CBSE Class 9 Mathematics Herons Formula Notes Set 02

Some Basic Terms

Area: The region covered by the base of a body is called the area of that body. Area is the same everywhere on the base.

 

Height: The perpendicular distance from a vertex to the opposite side (the base) is called the height. The side on which the foot of the perpendicular lies is called the base.

 

Examples on Height and Base

Example 1. ABCD is a parallelogram. If DE and BF are perpendiculars from D and B on the sides AB and DA respectively, then their bases are AB and DA respectively.

 

Example 2. ABC is an acute-angled triangle and AD is its height. The base corresponding to this height is BC.

 

Example 3. PQR is a triangle that is obtuse-angled at Q. The height of P from BC is PT, but the base is QR (not SR), so the base is extended to meet the foot of the perpendicular.

 

Area of a Triangle

Area of a triangle = \( \frac{1}{2} \) × base × height square units. We can use this formula when we can find or are given the height.

 

Note: Use \( \frac{1}{2} \) × (base) × (height) for the area of a right-angled triangle if any two sides are given.

 

Heron's Formula

If we have all three sides of a triangle and there is no way to find its height, then we use Heron's formula for the area of the triangle.

\[ \text{Area of } \Delta = \sqrt{s(s-a)(s-b)(s-c)} \]

where s is the semi-perimeter of the triangle and a, b, c are its sides:

\[ s = \frac{1}{2}(\text{sum of all sides}) = \frac{1}{2}(a+b+c) \]

 

Examples on Heron's Formula

Example 4. For the given figure, find s(s − a).

Answer:
Perimeter = 2s = 3 + 4 + 5 = 12 cm, so the semi-perimeter s = 12 ÷ 2 = 6 cm.
Here a = BC = 4 cm, so s − a = 6 − 4 = 2 cm.
∴ s(s − a) = 6 × 2 = 12 cm².

 

Example 5. If the semi-perimeter of a triangle is 60 cm and two of its sides are 45 cm and 40 cm, find the third side.
Answer:
Semi-perimeter = 60 cm, so perimeter = 2 × 60 = 120 cm.
Sum of all three sides = 120. Let the third side be x cm.
x + 45 + 40 = 120 ⇒ x + 85 = 120 ⇒ x = 120 − 85 = 35 cm.

 

Example 6. If the perimeter of an equilateral triangle is 96 cm, then find each side.
Answer:
All sides of an equilateral triangle are equal. Let each side be x cm.
x + x + x = 96 ⇒ 3x = 96 ⇒ x = 32 cm.

 

Example 7. If one side from two equal sides of a triangle is 14 cm and the semi-perimeter is 22.5 cm, then find the third side.
Answer:
Let the third side be x cm. Then x + 14 + 14 = 2 × 22.5 = 45.
x = 45 − 28 = 17 cm.

 

Example 8. Find the length of AD in the given figure if EC = 4 cm and AB = 5 cm.

Answer:
Area of \( \Delta ABC \) = \( \frac{1}{2} \)(AB × EC) = \( \frac{1}{2} \)(5 × 4) = 10 cm².
Also, area of \( \Delta ABC \) = \( \frac{1}{2} \)(BC × AD) = \( \frac{1}{2} \)(6 × AD) = 3AD cm².
∴ 3AD = 10 ⇒ AD = \( \frac{10}{3} \) = 3.33 cm.
Note: \( \sqrt{2}=1.41,\ \sqrt{3}=1.73,\ \sqrt{5}=2.23,\ \sqrt{6}=2.45,\ \sqrt{7}=2.64,\ \sqrt{8}=2.82,\ \sqrt{11}=3.31,\ \sqrt{15}=3.87 \)

 

Example 9. Find the area of a triangle whose sides are of lengths 52 cm, 56 cm and 60 cm respectively.
Answer:
Let a = 52 cm, b = 56 cm and c = 60 cm. Perimeter = 52 + 56 + 60 = 168 cm.
s = \( \frac{1}{2}(a+b+c) = \frac{1}{2} \times 168 \) = 84 cm.
s − a = 84 − 52 = 32 cm, s − b = 84 − 56 = 28 cm and s − c = 84 − 60 = 24 cm.
By Heron's formula, area = \( \sqrt{s(s-a)(s-b)(s-c)} \) = \( \sqrt{84 \times 32 \times 28 \times 24} \) cm²
= \( \sqrt{(14 \times 6) \times (16 \times 2) \times (14 \times 2) \times (6 \times 4)} \) cm² = (14 × 6 × 4 × 2 × 2) cm² = 1344 cm².

