Similar Triangles

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

(A) 9/4
(B) 12/5
(C) 16/5
(D) 15/4
(E) 20/3

I am stuck...

Since it's 3:4:5 right triangle, so QS is 5. divide it by 2 to get the base of the two small triangles which is 2.5
now we need the hight of one of the smaller triangles. So,
(2.5)^2+ h^2 = 4^2
h^2= 16- 6.25
h = sqrt 9.75
h approx = 3.1 which is 16/5 the answer should be C unless If i made some mistakes. pls correct me if I'm wrong

Abdulla wrote:

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

(A) 9/4
(B) 12/5
(C) 16/5
(D) 15/4
(E) 20/3

I am stuck...

Since it's 3:4:5 right triangle, so QS is 5. divide it by 2 to get the base of the two small triangles which is 2.5
now we need the hight of one of the smaller triangles. So,
(2.5)^2+ h^2 = 4^2
h^2= 16- 6.25
h = sqrt 9.75
h approx = 3.1 which is 16/5 the answer should be C unless If i made some mistakes. pls correct me if I'm wrong

I did it your way: h^2 = 4^2 - (5/2)^2 = 39/4 ---> h = sqrt 39/2 . Not sure if that is the same as 16/5.

In any case, we are both wrong because i don't think we can assume that PR bisects QS, although PR makes a right angle with QS.

From the picture that I edited above, you can clearly see that:

9-x^2=16-(5-x)^(2) By the pythagorean theorem

9-x^2=16-(25-10x+x^2)

9-x^2=-9+10x-x^2

9=-9+10x

x=9/5

PR= sqrt(9-(9/5)^(2))=sqrt(144/25)=12/5

Hope that helps.

I understand all this complicated but necessary solution. what I don't understand is why you assumed half the hypatenuse to be x ? did you assume that they are not halves and that the perpendicular PR does not bisect QS ?

Triangles PQS and PRS are Slr since they share a common angle, a common side and a Rt angle.

Ratio of the sides is Equal=

SR=x/4=3/5
x=12/5


Now PR/12/5=4/3
PR=16/5

C is the Right Choice/./

looking at the given figure the height of the traingle can not be greater that any of its side. So it has to be less than 3.
so is looks like choice C is the wrong answer.only choice a and b are less than 3

Triangles PQS and PRS are Slr since they share a common angle, a common side and a Rt angle.

Ratio of the sides is Equal=

SR=x/4=3/5
x=12/5


Now PR/12/5=4/3
PR=16/5

C is the Right Choice/./

GMatter2.0. Excellent answer.
I proved it correctly going through the cumbersome algebra. But, I couldn't get the thought of the triangles being similar out of my head. You did that for me!

Thanks!

I get B. I did it by thinking about area. The area of the big triangle is equal to the sum of the two smaller triangles.

Area of big = (1/2)*b*h = (1/2)*(3)*(4) = 6.

Area of PQR = (1/2) * x * h = (1/2) * (5/2) * h = 1/2xh
Area of PSR = (1/2) * (5-x) * h = (1/2) * (5/2) * h = (1/2)(5-x)h

You know triangles are similar, so apply pythagorean on both smaller triangles:

9 = x^2 + h^2
16 = (5-x)^2 + h^2
Substituting and solving for x yields:

9 - x^2 = 16 - (5-x)^2 --> 9 - x^2 = 16 - 25 + 10x - x^2, or x = 18/10 = 9/5
Area of big (PQS) = Area of PQR + Area of PSR
6 = (1/2)xh + (1/2)(5-x)h
Substitute x: 6 = (1/2)(9/5)h + (1/2)(5-(9/5))h
6 = 9/10h + 16/10h
h = 6*(10/25) = 12/5.

Choice B.

mp2437-

We haven't been told anywhere that PR bisects the triangle. In fact, we know that is most certainly NOT the case (since one leg is larger than the other and the line PR is perpendicular to the hypot).

Be careful when making assumptions on the exam.

Buckeye,

You are right. Just edited and resolved the question, and it is solved just like Truplayer's response.. Thanks for the correction...fortunately was able to pull it off.

This can be solved using knowledge about area
Area = 1/2(PS)(QP) = 6

Therefore, 1/2(QS)(PR) = 6, you know QS = 5 using 3,4,5 rule

Therefore PR = 12/5