Respuesta :

calculista

we know that

In the figure triangle ABC and triangle AQR are similar triangles

so

by proportion

[tex] \frac{QR}{AQ} =\frac{BC}{AB} \\ \\ \frac{2p+3}{4p} =\frac{6p-4}{8p} \\ \\ 2*(2p+3)=(6p-4)\\ \\ 4p+6=6p-4\\ \\ 6p-4p=6+4\\ \\ 2p=10\\ \\ p=5 [/tex]

[tex] AQ=4p\\ AQ=4*5\\ AQ=20\ units [/tex]

therefore

the answer is

AQ is [tex] 20\ units [/tex]

The value of AQ can be found using the mid-segment theorem, therefore, the value of AQ in ΔAQR is equal to 20 units.

Given to us

Points Q and R are midpoints of the sides of ΔABC.

What is the Mid-segment theorem?

The line segment joining the two midpoints of two adjacent sides of a triangle is half the length of the third side and is parallel to the third side.

What is the value of P?

We know about the mid-segment theorem, therefore,

[tex]BC = 2RQ\\\\(6p-4) = 2(2p+3)\\\\6p-4=4p+6\\\\6p-4p=4+6\\\\2p=10\\\\p=5[/tex]

Thus, the value of p is 5.

What is the value of AQ?

The value of AQ can be found using by substituting the value of p,

AQ = 4p

     = 4(5)

      = 20 units

Hence, the value of AQ in ΔAQR is equal to 20 units.

Learn more about Midsegment Theorem:

https://brainly.com/question/7826311

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