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3 years ago

Help with 21 would be much appreciated.

Mathematics

1 answer:

umka2103 [35]3 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:

$\dfrac{6}{y}=\dfrac{y}{8}$ cross multiply

$y^2=(6)(8)$

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

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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