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Find the area of a triangle whose two sides are 10 inches and 18 inches long, and has a perimeter of 42 inches.

A. 27.5 in[tex]$^2$[/tex]
B. [tex]$69.65$[/tex] in[tex]$^2$[/tex]
C. [tex]$155.74$[/tex] in[tex]$^2$[/tex]
D. [tex]$30.4$[/tex] in[tex]$^2$[/tex]

Sagot :

To find the area of the triangle, we need to follow these steps:

1. Identify the given information:
- Side \( a = 10 \) inches
- Side \( b = 18 \) inches
- Perimeter \( P = 42 \) inches

2. Determine the length of the third side \( c \):
- The perimeter formula for a triangle is \( P = a + b + c \).
- Rearranging to find \( c \), we get \( c = P - (a + b) \).
- So, \( c = 42 - (10 + 18) = 42 - 28 = 14 \) inches.

3. Calculate the semi-perimeter \( s \):
- The semi-perimeter \( s \) is half of the perimeter.
- \( s = \frac{P}{2} = \frac{42}{2} = 21 \) inches.

4. Use Heron's formula to find the area \( A \):
- Heron's formula states \( A = \sqrt{s(s-a)(s-b)(s-c)} \).
- Substituting the values, we have:
[tex]\[ A = \sqrt{21(21-10)(21-18)(21-14)} = \sqrt{21 \times 11 \times 3 \times 7} \][/tex]

5. Calculate the value inside the square root:
- Calculate \( 21 \times 11 = 231 \)
- Then \( 231 \times 3 = 693 \)
- Finally \( 693 \times 7 = 4851 \)

6. Take the square root of the product:
- \( A = \sqrt{4851} \approx 69.65 \)

Therefore, the area of the triangle is approximately \( 69.65 \) square inches.

The correct answer is:

B. [tex]\( 69.65 \text{ in}^2 \)[/tex]