How do you use Heron's formula to find the area of a triangle with sides of lengths #1 #, #5 #, and #5 #?

Question

How do you use Heron's formula to find the area of a triangle with sides of lengths #1  #, #5    #, and #5   #?

Aurora Ayala · Accepted Answer

#Area=2.48746# square units

Heron's formula for finding area of the triangle is given by #Area=sqrt(s(s-a)(s-b)(s-c))# Where #s# is the semi perimeter and is defined as #s=(a+b+c)/2# and #a, b, c# are the lengths of the three sides of the triangle. Here let #a=1, b=5# and #c=5# #implies s=(1+5+5)/2=11/2=5.5# #implies s=5.5# #implies s-a=5.5-1=4.5, s-b=5.5-5=0.5 and s-c=5.5-5=0.5# #implies s-a=4.5, s-b=0.5 and s-c=0.5# #implies Area=sqrt(5.5*4.5*0.5*0.5)=sqrt6.1875=2.48746# square units #implies Area=2.48746# square units

Violet Aldrich · Answer

undefined To use Heron's formula to find the area of a triangle with sides of lengths $a$, $b$, and $c$, where $s$ is the semiperimeter calculated as $\frac{a + b + c}{2}$, and $A$ is the area of the triangle:

$$A = \sqrt{s(s - a)(s - b)(s - c)}$$

For the given triangle with sides of lengths 1, 5, and 5:

$a = 1$, $b = 5$, $c = 5$

Calculate the semiperimeter:

$$s = \frac{1 + 5 + 5}{2} = \frac{11}{2}$$

Substitute the values into Heron's formula:

$$A = \sqrt{\frac{11}{2}\left(\frac{11}{2} - 1\right)\left(\frac{11}{2} - 5\right)\left(\frac{11}{2} - 5\right)}$$

$$= \sqrt{\frac{11}{2}\left(\frac{9}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)}$$

$$= \sqrt{\frac{11}{2} \times \frac{9}{2} \times \frac{1}{2} \times \frac{1}{2}}$$

$$= \sqrt{\frac{99}{8}}$$

$$= \frac{\sqrt{99}}{2\sqrt{2}}$$

$$= \frac{3\sqrt{11}}{4}$$

Therefore, the area of the triangle is $\frac{3\sqrt{11}}{4}$ square units.