What is the area of a triangle with sides of length 2, 4, and 5?

Question

Daniel Acosta · Accepted Answer

Area is #3.8# square units

If three sides of a triangle are #a,b# and #c#, then its area is #sqrt(s(s-a)(s-b)(s-c))#, where #s=(a+b+c)/2# Here sides are #2,4# and #5# and hence #s=(2+4+5)/2=11/2# and the triangle's area is #sqrt(11/2(11/2-2)(11/2-4)(11/2-5))# = #sqrt(11/2xx7/2xx3/2xx1/2)# = #1/4sqrt231=15.19868/4=3.79967# or say #3.8#

Delilah Aguilar · Answer

undefined To find the area of a triangle given the lengths of its sides, you can use Heron's formula. First, calculate the semi-perimeter $ s $ using the formula:

$$ s = \frac{a + b + c}{2} $$

Where $ a $, $ b $, and $ c $ are the lengths of the sides of the triangle. Then, use Heron's formula to find the area:

$$ \text{Area} = \sqrt{s \cdot (s - a) \cdot (s - b) \cdot (s - c)} $$

Substituting the given side lengths $ a = 2 $, $ b = 4 $, and $ c = 5 $:

$$ s = \frac{2 + 4 + 5}{2} = \frac{11}{2} $$

$$ \text{Area} = \sqrt{\frac{11}{2} \cdot \left(\frac{11}{2} - 2\right) \cdot \left(\frac{11}{2} - 4\right) \cdot \left(\frac{11}{2} - 5\right)} $$

$$ \text{Area} = \sqrt{\frac{11}{2} \cdot \frac{7}{2} \cdot \frac{3}{2} \cdot \frac{1}{2}} $$

$$ \text{Area} = \sqrt{\frac{11 \times 7 \times 3 \times 1}{2 \times 2 \times 2 \times 2}} $$

$$ \text{Area} = \sqrt{\frac{231}{16}} $$

$$ \text{Area} \approx \sqrt{14.4375} $$

$$ \text{Area} \approx 3.8 $$

Therefore, the area of the triangle with sides of length 2, 4, and 5 is approximately $ 3.8 $ square units.