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Ask QuestionPosted by Priyanshu Tendulkar 8 years, 3 months ago
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Posted by Anishka Arora 8 years, 3 months ago
- 1 answers
Soumya Ghoshal 8 years, 3 months ago
Height of equilateral triangle,
{tex}\frac{{√3a}}{2}=6{/tex}
or,√3a=12
or,a=12÷√3
or,a=4√3
Area of this triangle {tex}\frac{{√3a²}}{4}{/tex}
putting the value of a we get, {tex}\frac{{√3×4√3×4√3}}{4}{/tex}
=12√3cm²
Posted by Krishna Lau 8 years, 3 months ago
- 1 answers
Rashmi Bajpayee 8 years, 3 months ago
If x + y + z = 0, then x<font size="2">3</font> + y<font size="2">3</font> + z<font size="2">3</font> = 3xyz
Therefore, x<font size="2">3</font> + y<font size="2">3</font> + 8 = x<font size="2">3</font> + y<font size="2">3</font> + 2<font size="2">3</font> = 3(x)(y)(2) = 6xy
Posted by Krishna Lau 8 years, 3 months ago
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Posted by Alok Pandey 8 years, 3 months ago
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Posted by Kalpesh Soni 8 years, 3 months ago
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Posted by Aryant Mishra 8 years, 3 months ago
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Posted by Archie Bolya 8 years, 3 months ago
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Posted by Parush Uppal 8 years, 3 months ago
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Posted by Sharanya Sharu 8 years, 3 months ago
- 1 answers
Rashmi Bajpayee 8 years, 3 months ago
According to question,
Dimensions of new formed cuboid are
Length = 12 cm, Breadth = 6 cm and Height = 6 cm
Surface Area of Cuboid = 2(lb + bh + hl)
= 2(12 x 6 + 6 x 6 + 6 x 12)
= 2(72 + 36 + 72)
= 2 x 180
= 360 cm
Posted by Debaahish Sahoo 8 years, 3 months ago
- 4 answers
Anishka Arora 8 years, 3 months ago
Shashank Dwivedi 8 years, 3 months ago
Posted by Deepak Kumar 8 years, 3 months ago
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Posted by Jit Kumar 8 years, 3 months ago
- 1 answers
Sadhu Hiremath 8 years, 3 months ago
Posted by Thor Pacholi 8 years, 3 months ago
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Posted by Praveen Sheetal 8 years, 3 months ago
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Posted by Isha Thakur 8 years, 3 months ago
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Posted by Sid Banerjee 8 years, 3 months ago
- 2 answers
Arun Soni 8 years, 3 months ago
{tex}\eqalign{ & Here,a = 7 - 4\sqrt 3 \left( {Given} \right) \cr & Let{\text{ }}x = \sqrt a + \frac{1}{{\sqrt a }} \cr & \Rightarrow {x^2} = {\left( {\sqrt a + \frac{1}{{\sqrt a }}} \right)^2} \cr & \Rightarrow {x^2} = {\left( {\sqrt a } \right)^2} + {\left( {\frac{1}{{\sqrt a }}} \right)^2} + 2 \times \sqrt a \times \frac{1}{{\sqrt a }} \cr & \Rightarrow {x^2} = a + \frac{1}{a} + 2 \cr & \Rightarrow {x^2} = 7 - 4\sqrt 3 + \frac{1}{{7 - 4\sqrt 3 }} + 2 \cr & \Rightarrow {x^2} = 7 - 4\sqrt 3 + 7 + 4\sqrt 3 + 2\left[ {\because \frac{1}{{7 - 4\sqrt 3 }} \times \frac{{7 + 4\sqrt 3 }}{{7 + 4\sqrt 3 }} = 7 + 4\sqrt 3 } \right] \cr & \Rightarrow {x^2} = 16 \cr & \Rightarrow x = \pm 4 \cr & \Rightarrow \sqrt a + \frac{1}{{\sqrt a }} = \pm 4 \cr} {/tex}
Posted by Sadiq Khan 8 years, 3 months ago
- 1 answers
Goldi Kumari 8 years, 3 months ago
Posted by Rahul Tiwari 8 years, 3 months ago
- 1 answers
Arun Soni 8 years, 3 months ago
Yes,indeed,this is not necessary every time to mention the type of triangle.Sometime it is asked to check students application.
Posted by Preeti Thakur 8 years, 3 months ago
- 2 answers
Anuj Aryaman 8 years, 3 months ago
Posted by Preeti Thakur 8 years, 3 months ago
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Posted by English Nagora 8 years, 3 months ago
- 1 answers
Hans Raj 8 years, 3 months ago
{tex} \sqrt{}{/tex}367
{tex} \sqrt{}{/tex}361 = 19
{tex}\sqrt{400}{/tex}= 20
so {tex} \sqrt{367}{/tex} is between > 19 and < 20
{tex} \sqrt{367}{/tex} = {tex}\sqrt{}{/tex}19.157 x 19.157
{tex} \sqrt{367}{/tex} = 19.157
Posted by Vedant Vedant 8 years, 3 months ago
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Posted by Tanuj Maurya 8 years, 3 months ago
- 1 answers
Hans Raj 8 years, 3 months ago
a system of representing numbers of various types such as
decimal system 1, 2, 3, 4, 5, ......10 ,11,......88, 100, .......
Roman numerals I, II, II, IV, V , .......... X, XI, XII, ............
Binary system 1 , 0 , 10, 11, ..... 101, 111, ........ 1001.................(read as one, zero, one zero, one one. one zero one, one one one, one zero zero one ........ )
Posted by Mickey Bhardwaj 8 years, 3 months ago
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Posted by Chinmay Asodekar 8 years, 3 months ago
- 1 answers
Devendra Prajapati 8 years, 3 months ago
Posted by Micro Max Chamoli 8 years, 3 months ago
- 1 answers
Rashmi Bajpayee 8 years, 3 months ago
s = {tex}{{a + a + b} \over 2} = {{2a + b} \over 2}{/tex}
Now, Area of triangle = {tex}\sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} {/tex}
= {tex}\sqrt {{{2a + b} \over 2}\left( {{{2a + b} \over 2} - a} \right)\left( {{{2a + b} \over 2} - a} \right)\left( {{{2a + b} \over 2} - b} \right)} {/tex}
= {tex}\sqrt {{{2a + b} \over 2}\left( {{{2a + b - 2a} \over 2}} \right)\left( {{{2a + b - 2a} \over 2}} \right)\left( {{{2a + b - 2b} \over 2}} \right)} {/tex}
= {tex}\sqrt {{{2a + b} \over 2}\left( {{b \over 2}} \right)\left( {{b \over 2}} \right)\left( {{{2a - b} \over 2}} \right)} {/tex}
= {tex}\sqrt {{{4{a^2} - {b^2}} \over 4}{{\left( {{b \over 2}} \right)}^2}} {/tex}
= {tex}{b \over 4}\sqrt {4{a^2} - {b^2}} {/tex}
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Soumya Ghoshal 8 years, 3 months ago
We know that complementary angle means 90°
Therefore, 89°59'60'' -42°47'55''
= 47°12'5''
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