Notes
Questions
\(\hspace{-12pt}\small{\textbf{1)}}\)Find the area of this triangle.
\(\text{The area is }\approx 19.90\)
\(\text{Step 1: Find Semiperimeter}\)
\(\,\,\,\,\,\,s=\displaystyle \frac{a+b+c}{2}\)
\(\,\,\,\,\,\,s=\displaystyle \frac{5+8+9}{2}\)
\(\,\,\,\,\,\,s=\displaystyle 11\)
\(\text{Step 2: Find the Area}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{s(s-a)(s-b)(s-c)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{11(11-9)(11-8)(11-5)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{11(2)(3)(6)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{396}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle 6\sqrt{11}\)
\(\text{The area is }\approx 19.90\) units squared
\(\text{Step 1: Find Semiperimeter}\)
\(\,\,\,\,\,\,s=\displaystyle \frac{a+b+c}{2}\)
\(\,\,\,\,\,\,s=\displaystyle \frac{5+8+9}{2}\)
\(\,\,\,\,\,\,s=\displaystyle 11\)
\(\text{Step 2: Find the Area}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{s(s-a)(s-b)(s-c)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{11(11-9)(11-8)(11-5)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{11(2)(3)(6)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{396}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle 6\sqrt{11}\)
\(\text{The area is }\approx 19.90\) units squared
\(\hspace{-12pt}\small{\textbf{2)}}\) Find the area of this triangle.
The area is \(\approx13.42\) units squared
\(\text{Step 1: Find Semiperimeter}\)
\(\,\,\,\,\,\,s=\displaystyle \frac{a+b+c}{2}\)
\(\,\,\,\,\,\,s=\displaystyle \frac{9+7+4}{2}\)
\(\,\,\,\,\,\,s=\displaystyle 10\)
\(\text{Step 2: Find the Area}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{s(s-a)(s-b)(s-c)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{10(10-9)(10-7)(10-4)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{10(1)(3)(6)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{180}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle 6\sqrt{5}\)
\(\text{The area is }\approx13.42\) units squared
\(\text{Step 1: Find Semiperimeter}\)
\(\,\,\,\,\,\,s=\displaystyle \frac{a+b+c}{2}\)
\(\,\,\,\,\,\,s=\displaystyle \frac{9+7+4}{2}\)
\(\,\,\,\,\,\,s=\displaystyle 10\)
\(\text{Step 2: Find the Area}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{s(s-a)(s-b)(s-c)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{10(10-9)(10-7)(10-4)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{10(1)(3)(6)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{180}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle 6\sqrt{5}\)
\(\text{The area is }\approx13.42\) units squared
\(\hspace{-12pt}\small{\textbf{3)}}\)Find the area of this triangle.
\(\text{The area is }=54\)
\(\text{Step 1: Find Semiperimeter}\)
\(\,\,\,\,\,\,s=\displaystyle \frac{a+b+c}{2}\)
\(\,\,\,\,\,\,s=\displaystyle \frac{9+12+15}{2}\)
\(\,\,\,\,\,\,s=\displaystyle 18\)
\(\text{Step 2: Find the Area}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{s(s-a)(s-b)(s-c)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{18(18-9)(18-12)(18-15)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{18(9)(6)(3)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{2916}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle 54\)
\(\text{The area is }=54\) units squared
\(\text{Step 1: Find Semiperimeter}\)
\(\,\,\,\,\,\,s=\displaystyle \frac{a+b+c}{2}\)
\(\,\,\,\,\,\,s=\displaystyle \frac{9+12+15}{2}\)
\(\,\,\,\,\,\,s=\displaystyle 18\)
\(\text{Step 2: Find the Area}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{s(s-a)(s-b)(s-c)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{18(18-9)(18-12)(18-15)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{18(9)(6)(3)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{2916}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle 54\)
\(\text{The area is }=54\) units squared
\(\hspace{-12pt}\small{\textbf{4)}}\)Find the area of this triangle.
\(\text{The area is }\approx 35.14\)
\(\text{Step 1: Find Semiperimeter}\)
\(\,\,\,\,\,\,s=\displaystyle \frac{a+b+c}{2}\)
\(\,\,\,\,\,\,s=\displaystyle \frac{6+14+18}{2}\)
\(\,\,\,\,\,\,s=\displaystyle 19\)
\(\text{Step 2: Find the Area}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{s(s-a)(s-b)(s-c)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{19(19-6)(19-14)(19-18)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{19(13)(5)(1)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{1235}\)
\(\text{The area is }\approx 35.14\) units squared
\(\text{Step 1: Find Semiperimeter}\)
\(\,\,\,\,\,\,s=\displaystyle \frac{a+b+c}{2}\)
\(\,\,\,\,\,\,s=\displaystyle \frac{6+14+18}{2}\)
\(\,\,\,\,\,\,s=\displaystyle 19\)
\(\text{Step 2: Find the Area}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{s(s-a)(s-b)(s-c)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{19(19-6)(19-14)(19-18)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{19(13)(5)(1)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{1235}\)
\(\text{The area is }\approx 35.14\) units squared
\(\hspace{-12pt}\small{\textbf{5)}}\)Find the area of this triangle.
