Exercise 9.1 — Areas of Polygons
Problems based on areas of polygons.
Exercise 9.1 – Areas of Composite Figures: Trapeziums, Quadrilaterals & Irregular Fields
Exercise 9.1 from Chapter 9, "Areas of Plane Figures," of Class 8 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) puts the area formulas — for triangles, rectangles, squares, quadrilaterals, rhombuses, and trapeziums — into practical use. The exercise focuses on composite or irregular figures, which are split into simpler shapes whose areas are calculated separately and then added together to find the total area.
This exercise builds directly on the formulas covered in the chapter introduction. If you're not yet confident with the area formulas for a triangle, trapezium, or rhombus, it's worth reviewing those first, since every question here relies on combining two or more of those formulas.
Question 1: Dividing Shapes into Simple Figures
The first question asks students to divide given irregular shapes according to specific instructions — practising the "divide and conquer" strategy that underlies this entire chapter. The instructions include dividing shapes into 3 rectangles, 2 trapeziums, 2 triangles and a rectangle, and 3 triangles.
- An L-shaped or staircase-shaped figure can usually be split into multiple rectangles by extending the internal edges until they meet the boundary.
- An irregular hexagon or similar shape can often be divided into two trapeziums by drawing one line parallel to a pair of opposite sides.
- A house-shaped figure (rectangle with a triangular roof) naturally splits into a rectangle and two triangles (or one triangle, depending on the shape of the roof).
- Any polygon can be divided into triangles by drawing diagonals from a single vertex to every other non-adjacent vertex.
Question 2: Finding the Area Enclosed by Composite Figures
The figure is a combination of triangle ABC and square ACDE, where side AE = 4 cm. Since AE = 4 cm, the square ACDE has area:
Area of square ACDE = side × side = 4 × 4 = 16 sq.cmThe base of triangle ABC is AC = 4 cm (same as the square's side, since AC is a shared edge). The total height of the figure is 6 cm, so the height of the triangle alone is 6 − 4 = 2 cm.
Area of △ABC = ½ × base × height = ½ × 4 × 2 = 4 sq.cmTotal area = Area of △ABC + Area of square ACDE = 4 + 16 = 20 sq.cmThe figure combines square ABCF (side AB = 18 cm) with trapezium CDEF sitting on top of it.
Area of square ABCF = side × side = 18 × 18 = 324 sq.cmIn trapezium CDEF, the parallel sides are CF (= AB = 18 cm, the top edge of the square) and DE = 7 cm. The distance between these parallel sides is 8 cm.
Area of trapezium CDEF = ½ × (a + b) × h = ½ × (18 + 7) × 8 = 25 × 4 = 100 sq.cmTotal area = Area of square ABCF + Area of trapezium CDEF = 324 + 100 = 424 sq.cmThe figure combines rectangle ABCD (AB = CD = 20 cm, BC = 15 cm) with trapezium ADEF attached to its left side.
Area of rectangle ABCD = length × breadth = 20 × 15 = 300 sq.cmIn trapezium ADEF, the parallel sides are AD (= BC = 15 cm) and EF = 6 cm. The distance between them is 28 − 20 = 8 cm (the difference between the total outer length and the rectangle's length).
Area of trapezium ADEF = ½ × (a + b) × h = ½ × (15 + 6) × 8 = 21 × 4 = 84 sq.cmTotal area = Area of rectangle ABCD + Area of trapezium ADEF = 300 + 84 = 384 sq.cmQuestion 3: Area of a Quadrilateral Using a Diagonal
Here, the diagonal AC = 10 cm, and the perpendicular distances from the other two vertices, B and D, onto this diagonal are BM = 6 cm and DN = 5 cm respectively. Using the general quadrilateral area formula:
Area of ABCD = ½ × AC × (DN + BM) = ½ × 10 × (5 + 6) = 5 × 11 = 55 sq.cmThis question is a direct, single-step application of the formula introduced in the chapter introduction — the diagonal splits the quadrilateral into two triangles (ABC and ACD), and this formula adds their areas together in one go.
Question 4: Area of a Picture Frame (Shaded Region)
The picture frame has outer dimensions 28 cm × 24 cm and inner dimensions 20 cm × 16 cm. The shaded region (the frame border ABFE on one side) forms a trapezium. To find the area of this trapezium, the width of the frame ('x') must first be calculated.
Since the inner rectangle's height (16 cm) plus the frame widths on the top and bottom (x + x) must equal the outer rectangle's height (24 cm):
x + 16 + x = 24 ⇒ 2x = 8 ⇒ x = 4 cmThe trapezium ABFE has parallel sides AB = 28 cm and EF = 20 cm, with the distance between them being the frame width, x = 4 cm.
Area of trapezium ABFE = ½ × (AB + EF) × x = ½ × (28 + 20) × 4 = 48 × 2 = 96 sq.cmQuestion 5: Finding the Area of Irregular Fields
This question presents two large irregular pentagon/hexagon-shaped fields, each divided into a series of triangles and trapeziums by internal perpendicular lines. The total area of each field is found by calculating the area of every individual piece and adding them all together.
