CUET · MATHS · PYQ PAPER 2023

A right angled triangle has its longest side \(B C\) as the diameter of a circle of radius \(r\) and a vertex \(A\) lying on the circle. Maximum area that the triangle \(A B C\) can enclose is:

A \(r^2\)
B \(2 r^2\)
C \(\frac{r^2}{2}\)
D \(\frac{\pi}{2} r^2\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(r^2\)

Step-by-step Solution

Detailed explanation

The hypotenuse of the triangle is the diameter: \(BC = 2r\). The maximum height from vertex \(A\) to the base \(BC\) occurs when \(A\) is at the perpendicular bisector of \(BC\). This height is the radius: \(h = r\). Area…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Match List - I with List - II

List - I	List - II
(A) \(\int \frac{d x}{a^2-x^2}\)	(I) \(\frac{1}{2 a} \log \left\|\frac{x-a}{x+a}\right\|+C\)
(B) \(\int \frac{d x}{x^2-a^2}\)	(II) \(\frac{1}{2 a} \log \left\|\frac{a+x}{a-x}\right\|+C\)
(C) \(\int \frac{d x}{x^2+a^2}\)	(III) \(\sin ^{-1}\left(\frac{x}{a}\right)+C\)
(D) \(\int \frac{d x}{\sqrt{a^2-x^2}}\)	(IV) \(\frac{1}{a} \tan ^{-1} \frac{x}{a}+C\)

Choose the correct answer from the options given below:

CUET 2023 Easy

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