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In ΔPQS above, if PQ =3 and PS = 4, then PR =?

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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =?  [#permalink]

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New post 22 Dec 2018, 04:43
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Hey there,

I am equating areas as it is the exact same triangle. For example ... if there was a triangle ABC whose area was known to be 30. Also, you knew that the area of the triangle was 1/2*AB*BC, we would equate the two as it is triangle ABC. SO above we are talking about triangle QPS only and hence can equate. The perpendicular from the right angle to the opposite side ( hypotenuse) can be a bisector only if it is an isosceles right triangle ( which is not the case here as it is a 3-4-5 triangle.

Is it clear now?

We are just equating two different ways to write the area of triangle QPS which is 1/2 * QP * PS and 1/2* PR * QS ...

Best,
Gladi
dave13 wrote:

Gladiator59 thank you, so are equating areas of triangles based on the rule below ? correct ?

If the measures of the corresponding sides of two triangles are proportional then the triangles are similar. Likewise if the measures of two sides in one triangle are proportional to the corresponding sides in another triangle and the including angles are congruent then the triangles are similar.

source: https://www.mathplanet.com/education/ge ... /triangles

but how can I know that PR is not a median ?

have a look here https://www.khanacademy.org/math/geomet ... equal-area

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Regards,
Gladi



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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =?  [#permalink]

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New post 22 Dec 2018, 05:38
thanks Gladi, got it :)
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =? &nbs [#permalink] 22 Dec 2018, 05:38

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In ΔPQS above, if PQ =3 and PS = 4, then PR =?

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