 

Example 10. Using Heron's formula, find the area of an equilateral triangle of side a units.
Answer:
Here a = b = c = a, so s = \( \frac{1}{2}(a+a+a) = \frac{3a}{2} \).
s − a = s − b = s − c = \( \frac{3a}{2}-a = \frac{a}{2} \).
Area = \( \sqrt{s(s-a)(s-b)(s-c)} \) = \( \sqrt{\frac{3a}{2} \times \frac{a}{2} \times \frac{a}{2} \times \frac{a}{2}} \) = \( \frac{\sqrt{3}a^2}{4} \) sq units.
Hence the area of an equilateral triangle of side a is \( \frac{\sqrt{3}}{4}a^2 \) sq units.
Note: \( \Delta = \frac{\sqrt{3}}{4}a^2 = \frac{1}{2} \times a \times \frac{\sqrt{3}a}{2} = \frac{1}{2} \times \text{base} \times \text{height} \), so the height of an equilateral triangle is \( \frac{\sqrt{3}a}{2} \) units.

 

Example 11. Find the area of an isosceles triangle each of whose equal sides is 13 cm and whose base is 24 cm.
Answer:
Here a = 13 cm, b = 13 cm and c = 24 cm.
s = \( \frac{1}{2}(13+13+24) \) = 25 cm, so s − a = 12 cm, s − b = 12 cm and s − c = 1 cm.
Area = \( \sqrt{25 \times 12 \times 12 \times 1} \) cm² = (5 × 12) cm² = 60 cm².

 

Example 12. The perimeter of a triangular field is 450 m and its sides are in the ratio 13 : 12 : 5. Find the area of the triangle.
Answer:
Let a : b : c = 13 : 12 : 5, so a = 13x, b = 12x and c = 5x.
Perimeter = 450 ⇒ 13x + 12x + 5x = 450 ⇒ 30x = 450 ⇒ x = 15.
So the sides are a = 13 × 15 = 195 m, b = 12 × 15 = 180 m and c = 5 × 15 = 75 m.
2s = 450 ⇒ s = 225 m.
Area = \( \sqrt{225(225-195)(225-180)(225-75)} \) = \( \sqrt{225 \times 30 \times 45 \times 150} \) = \( \sqrt{5^2 \times 3^2 \times 3 \times 5 \times 2 \times 3^2 \times 5 \times 5^2 \times 2 \times 3} \) = 6750 m².

 

Example 13. Find the percentage increase in the area of a triangle if its each side is doubled.
Answer:
Let a, b, c be the sides of the old triangle and s its semi-perimeter, so s = \( \frac{1}{2}(a+b+c) \). Let its area be Δ.
The sides of the new triangle are 2a, 2b and 2c. Its semi-perimeter is s′ = \( \frac{1}{2}(2a+2b+2c) \) = a + b + c = 2s.
Δ′ = \( \sqrt{s'(s'-2a)(s'-2b)(s'-2c)} \) = \( \sqrt{2s(2s-2a)(2s-2b)(2s-2c)} \) = \( \sqrt{16\,s(s-a)(s-b)(s-c)} \) = 4Δ.
Increase in area = Δ′ − Δ = 4Δ − Δ = 3Δ.
Percentage increase = \( \frac{3\Delta}{\Delta} \times 100 \) = 300%.

 

Example 14. The lengths of the sides of a triangle are in the ratio 3 : 4 : 5 and its perimeter is 144 cm. Find (i) the area of the triangle and (ii) the height corresponding to the longest side.
Answer:
Sum of the ratio terms = 3 + 4 + 5 = 12. Let the sides be a, b and c.
a = \( 144 \times \frac{3}{12} \) = 36 cm, b = \( 144 \times \frac{4}{12} \) = 48 cm and c = \( 144 \times \frac{5}{12} \) = 60 cm.
s = \( \frac{1}{2}(36+48+60) \) = 72 cm, so s − a = 36 cm, s − b = 24 cm and s − c = 12 cm.
(i) Area = \( \sqrt{72 \times 36 \times 24 \times 12} \) cm² = (36 × 24) cm² = 864 cm².
(ii) The longest side is 60 cm. Let the corresponding height be h cm. Area = \( \frac{1}{2} \) × 60 × h = 30h cm².
∴ 30h = 864 ⇒ h = \( \frac{864}{30} \) = 28.8 cm.

