Find the area of an isocelas …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Micro Max Chamoli 8 years, 3 months ago
- 1 answers
Test
Related Questions
Posted by Sheikh Alfaz 1 month, 2 weeks ago
- 0 answers
Posted by Yash Pandey 6 months, 1 week ago
- 0 answers
Posted by Duruvan Sivan 6 months, 1 week ago
- 0 answers
Posted by Alvin Thomas 3 months ago
- 0 answers
Posted by Akhilesh Patidar 1 year, 4 months ago
- 0 answers
Posted by Savitha Savitha 1 year, 4 months ago
- 0 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
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}
0Thank You