Hello, and welcome to today's mathematics lesson. We are going to explore Chapter 23: Mensuration — the study of measuring the boundaries and spaces of flat shapes. By the end of this lesson, you will understand how to calculate perimeters, circumferences, and areas of rectangles, squares, triangles, and circles. You will also learn how units of length and area relate to each other.
Let us begin with the fundamentals. Mensuration deals with two main ideas for any closed plane figure. First, the perimeter — this is the total distance around the boundary of a shape. Imagine walking along the edge of a garden; the distance you cover is the perimeter. Second, the area — this is the region enclosed within that boundary, the actual space the shape covers on a flat surface.
Now, before we calculate, we must understand how units work. When we measure length, we use centimetres, metres, or kilometres. But area is two-dimensional — it involves length multiplied by length. This means units of area change much faster than units of length.
Consider this: one centimetre equals ten millimetres. But one square centimetre equals (10 mm × 10 mm), which gives one hundred square millimetres. Similarly, one metre equals one hundred centimetres, so one square metre equals (100 cm × 100 cm), which is ten thousand square centimetres, or 10⁴ square centimetres. And one kilometre equals one thousand metres, so one square kilometre equals (1000 m × 1000 m), which is one million square metres, or 10⁶ square metres. Remember: when converting to area units, always square the conversion factor.
Let us move to calculating perimeters of specific shapes.
For a rectangle, the perimeter is simply twice the sum of its length and breadth. The formula is: perimeter equals 2(l + b), where l stands for length and b stands for breadth. If you know the perimeter and one side, you can find the other. Length equals P/2 − b, and breadth equals P/2 − l.
For a square, where all four sides are equal, the perimeter is simply four times the side. The formula is: P = 4s. Conversely, if you know the perimeter, the side equals P/4.
For any triangle, the perimeter is the sum of all three sides: a + b + c.
Now we come to a special and beautiful shape — the circle. The perimeter of a circle is called its circumference. If the radius is r, then the circumference equals 2πr. Here, π is approximately 22/7 or about three point one four. Since diameter equals twice the radius, you can also write circumference as πd, where d is the diameter.
Here is a useful insight: if you bend a wire of any length into the largest possible circle, that wire's length becomes the circle's circumference. This principle helps solve many practical problems.
Let us work through a few examples to see these ideas in action.
Example one: a rectangular field has length two hundred thirty metres and breadth eighty-five metres. Using our formula, perimeter equals two times (230 + 85), which is two times three hundred fifteen, giving six hundred thirty metres.
Example two: a rectangle has perimeter two hundred forty centimetres, with length and breadth in the ratio five to three. Let length be five x and breadth be three x. Then two times (5x + 3x) equals two hundred forty. This simplifies to sixteen x equals two hundred forty, so x equals fifteen. Therefore, length is seventy-five centimetres and breadth is forty-five centimetres.
Example three: a rectangle has length fifteen metres and diagonal seventeen metres. Using Pythagoras theorem, breadth squared equals diagonal squared minus length squared: 17² − 15², which is two hundred eighty-nine minus two hundred twenty-five, giving sixty-four. So breadth is eight metres. The perimeter becomes two times (15 + 8), which is forty-six metres.
Example four: a circle has radius fourteen centimetres. Taking π as 22/7, the circumference equals two times 22/7 times fourteen. The sevens cancel, giving two times twenty-two times two, which equals eighty-eight centimetres.
Example five: two circles have radii six centimetres and nine centimetres. We want the radius of a circle whose circumference equals the sum of these two circumferences. Using the alternative method: 2πR equals 2π × 6 + 2π × 9. Dividing everything by 2π, we get R equals six plus nine, which is fifteen centimetres. Notice how the radii simply add up when circumferences are equal.
Now we turn to area — the space enclosed within shapes.
For a rectangle, area equals length multiplied by breadth: l × b. The diagonal, by Pythagoras theorem, equals √(l² + b²).
