Find x and y, given that KL and line MN are parallel. Show all work

Mathematics

2 answers:

iren [92.7K]3 years ago

5 0

X equals 9.5 y equals

Setler [38]3 years ago

4 0

$\dfrac{|MN|}{|NJ|}=\dfrac{|KL|}{|LJ|}\\\\\dfrac{22}{12+8}=\dfrac{y}{8}\\\\\dfrac{22}{20}=\dfrac{y}{8}\ \ \ |cross\ multiply\\\\20y=22\cdot8\\\\20y=176\ \ \ |:20\\\\\boxed{y=8.8}$

$\dfrac{|MK|}{|KJ|}=\dfrac{|NL|}{|LJ|}\\\\\dfrac{x-2}{5}=\dfrac{12}{8}\ \ \ \ |cross\ multiply\\\\8(x-2)=5\cdot12\\\\8x-16=60\ \ \ |+16\\\\8x=76\ \ \ |:8\\\\\boxed{x=9.5}$

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Answer:

Part a) The radii are segments AC and AD and the tangents are the segments CE and DE

Part b) $DE=4\sqrt{10}\ cm$

Step-by-step explanation:

Part a)

we know that

A <u>radius</u> is a line from any point on the circumference to the center of the circle

A <u>tangent</u> to a circle is a straight line which touches the circle at only one point. The tangent to a circle is perpendicular to the radius at the point of tangency.

In this problem

The radii are the segments AC and AD

The tangents are the segments CE and DE

Part b)

we know that

radius AC is perpendicular to the tangent CE

radius AD is perpendicular to the tangent DE

CE=DE

Triangle ACE is congruent with triangle ADE

Applying the Pythagoras Theorem

$AE^{2}=AC^{2}+CE^{2}$