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Calibri20Bactice...In the diagram below, EF is parallel to BC. If BC is twice the length ofED, BD = 8, and EF = 4, find the length of ED. Figures are notnecessarily drawn to scale. State your answer in simplest radical form, ifnecessary.PracticePracticePractice- Optio...EacticeBctice-18)

Calibri20BacticeIn The Diagram Below EF Is Parallel To BC If BC Is Twice The Length OfED BD 8 And EF 4 Find The Length Of ED Figures Are Notnecessarily Drawn To class=

Sagot :

The two triangles are simillar due to AAA theorem. This means that equivalent sides of the triangles are proportional, which means that we can create a relationship between them.

We were given the measurements for sides BD and EF. We were also told that BD is twice the length of ED, therefore we have:

[tex]\begin{gathered} \frac{BC}{EF}=\frac{BD}{ED} \\ \frac{2\cdot ED}{4}=\frac{8}{ED} \\ ED^2=2\cdot8 \\ ED^2=16 \\ ED=\sqrt[]{16} \\ ED=4 \end{gathered}[/tex]

The length of ED is equal to 4.

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