 

Example 15. Find the area of the shaded region in the figure.

Answer:
By Pythagoras theorem in \( \Delta ADB \): AB = \( \sqrt{AD^2+BD^2} \) = \( \sqrt{12^2+16^2} \) = \( \sqrt{400} \) = 20 cm.
For \( \Delta ABC \): sides 20 cm, 48 cm and 52 cm, so s = \( \frac{20+48+52}{2} \) = 60 cm.
Area of \( \Delta ABC \) = \( \sqrt{60(60-20)(60-48)(60-52)} \) = \( \sqrt{60 \times 40 \times 12 \times 8} \) = \( \sqrt{5^2 \times 8^2 \times 12^2} \) = 5 × 8 × 12 = 480 cm².
Area of \( \Delta ADB \) = \( \frac{1}{2} \) × AD × BD = \( \frac{1}{2} \) × 12 × 16 = 96 cm².
Shaded area = ar(\( \Delta ABC \)) − ar(\( \Delta ADB \)) = 480 − 96 = 384 cm².

 

Example 16. Find the area of an isosceles triangle of sides a cm, a cm and b cm.
Answer:
Semi-perimeter s = \( \frac{a+a+b}{2} = \frac{2a+b}{2} \) cm.
s − a = \( \frac{b}{2} \), s − b = \( \frac{2a-b}{2} \).
Area = \( \sqrt{\frac{2a+b}{2} \cdot \frac{b}{2} \cdot \frac{b}{2} \cdot \frac{2a-b}{2}} \) = \( \frac{b}{4}\sqrt{(2a+b)(2a-b)} \) = \( \frac{b}{4}\sqrt{4a^2-b^2} \) cm².

 

Example 17. A traffic signal board, indicating 'SCHOOL AHEAD', is an equilateral triangle with side 'a'. Find the area of the signal board using Heron's formula. If its perimeter is 180 cm, what will be the area of the signal board?
Answer:
Let 2s be the perimeter, so 2s = 3a and s = \( \frac{3a}{2} \). Then s − a = \( \frac{a}{2} \) for each side.
Area Δ = \( \sqrt{\frac{3a}{2} \cdot \frac{a}{2} \cdot \frac{a}{2} \cdot \frac{a}{2}} \) = \( \frac{\sqrt{3}}{4}a^2 \).
If the perimeter is 180 cm, then 3a = 180 ⇒ a = 60 cm.
∴ Δ = \( \frac{\sqrt{3}}{4} \) × (60)² = \( 900\sqrt{3} \) cm².

 

Area of a Quadrilateral

If all four sides and a diagonal of a quadrilateral are given, then the diagonal divides it into two triangles. By Heron's formula we can find the area of both triangles, and by adding them we get the area of the quadrilateral.

 

Area of a Rhombus

1. If both diagonals are given (or we can find their lengths), then area = \( \frac{1}{2} \) × (product of the diagonals) = \( \frac{1}{2} \)(AC × DB) sq units.

2. If we use Heron's formula, we find the area of one triangle made by two sides and a diagonal. Twice this area is the area of the rhombus: area = 2 × ar(\( \Delta ABC \)) sq units.

 

Examples on Rhombus and Quadrilateral

Example 18. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?

Answer:
Let ABCD be the rhombus with AB = BC = CD = DA = 30 m and diagonal AC = 48 m. The diagonal divides it into two congruent triangles with sides 30 m, 30 m and 48 m.
s = \( \frac{30+30+48}{2} \) = 54 m.
Area of the rhombus = 2 × ar(\( \Delta ABC \)) = 2\( \sqrt{54(54-30)(54-30)(54-48)} \) = 2\( \sqrt{9 \times 6 \times 24 \times 24 \times 6} \) = 2 × 3 × 6 × 24 = 864 m².
OR area = \( 2\left[\frac{b}{4}\sqrt{4a^2-b^2}\right] \) = \( 2 \times \frac{48}{4}\sqrt{4(30)^2-(48)^2} \) = 24\( \sqrt{1296} \) = 24 × 36 = 864 m².
Area required for 18 cows = 864 m², so the area required for 1 cow = \( \frac{864}{18} \) = 48 m².

