Question

question 10
using given a triangle with sides a = 30ft, b = 30ft, c = 23ft, find its area.
assume ∠a is opposite side a, ∠b is opposite side b, and ∠c is opposite side c.

• round your solution to the nearest tenth place.
• a solution write - up is required for this problem.

answer:
area =

Explanation:

Step1: Identify the triangle type

We have a triangle with sides \( a = 30\,\text{ft} \), \( b = 30\,\text{ft} \), \( c = 23\,\text{ft} \). This is an isosceles triangle with \( a = b \). We can use Heron's formula to find the area. First, find the semi - perimeter \( s=\frac{a + b + c}{2}\).
\( s=\frac{30 + 30+23}{2}=\frac{83}{2}=41.5\,\text{ft} \)

Step2: Apply Heron's formula

Heron's formula for the area \( A \) of a triangle with sides \( a,b,c \) and semi - perimeter \( s \) is \( A=\sqrt{s(s - a)(s - b)(s - c)} \)
Substitute \( s = 41.5 \), \( a = 30 \), \( b = 30 \), \( c = 23 \) into the formula:
\( A=\sqrt{41.5\times(41.5 - 30)\times(41.5 - 30)\times(41.5 - 23)} \)
\( 41.5-30 = 11.5 \), \( 41.5 - 23=18.5 \)
\( A=\sqrt{41.5\times11.5\times11.5\times18.5} \)
First, calculate \( 41.5\times18.5=(40 + 1.5)\times(40 - 1.5)=40^{2}-1.5^{2}=1600 - 2.25 = 1597.75 \)
and \( 11.5\times11.5 = 132.25 \)
Then \( A=\sqrt{1597.75\times132.25} \)
\( 1597.75\times132.25 = 1597.75\times(132+0.25)=1597.75\times132+1597.75\times0.25 \)
\( 1597.75\times132=(1600 - 2.25)\times132 = 1600\times132-2.25\times132=211200 - 297 = 210903 \)
\( 1597.75\times0.25=\frac{1597.75}{4}=399.4375 \)
\( 1597.75\times132.25=210903 + 399.4375=211302.4375 \)
\( A=\sqrt{211302.4375}\approx459.7\,\text{ft}^2 \) (rounded to the nearest tenth)

Answer:

The area of the triangle is approximately \( 459.7\,\text{square feet} \)