Question:
The sides of a rhombus ABCD measure 2 cm each, and the difference between two of its angles is \(90^{\circ}\). Then the area of the rhombus is:
The sides of a rhombus ABCD measure 2 cm each, and the difference between two of its angles is \(90^{\circ}\). Then the area of the rhombus is:
Show Hint
A rhombus has only two distinct angles that add up to 180 degrees; use their given difference to find each one, then apply Area = side squared times sine of the angle.
Updated On: Jul 13, 2026
- \(\sqrt{2}\) sq cm
- \(2\sqrt{2}\) sq cm
- \(3\sqrt{2}\) sq cm
- \(4\sqrt{2}\) sq cm
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Recall the angle property of a rhombus.
A rhombus has only two distinct interior angles, since opposite angles are equal and any two adjacent angles are supplementary (they add up to \(180^{\circ}\)). So if one angle is \(\theta\), the angle next to it is \(180^{\circ} - \theta\).
Step 2: Use the given angle difference to find \(\theta\).
The difference between the two distinct angles is \(90^{\circ}\).
\[ (180^{\circ} - \theta) - \theta = 90^{\circ} \]
\[ 180^{\circ} - 2\theta = 90^{\circ} \]
\[ 2\theta = 90^{\circ} \]
\[ \theta = 45^{\circ} \]
So the two angles of the rhombus are \(45^{\circ}\) and \(135^{\circ}\).
Step 3: Recall the area formula for a rhombus in terms of side and angle.
For a rhombus with side \(a\) and one interior angle \(\theta\), the area is
\[ \text{Area} = a^2 \sin\theta \]
This comes from splitting the rhombus into two triangles along a diagonal and using the triangle area formula \(\frac{1}{2}ab\sin\theta\), since both sides meeting at that angle equal \(a\).
Step 4: Substitute the known values.
Here \(a = 2\) and \(\theta = 45^{\circ}\) (using \(135^{\circ}\) instead gives the same result, since \(\sin 135^{\circ} = \sin 45^{\circ}\)).
\[ \text{Area} = 2^2 \times \sin 45^{\circ} = 4 \times \frac{\sqrt{2}}{2} = 2\sqrt{2} \]
Step 5: Rule out the other options.
Option (A) would come from forgetting to square the side length. Options (C) and (D) would come from using the wrong angle or a slip while simplifying \(4 \times \frac{\sqrt{2}}{2}\).
Final Answer:
The area of the rhombus is \(2\sqrt{2}\) sq cm. \[ \boxed{2\sqrt{2} \text{ sq cm}} \]
A rhombus has only two distinct interior angles, since opposite angles are equal and any two adjacent angles are supplementary (they add up to \(180^{\circ}\)). So if one angle is \(\theta\), the angle next to it is \(180^{\circ} - \theta\).
Step 2: Use the given angle difference to find \(\theta\).
The difference between the two distinct angles is \(90^{\circ}\).
\[ (180^{\circ} - \theta) - \theta = 90^{\circ} \]
\[ 180^{\circ} - 2\theta = 90^{\circ} \]
\[ 2\theta = 90^{\circ} \]
\[ \theta = 45^{\circ} \]
So the two angles of the rhombus are \(45^{\circ}\) and \(135^{\circ}\).
Step 3: Recall the area formula for a rhombus in terms of side and angle.
For a rhombus with side \(a\) and one interior angle \(\theta\), the area is
\[ \text{Area} = a^2 \sin\theta \]
This comes from splitting the rhombus into two triangles along a diagonal and using the triangle area formula \(\frac{1}{2}ab\sin\theta\), since both sides meeting at that angle equal \(a\).
Step 4: Substitute the known values.
Here \(a = 2\) and \(\theta = 45^{\circ}\) (using \(135^{\circ}\) instead gives the same result, since \(\sin 135^{\circ} = \sin 45^{\circ}\)).
\[ \text{Area} = 2^2 \times \sin 45^{\circ} = 4 \times \frac{\sqrt{2}}{2} = 2\sqrt{2} \]
Step 5: Rule out the other options.
Option (A) would come from forgetting to square the side length. Options (C) and (D) would come from using the wrong angle or a slip while simplifying \(4 \times \frac{\sqrt{2}}{2}\).
Final Answer:
The area of the rhombus is \(2\sqrt{2}\) sq cm. \[ \boxed{2\sqrt{2} \text{ sq cm}} \]
Was this answer helpful?
0
0
Top XAT Quantitative Ability Questions
- Last year, Mr Basu bought two scooters. This year, he sold both of them for Rs 30,000 each. On one scooter, he earned a profit of 20%, and on the other, he made a loss of 20%. What was his net profit or loss?
- In an examination, the average marks obtained by students who passed was \(x\%\), while the average of those who failed was \(y\%\). The average marks of all the students taking the exam was \(z\%\). Find, in terms of \(x\), \(y\) and \(z\), the percentage of students taking the exam who failed.
- Three circles A, B and C have a common centre O. A is the inner circle, B is the middle circle, and C is the outer circle. P is a point on the outer circle C, and the radius OP cuts the inner circle at X and the middle circle at Y, such that \(OX = XY = YP\). The ratio of the area of the region between the inner and middle circles to the area of the region between the middle and outer circles is:
- If \(S_n\) denotes the sum of the first \(n\) terms of an Arithmetic Progression, and \(S_1 : S_4 = 1 : 10\), then the ratio of the first term to the fourth term is:
- The curve \(y = 4x^2\) and \(y^2 = 2x\) meet at the origin O and at a point P, forming a loop. The straight line OP divides the loop into two parts. What is the ratio of the areas of the two parts of the loop?
View More Questions
Top XAT Questions
- Ashok, a maths teacher in the school, asked his students to guess his date of birth by giving the following hints.
Hint 1: He was born in 1995
Hint 2: The sum of the products of his birth date multiplied by 24 and his birthday month multiplied by 60 is 1284.
Find the sum of Ashok’s birth date, birth month, and birth year. - Solve for the number of integral values of \(x: \frac{(x^2-15x+47)(x-13)}{(x-8)}<0\).
- In an examination, Akhil scored 11.11% less than the pass marks and Akshay scored 12.5% more than the pass marks. Find the percentage of pass marks if the sum of the scores of Akhil and Akshay is 25% more than the total marks.
- The ratio of paint and oil in Tank A is 3: 4, whereas in Tank B, the respective ratio is 4: 5 . Then, 96 liters of a mixture consisting of paint and oil in the ratio of 5: 1 is added to Tank A, such that the respective ratio in Tank A becomes exactly the reverse of that in Tank B. Find the amount (in liters) of the mixture from Tank B that is poured into Tank A such that the amount of paint and water in Tank A become equal.
- The points (–4, 3) and (–1, –4) are two end points of a diagonal of a parallelogram. If the other diagonal has the equation 7y + 3x + c = 0, then find the value of c.
View More Questions