HERON’S FORMULA

EXERCISE 12.2

Q1 A park in the shape of quadrilateral ABCD , has C = 90, AB = 9m , BC = 12m, CD = 5m and AD = 8m. How much area does it occupy?

A

B C

D9m

8 m

5 m

12 m

SOLUTIONArea of Δ BCD = ½ × BC × CD

= ( ½ × 12 × 5 ) m²= 30 cm²

Using Pythagoras theorem, we haveBD² = BC² + CD²

BD² = 12² + 5² BD² = 144 + 25 BD² = 169

BD = √169 = 13 m

For ∆ ABD Let a = 13m, b = 8m and c = 9m Now, s = ½ ( a + b + c) = ½ ( 13 + 8 + 9) m = ½ × 30m = 15 m s – a = (15 -13) m = 2m s – b = (15 - 8) m = 7 m s − c = (15 − 9) m = 6 m and Area of ∆ABD = √s(s — a)(s —b)(s—c) = √15 × 2 × 7 × 6 m² = √3 × 5 × 2 × 7 × 2 × 3

m²

= √2 × 2 × 3 × 3 × 5 × 7 m²

= 2 × 3√35 m² (approx.)= 6 × 5.9 m² (approx.) Required area = ABD + BCD

= 35.4 m² + 30 m² = 65.4 m²

Q2 Find the area of a quadrilateral ABCD in which AB = 3 cm, CD = 4 cm, DA = 5 cm, and AC = 5 cm.

5 cm

5 c

m4 cm

4 cm

3 cm

D

A B

C

SOLUTIONArea of ∆ ABC = ½ × AB × BC = ( ½ × 3 × 4) cm² = 6 cm² For Δ ACD: Let a = 5 cm, b = 4 cm and c =

5 cm. then, s = ½ × (a + b + c) s = ½ ( 5 + 4 + 5 ) cm s = ½ × 14 cm s = 7 cm

Now, s – a = ( 7 – 5 ) = 2 cm s – b = ( 7 – 4 ) = 3 cm s – c = ( 7 – 5 ) = 2 cm Area of ∆ ACD = √s(s — a)(s —b)(s—c) = √ 7 (2) (3) (2) = √ 2 × 2 × 3 × 7 = 2 × √3 ×√7 = 2 × √21 = 2 × 4.58 = 9.167 /-

Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆ ACD

= 6 + (9.16) cm² = 15.2 cm² (approx.)

Q3 Radha made a picture of aeroplane with coloured paper as shown in Fig. 17.24. find the total area of the paper used.

I

II

6 cm 6 cm

5 cm

6.5 cm

1 cm 1 cm

2 cm

1.5 cmIV

V

III

SOLUTION

Area of region I : Region I is enclosed by a triangle of sides a = 5 cm, b = 5 cm and c =1 cm

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