This MCQ module is based on: Quadrilaterals Chapter Exercises
TOPIC 23 OF 23
Quadrilaterals Chapter Exercises
🎓 Class 8
Mathematics
CBSE
Theory
Ch 7 — Understanding Quadrilaterals
⏱ ~35 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Quadrilaterals Chapter Exercises
Targeting Class 8 level in General Mathematics, with Basic difficulty.
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Chapter 7 — Mixed Exercises (Figure it Out)
These questions draw from all of Chapter 7: Area — rectangles, triangles, quadrilaterals, parallelograms, rhombuses, trapeziums and real-life composite shapes?. Attempt each one with paper and pencil, then check the worked solution.
A — Rectangles & Paths
A1. A rectangular park EFGH measures 30 m × 20 m. A uniform path of width 2 m runs inside the park along its boundary. Find the area of the path using the formula: Area(path) = Area(outer rectangle) − Area(inner rectangle).
Inner rectangle dimensions = (30 − 4) × (20 − 4) = 26 × 16. Area(path) = 30 × 20 − 26 × 16 = 600 − 416 = 184 m².
A2. The side-length of a square is doubled. By what factor does the area increase?
If original side = s, new side = 2s. New area = (2s)² = 4s². The area becomes 4 times the original.
B — Triangles
B1. Find the area of (i) a triangle with base 6 cm and height 4 cm, (ii) a triangle with base 5.2 cm and height 3.2 cm.
(i) ½ × 6 × 4 = 12 cm². (ii) ½ × 5.2 × 3.2 = 8.32 cm².
B2. In ΔPQR, the altitude from P to QR has length 4 units and QR = 8 units. If PR = 6 units, find the altitude from Q to PR.
Area(ΔPQR) = ½ × 8 × 4 = 16 sq. units. Using base PR = 6: ½ × 6 × h = 16 ⇒ h = 16/3 ≈ 5.33 units.
B3. ΔSUB is isosceles with SE ⊥ UB, SE = 8 units, UB = 6 units. Find SB and the area of ΔSUB.
SE is also a median (isosceles + altitude), so EB = 3. Using Pythagoras in right ΔSEB: SB² = 8² + 3² = 64 + 9 = 73 ⇒ SB = √73 ≈ 8.54 units. Area = ½ × 6 × 8 = 24 sq. units.
B4. M and N are the midpoints of XY and XZ in triangle XYZ. What fraction of the area of ΔXYZ is the area of ΔXMN? (Hint: join NY.)
Since M is the midpoint of XY, ΔXMN has half the base of ΔXYN ⇒ Area(ΔXMN) = ½ Area(ΔXYN). Similarly N is the midpoint of XZ, so ΔXYN has half the base of ΔXYZ ⇒ Area(ΔXYN) = ½ Area(ΔXYZ). Therefore Area(ΔXMN) = ¼ Area(ΔXYZ).
C — Quadrilaterals & Parallelograms
C1. Find the area of a quadrilateral ABCD with AC = 22 cm, BM = 3 cm, DN = 3 cm, where BM ⊥ AC and DN ⊥ AC.
Area = ½ × AC × (BM + DN) = ½ × 22 × 6 = 66 cm².
C2. A regular hexagon divides neatly into one rectangle (in the middle), an equilateral triangle (on one end) and a rhombus (on the other) — see a honeycomb cell. If the rectangle has sides 6 cm × 10 cm, the equilateral triangle has side 6 cm (and height ≈ 5.2 cm), and the rhombus has diagonals 6 cm and 5.2 cm, find the ratio of their areas.
Rectangle = 60 cm². Triangle = ½ × 6 × 5.2 = 15.6 cm². Rhombus = ½ × 6 × 5.2 = 15.6 cm². Ratio = 60 : 15.6 : 15.6 ≈ 3.85 : 1 : 1. (The triangle and rhombus have equal areas.)
C3. Find the areas of two parallelograms: (i) base 7 cm, height 4 cm; (ii) base 5 cm, slant side 4.8 cm with 2 cm perpendicular height to the 5-cm base.
