5 Marks Questions

Question 15 Marks

A chord of a circle of radius 12 cm subtends an angle of $120^o$ at the center. Find the area of the corresponding segment of the circle. (Using $\pi$ = 3.14 and $\sqrt3$ = 1.73)

Answer

r = 12 cm, $\theta$ $= 120^o$
$\therefore$ Area of the corresponding sector of the circle = $\frac\theta{360^\circ}\mathrm{πr}^2\;=\;\frac{\displaystyle120}{\displaystyle360^\circ}\times3.14\;\times12\times12\;$= $150.72 cm^2$
Area of $\triangle$AOB
Draw OM $\perp$ AB
In right triangle OMA and OMB,
OA = OB ....... Radii of the same circle
OM = OM ........ Common side
$\therefore$ $\triangle$OMA $\cong$ $\triangle$OMB ........RHS congruence criterion
$\therefore$ AM = BM ....... CPCT
$\Rightarrow$ AM = BM = $\frac 12$AB
and $\angle$AOM = $\angle$BOM [CPCT]
$\Rightarrow$ $\angle$AOM = $\angle$BOM = $\frac 12$$\angle$AOB
= $\frac 12$ $\times$ $120^o = 60^o$
$\therefore$ In right triangle OMA,
$cos60^o$ = $\frac {OM}{OA}$
$\Rightarrow$ $\frac 12$= $\frac {OM}{12}$
$\Rightarrow$ OM = 6 cm
sin$60^o$= $\frac {AM}{OA}$
$\Rightarrow$ $\frac{\sqrt3}2$= $\frac {AM}{12}$
$\Rightarrow$ AM = 6$\sqrt3$ cm
$\Rightarrow$ 2AM = 12$\sqrt3$ cm
$\Rightarrow$ AB =12$\sqrt3$ cm
$\therefore$ Area of $\triangle$AOB =$\frac 12$ $\times$ AB $\times$ OM
= $\frac 12$ $\times$ 12$\sqrt3$ $\times$ 6 = 36$\sqrt3$ $cm^2$
= 36 $\times$ 1.73 $cm^2$ = 62.28 $cm^2$
So, Area of the corresponding segment of the circle = Area of the correspoding sector of circle - Area of $\triangle$AOB
= 150.72 - 62.28 = 88.44 $cm^2$

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Question 25 Marks

A chord of a circle of radius 15 cm subtends an angle of $60^o$ at the center. Find the areas of the corresponding minor and major segments of the circle. (Use $\pi$ = 3.14 and $\sqrt3$ = 1.73)

Answer

$r = 15 cm, \theta = 60^o$
Area of the minor sector = $\frac\theta{360^\circ}\mathrm{πr}^2\;=\;\frac{\displaystyle60^\circ}{\displaystyle360^\circ}\times3.14\;\times15\times15$ = $117.75 cm^2$
In $\triangle$AOB, draw OM $\perp$ AB
In right triangle OMA and OMB,
OA = OB .........Radii of the same circle
OM = OM .........Common side
$\therefore$ $\triangle$OMA $\cong$ $\triangle$OMB .........RHS congruence criterion
$\therefore$ AM = BM .......CPCT
$\Rightarrow$ AM = BM = $\frac 12$AB
$\angle$AOM = $\angle$BOM .......CPCT
$\Rightarrow$ $\angle$AOM = $\angle$BOM = $\frac 12$$\angle$AOB = $\frac 12$ $\times$ $60^o = 30^o$
$\therefore$ In right triangle OMA, $cos30^o$ = $\frac {OM}{OA}$
$\Rightarrow$ $\frac{\sqrt3}2$= $\frac {OM}{15}$
$\Rightarrow$ OM = $\frac{15\sqrt3}2$cm
$sin30^o$ = $\frac {AM}{OA}$
$\Rightarrow$ $\frac 12$= $\frac {AM}{15}$
$\Rightarrow$ AM = $\frac{15}2$cm
$\Rightarrow$ AB = 15 cm
$\therefore$ Area of $\triangle$AOB = $\frac 12$ $\times$ AB $\times$ OM
= $\frac 12$ $\times$ 15 $\times$ $\frac{15\sqrt3}2$ = $\frac{225\sqrt3}4$
= $\frac {225 × 1.73}4$ $= 97.3125 cm^2$
$\therefore$ Area of the corresponding minor segment of the circle = Area of minor sector - Area of $\triangle$AOB
$= 117.75 - 97.3125 = 20.4375 cm^2$
and, area of the corresponding major segment of the circle = $\pi$r$^2$ - area of the corresponding minor segment of the circle
= 3.14 $\times$ 15 $\times$ 15 - 20.4375
$= 706.5 - 20.4375 = 686.0625 cm^2$