For a square with side s, area equals s². The diagonal equals s√2. Interestingly, you can also find area from the diagonal: area equals half of diagonal squared, or ½d².
For triangles, we have several methods. When base and height are known, area equals half times base times height: ½bh.
For any triangle with three known sides, we use Heron's formula. First find the semi-perimeter: s = (a+b+c)/2. Then area equals the square root of s(s−a)(s−b)(s−c).
For an isosceles triangle with equal sides a and base b, area equals ¼b√(4a²−b²).
For an equilateral triangle with side s, area equals (√3/4)s².
For a circle, area equals πr². Since diameter is twice the radius, you can also write this as πd²/4.
Let us see area calculations in practice.
Example: find the area of a triangle with base thirty-two centimetres and height twenty centimetres. Area equals half times thirty-two times twenty, which is three hundred twenty square centimetres.
Another example: a triangle has sides thirty-nine, twenty-five, and fifty-six centimetres. Semi-perimeter equals (39+25+56)/2, which is sixty. Using Heron's formula: area equals square root of 60 × 21 × 35 × 4. This simplifies to square root of 2×2×3×5×3×7×5×7×2×2, giving two times three times five times seven times two, which equals four hundred twenty square centimetres.
For an equilateral triangle with side twenty centimetres, taking root three as one point seven three, area equals (1.73/4) × 20 × 20, which is one hundred seventy-three square centimetres.
For a circle with radius twenty-one centimetres, taking π as 22/7, area equals (22/7) × 21 × 21. The sevens cancel with twenty-one leaving three, so we get twenty-two times three times twenty-one, which equals one thousand three hundred eighty-six square centimetres.
Here is a clever problem: a circle has circumference eighty-eight centimetres. Find its area. First, from 2πr = 88, we get r = 88 × 7/(2×22), which is fourteen centimetres. Then area equals (22/7) × 14 × 14, giving six hundred sixteen square centimetres.
Let us also consider composite figures — shapes made by combining or removing simpler shapes.
For shaded regions, the approach is straightforward: calculate the area of the outer figure, subtract the area of the inner figure or cut-out portion. For example, if a rectangle forty by twenty centimetres has a smaller rectangle twenty-five by ten centimetres removed from its centre, the shaded area equals eight hundred minus two hundred fifty, which is five hundred fifty square centimetres.
Similarly, for a square thirty by thirty centimetres with a rectangle twenty by fifteen centimetres removed, shaded area equals nine hundred minus three hundred, giving six hundred square centimetres.
Now, a quick word about parallelograms and rhombuses, which you may encounter.
The area of a parallelogram equals base times height — the perpendicular distance between opposite sides. For a rhombus, where all sides are equal, area equals half the product of the diagonals: ½d₁d₂. These formulas extend your mensuration toolkit.
Let us recap the key takeaways from this lesson.
First, perimeter is the total boundary length of a closed figure, while area is the space enclosed within that boundary.
Second, when converting units of area, square the linear conversion factor — one square metre equals ten thousand square centimetres, not one hundred.
Third, for rectangles: perimeter equals 2(l+b) and area equals l×b. For squares: perimeter equals 4s and area equals s².
Fourth, for circles: circumference equals 2πr or πd, and area equals πr². Remember that π is approximately 22/7 or three point one four.
Fifth, for triangles: when base and height are known, use ½bh. When three sides are known, use Heron's formula with semi-perimeter s. For equilateral triangles, use (√3/4)s².
Sixth, for composite figures, find areas of individual parts and combine them appropriately — adding for combined shapes, subtracting for cut-out regions.
Mensuration is not just about memorising formulas. It is about understanding the geometry of shapes and applying logical reasoning to measure the world around you. Whether you are fencing a field, tiling a floor, or designing a garden, these principles will serve you well.
Keep practising, stay curious, and remember: every shape tells a story through its measurements. Until next time, happy learning!