Answer :
Let's solve the problem step-by-step.
### Step-by-Step Solution
We are given:
- Area of the isosceles triangle, [tex]\( A = 240 \text{ cm}^2 \)[/tex]
- Length of each of the equal sides, [tex]\( a = 26 \text{ cm} \)[/tex]
We need to find:
1. The length of the third side (the base) [tex]\( b \)[/tex].
2. The perimeter of the triangle.
#### Step 1: Calculate the length of the base ([tex]\( b \)[/tex])
The area [tex]\( A \)[/tex] of an isosceles triangle can be given by:
[tex]\[ A = \frac{1}{2} \times \text{base} \times \text{height} \][/tex]
To proceed, we need to find the height [tex]\( h \)[/tex]. The height can be calculated using the Pythagorean theorem in the right triangle formed by the height and half of the base, since the height bisects the base in an isosceles triangle.
From the Pythagorean theorem:
[tex]\[ h^2 + \left(\frac{b}{2}\right)^2 = a^2 \][/tex]
Rearranging it to solve for [tex]\( h \)[/tex]:
[tex]\[ h^2 = a^2 - \left(\frac{b}{2}\right)^2 \][/tex]
[tex]\[ h = \sqrt{a^2 - \left(\frac{b}{2}\right)^2} \][/tex]
Now, substitute [tex]\( h \)[/tex] back into the area formula:
[tex]\[ 240 = \frac{1}{2} \times b \times \sqrt{a^2 - \left(\frac{b}{2}\right)^2} \][/tex]
Solving for [tex]\( b \)[/tex]:
[tex]\[ 240 = \frac{1}{2} \times b \times \sqrt{26^2 - \left(\frac{b}{2}\right)^2} \][/tex]
[tex]\[ 480 = b \times \sqrt{676 - \frac{b^2}{4}} \][/tex]
[tex]\[ 480^2 = b^2 \times \left(676 - \frac{b^2}{4}\right) \][/tex]
[tex]\[ 230400 = b^2 \times \left(676 - \frac{b^2}{4}\right) \][/tex]
Let [tex]\( b^2 = x \)[/tex]:
[tex]\[ 230400 = x \times \left(676 - \frac{x}{4}\right) \][/tex]
[tex]\[ 230400 = 676x - \frac{x^2}{4} \][/tex]
[tex]\[ 4 \times 230400 = 4 \times 676x - x^2 \][/tex]
[tex]\[ 921600 = 2704x - x^2 \][/tex]
[tex]\[ x^2 - 2704x + 921600 = 0 \][/tex]
Solving this quadratic equation using the quadratic formula:
[tex]\[ x = \frac{2704 \pm \sqrt{2704^2 - 4 \times 921600}}{2} \][/tex]
[tex]\[ x = \frac{2704 \pm \sqrt{7311616 - 3686400}}{2} \][/tex]
[tex]\[ x = \frac{2704 \pm \sqrt{3625216}}{2} \][/tex]
[tex]\[ x = \frac{2704 \pm 1904}{2} \][/tex]
Thus, we have two potential solutions for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{2704 + 1904}{2} = 2304 \][/tex]
[tex]\[ x = \frac{2704 - 1904}{2} = 400 \][/tex]
We discard the negative value because it doesn't make physical sense in this context. Therefore:
[tex]\[ b^2 = 400 \][/tex]
[tex]\[ b = \sqrt{400} = 20 \text{ cm} \][/tex]
So, the length of the third side (the base) is [tex]\( 20 \text{ cm} \)[/tex].