 

Example 19. A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in the figure. How much paper of each shade has been used in it?

Answer:
ABCD is a square, so both diagonals are equal (32 cm each) and bisect each other at right angles: AC = 32 cm and BP = PD = 16 cm.
Area of region I = area of region II = \( \frac{1}{2} \) × AC × BP = \( \frac{1}{2} \) × 32 × 16 = 256 cm².
Region III is a triangle with sides 6 cm, 6 cm and 8 cm, so s = \( \frac{8+6+6}{2} \) = 10 cm.
Area of region III = \( \sqrt{10(10-8)(10-6)(10-6)} \) = \( \sqrt{10 \times 2 \times 4 \times 4} \) = 4\( \sqrt{20} \) = 8\( \sqrt{5} \) cm² = 8 × 2.236 = 17.88 cm².
Hence the paper used is 256 cm², 256 cm² and 17.88 cm² for regions I, II and III respectively.

 

Area of a Trapezium

Example 20. Find the area of a trapezium whose parallel sides are 25 cm and 13 cm and the other sides are 15 cm and 15 cm.

Answer:
Let ABCD be the trapezium with AB = 25 cm, CD = 13 cm, BC = 15 cm and AD = 15 cm. Draw CE || AD.
Now ADCE is a parallelogram in which AD || CE and AE || DC. ∴ AE = DC = 13 cm and BE = AB − AE = 25 − 13 = 12 cm.
In \( \Delta BCE \): the sides are 15 cm, 15 cm and 12 cm, so s = \( \frac{15+15+12}{2} \) = 21 cm.
Area of \( \Delta BCE \) = \( \sqrt{21(21-15)(21-15)(21-12)} \) = \( \sqrt{21 \times 6 \times 6 \times 9} \) = \( 18\sqrt{21} \) cm². ...(i)
Let h be the height of \( \Delta BCE \). Area = \( \frac{1}{2} \) × 12 × h = 6h cm². ...(ii)
From (i) and (ii), 6h = \( 18\sqrt{21} \) ⇒ h = \( 3\sqrt{21} \) cm. The height of the trapezium ABCD is the same as that of \( \Delta BCE \).
∴ Area of trapezium = \( \frac{1}{2} \)(AB + CD) × h = \( \frac{1}{2} \)(25 + 13) × \( 3\sqrt{21} \) = \( 57\sqrt{21} \) cm².

 

Example 21. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

Answer:
Let ABCD be the trapezium with AB = 25 m, CD = 10 m, BC = 14 m and AD = 13 m. From C, draw CE || DA. Then ADCE is a parallelogram with AE = DC = 10 m and CE = AD = 13 m.
∴ BE = AB − AE = 25 − 10 = 15 m. In \( \Delta BCE \): BC = 14 m, CE = 13 m and BE = 15 m, so s = \( \frac{14+13+15}{2} \) = 21 m.
Area of \( \Delta BCE \) = \( \sqrt{21(21-14)(21-13)(21-15)} \) = \( \sqrt{21 \times 7 \times 8 \times 6} \) = \( \sqrt{7^2 \times 3^2 \times 4^2} \) = 84 m².
Also, area of \( \Delta BCE \) = \( \frac{1}{2} \) × BE × CL, so 84 = \( \frac{1}{2} \) × 15 × CL ⇒ CL = \( \frac{168}{15} = \frac{56}{5} \) m.
Height of the parallelogram ADCE = CL = \( \frac{56}{5} \) m, so its area = base × height = 10 × \( \frac{56}{5} \) = 112 m².
Hence the area of the trapezium ABCD = area of parallelogram ADCE + area of \( \Delta BCE \) = 112 + 84 = 196 m².

