Heron's Formula

Exercise 12.2 Part 1

Question 1: A park, in the shape of a quadrilateral ABCD, has ∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

Answer: The following figure shows AB = 9 cm, BC = 12 cm, CD = 5 cm and AD = 8 cm. ∠C = 90°

In ΔBDC; BD² = BC² + DC²
= 12² + 5²
= 144 + 25 = 169
Or, BD = 13 cm

Area of ΔBCD `= ½ xx BC xx CD`

`= ½ xx 12 xx 5 = 30 sq cm`

For ΔABD; a = 9 cm, b = 13 cm and c = 8 cm
Now, `s = (9 + 13 + 8)/2 = 15`

Area `=sqrt(s(s-a)(s-b)(s-c))`

`=sqrt(15(15-9)(15-13)(15-8))`

`=sqrt(15xx6xx2xx7)=6sqrt35 sq cm`

Total Area = 30 + 36 = 66 sq cm (approx)

Question 2: Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

Answer: For ΔABC; `s = (3 + 4 + 5)/2 = 6`

Area `=sqrt(s(s-a)(s-b)(s-c))`

`=sqrt(6(6-3)(6-4)(6-5))`

`=sqrt(6xx3xx2xx1)=6 sq cm`

For Δ ADC; `s = (5 + 5 + 4)/2 = 7`

Area `=sqrt(s(s-a)(s-b)(s-c))`

`=sqrt(7(7-5)(7-5)(7-4))`

`=sqrt(7xx2xx2xx3)=2sqrt21=9 sq cm`(approx)

Total Area = 6 + 9 = 15 sq cm (approx)

Question 3: Radha made a picture of an aeroplane with coloured paper as shown in the figure. Find the total area of the paper used.

Answer: Part I (Triangle): a = 5 cm, b = 5 cm and c = 1 cm
So, `s = (5 + 5 + 1)/2 = 5.5`

Area`=sqrt(s(s-a)(s-b)(s-c) )`

`=sqrt(5.5(5.5-5)(5.5-5)(5.5-1))`

`=sqrt(5.5xx0.5xx0.5xx4.5)`

`=0.75sqrt11=2.488 sq cm` (approx)

Part II (Rectangle): l = 6.5 cm and b = 1 cm
Area `= 6.5 xx 1 = 6.5 sq cm`

Part III (Trapezium): It is composed of three equilateral triangles with sides = 1 cm
Area of equilateral triangle `= (sqrt3)/(4) xx (si\de)^2`

`= (sqrt3)/(4) xx 1^2`

Hence, area of part III `= (3sqrt3)/(4) = 1.3 sq cm `(approx)

Part IV and IV(Triangles): b = 6 cm and h = 1.5 cm
Area of triangle `= ½ xx h xx b`

`= ½ xx 1.5 xx 6 = 4.5 sq cm`

Area of Part IV + Part V = 9 sq cm
Total area `= 2.488 + 6.5 + 1.3 + 9 = 19.3 sq cm`