#### Step 2: Calculate the perimeter of the triangle
The perimeter [tex]\( P \)[/tex] of an isosceles triangle is given by:
[tex]\[ P = 2a + b \][/tex]
[tex]\[ P = 2 \times 26 + 20 \][/tex]
[tex]\[ P = 52 + 20 \][/tex]
[tex]\[ P = 72 \text{ cm} \][/tex]
### Summary
1. The length of the third side (the base) [tex]\( b \)[/tex] is [tex]\( 20 \text{ cm} \)[/tex].
2. The perimeter of the triangle is [tex]\( 72 \text{ cm} \)[/tex].
### Step-by-Step Solution
We are given:
- Area of the isosceles triangle, [tex]\( A = 240 \text{ cm}^2 \)[/tex]
- Length of each of the equal sides, [tex]\( a = 26 \text{ cm} \)[/tex]
We need to find:
1. The length of the third side (the base) [tex]\( b \)[/tex].
2. The perimeter of the triangle.
#### Step 1: Calculate the length of the base ([tex]\( b \)[/tex])
The area [tex]\( A \)[/tex] of an isosceles triangle can be given by:
[tex]\[ A = \frac{1}{2} \times \text{base} \times \text{height} \][/tex]
To proceed, we need to find the height [tex]\( h \)[/tex]. The height can be calculated using the Pythagorean theorem in the right triangle formed by the height and half of the base, since the height bisects the base in an isosceles triangle.
From the Pythagorean theorem:
[tex]\[ h^2 + \left(\frac{b}{2}\right)^2 = a^2 \][/tex]
Rearranging it to solve for [tex]\( h \)[/tex]:
[tex]\[ h^2 = a^2 - \left(\frac{b}{2}\right)^2 \][/tex]
[tex]\[ h = \sqrt{a^2 - \left(\frac{b}{2}\right)^2} \][/tex]
Now, substitute [tex]\( h \)[/tex] back into the area formula:
[tex]\[ 240 = \frac{1}{2} \times b \times \sqrt{a^2 - \left(\frac{b}{2}\right)^2} \][/tex]
Solving for [tex]\( b \)[/tex]:
[tex]\[ 240 = \frac{1}{2} \times b \times \sqrt{26^2 - \left(\frac{b}{2}\right)^2} \][/tex]
[tex]\[ 480 = b \times \sqrt{676 - \frac{b^2}{4}} \][/tex]
[tex]\[ 480^2 = b^2 \times \left(676 - \frac{b^2}{4}\right) \][/tex]
[tex]\[ 230400 = b^2 \times \left(676 - \frac{b^2}{4}\right) \][/tex]
Let [tex]\( b^2 = x \)[/tex]:
[tex]\[ 230400 = x \times \left(676 - \frac{x}{4}\right) \][/tex]
[tex]\[ 230400 = 676x - \frac{x^2}{4} \][/tex]
[tex]\[ 4 \times 230400 = 4 \times 676x - x^2 \][/tex]
[tex]\[ 921600 = 2704x - x^2 \][/tex]
[tex]\[ x^2 - 2704x + 921600 = 0 \][/tex]
Solving this quadratic equation using the quadratic formula:
[tex]\[ x = \frac{2704 \pm \sqrt{2704^2 - 4 \times 921600}}{2} \][/tex]
[tex]\[ x = \frac{2704 \pm \sqrt{7311616 - 3686400}}{2} \][/tex]
[tex]\[ x = \frac{2704 \pm \sqrt{3625216}}{2} \][/tex]
[tex]\[ x = \frac{2704 \pm 1904}{2} \][/tex]
Thus, we have two potential solutions for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{2704 + 1904}{2} = 2304 \][/tex]
[tex]\[ x = \frac{2704 - 1904}{2} = 400 \][/tex]
We discard the negative value because it doesn't make physical sense in this context. Therefore:
[tex]\[ b^2 = 400 \][/tex]
[tex]\[ b = \sqrt{400} = 20 \text{ cm} \][/tex]
So, the length of the third side (the base) is [tex]\( 20 \text{ cm} \)[/tex].
#### Step 2: Calculate the perimeter of the triangle
The perimeter [tex]\( P \)[/tex] of an isosceles triangle is given by:
[tex]\[ P = 2a + b \][/tex]
[tex]\[ P = 2 \times 26 + 20 \][/tex]
[tex]\[ P = 52 + 20 \][/tex]
[tex]\[ P = 72 \text{ cm} \][/tex]
### Summary
1. The length of the third side (the base) [tex]\( b \)[/tex] is [tex]\( 20 \text{ cm} \)[/tex].
2. The perimeter of the triangle is [tex]\( 72 \text{ cm} \)[/tex].