 

More Solved Examples

Example 22. Students of a school stage a rally for cleanliness campaign. They walked through the lanes in two groups. One group walked through the lanes AB, BC and CA, while the other through AC, CD and DA (see figure). Then they cleaned the area enclosed within their lanes. If AB = 9 m, BC = 40 m, CD = 15 m, DA = 28 m and ∠B = 90°, which group cleaned more area and by how much? Find the total area cleaned by the students.

Answer:
In \( \Delta ABC \), ∠B = 90°, so AC² = AB² + BC² = 9² + 40² = 1681 ⇒ AC = 41 m.
Area of \( \Delta ABC \) = \( \Delta_1 = \frac{1}{2} \) × BC × AB = \( \frac{1}{2} \) × 40 × 9 = 180 m².
For \( \Delta ACD \): the sides are AC = 41 m, CD = 15 m and DA = 28 m, so 2s = 41 + 15 + 28 = 84 ⇒ s = 42 m.
Area of \( \Delta ACD \) = \( \Delta_2 \) = \( \sqrt{42(42-41)(42-15)(42-28)} \) = \( \sqrt{42 \times 1 \times 27 \times 14} \) = 126 m².
So the first group cleaned 180 m² and the second group 126 m². Also Δ₁ − Δ₂ = 180 − 126 = 54 m².
Thus the first group cleaned 54 m² more than the second group, and the total area cleaned by all the students is 180 + 126 = 306 m².

 

Example 23. Radha made a picture of an aeroplane with coloured paper as shown in the figure. Find the total area of the paper used.

Answer:
Region I is a triangle with sides 5 cm, 5 cm and 1 cm. 2s = 5 + 5 + 1 ⇒ s = \( \frac{11}{2} \) cm.
Area = \( \sqrt{\frac{11}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{9}{2}} \) = \( \frac{3}{4}\sqrt{11} \) = \( \frac{3}{4} \) × 3.32 = 2.49 cm².
Region II is a rectangle of length 6.5 cm and breadth 1 cm. Area = 6.5 × 1 = 6.5 cm².
Region III is an isosceles trapezium (figure below), with AD = 2 cm, BC = 1 cm, AB = CD = 1 cm and AE = 0.5 cm. In \( \Delta ABE \), AB² = AE² + BE² ⇒ 1 = 0.25 + BE² ⇒ BE = \( \sqrt{0.75} = \frac{\sqrt{3}}{2} \) cm.

Area of region III = \( \frac{1}{2} \)(AD + BC) × BE = \( \frac{1}{2}(2+1) \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{4} \) cm² = 1.3 cm².
Region IV is a right triangle whose two sides are 6 cm and 1.5 cm. Area = \( \frac{1}{2} \) × 6 × 1.5 = 4.5 cm².
Region V is congruent to region IV, so its area is also 4.5 cm².
Hence the total area of the paper used = 2.49 + 6.5 + 1.3 + 4.5 + 4.5 = 19.29 cm².

 

Example 24. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

Answer:
The sides of one triangular piece are a = 20 cm, b = 50 cm and c = 50 cm. 2s = 20 + 50 + 50 = 120 ⇒ s = 60 cm.
Area of one piece = \( \sqrt{60(60-20)(60-50)(60-50)} \) = \( \sqrt{60 \times 40 \times 10 \times 10} \) = \( \sqrt{6 \times 4 \times 10 \times 10 \times 10 \times 10} \) = \( 200\sqrt{6} \) cm².
There are 10 pieces of two colours, so there are 5 pieces of each colour.
Area of cloth of each colour = 5 × \( 200\sqrt{6} \) = \( 1000\sqrt{6} \) cm².

 

Example 25. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). Find the cost of polishing the tiles at the rate of 50 paisa per cm².

Answer:
The sides of a tile are 28 cm, 9 cm and 35 cm, so s = \( \frac{28+9+35}{2} \) = 36 cm.
Area of one tile = \( \sqrt{36(36-28)(36-9)(36-35)} \) = \( \sqrt{36 \times 8 \times 27 \times 1} \) = \( 36\sqrt{6} \) cm².
Area of 16 tiles = 16 × \( 36\sqrt{6} \) = \( 576\sqrt{6} \) cm².
Cost of polishing at 50 paise, i.e. Rs. \( \frac{1}{2} \) per cm² = Rs. \( 576\sqrt{6} \times \frac{1}{2} \) = Rs. 288 × 2.4495 = Rs. 705.45 (approximately).

