Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Khushi Jain 5 years, 1 month ago
- 1 answers
Posted by Archi Kumari 5 years, 1 month ago
- 3 answers
Archi Kumari 5 years, 1 month ago
Bhavya Garg 5 years, 1 month ago
Posted by Dileep Dileep Chaudhary 5 years, 1 month ago
- 0 answers
Posted by Rajesh Poddar 5 years, 1 month ago
- 3 answers
Posted by Rinki Tyagi 5 years, 1 month ago
- 4 answers
Rohan Bamniya 5 years, 1 month ago
Ann Riya Dsouza 5 years, 1 month ago
Nikhil Pandey 5 years, 1 month ago
Posted by Angadh Kamboj 5 years, 1 month ago
- 1 answers
Posted by Aryan Chaudhary 5 years, 1 month ago
- 0 answers
Posted by Avantika Sharma 5 years, 1 month ago
- 2 answers
Tanushree Bhosale 5 years, 1 month ago
Posted by Pragati Yadav 5 years, 1 month ago
- 0 answers
Posted by Rohit Mohit 5 years, 1 month ago
- 1 answers
Posted by Urvashi Singh 5 years, 1 month ago
- 1 answers
Maakalyani Studio 5 years, 1 month ago
Posted by Lakshita Teotia 5 years, 1 month ago
- 3 answers
Sia ? 5 years, 1 month ago
Since, sides of coloured triangular wall are 15 m, 11 m and 6 m.
{tex}\therefore{/tex} Semi-perimeter of coloured triangular wall
S ={tex}\frac{15+11+6}{2}{/tex}={tex}\frac{32}{2}{/tex}=16 m
Now, Using Heron’s formula,
Area of coloured triangular wall
={tex}\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}{/tex}
={tex}\sqrt{16\left( 16-15 \right)\left( 16-11 \right)\left( 16-6 \right)}{/tex}
={tex}\sqrt{16\times 1\times 5\times 10}{/tex}
={tex}20\sqrt{2}{{m}^{2}}{/tex}
Hence area painted in blue colour = {tex}20\sqrt{2}{{m}^{2}}{/tex}
Angadh Kamboj 5 years, 1 month ago
Posted by Sunita Gupta 5 years, 1 month ago
- 2 answers
Posted by Ajit Kasotiya 5 years, 1 month ago
- 1 answers
Posted by Anu Dhingra 5 years, 1 month ago
- 0 answers
Posted by Hars H.T Tripathi 5 years, 1 month ago
- 3 answers
Posted by Divyanshu Bailwal 5 years, 1 month ago
- 1 answers
Tishya Jha 5 years, 1 month ago
Posted by Asmita Mukane Mukane 5 years, 1 month ago
- 3 answers
Asmita Mukane Mukane 5 years, 1 month ago
Asmita Mukane Mukane 5 years, 1 month ago
Posted by Vimal Nath 5 years, 1 month ago
- 1 answers
Aditya Tripathi 5 years, 1 month ago
Posted by Bhupesh Baloda 5 years, 1 month ago
- 1 answers
Aditya Tripathi 5 years, 1 month ago
Posted by Shrivardhan Sharma 5 years, 1 month ago
- 1 answers
Prince Jindal 5 years, 1 month ago
Posted by Hardeep Kaur 5 years, 1 month ago
- 1 answers
Posted by Rutvik Gupta 5 years, 1 month ago
- 1 answers
Rakshit Goel 5 years, 1 month ago
Posted by Prachi Srivastava 5 years, 1 month ago
- 2 answers
Tishya Jha 5 years, 1 month ago
Posted by Sneha Saini 5 years, 1 month ago
- 1 answers
Yogita Ingle 5 years, 1 month ago
Given, ∠ DAB + EBA = 180°. CA and CB are bisectors of ∠ DAB ∠ EBA respectively.
∴ ∠ DAC + ∠ CAB = 1/2 (∠ DAB).....(1)
⇒ ∠ EBC + ∠ CBA = 1/2 (∠ EBA)....(2)
⇒ ∠ DAB + ∠ EBA = 180°
⇒ 2 (∠ CAB) + 2 (∠ CBA) = 180° [using (1) and (2)]
⇒ ∠ CAB + ∠ CBA = 90°
In Δ ABC,
∠ CAB + ∠ CBA + ∠ ABC = 180° (Angle Sum property)
⇒ 90° + ∠ ABC = 180°
⇒ ∠ ABC = 180° - 90°⇒ ∠ ABC = 90°
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
Vinit Kumar Dubey 5 years, 1 month ago
0Thank You