We want to find the lengths of the sides of $△ C D E .$ We can see from the graph that the length of $\overline{C D}$ is $3$ units and the length of $\overline{D E}$ is $4$ units. We also know that angle $C D E$ is a right angle.

Since this is a right triangle, let's apply the Pythagorean Theorem to find the side length of

\overline{E C} .

\overline{C D}^{2} + \overline{D E}^{2} = \overline{E C}^{2}

We will substitute the values into the formula and solve for

\overline{E C} .

\overline{C D}^{2} + \overline{D E}^{2} = \overline{E C}^{2}

SubstituteII

$\overline{C D} = 3$ , $\overline{D E} = 4$

3^{2} + 4^{2} = \overline{E C}^{2}

▼

Simplify left-hand side

CalcPow

Calculate power

9 + 16 = \overline{E C}^{2}

AddTerms

Add terms

25 = \overline{E C}^{2}

SplitIntoFactors

Split into factors

5 \cdot 5 = \overline{E C}^{2}

WritePow

Write as a power

5^{2} = \overline{E C}^{2}

SqrtEqn

$LHS = RHS$

5^{2} = \overline{E C}^{2}

$a^{2} = \pm a$

\pm 5 = \overline{E C}

RearrangeEqn

Rearrange equation

\overline{E C} = \pm 5

Since

\overline{E C}

is the side length of the triangle and lengths cannot be negative, we know that the length of the side must be positive. We can write the side lengths of

△ C D E

as follows.

\overline{C D} = 3 units \overline{D E} = 4 units \overline{E C} = 5 units

Exercise