 

Example 26. Sanya has a piece of land which is in the shape of a rhombus. She wants her one daughter and one son to work on the land and produce different crops to suffice the needs of their family. She divided the land in two equal parts. If the perimeter of the land is 400 m and one of the diagonals is 160 m, how much area each of them will get?

Answer:
Let ABCD be the field, divided by the diagonal BD = 160 m into two equal triangles. Since ABCD is a rhombus of perimeter 400 m, AB = BC = CD = DA = \( \frac{400}{4} \) = 100 m.
In \( \Delta BCD \): s = \( \frac{100+100+160}{2} \) = 180 m.
Area of \( \Delta BCD \) = \( \sqrt{180(180-100)(180-100)(180-160)} \) = \( \sqrt{180 \times 80 \times 80 \times 20} \) = 4800 m².
Hence each of the two children will get an area of 4800 m².

 

Example 27. There is a slide in a park. One of its side walls has been painted in some colour with a message "KEEP THE PARK GREEN AND CLEAN" (see figure). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Answer:
The side wall is triangular with sides a = 15 m, b = 6 m and c = 11 m. 2s = 15 + 6 + 11 = 32 ⇒ s = 16 m.
So s − a = 1 m, s − b = 10 m and s − c = 5 m.
Area to be painted in colour = area of the side wall = \( \sqrt{16 \times 1 \times 10 \times 5} \) = \( 20\sqrt{2} \) m².

 

Example 28. A triangular park ABC has sides 120 m, 80 m and 50 m (see figure). A gardener Dhania has to put a fence all around it and also plant grass inside. How much area does she need to plant? Find the cost of fencing it with barbed wire at the rate of Rs. 20 per metre leaving a space 3 m wide for a gate on one side.

Answer:
The park is triangular with a = BC = 120 m, b = CA = 80 m and c = AB = 50 m. 2s = 120 + 80 + 50 ⇒ s = 125 m.
s − a = 5 m, s − b = 45 m and s − c = 75 m.
Area of the park = \( \sqrt{125 \times 5 \times 45 \times 75} \) = \( 375\sqrt{15} \) m².
Length of the wire needed for fencing = perimeter − width of the gate = 250 − 3 = 247 m.
Cost of fencing = Rs. (20 × 247) = Rs. 4940.

 

Example 29. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m. The advertisements yield an earning of Rs. 5000 per m² per year. A company hired both walls for 3 months. How much rent did it pay?
Answer:
The sides of the walls are 122 m, 22 m and 120 m. Since 122² = 120² + 22² (14884 = 14400 + 484), the walls are right-angled triangles with legs 120 m and 22 m.
Area of two walls = 2 × \( \frac{1}{2} \) × base × height = 2 × \( \frac{1}{2} \) × 120 × 22 = 2640 m².
Yearly rent = Rs. 5000 per m², so monthly rent = Rs. \( \frac{5000}{12} \) per m².
Rent paid by the company for 3 months = Rs. \( \frac{5000}{12} \times 3 \times 2640 \) = Rs. 33,00,000.

 

Example 30. The sides of a triangle are in the ratio 12 : 17 : 25 and its perimeter is 540 cm. Find its area.
Answer:
Let the sides be 12x cm, 17x cm and 25x cm. Then 12x + 17x + 25x = 540 ⇒ 54x = 540 ⇒ x = 10.
So the sides are 120 cm, 170 cm and 250 cm, and s = \( \frac{540}{2} \) = 270 cm.
Area = \( \sqrt{270(270-120)(270-170)(270-250)} \) = \( \sqrt{270 \times 150 \times 100 \times 20} \) = 9000 cm².

 

Example 31. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Answer:
Third side = 42 − 18 − 10 = 14 cm, and s = \( \frac{42}{2} \) = 21 cm.
Area = \( \sqrt{21(21-18)(21-10)(21-14)} \) = \( \sqrt{21 \times 3 \times 11 \times 7} \) = \( 21\sqrt{11} \) cm² = 21 × 3.31 = 69.51 cm² (approximately).