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Question 35 Marks

In a circle of radius 21 cm, an arc subtends an angle of $60^o$ at the center. Find:

the length of the arc.
area of the sector formed by the arc.
area of the segment formed by the corresponding chord

Answer

Given, r = 21 cm and $\theta = 60^o$

Length of arc = $=\;\frac\theta{360}2\mathrm{πr}\;=\frac{60}{360}\times2\times\frac{22}7\times21\;=\;22\;\mathrm{cm}$
Area of the sector = $=\;\frac\theta{360}\mathrm{πr}^2\;=\frac{60}{360}\times\frac{22}7\times21\times21\;=\;231\;\mathrm{cm}^2$
Area of segment formed by corresponding chord = Area of the sector - Area of $\triangle$OAB
= $\frac\theta{360}\mathrm{πr}^2$ - Area of $\triangle$OAB
⇒ Area of segment = 231 - Area of $\triangle$OAB ....... (1)
In right angled triangle OMA and OMB,
OM = OB [Radii of the same circle]
OM = OM [Common]
$\therefore$ $\triangle$OMA $\cong$ $\triangle$OMB [ RHS congruency]
$\therefore$ AM = BM [By CPCT]
$\therefore$ M is the mid-point of AB and $\angle$AOM = $\angle$BOM
$\Rightarrow$ $\angle$AOM = $\angle$BOM = $\frac{1}2$
$\angle$AOB =$\frac{1}2$ $\times 60^o = 30^o$
Therefore, in right angled triangle OMA,
cos$30^o$ = $\frac{OM}{OA}$
$\Rightarrow$ $\frac{\sqrt3}2$= $\frac{OM}{21}$
$\Rightarrow$ OM = $\frac{21\sqrt3}2$ cm
Also, sin $30^o$ = $\frac{AM}{OA}$
$\Rightarrow$ $\frac{1}2$= $\frac{AM}{21}$
$\Rightarrow$ AM = $\frac{21}2$ cm
$\therefore$ AB = 2 AM = 2 $\times$ $\frac{21}2$ = 21 cm
$\therefore$ Area of $\triangle$OAB = $\frac{1}2$ $\times$ AB $\times$ OM = $\frac{1}2$ $\times$ 21 $\times$ $\frac{21\sqrt3}2$ =$\frac{441\sqrt3}4$ $cm^2$
Using eq. (1),
Area of segment formed by corresponding chord = [231 - $\frac{441\sqrt3}4$] $cm^2$
$= 40.05 cm^2$

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Question 45 Marks

A round table cover has six equal designs as shown in figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per $cm^2$. (use $\sqrt3$ = 1.7)

Answer

r = 28 cm and θ = $\frac{360}6\;=60^\circ$
Area of minor sector = $\frac\theta{360}\mathrm{πr}^2\;=\frac{\displaystyle60}{\displaystyle360}\times\frac{22}7\times\;28\times\;28\;=\frac{1232}3$
$= 410.67 cm^2$
For, Area of $\triangle$AOB,

