\;2(80+x)=4\times60$$
$$\Rightarrow\; 80+x=120$$
$$\Rightarrow\;x=120-80=40$$
i.e. breadth of the rectangle $$=40\;m$$
Now, Area of the square $$=(60\times60)\;m^2$$
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=3600\;m^2$$.
and the area of the rectangle $$=(40\times80)\;m^2$$.
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=3200\;m^2$$.
Hence, the square field has a larger area.
Question 4
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The radius of a circle is $$14 m$$, then the circumference of a circle is

A
$$616 m$$

B
$$88 m$$

C
$$154 m$$

D
none of these

Solution

$$r =14 m$$
Circumference = $$2\pi r$$
$$=\, 2\, \times\, \displaystyle \dfrac {22}{7}\, \times14\, =\, 88\, m$$
Question 5
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The area of a parallelogram is $$120$$ $$cm^{2}$$ and its altitude is $$10$$ cm. The length of the base is

A
$$24$$ cm

B
$$12$$ cm

C
$$8$$ cm

D
$$4$$ cm

Solution

$$Area of parallelogram$$
= $$Base \times height = 120 cm^{2}$$.
$$Base=\cfrac{120}{10}$$
$$\therefore $$ Base = $$12 cm$$
Question 6
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Find the area of the unshaded portion of figure, the shaded portions being semi-circular regions.

A
$$100.5\;m^2$$.

B
$$101.5\;m^2$$.

C
$$201.5\;m^2$$.

D
$$101.8 \;m^2$$.

Solution

Area of unshaded portion $$=$$ Area of rectangle - Area of circle
$$=$$ $$l\times b-\pi { r }^{ 2 }$$
$$=$$ $$14$$$$\times 10-\dfrac { 22 }{ 7 } \times 3.5\times 3.5$$
$$=$$ $$140 - 38.5$$
$$=$$ $$101.5$$ $${ m }^{ 2 }$$
Question 7
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The circumference of a circle is $$44 m$$, then the area of the circle is

A
$$6084.5\, m^2$$

B
$$276.5\, m^2$$

C
$$154\, m^2$$

D
$$44\, m^2$$

Solution

Circumference of circle $$(C) = 2\pi r\, =\, 44$$

$$\therefore r\, =\, \displaystyle \frac {44}{2\, \times\, \displaystyle \frac {22}{7}}\, = \dfrac {44 \times 7}{2 \times 22} = 7\, m$$

Area of a circle $$=\, \pi r^2$$

$$=\, \displaystyle \frac {22}{7}\, \times\, 7\, \times\,7\, =\, 154\, m^2$$
Question 8
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The area of a circle is $$2464 m^2$$, then the diameter is

A
$$56 m$$

B
$$154 m$$

C
$$176 m$$

D
none of these

Solution

By given, Area =$$2464m^2$$
$$\displaystyle \frac {\pi d^2}{4}\, =\, 2464$$
$$d^2\, =\, \displaystyle \frac {2464\, \times\, 4}{\displaystyle \frac {22}{7}}\, =\, 3136$$
$$d\, =\, \sqrt {3136}\, =\, 56$$m
Question 9
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Abhay made a straight cut through a circular rubber band and then laid the rubber band flat, as shown in the figure.
Which measure corresponds to the length of the cut rubber band ?

A
Chord

B
Circumference

C
Diameter

D
Radius

Solution

Circumference of the rubber band is equal to the length of the cut rubber band .
Hence option B is answer
Question 10
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The inner diameter of a circular building is $$54 cm$$ and the base of the circular wall occupies a space of $$\displaystyle 2464cm^{2}$$. The thickness of the wall is:

A
$$1 cm$$

B
$$2 cm$$

C
$$4 cm$$

D
$$5 cm$$

Solution

Let the outer radius of the wall be $$R$$ cm.

Base of the wall is circular in shape with R as its radius.

Given, $$ \pi { R }^{ 2 } = 2464 $$

$$ => { R }^{ 2 } = \cfrac {2464 \times 7}{22} \\\ \ \ \ \ \ \ \ \ \ \ = 784 $$

$$ => R = 28 cm $$

Inner Radius, $$ r= \cfrac {54}{2} = 27 cm $$

Hence, thickness $$ = $$ Outer radius $$ - $$ Inner Radius
$$ = 28 - 27 \\= 1 cm $$