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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 
8 m 
D 
9m 
5 m 
B 12 m 
C
SOLUTION 
Area of Δ BCD = ½ × BC × CD 
= ( ½ × 12 × 5 ) m² 
= 30 cm² 
Using Pythagoras theorem, we have 
BD² = BC² + CD² 
BD² = 12² + 5² 
BD² = 144 + 25 
BD² = 169 
BD = √169 = 13 m
For Δ 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 ΔABD = √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 
C 
4 cm 
3 cm 
D 
A B
SOLUTION 
Area of Δ 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 Δ 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 ΔABC + 
Area of Δ 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. 
6 cm 6 cm 
I 
II 
5 cm 
6.5 cm 
1.5 cm 
1 cm 1 cm 
2 cm 
IV 
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
Heron's Formula and Area Word Problems
Heron's Formula and Area Word Problems
Heron's Formula and Area Word Problems

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Heron's Formula and Area Word Problems

  • 2. 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 8 m D 9m 5 m B 12 m C
  • 3. SOLUTION Area of Δ BCD = ½ × BC × CD = ( ½ × 12 × 5 ) m² = 30 cm² Using Pythagoras theorem, we have BD² = BC² + CD² BD² = 12² + 5² BD² = 144 + 25 BD² = 169 BD = √169 = 13 m
  • 4. For Δ 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 ΔABD = √s(s — a)(s —b)(s—c) = √15 × 2 × 7 × 6 m² = √3 × 5 × 2 × 7 × 2 × 3 m²
  • 5. = √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
  • 6. 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 C 4 cm 3 cm D A B
  • 7. SOLUTION Area of Δ 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
  • 8. Now, s – a = ( 7 – 5 ) = 2 cm s – b = ( 7 – 4 ) = 3 cm s – c = ( 7 – 5 ) = 2 cm Area of Δ 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 ΔABC + Area of Δ ACD = 6 + (9.16) cm² = 15.2 cm² (approx.)
  • 9. Q3 Radha made a picture of aeroplane with coloured paper as shown in Fig. 17.24. find the total area of the paper used. 6 cm 6 cm I II 5 cm 6.5 cm 1.5 cm 1 cm 1 cm 2 cm IV V III
  • 10. SOLUTION Area of region I : Region I is enclosed by a triangle of sides a = 5 cm, b = 5 cm and c =1 cm