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Given that point P is equidistant from the vertices of TUV , find VU. A. VU = 4 B. VU = 5 C. VU = 8 D. VU = 10

Given that point P is equidistant from the vertices of TUV , find VU. A. VU = 4 B-example-1
User Morteza R
by
9.5k points

2 Answers

1 vote
1) Find x

3x - 1 = x + 3
x=2

2) Plug in for x

3(2)-1=5

2+3=5

3) Add up and that's your answer!
User Nico Mee
by
8.2k points
7 votes

Answer:

The length of VU is 10 unit.

Explanation:

Consider the provided information.

The point P is equidistant from the vertices of ΔTUV, thus the segment PU equals to the segment PV, then these two segments are congruent:

PU=PV

Then ΔUVP is an isosceles triangle, because PU=PV.

Now, observe the provided figure, PR divides the ΔTUV in two triangle (ΔPUR and ΔPRV)

Where, ∠P=∠P, PR=PR and PV=PU. Thus ΔPUR congruent to ΔPRV and the legs UR and RV must be congruent:

UR = RV

3x - 1 = x + 3

Subtracting x from both side and add 1 to both sides of the above equation:

3x - 1 - x + 1 = x + 3 - x + 1

2x = 4

Divide both sides of the equation by 2:

x=2

VU = UR + RV

Where the length of UR is x + 3 and length of RV is 3x - 1.

Therefore, the length of UR is, 2+3=5

The length of RV is, 3(2)-1=5

Thus, the length of VU is 5+5=10.

Hence, the length of VU is 10 unit.

User Feupeu
by
8.7k points
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