**Title: Mastering Triangle Area Calculations: A Guide to Heron’s Formula**
When it comes to calculating the area of a triangle, most are familiar with the base times height method. However, when the triangle is irregular and we only know the lengths of its sides, we turn to Heron’s Formula—a gem in the field of geometry.
**Understanding Heron’s Formula**
Heron’s Formula, named after Hero of Alexandria, a Greek mathematician and engineer, is given as:
\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]
where \( A \) is the area of the triangle, \( a \), \( b \), and \( c \) are the lengths of the sides, and \( s \) is the semi-perimeter of the triangle, which is calculated as:
\[ s = \frac{a + b + c}{2} \]
The beauty of Heron’s Formula lies in its simplicity and elegance—it requires no complex geometrical understanding, just the side lengths.
**Applying Heron’s Formula — 20 Problems and Solutions**
Let’s apply Heron’s Formula with 20 sample problems and detailed solutions.
**Problem 1:**
Given a triangle with sides of length 7, 24, and 25 units, calculate the area.
**Solution 1:**
\[
s = \frac{7 + 24 + 25}{2} = 28
\]
\[
A = \sqrt{28(28-7)(28-24)(28-25)} = \sqrt{28(21)(4)(3)} = \sqrt{7056} = 84 \text{ square units}
\]
**Problem 2:**
Find the area of a triangle with side lengths 8, 15, and 17 units.
**Solution 2:**
\[
s = \frac{8 + 15 + 17}{2} = 20
\]
\[
A = \sqrt{20(20-8)(20-15)(20-17)} = \sqrt{20(12)(5)(3)} = \sqrt{3600} = 60 \text{ square units}
\]
**Problem 3:**
A triangle has sides measuring 9, 10, and 11 units. Determine its area.
**Solution 3:**
\[
s = \frac{9 + 10 + 11}{2} = 15
\]
\[
A = \sqrt{15(15-9)(15-10)(15-11)} = \sqrt{15(6)(5)(4)} = \sqrt{1800} = 30 \sqrt{2} \text{ square units}
\]
**Problem 4:**
Calculate the area of a triangle with sides 13, 14, and 15 units in length.
**Solution 4:**
\[
s = \frac{13 + 14 + 15}{2} = 21
\]
\[
A = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21(8)(7)(6)} = \sqrt{7056} = 84 \text{ square units}
\]
**Problem 5:**
A triangle’s sides measure 5, 12, and 13 units. What’s the area?
**Solution 5:**
\[
s = \frac{5 + 12 + 13}{2} = 15
\]
\[
A = \sqrt{15(15-5)(15-12)(15-13)} = \sqrt{15(10)(3)(2)} = \sqrt{900} = 30 \text{ square units}
\]
**Problem 6:**
Calculate the area for a triangle with 6, 8, and 10 unit sides.
**Solution 6:**
\[
s = \frac{6 + 8 + 10}{2} = 12
\]
\[
A = \sqrt{12(12-6)(12-8)(12-10)} = \sqrt{12(6)(4)(2)} = \sqrt{576} = 24 \text{ square units}
\]
**Problem 7:**
Find the area for a triangle with 13, 37, and 30 units as side lengths.
**Solution 7:**
\[
s = \frac{13 + 37 + 30}{2} = 40
\]
\[
A = \sqrt{40(40-13)(40-37)(40-30)} = \sqrt{40(27)(3)(10)} = \sqrt{32400} = 180 \text{ square units}
\]
**Problem 8:**
A triangle has sides of 20, 21, and 29 units. Determine its area.
**Solution 8:**
\[
s = \frac{20 + 21 + 29}{2} = 35
\]
\[
A = \sqrt{35(35-20)(35-21)(35-29)} = \sqrt{35(15)(14)(6)} = \sqrt{44100} = 210 \text{ square units}
\]
**Problem 9:**
Given a triangle with sides 17, 25, and 26 units, calculate its area.
**Solution 9:**
\[
s = \frac{17 + 25 + 26}{2} = 34
\]
\[
A = \sqrt{34(34-17)(34-25)(34-26)} = \sqrt{34(17)(9)(8)} = \sqrt{39168} = 198 \text{ square units}
\]
**Problem 10:**
Calculate the area of a triangle with side lengths 7, 15, and 20 units.