(i) 7 × 4 = 28 cm². (ii) 5 × 2 = 10 cm². (Slant side is irrelevant — only the perpendicular height matters.)
D — Rhombus, Trapezium & Composite
D1. Find the area of a rhombus whose diagonals are 20 cm and 15 cm.
Area = ½ × 20 × 15 = 150 cm².
D2. A trapezium has parallel sides 24 m and 14 m with perpendicular distance 7 m between them. Find the area.
Area = ½ × 7 × (24 + 14) = ½ × 7 × 38 = 133 m².
D3. A trapezium-shaped park has the two parallel sides 36 m and 18 m with perpendicular distance 12 m. The park also contains a rectangular pond of dimensions 10 m × 6 m in the middle. Find the grass-covered area of the park.
Park area = ½ × 12 × (36 + 18) = 6 × 54 = 324 m². Pond area = 10 × 6 = 60 m². Grass area = 324 − 60 = 264 m².
D4. Find the area of a composite figure consisting of a 10 m × 7 m rectangle with a semicircle of diameter 7 m attached to one short side. (Use π ≈ 22/7.)
Rectangle = 70 m². Semicircle radius = 3.5 m. Semicircle area = ½ × π × r² = ½ × (22/7) × 12.25 = 19.25 m². Total = 89.25 m².
D5. Suggest two distinct methods to convert a given triangle into a rectangle of equal area using dissection alone.
Method 1: Find the midpoints of two sides. The segment joining them is parallel to the third side and half its length. Cut off the small triangle above and flip it down to form a rectangle of the same base and half the height.
Method 2: Drop an altitude inside the triangle; it splits the triangle into two right-triangles. Slide each into a rectangle half its size, then combine.
Method 2: Drop an altitude inside the triangle; it splits the triangle into two right-triangles. Slide each into a rectangle half its size, then combine.
D6. Gopal needs to carry water from a river to his water tank. He starts from his house. What is the shortest path he can take from his house to the river and then to the water tank? Explain the reflection trick.
Reflect the water-tank's position across the river line to get a mirror-image point T'. Draw a straight line from the house to T'. The point where this line crosses the river is where Gopal should fill water. House → river point → tank is the shortest — because a straight line is the shortest distance, and the reflection preserves the river-to-tank leg length.
Exercise D3 — trapezium-shaped park containing a rectangular pond.
Activity: Estimate the Area of Your Classroom Floor
L3 Apply
Materials: Measuring tape, graph paper, calculator.
Predict: Is your classroom floor larger or smaller than an A4 sheet by what factor?
- Measure the length and breadth of the classroom to the nearest 10 cm.
- Compute the floor area in m².
- An A4 sheet ≈ 21 × 29.7 cm = 623.7 cm² = 0.06237 m². Divide your answer by 0.06237 to find how many A4 sheets fit on the floor.
- Also convert your answer to ft² using 1 m² ≈ 10.764 ft². Compare your two results.
A typical Indian classroom is roughly 8 m × 6 m = 48 m² ≈ 770 A4 sheets ≈ 516 ft². Comparing units builds intuition: metric and Imperial give the same area, just a different number attached.
Competency-Based Questions
Scenario: A municipal park committee is planning an entrance area made of three shapes placed side-by-side: a rectangular walking track (20 m × 3 m), a triangular flower bed (base 10 m, height 4 m), and a trapezium-shaped lawn with parallel sides 15 m & 9 m and height 8 m. The committee must buy turf priced at ₹85/m² for the lawn and at ₹110/m² for the flower bed.
Q1. Compute the total paved area (rectangle) and the trapezium lawn's area.
L3 Apply(a) 60 m² and 96 m². Rect = 20 × 3 = 60. Trap = ½ × 8 × (15 + 9) = 4 × 24 = 96.
Q2. Compute the total budget for turfing the lawn and the flower bed. Break your answer into two line items.
L4 AnalyseFlower bed area = ½ × 10 × 4 = 20 m². Flower-bed turf = 20 × 110 = ₹2200. Lawn turf = 96 × 85 = ₹8160. Total = ₹10 360.
