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Nastasia [14]
2 years ago
10

Help with 21 would be much appreciated.

Mathematics
1 answer:
umka2103 [35]2 years ago
3 0

Answer:

\large\boxed{x=2\sqrt{21}\ and\ y=4\sqrt3}

Step-by-step explanation:

Look at the picture.

ΔADC and ΔCDB are similar. Therefore the sides are in proportion:

\dfrac{AD}{CD}=\dfrac{CD}{DB}

We have

AD=14-8=6\\CD=y\\DB=8

Substitute:

\dfrac{6}{y}=\dfrac{y}{8}         <em>cross multiply</em>

y^2=(6)(8)

y^2=48\to y=\sqrt{48}\\\\y=\sqrt{16\cdot3}\\\\y=\sqrt{16}\ cdot\sqrt3\\\\\boxed{y=4\sqrt3}

For x use the Pythagorean theorem:

x^2=6^2+(4\sqrt3)^2\\\\x^2=36+48\\\\x^2=84\to x=\sqrt{84}\\\\x=\sqrt{4\cdot21}\\\\x=\sqrt4\cdot\sqrt{21}\\\\\boxed{x=2\sqrt{21}}

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