Chapter 9 · Class 9 CBSE · MCQ Test
Areas of Parallelograms & Triangles MCQ Test — Class 9 CBSE
Practice 10 multiple-choice questions with instant answer reveal and explanations.
Areas of Parallelograms & Triangles — MCQ Questions
1For two planar figures to be considered 'between the same parallels', what is the essential condition they must satisfy?
Show Answer+
Answer: A
Hint: Recall the definition of figures lying on the same base and between the same parallels. It involves specific placement relative to two parallel lines.
Solution:
The definition states that if two figures have a common base (or equal bases) and the vertices opposite to the base of each figure lie on a line parallel to the base, then they are said to be on the same base and between the same parallels.
Option A directly captures this definition, explaining the geometric arrangement required.
2In a figure, parallelogram ABCD and triangle EBC are drawn such that point E lies on the side AD. For these two figures to be on the same base and between the same parallels, which condition is necessary?
Show Answer+
Answer: B
Hint: Consider BC as the common base. For the figures to be between the same parallels, the line containing the vertices opposite to BC must be parallel to BC.
Solution:
The common base for both parallelogram ABCD and triangle EBC is BC.
For them to be between the same parallels, the line containing the vertex opposite to the base (AD for the parallelogram, and E for the triangle, which lies on AD) must be parallel to the base BC.
Therefore, the line AD must be parallel to BC.
3Parallelogram PQRS and parallelogram MNRS are on the same base SR and between the same parallels SR and PM. If Area(PQRS) = 45 cm², what is Area(MNRS)?
Show Answer+
Answer: B
Hint: Recall the theorem about the areas of parallelograms that share the same base and are situated between the same parallel lines.
Solution:
According to Theorem 9.1, parallelograms on the same base and between the same parallels are equal in area.
Given that PQRS and MNRS are on the same base SR and between the same parallels SR and PM, their areas must be equal.
Therefore, Area(MNRS) = Area(PQRS).
Area(MNRS) = 45 cm².
4Which of the following conditions is *not necessarily true* for two triangles on the same base and between the same parallels?
Show Answer+
Answer: C
Hint: While areas are equal, think about what determines the perimeter of a triangle. Does having the same base and height guarantee all side lengths are identical?
Solution:
If two triangles are on the same base and between the same parallels, it implies they have the same base length and the same height corresponding to that base.
Based on the formula Area = 1/2 × base × height, their areas will be equal (Option B is true). Their heights are indeed equal (Option A is true). And by definition, their opposite vertices lie on the same parallel line (Option D is true).
However, having the same base and height does not guarantee that the other two sides of the triangles are equal. Therefore, their perimeters are not necessarily equal.
5A triangle and a parallelogram are on the same base and between the same parallels. If the area of the parallelogram is 60 cm², what is the area of the triangle?
Show Answer+
Answer: A
Hint: Remember the relationship between the area of a triangle and a parallelogram when they share the same base and are between the same parallel lines.
Solution:
According to a key theorem, if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.
Given Area(parallelogram) = 60 cm².
Area(triangle) = 1/2 × Area(parallelogram).
Area(triangle) = 1/2 × 60 cm² = 30 cm².
6Ravi claims that if two triangles have the same base, their areas must be equal. Which of the following statements *best explains* why Ravi's claim is incorrect?
Show Answer+
Answer: D
Hint: Think about the formula for the area of a triangle. What two components determine its area?
Solution:
The formula for the area of a triangle is 1/2 × base × height.
If two triangles have the same base, their areas will only be equal if their heights corresponding to that base are also equal.
Ravi's claim is incorrect because he overlooks the crucial role of height. Two triangles can have the same base but different heights, leading to different areas.
7If two triangles have the same base and equal areas, then their corresponding vertices opposite to the common base must lie on a line __________ to the base.
Show Answer+
Answer: C
Hint: Consider the converse of the theorem about triangles on the same base and between the same parallels. Equal area with the same base implies something about their heights.
Solution:
The area of a triangle is given by 1/2 × base × height.
If two triangles have the same base and equal areas, it implies that their heights corresponding to that base must also be equal.
If the heights from the opposite vertices to the common base are equal, then these vertices must lie on a line parallel to the base (as the perpendicular distance between two parallel lines is constant).
8In ΔPQR, if PS is a median to side QR, and Area(ΔPQS) = 35 cm², what is Area(ΔPQR)?
Show Answer+
Answer: A
Hint: A median of a triangle divides it into two triangles. What can you say about the bases and heights of these two triangles relative to the original triangle?
Solution:
A median of a triangle divides it into two triangles of equal area.
In ΔPQR, PS is the median to side QR, dividing ΔPQR into ΔPQS and ΔPSR.
Therefore, Area(ΔPQS) = Area(ΔPSR).
Given Area(ΔPQS) = 35 cm², then Area(ΔPSR) = 35 cm².
Area(ΔPQR) = Area(ΔPQS) + Area(ΔPSR) = 35 cm² + 35 cm² = 70 cm².
9A parallelogram ABCD has side AB = 10 cm and side BC = 6 cm. If the height corresponding to base AB is 4 cm, what is the height corresponding to base BC?
Show Answer+
Answer: C
Hint: The area of a parallelogram is constant regardless of which side is chosen as the base. Use the known area to find the unknown height.
Solution:
The area of a parallelogram is given by Base × Height.
Using base AB, Area(ABCD) = AB × height_AB = 10 cm × 4 cm = 40 cm².
Now, using base BC, the area remains the same: Area(ABCD) = BC × height_BC.
So, 40 cm² = 6 cm × height_BC. Solving for height_BC: height_BC = 40 / 6 cm = 20 / 3 cm ≈ 6.67 cm.
10In a triangle ABC, point D is on BC such that AD is the altitude to BC. If we consider AB as the base, which point's perpendicular distance from the line containing AB would represent the height?
Show Answer+
Answer: C
Hint: The height of a triangle corresponding to a particular base is the perpendicular distance from the opposite vertex to the line containing that base.
Solution:
In any triangle, the height (or altitude) corresponding to a chosen base is the perpendicular distance from the vertex opposite to that base, to the line containing the base.
If AB is considered the base, the vertex opposite to AB is Point C.
Therefore, the height corresponding to base AB would be the perpendicular distance from Point C to the line containing AB.
Want more questions?
Practice 60+ questions with AI-powered doubt clearing and step-by-step solutions.
Tips for Areas of Parallelograms & Triangles MCQs
- 1Read each question carefully and identify what is being asked before looking at the options.
- 2Try to solve the problem mentally or on paper first, then match your answer with the options.
- 3Use elimination — rule out clearly wrong options to improve your chances even when unsure.
- 4Check units, signs, and edge cases — these are common traps in Areas of Parallelograms & Triangles MCQs.
- 5Review your mistakes after completing the test to build lasting understanding.
Master Areas of Parallelograms & Triangles on SparkEd
Go beyond MCQs. Practice at three difficulty levels with instant feedback, solutions, and an AI coach to clear every doubt.
Start PractisingSparkEd Maths offers free MCQ tests for Class 1-10 across 7 education boards. All questions are aligned to the 2025-26 syllabus with step-by-step solutions and AI-powered doubt clearing.