Field (i) is divided into two trapeziums and three triangles. Each piece's area is calculated using its respective formula:
| Part | Shape | Calculation | Area (sq.m) |
|---|---|---|---|
| Trapezium ABCH | Trapezium | ½ × (30 + 40) × 80 = 70 × 40 | 2800 |
| △CHD | Triangle | ½ × 40 × 80 = 20 × 80 | 1600 |
| △DIE | Triangle | ½ × 40 × 60 = 20 × 60 | 1200 |
| Trapezium EIGF | Trapezium | ½ × (60 + 50) × 70 = 110 × 35 | 3850 |
| △AGF | Triangle | ½ × 50 × 50 = 25 × 50 | 1250 |
Total area = 2800 + 1600 + 1200 + 3850 + 1250 = 10700 sq.mField (ii) is divided into three trapeziums and three triangles:
| Part | Shape | Calculation | Area (sq.m) |
|---|---|---|---|
| △ABK | Triangle | ½ × 30 × 50 = 15 × 50 | 750 |
| Trapezium BCIK | Trapezium | ½ × (30 + 40) × 60 = 70 × 30 | 2100 |
| Trapezium CDEI | Trapezium | ½ × (40 + 50) × 80 = 90 × 40 | 3600 |
| △EHF | Triangle | ½ × 40 × 20 = 20 × 20 | 400 |
| Trapezium FGJH | Trapezium | ½ × (20 + 40) × 80 = 60 × 40 | 2400 |
| △GJA | Triangle | ½ × 40 × 70 = 20 × 70 | 1400 |
Total area = 750 + 2100 + 3600 + 400 + 2400 + 1400 = 10650 sq.mQuestion 6: Finding Unknown Trapezium Sides from Area
The ratio of the parallel sides of a trapezium is 5 : 3, the distance between them is 16 cm, and the area is 960 sq.cm. Let the parallel sides be 5x and 3x.
- Step 1: Apply the trapezium area formula: 960 = ½ × (5x + 3x) × 16
- Step 2: Simplify the sum of the parallel sides: 960 = ½ × 8x × 16 = 8x × 8 = 64x
- Step 3: Solve for x: x = 960 ÷ 64 = 15
- Step 4: Substitute back: parallel sides = 5x = 5 × 15 = 75 cm, and 3x = 3 × 15 = 45 cm
Lengths of parallel sides = 75 cm and 45 cmQuestion 7: Real-World Application — Cost of Rhombus-Shaped Tiles
A floor uses 3000 rhombus-shaped tiles, each with diagonals 45 cm and 30 cm. The cost is ₹20 per square metre. Since the cost is given per square metre but the diagonals are in centimetres, the first step is converting units.
- Step 1 — Convert to metres: 45 cm = 0.45 m, and 30 cm = 0.30 m (since 100 cm = 1 m).
- Step 2 — Area of one tile: Area = ½ × d₁ × d₂ = ½ × 0.45 × 0.30 = 0.0675 sq.m
- Step 3 — Area of all 3000 tiles: 3000 × 0.0675 = 202.5 sq.m
- Step 4 — Total cost: 202.5 × 20 = ₹4050
Total cost of flooring = ₹4050Question 8: Two Different Ways to Find the Area of the Same Pentagon
A house-shaped pentagon has a square base of side 15 cm with a triangular "roof" making the total height 30 cm. Two students, Jyothi and Rashida, divide this same pentagon in two completely different ways to find its area.
Jyothi's Method: Two Trapeziums
Jyothi divides the pentagon ACDEF into two trapeziums, ABEF and BCDE, by drawing a vertical line EB from the roof peak down to the base. Since the roof is symmetrical (EF = ED), this line bisects AC, so AB = BC = AC ÷ 2 = 15 ÷ 2 = 7.5 cm.
In trapezium ABEF, the parallel sides are AF = 15 cm and BE = 30 cm (the total height), with the distance between them being AB = 7.5 cm.
Area of trapezium ABEF = ½ × (AF + BE) × AB = ½ × (15 + 30) × 7.5 = 168.75 sq.cmBy symmetry, trapezium BCDE has exactly the same area.
Area of pentagon (Jyothi's method) = 168.75 + 168.75 = 337.5 sq.cmRashida's Method: Square Plus Triangle
Rashida divides the same pentagon ABCDE into a square ABCD (the house's body) and a triangle DEC (the roof).
Area of square ABCD = side × side = 15 × 15 = 225 sq.cmThe base of triangle DEC is DC = 15 cm, and its height EF = 30 − 15 = 15 cm (the roof height is the total height minus the square's side).
Area of △DEC = ½ × DC × EF = ½ × 15 × 15 = 112.5 sq.cmArea of pentagon (Rashida's method) = 225 + 112.5 = 337.5 sq.cmCommon Mistakes to Avoid in Exercise 9.1
- Using the wrong "height" for a piece: When a composite figure is split into parts, make sure the height/distance used for each part's formula is the perpendicular measurement for that specific piece, not the total figure's height.
- Forgetting to subtract dimensions: In figures like Question 2(iii), the height of the trapezium part (8 cm) had to be found by subtracting the rectangle's length from the total length (28 − 20). Always check whether a given dimension is for the whole figure or just one part.
- Skipping unit conversion: As in Question 7, mixing centimetres and metres without converting leads to answers that are off by a factor of 100 or 10,000.
- Not setting up a variable for ratio problems: For Question 6, jumping straight to numbers without using "5x" and "3x" makes the equation much harder to set up correctly.
- Assuming there's only one "correct" way to split a figure: Question 8 shows that different decompositions can both be valid — don't worry if your division of a shape looks different from someone else's, as long as the final area matches.
What Exercise 9.1 Prepares You For
Exercise 9.1 is the practical heart of this chapter — it shows how the formulas from the chapter introduction combine to solve real measurement problems involving fields, frames, and tiled floors. The skill of breaking a complex shape into trapeziums and triangles will reappear in later exercises of this chapter involving even more complex composite figures.
These area concepts also connect to topics in symmetry and geometric figures, since recognising symmetrical shapes (like the symmetric roof in Question 8) often makes area calculations simpler. In Class 9 and Class 10, these same decomposition techniques are used in coordinate geometry to find the area of polygons using vertex coordinates, and in mensuration chapters dealing with surface areas and volumes of solid shapes.