The base of an isoceles of …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Ash J 5 years, 3 months ago
- 2 answers
Manoj Kumar 5 years, 3 months ago
Related Questions
Posted by Gnani Yogi 3 months, 2 weeks ago
- 0 answers
Posted by Akhilesh Patidar 3 months, 2 weeks ago
- 0 answers
Posted by Alvin Thomas 3 months, 2 weeks ago
- 0 answers
Posted by T Prudhvi 4 weeks ago
- 0 answers
Posted by Savitha Savitha 3 months, 3 weeks 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
Sia ? 5 years, 3 months ago
Here, base, b = 24 cm, and let each equal side be a cm. Then,
area = {tex}\frac { 1 } { 4 } b \sqrt { 4 a ^ { 2 } - b ^ { 2 } } \text { sq units } = \frac { 1 } { 4 } \times 24 \times \sqrt { 4 a ^ { 2 } - 576 } \mathrm { cm } ^ { 2 }{/tex}
={tex}12 \times \sqrt { a ^ { 2 } - 144 } \mathrm { cm } ^ { 2 }{/tex}
But, area = 192 cm2 [given].
{tex}\therefore \quad 12 \times \sqrt { a ^ { 2 } - 144 } = 192 \Rightarrow \sqrt { a ^ { 2 } - 144 } = 16{/tex}
{tex}\Rightarrow a ^ { 2 } - 144 = 256 \Rightarrow a ^ { 2 } = 400 \Rightarrow a = 20{/tex}
{tex}\therefore\quad {/tex}perimeter of the triangle = (2a + b) cm
= (2 {tex}\times{/tex} 20 + 24) cm = 64 cm.
1Thank You