Q3. A vendor offers a flat 10% discount if all three sections (including the walking track) are bought from them at ₹90/m² uniform rate. Evaluate whether the committee saves money compared with the mixed pricing above.
L5 EvaluateTotal area = 60 + 20 + 96 = 176 m². Flat cost = 176 × 90 = ₹15 840 → with 10% discount = ₹14 256. Original mixed plan (track not bought; only turf: 20 + 96) = ₹10 360 plus paving cost separately. If paving is ₹90/m² elsewhere, separate track = 60 × 90 = ₹5400 ⇒ total ₹15 760, versus bundled ₹14 256. The bundle saves ~₹1500 — accept the offer.
Q4. Design a new entrance layout of total area exactly 200 m² using at most three shapes drawn from: rectangle, triangle, rhombus, trapezium. Label all dimensions; justify your calculation.
L6 CreateOne valid design: Rectangle 10 × 12 = 120 m²; Triangle base 10 height 8 = 40 m²; Rhombus with diagonals 10 and 8 = 40 m². Total = 120 + 40 + 40 = 200 m². Many other designs work — the student should show each formula and sum.
Assertion–Reason Questions
Assertion (A): Doubling the side of a square multiplies its area by 4.
Reason (R): Area varies as the square of the side length.
Reason (R): Area varies as the square of the side length.
Answer: (a) — (2s)² = 4s², and R precisely captures why this happens.
Assertion (A): A triangle whose midpoints are joined to form a smaller triangle has area ¼ of the original triangle.
Reason (R): The segment joining midpoints of two sides is parallel to the third side and half its length.
Reason (R): The segment joining midpoints of two sides is parallel to the third side and half its length.
Answer: (a) — The midpoint triangle is similar with ratio 1:2, so its area is (1/2)² = 1/4 of the original. R is exactly the mid-point theorem, which is the reason.
Chapter 7 — Summary
Triangle
\( \text{Area} = \tfrac{1}{2} \times \text{base} \times \text{height} \)
Any Polygon
Break it into triangles along diagonals from one vertex, add up the triangle areas.
Parallelogram
\( \text{Area} = \text{base} \times \text{height} \)
Height is the perpendicular distance between the two parallel sides.
Rhombus
\( \text{Area} = \tfrac{1}{2} \times d_1 \times d_2 \) where \(d_1,d_2\) are the two diagonals.
Trapezium
\( \text{Area} = \tfrac{1}{2} \times \text{height} \times (\text{sum of parallel sides}) \)
Rectangle & Square
Rectangle: length × width. Square: side × side.
Key Principles
(i) Area is additive: the area of a composite figure is the sum of the areas of its non-overlapping pieces.(ii) Area is preserved under dissection: cutting and rearranging never changes total area.
(iii) Perimeter and area are independent — two shapes may have equal perimeters but different areas.
(iv) Triangles on the same base between the same parallels have equal areas.
Units & Conversions
\(1\ \text{m}^2 = 10\,000\ \text{cm}^2\) · \(1\ \text{in}^2 = 6.4516\ \text{cm}^2\) · \(1\ \text{ft}^2 = 144\ \text{in}^2\) · \(1\ \text{hectare} = 10\,000\ \text{m}^2\) · \(1\ \text{acre} \approx 4047\ \text{m}^2\).
Frequently Asked Questions
How do you find an unknown angle in a quadrilateral?
Use the property that the four interior angles of any quadrilateral sum to 360 degrees. Subtract the known angles from 360 to find the unknown angle.
How do you compute the area of a parallelogram?
Area equals base times height, where height is the perpendicular distance from the base to the opposite side. Side length alone is not enough without the perpendicular height.
What is the summary of Chapter 7 Quadrilaterals?
The chapter classifies quadrilaterals into special types (parallelogram, rectangle, rhombus, square, trapezium), studies diagonal properties, angle-sum rules and computes areas.
How do construction problems appear in exercises?
Students are asked to construct parallelograms and rhombuses given sides, diagonals or a diagonal and an angle, using ruler-and-compass techniques.
What is the exterior angle sum of a polygon?
The sum of exterior angles of any convex polygon is 360 degrees regardless of the number of sides, a key result used in exercises.
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