**Solution 10:**
\[
s = \frac{7 + 15 + 20}{2} = 21
\]
\[
A = \sqrt{21(21-7)(21-15)(21-20)} = \sqrt{21(14)(6)(1)} = \sqrt{1764} = 42 \text{ square units}
\]
**Problem 11:**
A triangle has sides measuring 8, 29, and 35 units. Find the area.
**Solution 11:**
\[
s = \frac{8 + 29 + 35}{2} = 36
\]
\[
A = \sqrt{36(36-8)(36-29)(36-35)} = \sqrt{36(28)(7)(1)} = \sqrt{7056} = 84 \text{ square units}
\]
**Problem 12:**
What is the area of a triangle with sides 9, 40, and 41 units long?
**Solution 12:**
\[
s = \frac{9 + 40 + 41}{2} = 45
\]
\[
A = \sqrt{45(45-9)(45-40)(45-41)} = \sqrt{45(36)(5)(4)} = \sqrt{32400} = 180 \text{ square units}
\]
**Problem 13:**
Find the area of a triangle with sides 11, 60, and 61 units.
**Solution 13:**
\[
s = \frac{11 + 60 + 61}{2} = 66
\]
\[
A = \sqrt{66(66-11)(66-60)(66-61)} = \sqrt{66(55)(6)(5)} = \sqrt{108900} = 330 \text{ square units}
\]
**Problem 14:**
Determine the area of a triangle with sides 3, 4, and 5 units.
**Solution 14:**
\[
s = \frac{3 + 4 + 5}{2} = 6
\]
\[
A = \sqrt{6(6-3)(6-4)(6-5)} = \sqrt{6(3)(2)(1)} = \sqrt{36} = 6 \text{ square units}
\]
**Problem 15:**
A triangle has side lengths 22, 120, and 122 units. What is its area?
**Solution 15:**
\[
s = \frac{22 + 120 + 122}{2} = 132
\]
\[
A = \sqrt{132(132-22)(132-120)(132-122)} = \sqrt{132(110)(12)(10)} = \sqrt{1742400} = 1320 \text{ square units}
\]
**Problem 16:**
Calculate the area of a triangle with sides 24, 45, and 51 units.
**Solution 16:**
\[
s = \frac{24 + 45 + 51}{2} = 60
\]
\[
A = \sqrt{60(60-24)(60-45)(60-51)} = \sqrt{60(36)(15)(9)} = \sqrt{291600} = 540 \text{ square units}
\]
**Problem 17:**
Given a triangle with sides 26, 28, and 30 units, find its area.
**Solution 17:**
\[
s = \frac{26 + 28 + 30}{2} = 42
\]
\[
A = \sqrt{42(42-26)(42-28)(42-30)} = \sqrt{42(16)(14)(12)} = \sqrt{112896} = 336 \text{ square units}
\]
**Problem 18:**
Find the area for a triangle with sides 14, 48, and 50 units long.
**Solution 18:**
\[
s = \frac{14 + 48 + 50}{2} = 56
\]
\[
A = \sqrt{56(56-14)(56-48)(56-50)} = \sqrt{56(42)(8)(6)} = \sqrt{112896} = 336 \text{ square units}
\]
**Problem 19:**
A triangle has sides measuring 33, 44, and 55 units. Determine its area.
**Solution 19:**
\[
s = \frac{33 + 44 + 55}{2} = 66
\]
\[
A = \sqrt{66(66-33)(66-44)(66-55)} = \sqrt{66(33)(22)(11)} = \sqrt{532956} = 730 \text{ square units}
\]
**Problem 20:**
Calculate the area of a triangle with side lengths 16, 63, and 65 units.
**Solution 20:**
\[
s = \frac{16 + 63 + 65}{2} = 72
\]
\[
A = \sqrt{72(72-16)(72-63)(72-65)} = \sqrt{72(56)(9)(7)} = \sqrt{254016} = 504 \text{ square units}
\]
Through these sample problems, it is evident that Heron’s Formula is a versatile tool that simplifies the calculation of a triangle’s area when only the side lengths are known. Its application is particularly useful in trigonometry, surveying, and architecture, where precise measurements are crucial. Hence, understanding and utilizing Heron’s Formula is a fundamental skill for students and professionals alike.
