This MCQ module is based on: Rectangles, Squares and Triangles Revisited
TOPIC 21 OF 23
Rectangles, Squares and Triangles Revisited
🎓 Class 8
Mathematics
CBSE
Theory
Ch 7 — Understanding Quadrilaterals
⏱ ~35 min
🌐 Language: [gtranslate]
🧠 AI-Powered MCQ Assessment
▲
📐 Maths Assessment
▲
This mathematics assessment will be based on: Rectangles, Squares and Triangles Revisited
Targeting Class 8 level in General Mathematics, with Basic difficulty.
Upload images, PDFs, or Word documents to include their content in assessment generation.
7.1 Rectangles and Squares — Revisiting Area
You have used area? before to measure how much of a surface a flat figure covers. In this chapter we will build on that idea and derive formulas for the area of triangles, parallelograms, rhombuses, trapeziums and any polygon — using nothing more than cutting, rearranging and simple algebra.
Let us start with a playful question: how many different ways can you divide a square into four parts of equal area? Infinitely many! One way is a cross-cut into four small squares; another is to carve a wiggly stepped boundary down the middle — if each part balances what it gives up against what it gains on the other side, the four pieces still have the same area.
Three of the infinitely many ways to split a square into 4 equal-area parts.
In each cut, area is simply compressed along one edge and expanded along another — if the compression and expansion have the same size, all four pieces still have the same area.
Counting Squares — Unit Squares as the Measure
Rangoli patterns are a beautiful use of coloured rectangles. To compare two rectangles — say a 7 cm × 4 cm one and an 8 cm × 3 cm one — we can count the number of non-overlapping unit squares? of side 1 cm that fit inside each rectangle.
The 7 cm × 4 cm rectangle has 28 unit squares; the 8 cm × 3 cm has only 24. The first needs more powder.
Key Formula
Area of a rectangle = length × widthWritten with units: 7 cm × 4 cm = 28 sq. cm (also 28 cm²).
When one diagonal of a rectangle is drawn, it splits the rectangle into two congruent? right triangles — each triangle has exactly half the rectangle's area.
Diagonal divides a rectangle into two congruent right triangles.
Why Perimeter Cannot Measure Area
Perimeter measures only the boundary length. Two regions with the same perimeter can enclose very different amounts of space.
Counter-example
Region 1: a 1 × 5 rectangle — perimeter = 12, area = 5.Region 2: a 2 × 4 rectangle — perimeter = 12, area = 8.
Same perimeter, different areas. So perimeter is not a measure of area.
Activity: Same Perimeter, Different Areas
L3 Apply
Materials: Squared paper, pencil, ruler.
Predict: If two rectangles share a perimeter of 20 cm, which one encloses the most unit squares?
- On squared paper, draw rectangles whose perimeter is 20 cm but with different length–width pairs: 1×9, 2×8, 3×7, 4×6, 5×5.
- Count unit squares inside each to record the area.
- Plot (length, area) as a small bar chart.
- Which shape has the largest area? Which has the smallest?
Areas: 9, 16, 21, 24, 25. The 5×5 square has the largest area. Among rectangles of fixed perimeter, the square encloses the greatest area — a fact farmers, gardeners and architects use every day.
Figure it Out (Section 7.1 — part)
Q1. Find the missing side-lengths in each rectangle: (i) 28 m² region with one side 4 m; (ii) an L-shape broken into pieces of side-lengths 7 m, 3 m, 4 m with total area 29 m².
(i) Unknown side = 28 ÷ 4 = 7 m. (ii) Area of L-shape = 29 m² ⇒ the missing strip's width comes out to 4 m after subtracting the rectangle piece 7 × 3 = 21 m² from the whole.
Q2. A path of uniform width is laid around a rectangular park EFGH. Write a formula for the area of the path as: Area of EFGH + any extra pieces − Area of inner park. Use your formula on a 20 m × 15 m park surrounded by a 1 m path.
Outer dimensions become 22 m × 17 m. Area of path = 22 × 17 − 20 × 15 = 374 − 300 = 74 m².
Q3. A square plot has sides of 14 m. A cross-shaped path of width 2 m runs through its centre (one arm parallel to each pair of sides). Find the area of the path.
Each arm of the cross is 14 × 2 = 28 m². The two arms overlap in the central 2 × 2 = 4 m² square. Path area = 28 + 28 − 4 = 52 m².
7.2 Triangles — From Rectangles to a General Formula
Suppose two triangles, \(\triangle XDC\) and \(\triangle YDC\), share the same base DC but sit inside the same rectangle ABCD. Which has the larger area? By dropping altitudes? from X and Y to DC, we see that both triangles equal exactly half the enclosing rectangle — so their areas are equal!
Fig. 7.1 — Any triangle on base DC with apex on line AB has the same area = ½ × (base DC) × (height).
Why half the rectangle?
Drop the altitude from the apex X to base DC. The rectangle is split into two smaller rectangles by this altitude. The diagonal of each small rectangle halves it — so \(\triangle XDC\) itself is the sum of two half-rectangles = half of the whole.
The Area Formula for a Triangle
Place any triangle inside a rectangle whose base equals the triangle's base and whose height equals the triangle's height. The triangle always occupies exactly half the rectangle — even if the apex falls outside the base (the obtuse case), because the "outer" sliver can be written as the difference of two half-rectangles.
Formula — Triangle
\[ \text{Area of a triangle} \;=\; \tfrac{1}{2}\,\times\,\text{base}\,\times\,\text{height} \]
Here height means the perpendicular distance from the apex to the line containing the base — the altitude.
