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Yogita Ingle 6 years, 2 months ago
The annual salary received by Subba Rao in the years 1995,1996,1997..... is 5000, 5200, 5400,...... 7000
The incomes that Subba Rao obtained in various years are in A.P. as every year, his salary is increased by Rs 200.
Therefore, the salaries of each year after 1995 are
5000, 5200, 5400, …
Given:
a = 5000
d = 200
Let after nth year, his salary be Rs 7000.
an = a + (n − 1) d
7000 = 5000 + (n − 1) 200
200(n − 1) = 2000
(n − 1) = 10
n = 11
Thus in 11th year, his salary will be Rs 7000.
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Sia ? 6 years, 2 months ago
Here, we have
r = 15cm, {tex}\theta{/tex} = 60o
Let OACBO be the given sector and OAB is a triangle. Then
(i). Area of the sector (OACBO)
{tex}=\pi r^{2} \times \frac{\theta}{360}{/tex}
{tex}=\left(3.14 \times 15 \times 15 \times \frac{60}{360}\right) \mathrm{cm}^{2}{/tex}
= 117.75 cm2
(ii) Area of the triangle (AOB)
{tex}=\frac{1}{2} r^{2} \sin \theta{/tex}
{tex}=\frac{1}{2} \times 15 \times 15 \times \frac{\sqrt{3}}{2}{/tex}
{tex}=\frac{15 \times 15 \times 1.73}{4}=97.313 \mathrm{cm}^{2}{/tex}
Now,
Area of MInor segment (ACBA)
= Area of sector (OACBO) - Area of triangle (AOB)
= (117.75 - 97.313) cm2
= 20.437 cm2
and Area of Major segment (ABDA)
= Area of circle - Area of Minor segment
= ({tex}\pi {/tex}r3 - 20.437) cm2
= (3.14 {tex}\times{/tex}15 {tex}\times{/tex} 15 - 20.437) cm2
= (706.5 - 20.437) cm2
= 686.063 cm2
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TAKE RATIO OF EACH I HOPE U KNOW HOW TO CALULATE AREA....
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Sahil ???? 6 years, 2 months ago
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