Question

ABCD is a cyclic rectangle. Prove that the centre of the circle through A, B, C
and D is the point of intersection of the diagonals of the rectangle.

Answer 1

Answer:

Hello............. ^_^

Q. (6)

Let Principal be 'P'

and Rate of interest be 'R'

Time (n) = 2 years

Simple Interest (S.I) = Rs. 400

Compound Interest (C.I) = Rs. 410

$\begin{gathered}S.I = \frac{P \times R \times T}{100} \\ \\ = \frac{P \times R \times T}{100} = 400 \\ \\ = \frac{PR \times 2}{100} = 400 \\ \\ = \frac{PR}{50} = 400 \\ \\ = PR = 400 \times 50 \\ = PR = 20000\end{gathered}$S.I=100P×R×T=100P×R×T=400=100PR×2=400=50PR=400=PR=400×50=PR=20000

$\begin{gathered}C.I = P( \: \: ( {1 + \frac{R}{100} })^{n} - 1) \\ \\ = P( \: \: ( {1 + \frac{R}{100} })^{n} - 1) = 410 \\ \\ = P( \: \: ( {1 + \frac{R}{100} })^{2} - 1) = 410 \\ \\ = P( \: \: {(1)}^{2} + 2 \times 1 \times \frac{R}{100} + { (\frac{R}{100}) }^{2} - 1 ) = 410 \\ \\ = P( \: \: 1 - 1 \times \frac{R}{50} + { (\frac{R}{100}) }^{2} \: \: \: ) = 410 \\ \\ = P( \: \frac{R}{100} + \frac{ {R}^{2} }{10000} \: \: ) = 410 \\ \\ (take \: \: common \: R \: ) \\ \\ = PR ( \frac{1}{50} + \frac{R}{10000} ) = 410 \\ (put \: the \: value \: of \: PR \: = 20000) \\ \\ = 20000( \frac{1}{50} + \frac{R}{10000} ) = 410 \\ \\ = 20000( \frac{200 + R }{10000} ) = 410 \\ \\ = \frac{R + 200}{10000} = \frac{410}{20000 } \\ \\ = \frac{R + 200}{10000} = \frac{41}{2000 } \\ \\ = R + 200 = \frac{41}{2000} \times 10000 \\ \\ = R + 200 = 41 \times 5 \\ = R + 200 = 205 \\ = R = 205 - 200 \\ = R = 5\end{gathered}$

Rate of interest = 5%

Now Principal,

as PR

$\begin{gathered}Now Principal, \\ as P \times R = 20000 \\ = > P \times 5 = 20000 \\ = > P = \frac{20000}{5} \\ \\ = > P = 4000\end{gathered}$NowPrincipal,asP×R=20000=>P×5=20000=>P=520000=>P=4000

Principal = Rs. 4000

Q. (7)

A man invested Rs. 1000 for 3 years at 11% Simple Interest .....................(CASE 1)

Principal (P) = Rs. 1000

Rate of interest (R) = 11% per annum

Time (n) = 3 years

$\begin{gathered}S.I = \frac{P \times R \times T}{100} \\ \\ = \frac{1000 \times 11 \times 3}{100} \\ \\ = 10 \times 11 \times 3 \\ = 330\end{gathered}$S.I=100P×R×T=1001000×11×3=10×11×3=330

S.I = Rs. 330

He also invested Rs. 1000 at 10% compound interest per annum compounded annually for 3 years .........................( CASE 2 )

Principal (P) = Rs. 1000

Rate of interest (R) = 10% per annum

Time (n) = 3 years

$\begin{gathered}C.I = P( \: \: {(1 + \frac{R}{100}) }^{n} - 1) \\ \\ = 1000( \: \: {(1 + \frac{10}{100}) }^{3} - 1) \\ \\ = 1000( \: \: {(1 + \frac{1}{10}) }^{3} - 1) \\ \\ = 1000( \: \: { (\frac{10 + 1}{10} )}^{3} - 1 ) \\ \\ = 1000( \: { (\frac{11}{10}) }^{3} - 1 ) \\ \\ = 1000 \times (\frac{1331}{1000} - 1) \\ \\ = 1000 \times ( \frac{1331 - 1000}{1000} ) \\ \\ = 1000 \times \frac{331}{1000} \\ \\ = 331\end{gathered}$C.I=P((1+100R)n−1)=1000((1+10010)3−1)=1000((1+101)3−1)=1000((1010+1)3−1)=1000((1011)3−1)=1000×(10001331−1)=1000×(10001331−1000)=1000×1000331=331

C.I = Rs. 331

(S.I =330 < C.I = 331)

(CASE 1) < (CASE 2)

S.I < C.I

So, investment at compound Interest is better.

Q. (8)

Principal (P) = Rs. 400000

Rate of interest = 16% per annum

= 16/2 = 8% per half yearly (R)

Time (n) = 1 year = 2 half years

$\begin{gathered}Amount = P( {1 + \frac{R}{100} ) }^{n} \\ \\ = 400000( {1 + \frac{8}{100} ) }^{2} \\ \\ = 400000( {1 + \frac{2}{25} ) }^{2} \\ \\ = 400000( {\frac{25 + 2}{25} ) }^{2} \\ \\ = 400000 \times { (\frac{27}{25}) }^{2} \\ \\ = 400000 \times \frac{27 \times 27}{25 \times 25} \\ \\ = 400000 \times \frac{729}{625} \\ \\ = 640 \times 729 \\ = 466560\end{gathered}$Amount=P(1+100R)n=400000(1+1008)2=400000(1+252)2=400000(2525+2)2=400000×(2527)2=400000×25×2527×27=400000×625729=640×729=466560

So, they earn Rs. 466560

...................^_^

Answer 2

the answer is 466560.

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