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Segments BF, CD, and AE are medians for triangle ABC. Point K is the intersection of all three medians. The measure of segment FK is 3.40 cm and segment EK = 1.63 cm. Find the sum of the lengths of segments AK and BK.

Respuesta :

The point of concurrency of the three median is the centroid of the triangle ABC.

  • The sum of the lengths of AK and BK is 10.06

Reasons:

The given parameters are;

The medians of the triangle ABC = Segments BF, CD, and AE

The  length of segment FK = 3.40

Length of segment EK = 1.63

Required:

The sum of the lengths of segment AK and BK

Solution:

The ratios of the lengths of AK to KE = 2:1

Therefore, we have;

[tex]\dfrac{AK}{KE} = \dfrac{2}{1} = \mathbf{\dfrac{AK}{1.63}}[/tex]

1.63 × 2 = 1 × AK

AK = 3.26

Similarly, we have;

[tex]\dfrac{BK}{KF} = \dfrac{2}{1} = \mathbf{\dfrac{BK}{3.40}}[/tex]

3.40 × 2 = BK

6.80 = BK

Which gives;

AK + BK = 3.26 + 6.80 = 10.06

The sum of the lengths of AK and BK = 10.06

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https://brainly.com/question/7856454

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