\(\text{The area is }=120 \)
\(\text{Step 1: Find Semiperimeter}\)
\(\,\,\,\,\,\,s=\displaystyle \frac{a+b+c}{2}\)
\(\,\,\,\,\,\,s=\displaystyle \frac{10+24+26}{2}\)
\(\,\,\,\,\,\,s=\displaystyle 30\)
\(\text{Step 2: Find the Area}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{s(s-a)(s-b)(s-c)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{30(30-10)(30-24)(30-26)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{30(20)(6)(4)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{14{,}400}\)
\(\text{The area is }=120 \) units squared
\(\text{Step 1: Find Semiperimeter}\)
\(\,\,\,\,\,\,s=\displaystyle \frac{a+b+c}{2}\)
\(\,\,\,\,\,\,s=\displaystyle \frac{10+24+26}{2}\)
\(\,\,\,\,\,\,s=\displaystyle 30\)
\(\text{Step 2: Find the Area}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{s(s-a)(s-b)(s-c)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{30(30-10)(30-24)(30-26)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{30(20)(6)(4)}\)
\(\,\,\,\,\,\,\text{Area}=\displaystyle \sqrt{14{,}400}\)
\(\text{The area is }=120 \) units squared
See Related Pages\(\)
\(\bullet\text{ Geometry Homepage}\)
\(\,\,\,\,\,\,\,\,\text{All the Best Topics…}\)
\(\bullet\text{ Triangle Calculator (Calculator.net)}\)
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\(\bullet\text{ Right Triangle Trigonometry}\)
\(\,\,\,\,\,\,\,\,\sin{(x)}=\displaystyle\frac{\text{opp}}{\text{hyp}}…\)
\(\bullet\text{ Angle of Depression and Elevation}\)
\(\,\,\,\,\,\,\,\,\text{Angle of Depression}=\text{Angle of Elevation}…\)
\(\bullet\text{ Convert to Radians and to Degrees}\)
\(\,\,\,\,\,\,\,\,\text{Radians} \rightarrow \text{Degrees}, \times \displaystyle \frac{180^{\circ}}{\pi}…\)
\(\bullet\text{ Degrees, Minutes and Seconds}\)
\(\,\,\,\,\,\,\,\,48^{\circ}34’21”…\)
\(\bullet\text{ Coterminal Angles}\)
\(\,\,\,\,\,\,\,\,\pm 360^{\circ} \text { or } \pm 2\pi n…\)
\(\bullet\text{ Reference Angles}\)
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\(…\)
\(\bullet\text{ Find All 6 Trig Functions}\)
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\(…\)
\(\bullet\text{ Unit Circle}\)
\(\,\,\,\,\,\,\,\,\sin{(60^{\circ})}=\displaystyle\frac{\sqrt{3}}{2}…\)
\(\bullet\text{ Law of Sines}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{\sin{A}}{a}=\frac{\sin{B}}{b}=\frac{\sin{C}}{c}\) 
\(…\)
\(\bullet\text{ Area of SAS Triangles}\)
\(\,\,\,\,\,\,\,\,\text{Area}=\frac{1}{2}ab \sin{C}\) \(…\)
\(\bullet\text{ Law of Cosines}\)
\(\,\,\,\,\,\,\,\,a^2=b^2+c^2-2bc \cos{A}\) \(…\)
\(\bullet\text{ Area of SSS Triangles (Heron’s formula)}\)
\(\,\,\,\,\,\,\,\,\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}\) \(…\)
\(\bullet\text{ Geometric Mean}\)
\(\,\,\,\,\,\,\,\,x=\sqrt{ab} \text{ or } \displaystyle\frac{a}{x}=\frac{x}{b}…\)
\(\bullet\text{ Geometric Mean- Similar Right Triangles}\)
\(\,\,\,\,\,\,\,\,\)
\(…\)
\(\bullet\text{ Inverse Trigonmetric Functions}\)
\(\,\,\,\,\,\,\,\,\sin {\left(cos^{-1}\left(\frac{3}{5}\right)\right)}…\)
\(\bullet\text{ Sum and Difference of Angles Formulas}\)
\(\,\,\,\,\,\,\,\,\sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B}…\)
\(\bullet\text{ Double-Angle and Half-Angle Formulas}\)
\(\,\,\,\,\,\,\,\,\sin{(2A)}=2\sin{(A)}\cos{(A)}…\)
\(\bullet\text{ Trigonometry-Pythagorean Identities}\)
\(\,\,\,\,\,\,\,\,\sin^2{(x)}+\cos^2{(x)}=1…\)
\(\bullet\text{ Product-Sum Identities}\)
\(\,\,\,\,\,\,\,\,\cos{\alpha}\cos{\beta}=\left(\displaystyle\frac{\cos{(\alpha+\beta)}+\cos{(\alpha-\beta)}}{2}\right)…\)
\(\bullet\text{ Cofunction Identities}\)
\(\,\,\,\,\,\,\,\,\sin{(x)}=\cos{(\frac{\pi}{2}-x)}…\)
\(\bullet\text{ Proving Trigonometric Identities}\)
\(\,\,\,\,\,\,\,\,\sec{x}-\cos{x}=\displaystyle\frac{\tan^2{x}}{\sec{x}}…\)
\(\bullet\text{ Graphing Trig Functions- sin and cos}\)
\(\,\,\,\,\,\,\,\,f(x)=A \sin{B(x-c)}+D \) 
\(…\)
\(\bullet\text{ Solving Trigonometric Equations}\)
\(\,\,\,\,\,\,\,\,2\cos{(x)}=\sqrt{3}…\)
\(\bullet\text{ Andymath Homepage}\)
In Summary
Heron’s formula is a process for finding the area of any triangle where all 3 sides are known. It works on right-angled, obtuse and acute triangles.
It’s named after an ancient Greek mathematician Heron of Alexandria.
Heron’s formula is typically introduced in a high school geometry course while learning about triangles.
One common mistake students make when working with Heron’s formula is to forget to divide the sum of the sides by 2 to get the semiperimeter. The formula doesn’t work if you use the entire perimeter.