Answer :
To find the area of a triangle with side lengths of 6, 10, and 12 units using Heron's Formula, we will follow these steps:
1. Calculate the semi-perimeter [tex]\( s \)[/tex]
The semi-perimeter [tex]\( s \)[/tex] is given by:
[tex]\[ s = \frac{a + b + c}{2} \][/tex]
Here, [tex]\( a = 6 \)[/tex], [tex]\( b = 10 \)[/tex], and [tex]\( c = 12 \)[/tex]. Thus,
[tex]\[ s = \frac{6 + 10 + 12}{2} = \frac{28}{2} = 14 \][/tex]
2. Apply Heron's Formula to find the area
Heron's Formula for the area [tex]\( A \)[/tex] of a triangle is:
[tex]\[ A = \sqrt{s(s - a)(s - b)(s - c)} \][/tex]
Substituting the values, we have:
[tex]\[ A = \sqrt{14(14 - 6)(14 - 10)(14 - 12)} \][/tex]
Simplifying inside the square root:
[tex]\[ A = \sqrt{14 \times 8 \times 4 \times 2} \][/tex]
3. Break down the multiplication inside the square root
[tex]\[ 14 \times 8 \times 4 \times 2 = 14 \times 8 = 112 \][/tex]
[tex]\[ 112 \times 4 = 448 \][/tex]
[tex]\[ 448 \times 2 = 896 \][/tex]
4. Expressing the area in radical form
[tex]\[ A = \sqrt{896} \][/tex]
The area in radical form can be expressed further. Notice that 896 can be factored to find a perfect square:
[tex]\[ 896 = 64 \times 14 \quad \text{(as } 64 = 8^2 \text{ and } 64 \text{ is a perfect square)} \][/tex]
Hence,
[tex]\[ \sqrt{896} = \sqrt{64 \times 14} = \sqrt{64} \times \sqrt{14} = 8\sqrt{14} \][/tex]
Thus, the area of a triangle with side lengths of 6, 10, and 12 units, in radical form, is:
[tex]\[ 8\sqrt{14} \][/tex]
1. Calculate the semi-perimeter [tex]\( s \)[/tex]
The semi-perimeter [tex]\( s \)[/tex] is given by:
[tex]\[ s = \frac{a + b + c}{2} \][/tex]
Here, [tex]\( a = 6 \)[/tex], [tex]\( b = 10 \)[/tex], and [tex]\( c = 12 \)[/tex]. Thus,
[tex]\[ s = \frac{6 + 10 + 12}{2} = \frac{28}{2} = 14 \][/tex]
2. Apply Heron's Formula to find the area
Heron's Formula for the area [tex]\( A \)[/tex] of a triangle is:
[tex]\[ A = \sqrt{s(s - a)(s - b)(s - c)} \][/tex]
Substituting the values, we have:
[tex]\[ A = \sqrt{14(14 - 6)(14 - 10)(14 - 12)} \][/tex]
Simplifying inside the square root:
[tex]\[ A = \sqrt{14 \times 8 \times 4 \times 2} \][/tex]
3. Break down the multiplication inside the square root
[tex]\[ 14 \times 8 \times 4 \times 2 = 14 \times 8 = 112 \][/tex]
[tex]\[ 112 \times 4 = 448 \][/tex]
[tex]\[ 448 \times 2 = 896 \][/tex]
4. Expressing the area in radical form
[tex]\[ A = \sqrt{896} \][/tex]
The area in radical form can be expressed further. Notice that 896 can be factored to find a perfect square:
[tex]\[ 896 = 64 \times 14 \quad \text{(as } 64 = 8^2 \text{ and } 64 \text{ is a perfect square)} \][/tex]
Hence,
[tex]\[ \sqrt{896} = \sqrt{64 \times 14} = \sqrt{64} \times \sqrt{14} = 8\sqrt{14} \][/tex]
Thus, the area of a triangle with side lengths of 6, 10, and 12 units, in radical form, is:
[tex]\[ 8\sqrt{14} \][/tex]