Verified Solution • 2025–26 Session moderate

QJEEM22Ma1365: JEE Main 2022 Mathematics - Class 11 JEEM22Ma1365 Solution: Let and be vertices of a . If is the circumcentre of , then…

Let A(α,−2),B(α,6)\mathrm{A}(\alpha,-2), \mathrm{B}(\alpha, 6) and C(α4,−2)\mathrm{C}\left(\frac{\alpha}{4},-2\right) be vertices of a △ABC\triangle \mathrm{ABC} . If (5,α4)\left(5, \frac{\alpha}{4}\right) is the circumcentre of △ABC\triangle \mathrm{ABC} , then which of the following is NOT correct about △ABC\triangle \mathrm{ABC} ?

Multiple Choice Options

  • A area is 24
  • B perimeter is 25
  • C circumradius is 5
  • D inradius is 2

Step-by-Step Verified Solution

JEE Main 2022 (Online) 29th July Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 46 English Explanation

Circumcentre of △ABC\triangle A B C

amp;=(α+α42,6−22)amp;=(5α8,2)amp;=(5,α4)amp;⇒α=8\begin{aligned}&=\left(\frac{\alpha+\frac{\alpha}{4}}{2}, \frac{6-2}{2}\right) \\\\&=\left(\frac{5 \alpha}{8}, 2\right) \\\\&=\left(5, \frac{\alpha}{4}\right) \\\\&\Rightarrow \alpha=8\end{aligned}

area⁡(△ABC)=12⋅3α4×8=24\operatorname{area}(\triangle A B C)=\frac{1}{2} \cdot \frac{3 \alpha}{4} \times 8=24 sq. units

 Perimeter amp;=8+3α4+82+(3α4)2amp;=8+6+10=24\begin{aligned}\text { Perimeter } &=8+\frac{3 \alpha}{4}+\sqrt{8^{2}+\left(\frac{3 \alpha}{4}\right)^{2}} \\\\&=8+6+10=24\end{aligned}

Circumradius =102=5=\frac{10}{2}=5

inradius (r)=Δs=2412=2(r)=\frac{\Delta}{s}=\frac{24}{12}=2

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring