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IRISSAK [1]
3 years ago
10

Points A, B, C, and D are collinear and positioned in that order. Solve for x then find the length of: AB, BC, and CD, if AB=x+5

, BC=2x+10, CD=4x, and AD=78
Mathematics
1 answer:
fenix001 [56]3 years ago
4 0

Step-by-step explanation:

we can solve this by Adding all points so you said that all are collinear points which means all points are in one line so this is solved as

= X+5+2X+10+4X=78

=7X+15=78. ....collecting like terms

=7x=78-15

=7x=63. divide both sides with 7

=x=9

So now All sides are solved

●AB=x+5

AB=9+5=14

●BC=2X+10=2(9)+10

BC=28

●CD=4X

CD=4(9)

CD=36

so AB+BC+CD=AD

14+28+36=78

True!!

Thanks in Advance

Nahom Wondale

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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

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Triangle ACE is congruent with triangle ADE

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

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CE^{2}=14^{2}-6^{2}

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CE=4\sqrt{10}\ cm

remember that

CE=DE

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DE=4\sqrt{10}\ cm

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