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Ask QuestionPosted by Sitanshu Tripathi 6 years, 10 months ago
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Posted by Priyanshee Maurya 6 years, 10 months ago
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Sia ? 6 years, 4 months ago
{tex}x^2 + 6x - 16 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}x^2 + 6x = 16{/tex}
{tex}\Rightarrow{/tex} {tex}x^2 + 6x + 9 = 16 + 9{/tex} [Adding on both sides square of coefficient of x, i.e. ({tex}\frac{6}{2}{/tex})2]
{tex}\Rightarrow{/tex} {tex}(x + 3)^2 = 25{/tex}
{tex}\Rightarrow{/tex} x + 3 = {tex}\pm{/tex}{tex}\sqrt{25}{/tex}
{tex}\Rightarrow{/tex} x + 3 =5 or x + 3 = -5
{tex}\Rightarrow{/tex} x = 2 or x = -8
Posted by Shriya Mathew 6 years, 10 months ago
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Posted by Khaja Moinuddin 6 years, 10 months ago
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Mehak Rathore 6 years, 10 months ago
Posted by Satish Kumar 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
P(4,-2), Q(7,2), R(0,9), S(-3,5) forms a parallelogram
Let the height of parallelogram taking PQ as base be h.
{tex}\therefore \quad PQ = \sqrt { ( 7 - 4 ) ^ { 2 } + ( 2 + 2 ) ^ { 2 } }{/tex}
{tex}= \sqrt { 3 ^ { 2 } + 4 ^ { 2 } } = \sqrt { 9 + 16 }{/tex}
{tex}=\sqrt{25}{/tex}
= 5 units
Area of triangle PQR
{tex}= \frac { 1 } { 2 } \left[ x _ { 1 } \left( y _ { 2 } - y _ { 3 } \right) + x _ { 2 } \left( y _ { 3 } - y _ { 1 } \right) + x _ { 3 } \left( y _ { 1 } - y _ { 2 } \right) \right]{/tex}
{tex}= \frac { 1 } { 2 } [ 4 ( 2 - 9 ) + 7 ( 9 + 2 ) + 0 ( 2 - 2 ) ]{/tex}
{tex}= \frac{1}{2} [4(-7)+7(11)]{/tex}
{tex}= \frac { 1 } { 2 } \times 49 = \frac { 49 } { 2 } {/tex}sq. units
Now, {tex}\frac { 1 } { 2 } \times PQ \times h = \frac { 49 } { 2 }{/tex}
{tex}\Rightarrow{/tex} {tex}\frac { 1 } { 2 } \times 5 \times h = \frac {49}{2}{/tex}
{tex}\Rightarrow{/tex} h = 9.8 units
Posted by Pawanyadav Pawan 6 years, 10 months ago
- 6 answers
Posted by Karma Desang 6 years, 10 months ago
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Posted by Grusha Reddy 6 years, 10 months ago
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Aagam Jain 6 years, 10 months ago
Posted by Afna Afna 6 years, 10 months ago
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Posted by Shivam Thakur 6 years, 10 months ago
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Isha Kapoor 6 years, 10 months ago
Posted by Thunder ?Yashwanth??? 6 years, 10 months ago
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Posted by Ritesh Kumar 6 years, 10 months ago
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Mehak Rathore 6 years, 10 months ago
Posted by Krishna Antil 6 years, 10 months ago
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Posted by Anil Kumar 6 years, 10 months ago
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Posted by Anil Kumar 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
LHS {tex} = 1 + \frac{{{{\cot }^2}\theta }}{{1 + \cos ec\theta }}{/tex}
{tex} = 1 + \frac{{\cos e{c^2}\theta - 1}}{{1 + \cos ec\;\theta }}{/tex} {tex}\left[ {\because {{\cot }^2}\theta = \cos e{c^2}\theta - 1} \right]{/tex}
{tex} = 1 + \frac{{(\cos ec\theta + 1)(\cos ec\theta - 1)}}{{(1 + \cos ec\theta )}}{/tex} {tex}\left[ {\because {a^2} - {b^2} = (a + b)(a - b)} \right]{/tex}
{tex} = 1 + \cos ec\;\theta - 1{/tex}
{tex} = \cos ec\theta {/tex}
= RHS
Hence proved.
Posted by Ann Maria 6 years, 10 months ago
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Posted by Krishnan Verma 6 years, 10 months ago
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Posted by Santosh Raj 6 years, 10 months ago
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Posted by Santosh Raj 6 years, 10 months ago
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Posted by Satyam Raja 6 years, 10 months ago
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Posted by Manpreet Kaur 6 years, 10 months ago
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Vipul Mishra 6 years, 10 months ago
3Thank You