Class 9 Maths

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