Worked Examples
Example A. Find BY in \(\triangle ABC\) given AX = 4, BC = 6, AB = 5, AC = ?, with area \(\triangle ABC = \tfrac{1}{2}\cdot AX\cdot BC = 12\) sq units, and BY the altitude to AC with AC = 32/5. Find BY.
Area = ½ × AX × BC = ½ × 4 × 6 = 12 sq units. Using the other base AC = 32/5, we also have area = ½ × AC × BY, so ½ × (32/5) × BY = 12 ⇒ BY = 60/16 = 3.75 units.
Example B. A triangle has base 10 cm and height 6 cm. What is its area? What is the area of the rectangle of the same base and height?
Triangle area = ½ × 10 × 6 = 30 cm². Rectangle area = 10 × 6 = 60 cm². The triangle is exactly half — as the formula predicts.
Example C. The diagonals of a rectangle split it into four triangles. Are the four triangles always congruent? Do any pairs of them have equal area?
They are not all congruent, but the diagonals of a rectangle bisect each other and the rectangle has two pairs of equal halves, so the four triangles form two pairs of equal area (the "top/bottom" pair and the "left/right" pair). In a square, all four are congruent.
Triangles Between Parallel Lines on a Common Base
If a line \(\ell\) is drawn parallel to base BC and we slide the apex along \(\ell\), the height of every such triangle is the same perpendicular distance between the two lines. So every triangle on base BC with apex on \(\ell\) has the same area. They differ only in perimeter.
All triangles on base BC with apex on line ℓ share the same area.
🔵 In-text question: Which such triangle has the least perimeter? The one whose apex sits on the perpendicular bisector of BC — the isosceles one. (It uses the reflection / mirror-line argument: the shortest path from B to the line ℓ and then to C reflects B across ℓ and draws a straight line.)
Competency-Based Questions
Scenario: A school is designing a triangular flower bed in the corner of a rectangular 12 m × 8 m lawn. The gardener wants to cover exactly half the lawn with flowers and keep the rest as grass.
Q1. If the flower bed is a right triangle using two sides of the lawn as legs, what is its area?
L3 Apply(b) 48 m². Area = ½ × 12 × 8 = 48 m², which is exactly half of 96 m² — what the gardener wants.
Q2. The gardener now proposes a scalene triangle using the full 12 m side as the base and a vertex on the opposite side of the lawn. Analyse whether this triangle can also cover exactly half the lawn, and where the vertex must sit.
L4 AnalyseYes. Any triangle with base 12 m and apex on the opposite 12 m side has height 8 m ⇒ area = ½ × 12 × 8 = 48 m². The vertex can sit anywhere on the far side — all such triangles have the same area.
Q3. A contractor claims: "Doubling both base and height quadruples the triangle's area; doubling just one of them doubles it." Evaluate this claim with a short calculation.
L5 EvaluateTrue. Original area = ½bh. Doubling both ⇒ ½(2b)(2h) = 4·½bh = 4 × original. Doubling only base ⇒ ½(2b)h = 2·½bh. The claim is correct because area is linear in each dimension separately and quadratic in both together.
Q4. Design a flower-bed layout that covers exactly 30 m² on the same 12 m × 8 m lawn using a single triangle. Describe the triangle (base + height) and sketch one possibility.
L6 CreateMany answers possible. Example 1: base 12 m, height 5 m ⇒ area = ½ × 12 × 5 = 30 m². Example 2: base 10 m, height 6 m ⇒ area = 30 m². Place apex anywhere along a line parallel to the base at the required height.
Assertion–Reason Questions
Assertion (A): Two triangles on the same base and between the same two parallel lines have equal areas.
Reason (R): Triangles with the same base and the same height have equal areas.
Reason (R): Triangles with the same base and the same height have equal areas.
Answer: (a) — Between the same parallels the perpendicular distance (height) is fixed, so both triangles have the same height. R explains A.
Assertion (A): Two rectangles with the same perimeter must have the same area.
Reason (R): Perimeter and area both describe how big a figure is.
Reason (R): Perimeter and area both describe how big a figure is.
Answer: (d) — A is false: 1×5 and 2×4 have equal perimeters but different areas. R is partially valid as a loose description but doesn't justify A.
Assertion (A): A diagonal of a rectangle divides it into two triangles of equal area.
Reason (R): The diagonal of a rectangle always bisects its angles.
Reason (R): The diagonal of a rectangle always bisects its angles.
Answer: (c) — A is true (two congruent right triangles). R is false in general — diagonals bisect angles only in a square.
Frequently Asked Questions
How do we know the diagonals of a rectangle are equal?
By using the Pythagoras theorem on the two right triangles formed by a diagonal, both diagonals have length equal to the square root of (length squared plus breadth squared).
What extra property does a square have over a rectangle?
A square has all sides equal and its diagonals are perpendicular bisectors of each other, each making 45-degree angles with the sides.
What makes a triangle special among polygons?
A triangle is rigid: given three side lengths, its shape is uniquely determined. No other polygon has this rigidity without diagonals.
How is this chapter different from Class 6 or 7 treatment?
Class 8 adds deeper reasoning, formal properties of diagonals, more rigorous constructions and links to coordinate geometry in later chapters.
Why revisit basic shapes in Class 8?
Revisiting consolidates the foundation and prepares students for formal proofs in Class 9 and mensuration and coordinate geometry in Class 10.
AI Tutor
Mathematics Class 8 — Ganita Prakash Part II
Ready