Question

Q.4: Find the area of the segment AYB shown in the figure, if the radius of the circle is 21 cm and ∠ AOB = 120°. (Use π = 22/7).

Q.5: Find the area of the shaded design in given figure, where ABCD is a square of side 10 cm and semicircles are drawn with each side of the square as diameter. (Use π = 3.14).

Q.6: A round table cover has six equal designs as shown in the figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs. 0.35 per cm² (Use 3 = 1.7)

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Answer 1

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In ∆AOB

Draw a perpendicular line from O which intersect AB at M.

In ∆AOM

• AMO = 90
• OAM = 30

cos 30 = AM/AO

√3/2 = AM/21

AM = 21×√3/2

AB = 2(AM)

=2(21×√3/2)

=21√3

OM^2 = AO^2-AM^2

=21^2-(21√3/2)^2

=441-330.51

=110.48

OM =√110.48

OM =10.51

OM = 10.51cm

Area of ∆AOM

= 1/2 AB × OM

=1/2 ×21√3 ×10.51

=191.14cm^2

Area of sector AOBY = 120πr^2/360

=120×21×21×22/2520

=462cm^2

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

=462-191.14

=270.86

Area of segment AYB is 270.86cm^2.

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Let Unshaded regions be 1, 2, 3 and 4

 Area of 1 + Area of 3= Area of ABCD – Areas of two semicircles of each of radius 5 cm 

Area of 1 and 3 = ( 10 * 10 - 2 * 1/2 * 3.14 * 5 *5)  [Area of semi circle                                                                                         =  1/2  pie r²]

                          = (100 - 3.14 * 25)

                          = (100 - 78.5)

                          =21.5 cm²

So,

Even the Area of 2 and 4 is equal to 21.5cm²

So,

Area of shaded region = Area of ABCD - Area 0f( 1+2+3+4)

                                       = 100 - (21.5 + 21.5)

                                       = 100 - 43

Area of shaded region = 57cm²



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Given

• Number of equal designs = 6
• The radius of round table cover = 28 cm
• Cost of making design = 0.35 per cm2
• ∠O = 360°/6 = 60°

Solution:

ΔAOB is isosceles as two sides are equal.

(Radius of the circle)

∴ ∠A = ∠B

Sum of all angles of triangle = 180°

$\tt \: ∠A + ∠B + ∠O = 180° \\ \tt \: ⇒ 2 ∠A = 180° – \: 60° \\ \tt \: ⇒ ∠A = \frac{120}{2} \\ \tt \: ⇒ ∠A = 60° \\ \\ \tt \: ∠A = ∠B = ∠C = 60° \\ \\ \tt \: Area \: of \: equilateralm ΔAOB \:\\ \tt \: = √3/4 × (OA)2 \\ \tt \:= √3/4 × 282\\ \tt \: = 333.2 cm {}^{2} \\ \\ \tt \:Area \: of \: sector \: ACB\\ \tt \: = ( \frac{60}{360} ) × π r {}^{2} cm {}^{2} \\ = \frac{1}{6} \times \frac{22}{7} × 28 × 28 \\ \tt \:= 410.66 cm {}^{2} \\ \\ \tt \:Area \: of \: design\\ \tt \: = Area \: of \: sector \: ACB – Area \: of \: equilateral \: ΔAOB\\ \tt \:</p><p>= 410.66 cm {}^{2} – 333.2 cm {}^{2} \\ \tt \:= 77.46 cm {}^{2} \\ \\ \tt \:Area \: of \: 6 \: design\\ \tt \: = 6 × 77.46 cm {}^{2} \\ \tt \:= 464.76 cm {}^{2} </p><p> \\ \\ \tt \: Total \: cost \: of \: making \: design\\ \tt \: = 464.76cm {}^{2} × 0.35 \: per \: cm {}^{2} \\ \tt \:= 162.66$

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