Sure, I’d be happy to help you with this step-by-step solution.
We are given the following values for the isosceles triangle:
- The height [tex]\( h \)[/tex] is 4 cm
- The equal side lengths [tex]\( s \)[/tex] are 7 cm
### Step 1: Find the length of half the base
To find the length of the base, we can use the Pythagorean theorem. Since the height divides the isosceles triangle into two right triangles, we can consider one of those right triangles. In that right triangle:
- The height [tex]\( h \)[/tex] is 4 cm
- The hypotenuse (one of the equal sides) [tex]\( s \)[/tex] is 7 cm
- Let [tex]\( b/2 \)[/tex] be half the base.
We can use the Pythagorean theorem:
[tex]\[ s^2 = (b/2)^2 + h^2 \][/tex]
[tex]\[ 7^2 = (b/2)^2 + 4^2 \][/tex]
[tex]\[ 49 = (b/2)^2 + 16 \][/tex]
[tex]\[ (b/2)^2 = 49 - 16 \][/tex]
[tex]\[ (b/2)^2 = 33 \][/tex]
[tex]\[ b/2 = \sqrt{33} \][/tex]
Simplifying, we obtain:
[tex]\[ b/2 \approx 5.744562646538029 \, \text{cm} \][/tex]
### Step 2: Find the full length of the base
Since [tex]\( b/2 \)[/tex] is half of the base, the full length of the base [tex]\( b \)[/tex] is:
[tex]\[ b = 2 \times 5.744562646538029 \][/tex]
[tex]\[ b \approx 11.489125293076057 \, \text{cm} \][/tex]
### Step 3: Calculate the area of the triangle
The area [tex]\( A \)[/tex] of a triangle is given by:
[tex]\[ A = \frac{1}{2} \times \text{base} \times \text{height} \][/tex]
Substituting in the values:
[tex]\[ A = \frac{1}{2} \times 11.489125293076057 \times 4 \][/tex]
[tex]\[ A \approx 22.978250586152114 \, \text{cm}^2 \][/tex]
### Step 4: Calculate the perimeter of the triangle
The perimeter [tex]\( P \)[/tex] is the sum of all the sides of the triangle:
[tex]\[ P = 2 \times \text{side length} + \text{base} \][/tex]
[tex]\[ P = 2 \times 7 + 11.489125293076057 \][/tex]
[tex]\[ P \approx 25.489125293076057 \, \text{cm} \][/tex]
To summarize:
- Half the base is approximately [tex]\( 5.74 \, \text{cm} \)[/tex]
- The full base is approximately [tex]\( 11.49 \, \text{cm} \)[/tex]
- The area is approximately [tex]\( 22.98 \, \text{cm}^2 \)[/tex]
- The perimeter is approximately [tex]\( 25.49 \, \text{cm} \)[/tex]
