The distances when put together create a right triangle.  

The distance between them will be the hypotenuse or the diagonal side.  

You use Pythagorean Theorem or \(\displaystyle a^{2}+b^{2}=c^{2}\) to find the length.  

So you plug \(\displaystyle 30\) and \(\displaystyle 40\) for \(\displaystyle a\) and \(\displaystyle b\) which gives you,

 \(\displaystyle 900+1600=c^{2}\) or \(\displaystyle 2500=c^{2}\).  

Then you find the square root of each side and that gives you your answer of \(\displaystyle c=50\).

The triangle in this problem is a variation of the 3, 4, 5 right triangle. However, it is 4 times bigger. We know this because \(\displaystyle 5\cdot4=20\) (the length of the hypotenuse) and \(\displaystyle 3\cdot4=12\) (the length of one of the legs). 

Therefore, the length of the other leg will be equal to:

\(\displaystyle 4\cdot4=16\)

For a given right triangle with base \(\displaystyle b\) and height \(\displaystyle h\), the area \(\displaystyle A\) can be defined by the formula \(\displaystyle A=\frac{1}{2}bh\). If one leg of the right triangle is taken as the base, then the other leg is the height.  

Therefore, to find the height \(\displaystyle h\), we restructure the formula for the area \(\displaystyle A\) as follows:

\(\displaystyle A=\frac{1}{2}bh\)

\(\displaystyle 2A=bh\)

\(\displaystyle \frac{2A}{b}=h\)

Plugging in our values for \(\displaystyle A\) and \(\displaystyle b\):

\(\displaystyle \frac{2(45cm^{2})}{9cm}=h\)

\(\displaystyle \frac{90cm^{2}}{9cm}=h\)

\(\displaystyle 10cm=h\)

For a given right triangle with base \(\displaystyle b\) and height \(\displaystyle h\), the area \(\displaystyle A\) can be defined by the formula \(\displaystyle A=\frac{1}{2}bh\). If one leg of the right triangle is taken as the base, then the other leg is the height.  

Therefore, to find the height \(\displaystyle h\), we restructure the formula for the area \(\displaystyle A\) as follows:

\(\displaystyle A=\frac{1}{2}bh\)

\(\displaystyle 2A=bh\)

\(\displaystyle \frac{2A}{b}=h\)

Plugging in our values for \(\displaystyle A\) and \(\displaystyle b\):

\(\displaystyle \frac{2(100cm^{2})}{10cm}=h\)

\(\displaystyle \frac{200cm^{2}}{10cm}=h\)

\(\displaystyle 20cm=h\)

For a given right triangle with base \(\displaystyle b\) and height \(\displaystyle h\), the area \(\displaystyle A\) can be defined by the formula \(\displaystyle A=\frac{1}{2}bh\). If one leg of the right triangle is taken as the base, then the other leg is the height. 

However, we have not been given a base or leg length for the right triangle, only the length of the hypotenuse and the area. We therefore do not have enough information to solve for the height \(\displaystyle h\). 

To find the height, plug what is given in the question into the formula used to find the area of a triangle.

\(\displaystyle \text{Area}=\frac{base\times height}{2}\)

Use the information given in the question:

\(\displaystyle \text{Base}=25\)

\(\displaystyle \text{Area}=100\)

\(\displaystyle 100=\frac{25\times height}{2}\)

Now, solve for the height.

\(\displaystyle 25\times height=200\)

\(\displaystyle height=8\)

To find the height, plug what is given in the question into the formula used to find the area of a triangle.

\(\displaystyle \text{Area}=\frac{base\times height}{2}\)

Use the information given in the question:

\(\displaystyle \text{Base}=15\)

\(\displaystyle \text{Area}=45\)

\(\displaystyle 45=\frac{15\times height}{2}\)

Now, solve for the height.

\(\displaystyle 15\times height=90\)

\(\displaystyle height=6\)

To find the height, plug what is given in the question into the formula used to find the area of a triangle.

\(\displaystyle \text{Area}=\frac{base\times height}{2}\)

Use the information given in the question:

\(\displaystyle \text{Base}=10\)

\(\displaystyle \text{Area}=50\)

\(\displaystyle 50=\frac{10\times height}{2}\)

Now, solve for the height.

\(\displaystyle 10\times height=100\)

\(\displaystyle height=10\)

The area of a right triangle is half the product of its legs. In each case, we know the length of one leg and the hypotenuse, so we need to apply the Pythagorean Theorem to find the second leg, then take half the product of the legs:

Right Triangle A:

The length of the second leg is

\(\displaystyle \sqrt{25^{2} -24 ^{2}} = \sqrt{625 - 576} = \sqrt{49} = 7\) inches.

The area is 

\(\displaystyle \frac{1}{2} \times 7 \times 24 = 84\) square inches.

Right Triangle B:

The length of the second leg is

\(\displaystyle \sqrt{15^{2} -9 ^{2}} = \sqrt{225 - 81} = \sqrt{144} = 12\) inches.

The area is 

\(\displaystyle \frac{1}{2} \times 9 \times 12 = 54\) square inches.

The sum of the areas is \(\displaystyle 84 + 54 = 138\) square inches.

The area of a rectangle is the product of its length and its height. Therefore, the height is the quotient of the area and the length, which, for Rectangle C, is \(\displaystyle 138 \div 16 = 8 \frac{5}{8}\) inches.

The area of a right triangle is half the product of its legs. The area of Right Triangle A is equal to \(\displaystyle \frac{1}{2} \times 10 \times 14 = 70\) square inches; that of Right Triangle B is equal to \(\displaystyle \frac{1}{2} \times 20 \times 13 = 130\) square inches. The sum of the areas is \(\displaystyle 70 + 130 = 200\) square inches, which is the area of Rectangle C.

The area of a rectangle is the product of its length and its height. Therefore, the height is the quotient of the area and the length, which, for Rectangle C, is \(\displaystyle 200 \div 30 = 6 \frac{2}{3}\) inches.