Find the area of a right …

CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Aman Gupta 8 years, 5 months ago
- 1 answers
Related Questions
Posted by Duruvan Sivan 6 months, 1 week ago
- 0 answers
Posted by Akhilesh Patidar 1 year, 4 months ago
- 0 answers
Posted by Yash Pandey 6 months, 1 week ago
- 0 answers
Posted by Sheikh Alfaz 1 month, 2 weeks ago
- 0 answers
Posted by Savitha Savitha 1 year, 4 months ago
- 0 answers
Posted by Alvin Thomas 3 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
myCBSEguide
Sahdev Sharma 8 years, 5 months ago
Using Pythagoras Theorm
{tex}Height = \sqrt {(hypotenuse)^2-(base)^2}{/tex}
{tex}= \sqrt{625-49}{/tex}
{tex}= \sqrt{576}= 24 {/tex} cm
Area of ∆ = {tex}{1\over 2}\times base\times height{/tex}
= {tex}{1\over2}\times 7\times 24= 84\ cm^2{/tex}
0Thank You