CBSE Class 10 Mathematics Perimeter and Area of Plane Figures Notes Set 06

Perimeter and Area of Basic Figures

Square of side a: perimeter 4a, area a², diagonal a√2. Rectangle l × b: perimeter 2(l + b), area lb, diagonal √(l² + b²). Parallelogram: area = base × height. Rhombus: area = ½ × d1 × d2 and side = ½√(d1² + d2²). Trapezium: area = ½ × (sum of parallel sides) × height.
Triangle: area = ½ × base × height; Heron's formula area = √(s(s − a)(s − b)(s − c)) with s as the semi-perimeter. Equilateral triangle of side a: area = (√3/4)a² and height = (√3/2)a. A triangle and a parallelogram on the same base between the same parallels have areas in the ratio 1 : 2.

Percentage Changes in Area

If a length changes by x%, a square's area changes by a factor (1 + x/100)². For a rectangle whose sides change by factors p and q, the area changes by the factor pq. To keep the area fixed when one side is multiplied by k, multiply the other side by 1/k.

Regular Polygons and Circles

A regular hexagon of side a splits into six equilateral triangles, so its area is (3√3/2)a² and its circumradius equals a. The inradius of a regular hexagon is (√3/2)a. The largest square inscribed in a circle of radius r has diagonal 2r and area 2r². For a square of side s, circumradius = s/√2 and inradius = s/2.

Worked Examples

Example 1. The sides of a rectangle are in the ratio 5 : 3 and its perimeter is 64 m. Find its area.
Solution: Let the sides be 5k and 3k. 2(8k) = 64, so k = 4. Sides 20 m and 12 m, area = 240 m².

Example 2. A rhombus has diagonals 24 cm and 10 cm. Find its side and area.
Solution: Half diagonals 12 and 5, so side = √(144 + 25) = 13 cm. Area = ½ × 24 × 10 = 120 cm².

Example 3. The side of a square is increased by 25%. By what percent does the area increase?
Solution: New area = 1.25² = 1.5625 times, an increase of 56.25%.

Example 4. The length of a rectangle is increased by 20% and the breadth is decreased by 20%. Find the change in area.
Solution: Area factor = 1.2 × 0.8 = 0.96, so the area decreases by 4%.

Example 5. Find the area of an equilateral triangle whose altitude is 6√3 cm.
Solution: Altitude = (√3/2)a gives a = 12 cm. Area = (√3/4) × 144 = 36√3 cm².

Example 6. The sides of a triangle are 13 cm, 14 cm and 15 cm. Find its area and the altitude on the side 14 cm.
Solution: s = 21; area = √(21 × 8 × 7 × 6) = 84 cm². Altitude = 2 × 84 ÷ 14 = 12 cm.

Example 7. A rectangular field is 48 m long and 14 m wide. A path of 2 m width runs all around it on the outside. Find the area of the path.
Solution: Outer rectangle = 52 × 18 = 936 m²; field = 48 × 14 = 672 m². Path = 264 m².

Example 8. Find the perimeter of a regular hexagon inscribed in a circle of radius 9 cm.
Solution: Side of the hexagon = radius = 9 cm, perimeter = 54 cm.

Example 9. The diagonal of a square is 10 cm. Find its area.
Solution: Area = ½ × diagonal² = ½ × 100 = 50 cm².

Example 10. A triangle and a parallelogram have the same base and equal areas. If the triangle's height is 18 cm, find the height of the parallelogram.
Solution: ½ × b × 18 = b × h gives h = 9 cm.