How do you use Heron's formula to find the area of a triangle with sides of lengths #6 #, #6 #, and #7 #?

Question

How do you use Heron's formula to find the area of a triangle with sides of lengths #6  #, #6    #, and #7   #?

Aaliyah Anderson · Accepted Answer

#17.06 \ "units"^2#

According to Heron's formula, the area of any triangle is equal to:

#A=sqrt(s(s-a)(s-b)(s-c))#

#s# is the semiperimeter of the triangle, given by #s=(a+b+c)/2#.

#a,b,c# are the sides of the triangle

Here, #s=(6+6+7)/2=19/2=9.5# units.

Thus, the region is:

#A=sqrt(9.5(9.5-6)(9.5-6)(9.5-7))#

#=sqrt(9.5*3.5*3.5*2.5)#

#=sqrt(290.9375)#

#~~17.06#

Claire Akin · Answer

Archimedes' Theorem is a modern form of Heron's Formula. We set #a=c=6, b=7# in

# 16A^2 = 4a^2b^2 - (c^2-a^2-b^2)^2 = 4655#

# A = {7 sqrt(95)}/4 # Archimedes' Theorem is usually superior to Heron's Formula. Given a triangle with sides #a,b,c# and area #A#, we have # 16A^2 = 4a^2b^2 - (c^2-a^2-b^2)^2 # # = (a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) # #= (a+b+c)(-a+b+c)(a-b+c)(a+b-c)# No semiperimeter, no multiple square roots when given coordinates. We're free to choose whichever form is convenient and assign #a,b,c# however we see fit. For an isosceles triangle, #a=c# gives nice cancellation in the first form. Let's set #a=c=6, b=7# in #16 A^2 = 4(6^2)(7^2) - 7^4 = 4655# # A = \sqrt{4744/16} = {7 sqrt(95)}/4 #

Savannah Aguilar · Answer

undefined To use Heron's formula to find the area of a triangle with side lengths $a$, $b$, and $c$, where $s$ is the semi-perimeter, you first calculate $s$ using the formula:

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

Then, you can use Heron's formula to find the area $A$ of the triangle:

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

For the given triangle with side lengths $a = 6$, $b = 6$, and $c = 7$, we first calculate $s$:

$$s = \frac{6 + 6 + 7}{2} = \frac{19}{2}$$

Then, we substitute $s$ and the side lengths into Heron's formula:

$$A = \sqrt{\frac{19}{2} \left(\frac{19}{2} - 6\right) \left(\frac{19}{2} - 6\right) \left(\frac{19}{2} - 7\right)}$$

$$= \sqrt{\frac{19}{2} \times \frac{7}{2} \times \frac{7}{2} \times \frac{5}{2}}$$

$$= \sqrt{\frac{19 \times 7 \times 7 \times 5}{2 \times 2 \times 2 \times 2}}$$

$$= \sqrt{\frac{9315}{16}}$$

Now, you can simplify the square root if possible:

$$A \approx \sqrt{582.1875}$$

Finally, you find the square root:

$$A \approx 24.1406$$

So, the area of the triangle with side lengths 6, 6, and 7 is approximately 24.1406 square units.