Let A(α, -2), B(α, 6) and C(\(\frac{\alpha}{4}\), -2) be vertices of a ΔABC. If (5, \(\frac{\alpha}{4}\)) is the circumcentre of ΔABC, then which of the following is NOT correct about ΔABC?
- Area is 24
- Perimeter is 25
- Circumradius is 5
- Inradius is 2
The Correct Option is B
Approach Solution - 1
To solve this problem, we need to analyze the given information about the vertices and circumcentre of ΔABC.
The vertices of ΔABC are A(α, -2), B(α, 6), and C(\(<\alpha/4\), -2). The circumcentre is given as (5, \(\alpha/4\)).
Step 1: Understanding the Circumcentre
The circumcentre of a triangle is equidistant from all the vertices. Hence, we have the following equations for the circumradius (R), considering the point (5, \(\alpha/4\)) as the circumcentre:
- Distance from (5, \(\alpha/4\)) to A(α, -2):
- Distance from (5, \(\alpha/4\)) to B(α, 6):
- Distance from (5, \(\alpha/4\)) to C(\(\frac{\alpha}{4}\), -2):
Step 2: Equating Distances
Equating any two distances will give us the value of α. Solving these equations will lead us to α = 4.
Step 3: Calculating the Lengths
- Coordinates of A, B, and C for α = 4 are A(4, -2), B(4, 6), and C(1, -2).
- Using the distance formula, calculate the lengths of sides AB, BC, and CA:
Step 4: Verifying the Options
- Area: Use the determinant method to find the area of a triangle with the given vertices:
- Perimeter: Add the lengths of all sides:
- Circumradius: From the coordinates of circumcentre and vertex distances:
- Inradius: Calculate using the formula:
Conclusion:
The perimeter option stating "Perimeter is 25" is incorrect because the calculated perimeter is 16.
Approach Solution -2
=(\(\alpha+\frac{\frac{\alpha}{4}}{2}\),\(\frac{6-2}{2}\))
=(\(\frac{5\alpha}{8}\), 2)
=(5,\(\frac{\alpha}{4}\))
⇒α=8
Area (ΔABC)=\(\frac{1}{2}\)⋅3α4×8=24 sq. units
Perimeter =8+\(\frac{3\alpha}{4}\)+\(\sqrt{82}\)+(\(\frac{3\alpha}{4}^2\))
=8+6+10=24
Circumradius=\(\frac{10}{2}\)=5
r=\(\frac{\Delta}{s}\)
r=\(\frac{24}{12}\)
r=2
So, the correct option is (B): Perimeter is 25.
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
Top JEE Main Horizontal and vertical lines Questions
- Let \((\alpha, \beta, y)\) be the foot of perpendicular form the point \((1,2,3)\) on the line \(\bigg(\frac{x + 3}{5}\) = \(\frac{y - 1}{2}\) = \(\frac{z + 4}{3}\) \(\bigg)\)then \(19 (\alpha + \beta + y)\)
- Let \(\alpha, \beta, \gamma\) be the foot of perpendicular from the point \((1, 2, 3)\) on the line \(\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}\). Then \(19(\alpha + \beta + \gamma)\) is equal to:
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
Concepts Used:
Horizontal and vertical lines
Horizontal Lines:
- A horizontal line is a sleeping line that means "side-to-side".
- These are the lines drawn from left to right or right to left and are parallel to the x-axis.
Equation of the horizontal line:
In all cases, horizontal lines remain parallel to the x-axis. It never intersects the x-axis but only intersects the y-axis. The value of x can change, but y always tends to be constant for horizontal lines.
Vertical Lines:
- A vertical line is a standing line that means "up-to-down".
- These are the lines drawn up and down and are parallel to the y-axis.
Equation of vertical Lines:
The equation for the vertical line is represented as x=a,
Here, ‘a’ is the point where this line intersects the x-axis.
x is the respective coordinates of any point lying on the line, this represents that the equation is not dependent on y.
⇒ Horizontal lines and vertical lines are perpendicular to each other.