Draw OM $\perp$ AB.
In right triangles OMA and OMB,
OA = OB [Radii of same circle]
OM = OM [Common]
$\therefore$ $\triangle$OMA $\cong$ OMB [RHS congruency]
$\therefore$ AM = BM [By CPCT]
$\Rightarrow$ AM = BM = $\frac12$AB and $\angle$AOM = $\angle$BOM = $\frac12$AOB = $\frac12$ $\times$ $60^o = 30^o$
In right angled triangle OMA, $cos30^o$ = $\frac{OM}{OA}$
$\Rightarrow$ $\frac{\sqrt3}2$ = $\frac{OM}{28}$
$\Rightarrow$ OM = 14$\sqrt3 $ cm
Also, sin30° = $\frac{AM}{OA}$
$\Rightarrow$ $\frac12\;=\frac{AM}{28}$
$\Rightarrow$ AM = 14 cm
$\Rightarrow$ 2 AM = 2 $\times$ 14 = 28 cm
$\Rightarrow$ AB = 28 cm
∴ Area of $\triangle$AOB = $\frac12$$\times$ AB $\times$ OM = $\frac12$ $\times$ 28 $\times$ 14$\sqrt3 $ = 196 $\sqrt3 $ = 196 $\times$ $1.7 = 333.2 cm^2$
∴ Area of minor segment = Area of minor sector - Area of $\triangle$AOB
$= 410.67 - 333.2 = 77.47 cm^2$
∴ Area of one design $= 77.47 cm^2$
∴ Area of six designs = 77.47 $\times$ 6 $= 464.82 cm^2$
Cost of making designs = 464.82 $\times$ 0.35 = Rs. 162.68

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Question 55 Marks

Find the area of the segment AYB shown in Figure, if the radius of the circle is 21 cm and $\angle$AOB = 120°. (Use $\pi = \frac{22}{7}$).

Answer

Area of the segment AYB = Area of sector OAYB - Area of ∆ OAB

Now, area of the sector OAYB = $\frac{120}{360} \times \frac{22}{7} \times 21 \times 21 \mathrm{cm}^{2}=462 \mathrm{cm}^{2}$

For finding the area of ∆ OAB, draw OM ⊥ AB as shown in Fig.

Note that OA = OB. Therefore, by RHS congruence, ∆ AMO ≅ ∆ BMO.

So, M is the mid-point of AB and $\angle \mathrm{AOM}=\angle \mathrm{BOM}=\frac{1}{2} \times 120^{\circ}=60^{\circ}$

Let OM = x cm

So, from ∆ OMA, $\frac{\mathrm{OM}}{\mathrm{OA}}=\cos 60^{\circ}$

or, $\frac{x}{21}=\frac{1}{2} \quad\left(\cos 60^{\circ}=\frac{1}{2}\right)$

or, $x=\frac{21}{2}$

So, $\mathrm{OM}=\frac{21}{2} \mathrm{cm}$

Also, $\frac{\mathrm{AM}}{\mathrm{OA}}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}$

So, $\mathrm{AM}=\frac{21 \sqrt{3}}{2} \mathrm{cm}$

Therefore, $A B=2 A M=\frac{2 \times 21 \sqrt{3}}{2} \mathrm{cm}=21 \sqrt{3} \mathrm{cm}$

So, area of ∆ OAB $=\frac{1}{2} \mathrm{AB} \times \mathrm{OM}=\frac{1}{2} \times 21 \sqrt{3} \times \frac{21}{2} \mathrm{cm}^{2}$

$=\frac{441}{4} \sqrt{3} \mathrm{cm}^{2}$

Therefore, area of the segment AYB = $\left(462-\frac{441}{4} \sqrt{3}\right) \mathrm{cm}^{2}$ [From (1), (2) and (3)]

$=\frac{21}{4}(88-21 \sqrt{3}) \mathrm{cm}^{2}$

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Question 65 Marks

Find the area of the sector of a circle with radius 4 cm and of angle $30^\circ$. Also, find the area of the corresponding major sector.(use $ \pi =3.14 )$

Answer

Here, $\theta$ = 30° and r = 4 cm
Area of sector OAPB =$\frac { \theta } { 360 } \times \pi r ^ { 2 }$
Let 'A' be the area of corresponding major sector.
Then, A = Area of sector OAQB
$\Rightarrow$A = Area of the circle - Area of the corresponding minor sector
$\Rightarrow A = \pi r ^ { 2 } - \frac { \theta } { 360 } \times \pi r ^ { 2 }$
$\Rightarrow A = \pi r ^ { 2 } \left( 1 - \frac { \theta } { 360 } \right)$
$\Rightarrow A = 3.14 \times 4 \times 4 \left( 1 - \frac { 30 } { 360 } \right) \mathrm { cm } ^ { 2 }$
$\Rightarrow A = 3.14 \times 4 \times 4 \times \frac { 11 } { 12 } \mathrm { cm } ^ { 2 }$$= \frac { 3.14 \times 44 } { 3 } \mathrm { cm } ^ { 2 }$= 46.05 $cm^2$

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