Area and Perimeter of Plane Figures
ICSE Class 9 · Mathematics · Chapter 19
What this chapter covers
- Perimeter, area and their units
- Area and perimeter of a triangle using base and height
- Heron's formula for the area of a triangle
- Rectangles and squares
- Parallelograms and rhombuses
- Trapeziums
- Area of any quadrilateral
- Circumference of a circle
- Area of a circle, semicircle and quadrant
- Inner and outer area: rings, paths and combined figures
Key points
- Perimeter is the length of the boundary (linear units); area is the region enclosed (square units).
- 1 m2 = 10 000 cm2; 1 are = 100 m2; 1 hectare = 10 000 m2.
- Triangle: Area = (1)/(2) × base × height; any height = (2 × Area)/(base).
- Right-angled triangle: Area = (1)/(2) × product of the sides containing the right angle.
- Equilateral triangle of side a: height (√(3))/(2)a, Area (√(3))/(4)a2, perimeter 3a.
- Isosceles triangle with equal sides a and base b: Area = (b)/(4)√(4a2 - b2).
- Heron's formula: Area = √(s(s - a)(s - b)(s - c)) with s = (a + b + c)/(2); the three differences add up to s.
- Rectangle: P = 2(l + b), A = lb, diagonal √(l2 + b2). Square: P = 4a, A = a2 = (d2)/(2), diagonal a√(2).
- Path or border: area of path = outer area - inner area; a path of width w changes each dimension by 2w.
- Parallelogram: Area = base × corresponding height; a diagonal splits it into two congruent triangles.
- Rhombus: Area = (1)/(2)d1 d2; side = √(((d1)/(2))2 + ((d2)/(2))2).
- Trapezium: Area = (1)/(2)(a + b)h; with four sides given, cut off a triangle and use Heron's formula for the height.
- Any quadrilateral: Area = (1)/(2)d(h1 + h2), or the sum of two triangles formed by a diagonal.
- Circle: C = 2π r = π d, A = π r2; revolutions of a wheel = distance ÷ circumference.
- Semicircle: A = (1)/(2)π r2, P = π r + 2r. Quadrant: A = (1)/(4)π r2, P = (1)/(2)π r + 2r.
Questions answered
Define the perimeter and the area of a plane figure, and state the kind of unit each is measured in.
The perimeter is the total length of the boundary of a closed figure, measured in units of length such as m. The area is the measure of the region enclosed by the boundary, measured in square units such as m2.
Express 5 hectares in square metres and 2.5 m2 in square centimetres.
5 × 10 000 = 50 000 m2; 2.5 × 10 000 = 25 000 cm2.
State Heron's formula and explain what s stands for.
Area = √(s(s - a)(s - b)(s - c)), where a, b, c are the sides and s = (a + b + c)/(2) is the semi-perimeter.
Find the area of a triangle with sides 18 cm, 24 cm and 30 cm, and the height on its longest side.
s = 36; Area = √(36 × 18 × 12 × 6) = √(46 656) = 216 cm2. Height on the 30 cm side = (432)/(30) = 14.4 cm.
Explain, using Heron's formula, why lengths of 4 cm, 5 cm and 10 cm cannot be the sides of a triangle.
s = 9.5 and s - 10 = -0.5, which is negative. The product under the root is negative, because 4 + 5 is less than 10 and the triangle inequality fails.
Find the area of an equilateral triangle of side 10 cm, taking √(3) = 1.732.
(√(3))/(4) × 100 = 25√(3) = 43.3 cm2.
A rectangle is 24 cm long and 10 cm wide. Find its perimeter, area and diagonal.
Perimeter = 68 cm, area = 240 cm2, diagonal = √(576 + 100) = 26 cm.
A rectangular garden 20 m by 15 m has a path 2 m wide all round it outside. Find the area of the path.
Outer rectangle 24 × 19 = 456 m2; path = 456 - 300 = 156 m2.
Past board exam questions
ICSE 2019 sample paper · 4 marks
AEFD is a rectangle. B is a point on AE and C is a point on DF such that BC is perpendicular to AE, and AC is joined. A semicircle with BC as diameter is drawn inside rectangle BCFE, touching EF. The shaded region is the part of AEFD not covered by △ABC and the semicircle. AB = (1)/(2) BC, where BC = 14 cm (Use π = (22)/(7)). Find: (i) Area of quad. AEFD (ii) Area of △ABC (iii) Area of semicircle. Hence find the area of shaded region.
Show answer
BC = 14 cm, AB = (1)/(2) × 14 = 7 cm. The semicircle on diameter BC has radius 7 cm and touches EF, so BE = 7 cm and AE = AB + BE = 14 cm; EF = BC = 14 cm.
(i) Area of AEFD = 14 × 14 = 196 cm²
(ii) Area of △ABC = (1)/(2) × AB × BC = (1)/(2) × 7 × 14 = 49 cm²
(iii) Area of semicircle = (1)/(2) × (22)/(7) × 72 = 77 cm²
Area of shaded region = 196 − 49 − 77 = 70 cm²