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Ask QuestionPosted by Vivek Raj 6 years, 11 months ago
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Posted by Baap Of All. 6 years, 11 months ago
- 3 answers
Gaurav Seth 6 years, 11 months ago
Since ABCD is a parallelogram,
AB = CD ---- i)
BC = AD ---- ii)
It can be observed that
DR = DS (Tangents on the circle from point D)
CR = CQ (Tangents on the circle from point C)
BP = BQ (Tangents on the circle from point B)
AP = AS (Tangents on the circle from point A)
Adding all these equations, we obtain
DR + CR + BP + AP = DS + CQ + BQ + AS
(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
CD + AB = AD + BC
On putting the values of equations (1) and (2) in this equation, we obtain
2AB = 2BC
AB = BC …(3)
Comparing equations (1), (2), and (3), we obtain
AB = BC = CD = DA
Hence, ABCD is a rhombus.
Baap Of All. 6 years, 11 months ago
Yaar 6 years, 11 months ago
Posted by Nikhil Nikhil 6 years, 11 months ago
- 3 answers
Anushka Jugran ? 6 years, 11 months ago
Posted by Naman Gupta 6 years, 11 months ago
- 1 answers
Posted by John Masih 6 years, 11 months ago
- 1 answers
Posted by Surya Prakash 6 years, 11 months ago
- 0 answers
Posted by Nitin Sharma 6 years, 11 months ago
- 1 answers
Posted by Muskan Chhilar 6 years, 11 months ago
- 3 answers
Baap Of All. 6 years, 11 months ago
Posted by Mayank Arya 6 years, 11 months ago
- 2 answers
. . 6 years, 11 months ago
Posted by Keshav Nevatiya 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
Let a be the first term and d be the common difference of the given AP. Therefore, the sum of first n terms is given by
{tex}S _ { n } = \frac { n } { 2 } \cdot \{ 2 a + ( n - 1 ) d \}{/tex}
{tex}\therefore{/tex} S10 = {tex}\frac{{10}}{2}{/tex}{tex}\cdot{/tex}(2a+9d) {tex}\Rightarrow{/tex}5(2a+9d)=210
{tex}\Rightarrow{/tex}2a+9d=42. ...(i)
Sum of last 15 terms = (S50 - S35).
{tex}\therefore{/tex} (S50 - S35) = 2565
{tex}\Rightarrow{/tex} {tex}\frac{{50}}{2}{/tex}(2a+49d)- {tex}\frac{{35}}{2}{/tex}(2a+34d)=2565
{tex}\Rightarrow{/tex} 25(2a+49d)-35(a+17d)=2565
{tex}\Rightarrow{/tex} (50a-35a)+(1225d-595d)=2565
{tex}\Rightarrow{/tex} 15a+630d = 2565 {tex}\Rightarrow{/tex} a + 42d = 171 ...... (ii)
Therefore, on solving (i) and (ii), we get a=3 and d=4.
Hence, the required AP is 3,7,11,15,19.....
Posted by Md Shahjeb 6 years, 11 months ago
- 0 answers
Posted by Abhishek Singh 6 years, 11 months ago
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Posted by Vinod Meena 6 years, 11 months ago
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Posted by Ansh Sharma 6 years, 11 months ago
- 2 answers
Sahitha?? Sharma 6 years, 11 months ago
Posted by Rizwan Sid 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
We have, Radius of hemispherical tank = {tex} \frac { 3 } { 2 } m{/tex}
{tex} \therefore{/tex} Volume of the tank ={tex} \frac { 2 } { 3 } \times \frac { 22 } { 7 } \times \left( \frac { 3 } { 2 } \right) ^ { 3 } \mathrm { m } ^ { 3 } = \frac { 99 } { 14 } \mathrm { m } ^ { 3 }{/tex}
Volume of the water to be emptied = {tex} \frac { 1 } { 2 } \times \frac { 99 } { 14 } \mathrm { m } ^ { 3 } = \frac { 99 } { 28 } \mathrm { m } ^ { 3 } = \frac { 99 } { 28 } \times 1000 \text { litres } = \frac { 99000 } { 28 }{/tex}litres
Since {tex} \frac { 25 } { 7 }{/tex} litres of water is emptied in one second. Therefore,
Total time taken to empty half the tank i.e {tex} \frac { 99000 } { 28 }{/tex} litres of water = {tex} = \frac { 99000 } { 28 } \div \frac { 25 } { 7 }{/tex}seconds
{tex} = \frac { 99000 } { 28 } \times \frac { 7 } { 25 }{/tex}seconds
{tex} = \frac { 99000 } { 28 } \times \frac { 7 } { 25 } \times \frac { 1 } { 60 }{/tex}minutes
= 16.5 minutes
Posted by Rahul Wadhwa 6 years, 11 months ago
- 1 answers
Puja Sahoo? 6 years, 11 months ago
Posted by Swetha Selvam 6 years, 11 months ago
- 3 answers
Puja Sahoo? 6 years, 11 months ago
Posted by Shivanand Shahi 6 years, 11 months ago
- 1 answers
Posted by Gurjar Kabir (बैंसला) 6 years, 11 months ago
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Neeran Barman 6 years, 11 months ago
Posted by Aj Raghav 6 years, 11 months ago
- 3 answers
Posted by Chitragupt Maurya 6 years, 11 months ago
- 3 answers
Posted by Mantasha Rahmani 6 years, 11 months ago
- 1 answers
Yogita Ingle 6 years, 11 months ago
Given series is 3, 8, 13, ...... 253
here, first term , a = 3
common difference , d = 5
Let us find total number of terms at first.
use an = a+(n-1)d}
⇒ 253 = 3 + (n - 1)5
⇒ 250 = 5(n - 1)
⇒ n - 1 = 50
⇒ n = 51
so, there are 51 terms in given series.
now we know, mth term from last =last term - mth term + 1
so, 20th term from last = 51 - 20 + 1 = 32
hence, 20th from last = 32th term from first
use , a32=a+(32-1)d}
= 3 + 31 × 5
= 3 + 155
= 158
hence , 20th term from last = 158
Posted by Jaya Rai 6 years, 11 months ago
- 4 answers
Lionel Messi⚽️ 6 years, 11 months ago
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Yaar 6 years, 11 months ago
1Thank You