Chapter 9.2 — Exercise 9.1 — Areas of Polygons

Practice at Spotting Where the Cuts Should Go

The first question here isn't a calculation at all; it's practice at spotting where those cuts should go: an L-shaped figure usually splits into three rectangles by extending its internal edges to the boundary, a hexagon-like outline often splits into two trapeziums with one well-placed parallel line, a house shape (box plus roof) splits into a rectangle and one or two triangles, and any polygon at all can be cut into triangles by drawing every diagonal from a single vertex.

Two Shapes Stacked, One Answer

The next set of figures each combine exactly two familiar shapes glued along a shared edge. In the first, a triangle ABC sits on top of a square ACDE, where AE = 4 cm is the square's side and the combined figure's total height is 6 cm.

The second combines square ABCF (side 18 cm) with trapezium CDEF sitting above it, where CF (= 18 cm, the square's own top edge) and DE = 7 cm are the trapezium's parallel sides, 8 cm apart:

The third combines rectangle ABCD (20 cm × 15 cm) with trapezium ADEF, whose parallel sides are AD (= 15 cm, matching the rectangle's breadth) and EF = 6 cm, separated by a height found by subtraction: the whole figure's outer length is 28 cm, and the rectangle already accounts for 20 cm of that, leaving 28 − 20 = 8 cm for the trapezium's height.

All three problems follow the identical two-step pattern — identify which two shapes were glued together, work out any dimension that isn't handed to you directly (a height by subtraction, a shared side by matching it to its twin), then add the two areas.

One Diagonal, Two Given Offsets

Quadrilateral ABCD has diagonal AC = 10 cm, with perpendiculars from B and D onto that diagonal measuring BM = 6 cm and DN = 5 cm. This is the general quadrilateral formula from the chapter introduction applied with no extra work at all:

Nothing here needs decomposing further — the diagonal and both offsets are already given directly, so the formula applies in a single line.

Finding a Frame's Width Before Its Area

A picture frame has outer dimensions 28 cm × 24 cm and inner (photo-opening) dimensions 20 cm × 16 cm, with every section of the border the same width. Unlike the previous problems, the frame's width — call it x — isn't given directly; it has to be worked out first from how the inner and outer rectangles relate along one side:

With the width known, one border strip is a trapezium with parallel sides AB = 28 cm (outer) and EF = 20 cm (inner), separated by that 4 cm width:

The border reads as a trapezium rather than a rectangle precisely because its two parallel edges — one from the outer rectangle, one from the inner one — are different lengths, while the width connecting them stays constant all the way around.

Two Fields, Five and Six Pieces

Real land surveys rarely hand you a single clean shape, so this question presents two irregular fields, each already marked up with the internal perpendiculars needed to split them into trapeziums and triangles. Field (i) breaks into five pieces:

Adding the five gives 2800 + 1600 + 1200 + 3850 + 1250 = 10,700 sq.m. Field (ii) is subdivided differently, into six pieces:

That total comes to 750 + 2100 + 3600 + 400 + 2400 + 1400 = 10,650 sq.m. Laying each field's pieces out in a table rather than a running list of numbers is what actually prevents a piece from being missed or double-counted once there are five or six of them to track.

Turning a Ratio Into an Equation

A trapezium's parallel sides are in the ratio 5 : 3, the distance between them is 16 cm, and its area is 960 sq.cm — the lengths themselves aren't given, only the ratio and the total area. Writing the sides as 5x and 3x turns the area formula into a solvable equation instead of two unknowns:

So the parallel sides are 5 × 15 = 75 cm and 3 × 15 = 45 cm. Whenever a problem gives a ratio rather than exact numbers, writing the unknowns as multiples of a single variable is what converts an otherwise underdetermined question into an ordinary linear equation.

Tiling a Floor, Rhombus by Rhombus

3000 rhombus-shaped tiles, each with diagonals 45 cm and 30 cm, cover a floor at a cost of ₹20 per square metre. Since the rate is per square metre but the diagonals are given in centimetres, converting units comes before anything else: 45 cm = 0.45 m and 30 cm = 0.30 m.

Skipping the centimetre-to-metre conversion here wouldn't just introduce a small rounding error — it would inflate the final answer by a factor of 10,000, since area scales with the square of a length conversion, not the length itself.

One Pentagon, Two Honest Ways to Cut It

A house-shaped pentagon — a square base of side 15 cm topped by a triangular roof, 30 cm tall overall — gets divided two completely different ways by two students, Jyothi and Rashida, to see whether the final answer depends on the choice of cut.

Jyothi's cut: two trapeziums, ABEF and BCDE, split by a vertical line EB from the roof peak to the base. Since the roof is symmetric (EF = ED), this line bisects AC, giving AB = BC = 15 ÷ 2 = 7.5 cm. Trapezium ABEF has parallel sides AF = 15 cm and BE = 30 cm, 7.5 cm apart:

Rashida's cut: a square ABCD (the house body) plus a triangle DEC (the roof), with the triangle's base DC = 15 cm and its height EF = 30 − 15 = 15 cm:

Both cuts land on exactly 337.5 sq.cm. That agreement isn't a coincidence specific to this pentagon — as long as a shape is divided into pieces that exactly tile it with no gap and no overlap, the sum of the pieces' areas always equals the area of the whole, no matter which internal lines were chosen to make the cut.

Reading This Exercise Alongside the Introduction

Every calculation above reuses one of the seven formulas from the chapter introduction — the new skill here isn't a new formula but the judgment to see which shapes are hiding inside a composite figure and which dimension needs to be found before a formula can even be applied. That same judgment carries directly into Exercise 9.2, where circles, rings, and sectors get added to the same composite-figure toolkit.