# How do you use the Pythagorean Theorem to find the missing side of the right triangle with the given measures given c is the hypotenuse and we have b=3x,c=7x?

Mar 25, 2016

$a = 2 x \sqrt{10}$

#### Explanation:

The Pythagorean Theorem states that

${a}^{2} + {b}^{2} = {c}^{2}$

in a triangle with legs $a , b$ and hypotenuse $c$, as you've already described in the problem.

With $b = 3 x$ and $c = 7 x$, we have the relation:

${a}^{2} + {\left(3 x\right)}^{2} = {\left(7 x\right)}^{2}$

Now, recall that when we have something like ${\left(3 x\right)}^{2}$, we have to square both the $3$ and the $x$:

${\left(3 x\right)}^{2} = {3}^{2} \cdot {x}^{2} = 9 {x}^{2}$

Similarly, for ${\left(7 x\right)}^{2}$:

${\left(7 x\right)}^{2} = {7}^{2} \cdot {x}^{2} = 49 {x}^{2}$

Substituting these back in to the Pythagorean Theorem equation, we see that

${a}^{2} + 9 {x}^{2} = 49 {x}^{2}$

Subtract $9 {x}^{2}$ from both sides of the equation.

${a}^{2} = 40 {x}^{2}$

Take the square root of both sides.

$a = \sqrt{40 {x}^{2}}$

We can rewrite $\sqrt{40 {x}^{2}}$ as a product of mostly squared terms in order to simplify. For example, it's important to note that $40 = 4 \times 10$.

$a = \sqrt{4} \cdot \sqrt{{x}^{2}} \cdot \sqrt{10}$

$a = 2 x \sqrt{10}$

Mar 25, 2016

$A = 2 x \sqrt{10}$

#### Explanation: Using the principle of proportionality disregard the $x ' s$ for now.
Think if it as working on a triangle that has been reduced in scale but is of the same ratio.

By Pythagoras ${A}^{2} + {B}^{2} = {C}^{2}$

So $A \to \sqrt{{7}^{2} - {3}^{2}} \text{ "=" } \sqrt{49 - 9}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Technically we could write $A = x \sqrt{{7}^{2} - {3}^{3}}$
It is simpler just to leave it out for now but incorporate it at the end.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$A \to \sqrt{{2}^{2} \times 10}$

$A \to 2 \sqrt{10}$

Scaling back up we have

$A = 2 x \sqrt{10}$