Question:
A rectangle ABCD with ABCD with AB = 2 and BC = 4 is inscribed in rectangle PQRS such that vertices of ABCD lie on sides of PQRS then maximum possible area(in sq. unit) of rectangle PQRS is :
A rectangle ABCD with ABCD with AB = 2 and BC = 4 is inscribed in rectangle PQRS such that vertices of ABCD lie on sides of PQRS then maximum possible area(in sq. unit) of rectangle PQRS is :
Updated On: Apr 6, 2024
- 9
- 20
- 18
- 12
Hide Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
The Correct answer is option is (C) : 18
Was this answer helpful?
0
0
Top Questions on Area of a Triangle - by Heron’s Formula
- The area of the region bounded by the curves \( x(1 + y^2) = 1 \) and \( y^2 = 2x \) is:
- JEE Main - 2025
- Mathematics
- Area of a Triangle - by Heron’s Formula
- Consider the region \[ R = \left\{ (x, y): x \leq y \leq 9 - \frac{11}{3} x^2, x \geq 0 \right\}. \] The area of the largest rectangle of sides parallel to the coordinate axes and inscribed in \( R \) is:
- JEE Main - 2025
- Mathematics
- Area of a Triangle - by Heron’s Formula
- Let $f: [0, 3] \to A$ be defined by $f(x) = 2x^3 - 15x^2 + 36x + 7$ and $g: [0, \infty) \to B$ be defined by $g(x) = \frac{x}{x^{2025} + 1}$. If both functions are onto and $S = \{ x \in \mathbb{Z} : x \in A \text{ or } x \in B \}$, then $n(S)$ is equal to:
- JEE Main - 2025
- Mathematics
- Area of a Triangle - by Heron’s Formula
Let \( A = \begin{bmatrix} \frac{1}{\sqrt{2}} & -2 \\ 0 & 1 \end{bmatrix} \) and \( P = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}, \theta > 0. \) If \( B = P A P^T \), \( C = P^T B P \), and the sum of the diagonal elements of \( C \) is \( \frac{m}{n} \), where gcd(m, n) = 1, then \( m + n \) is:
- JEE Main - 2025
- Mathematics
- Area of a Triangle - by Heron’s Formula
- Let A, B, C be three points in the xy-plane, whose position vectors are given by \[ \sqrt{3} \hat{i} + \hat{j}, \quad \hat{i} + \sqrt{3} \hat{j}, \quad \text{and} \quad a\hat{i} + (1-a) \hat{j} \] respectively with respect to the origin \( O \). If the distance of the point \( C \) from the line bisecting the angle between the vectors \( \overrightarrow{OA} \) and \( \overrightarrow{OB} \) is \( \frac{9}{\sqrt{2}} \), then the sum of all possible values of \( a \) is:
- JEE Main - 2025
- Mathematics
- Area of a Triangle - by Heron’s Formula
View More Questions
Questions Asked in JEE Main exam
- 2.8 \( \times 10^{-3} \) mol of \( \text{CO}_2 \) is left after removing \( 10^{21} \) molecules from its ‘\( x \)’ mg sample. The mass of \( \text{CO}_2 \) taken initially is: Given: \( N_A = 6.02 \times 10^{23} \, \text{mol}^{-1} \)
- JEE Main - 2025
- Mole concept and Molar Masses
- A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field \( B \) exists into the page. The bar starts to move from the vertex at time \( t = 0 \) with a constant velocity. If the induced EMF is \( E \propto t^n \), then the value of \( n \) is ________________________.
- JEE Main - 2025
- Electromagnetic induction
In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by:
- JEE Main - 2025
- Electromagnetic Field (EMF)
- The value of \( (\sin 70^\circ)(\cot 10^\circ \cot 70^\circ - 1) \) is:
- JEE Main - 2025
- Trigonometric Identities
- Ice at \( -5^\circ C \) is heated to become vapor with temperature of \( 110^\circ C \) at atmospheric pressure. The entropy change associated with this process can be obtained from:
- JEE Main - 2025
- Thermodynamics
View More Questions