An isosceles triangle has a 10 inches base and two 13 inches sides .What other value can the base have and still yields a triangle with the same areas ?

Options

A. 18 inches
B. 19 inches
C. 24 inches
D. 27 inches

Answer

To solve the problem of finding another base for the isosceles triangle that yields the same area, we can follow these steps:

### Step 1: Calculate the height of the original triangle
We have an isosceles triangle with a base of 10 inches and two equal sides of 13 inches. To find the height, we can drop a perpendicular from the apex to the midpoint of the base. This creates two right triangles.

- The base of each right triangle is \( \frac{10}{2} = 5 \) inches.
- The hypotenuse (the equal sides) is 13 inches.

Using the Pythagorean theorem:
\[ h^2 + 5^2 = 13^2 \]
\[ h^2 + 25 = 169 \]
\[ h^2 = 169 - 25 = 144 \]
\[ h = \sqrt{144} = 12 \text{ inches} \]

### Step 2: Calculate the area of the original triangle
The area \( A \) of a triangle is given by:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
For our triangle:
\[ A = \frac{1}{2} \times 10 \times 12 = 60 \text{ square inches} \]

### Step 3: Set up the equation for the new triangle
Let the new base be \( 2y \) (since it is still an isosceles triangle with equal sides of 13 inches). The height remains the same at \( h \).

Using the area formula again:
\[ A = \frac{1}{2} \times 2y \times h \]
Setting this equal to the area of the original triangle:
\[ \frac{1}{2} \times 2y \times 12 = 60 \]
This simplifies to:
\[ 12y = 60 \]
\[ y = 5 \]

### Step 4: Calculate the new base
The new base \( b \) is:
\[ b = 2y = 2 \times 5 = 10 \text{ inches} \]

### Step 5: Find the other possible base
To find another base that yields the same area, we can use the Pythagorean theorem again for the new triangle:
\[ h^2 + y^2 = 13^2 \]
Substituting \( h = 12 \):
\[ 12^2 + y^2 = 169 \]
\[ 144 + y^2 = 169 \]
\[ y^2 = 169 - 144 = 25 \]
\[ y = 5 \]

Now, we can find the new base:
\[ \text{New base} = 2y = 2 \times 5 = 10 \text{ inches} \]

However, we also need to consider the possibility of a larger base that still maintains the same area. We can set up the equation:
\[ \frac{1}{2} \times 2y \times h = 60 \]
And solve for \( y \) again, leading to:
\[ y^4 - 169y^2 + 3600 = 0 \]
This can be factored or solved using the quadratic formula, yielding:
\[ y = 12 \text{ or } 5 \]
Thus, the new base can be:
\[ b = 2y = 2 \times 12 = 24 \text{ inches} \]

### Final Answer
The other possible base for the triangle that yields the same area is **24 inches**.