Let the width (base) of the triangle be denoted as \( b \) and the height as \( h \). According to the problem, the height is half of the width, which can be expressed as:
\[ h = \frac{1}{2}b \]
The area \( A \) of a triangle is given by the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Substituting the values for the area and height, we have:
\[ 25 = \frac{1}{2} \times b \times \left(\frac{1}{2}b\right) \]
Simplifying the right side:
\[ 25 = \frac{1}{2} \times b \times \frac{1}{2}b = \frac{1}{4} b^2 \]
Now, multiply both sides by 4 to eliminate the fraction:
\[ 100 = b^2 \]
Taking the square root of both sides gives:
\[ b = 10 , \text{cm} \]
Now, substituting \( b \) back into the equation for height:
\[ h = \frac{1}{2}b = \frac{1}{2} \times 10 = 5 , \text{cm} \]
Thus, the measures of the base and height of the triangle are:
\[ \text{Base} = 10 , \text{cm} \] \[ \text{Height} = 5 , \text{cm} \]