Answer:
[tex]\dfrac{24}{7}\;\sf inches[/tex]
Step-by-step explanation:
When a square is inscribed in a right triangle, each vertex of the square touches a side of the triangle.
If the square and the right triangle share a common right angle, it means that the right angle of the triangle is also one of the corners of the square. So, two adjacent sides of the square align with the legs of the triangle, and one vertex of the square touches the hypotenuse of the right triangle.
The large right triangle is now composed of the square and two smaller right triangles. So the sum of the areas of these three shapes is equal to the area of the large triangle.
Let x be the side length of the square.
Given that the legs of the large right triangle are 6 in and 8 in, the dimensions of the two smaller right triangles are:
Triangle 1:
- Base = x inches
- Height = (6 - x) inches
Triangle 2:
- Base = (8 - x) inches
- Height = x inches
Now, we can create an equation that equates the sum of the areas of the 3 shapes with the area of the large triangle, keeping in mind that the area of a square is the square of its side length, and the area of a triangle is half the product of its base and height:
[tex]x^2+\left(\dfrac{1}{2} \cdot x \cdot (6-x)\right)+\left(\dfrac{1}{2}\cdot (8-x) \cdot x\right)=\dfrac{1}{2}\cdot 8 \cdot 6[/tex]
To find the length of the side of the square, simply solve the equation for x:
[tex]\begin{aligned}x^2+\dfrac{1}{2} (6x-x^2)+\dfrac{1}{2}(8x-x^2)&=24\\\\2x^2+6x-x^2+8x-x^2&=48\\\\14x&=48\\\\x&=\dfrac{48}{14}\\\\x&=\dfrac{24}{7}\end{aligned}[/tex]
Therefore, the side length of the square is:
[tex]\Large\boxed{\boxed{\dfrac{24}{7}\;\sf inches}}[/tex]
