Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Let <img alt="\mathrm{\mathrm{A}(\alpha,-2), \mathrm{B}(\alpha, 6) \: and\: \mathrm{C}\left(\frac{\alpha}{4},-2\right)}" src="/latex-image/?%5Cmathrm%7B%5Cmathrm%7BA%7D%28%5Calpha%2C-2%29%2C%20%5Cm

Let \mathrm{\mathrm{A}(\alpha,-2), \mathrm{B}(\alpha, 6) \: and\: \mathrm{C}\left(\frac{\alpha}{4},-2\right)} be vertices of a \mathrm{\triangle \mathrm{ABC}. If \left(5, \frac{\alpha}{4}\right)} is the circumcentre of \mathrm{\triangle \mathrm{ABC}}, then which of the following is NOT correct about \mathrm{\triangle \mathrm{ABC}} ?
 

Option: 1

area is 24
 


Option: 2

perimeter is 25
 


Option: 3

circumradius is 5
 


Option: 4

inradius is 2


Answers (1)

best_answer

\begin{aligned} & \text{clearly A is right angle and this } \mathrm{D\left(5, \frac{\alpha}{4}\right) } \text{ is mid-point of BC.}\\ &\mathrm{\quad \therefore \quad \frac{6-2}{2}=\frac{\alpha}{4} \Rightarrow \alpha=8 }\\ &\mathrm{\therefore A B=8, \quad A C=6} \\ &\text { Area }\mathrm{=\frac{1}{2} \times 8 \times 6=24} \\ &\text { Perimeter }\mathrm{=8+6+\sqrt{8^{2}+6^{2}}=24} \\ &\therefore \text { option (B) } \end{aligned}

Posted by

Pankaj

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2027 exam dates out; session 1 from January 22
anam-khan

NTA Exam Calendar 2027 Out: JEE Main from January 22, CUET PG in March

Latest Question