Geometry Area

[matthew042]

I need help finding the area [OQS] in the picture attached.[attachment=0:1vzj8egv]Picture.jpg[/attachment:1vzj8egv]

 

[Denis]

HINT: x = PQ, p = OS
p^2 = 4^2 + (x+6)^2

 

[matthew042]

Can someone please provide me with more details on how to solve this? All I need is OS. I went nowhere with the hint. I see the concept of using the a^2+b^=c^ formula though. Please someone help me. Thanks.

 

[Mrspi]

I need help finding the area [OQS] in the picture attached.[attachment=0:u04mj529]Picture.jpg[/attachment:u04mj529]

Look at triangles QRS and OPS.....each is a right triangle (so that gives you ONE pair of congruent angles) and each contains angle S (which gives you a second pair of congruent angles). If two angles in one triangle are congruent to two angles in a second triangle, the two triangles are similar.

So, triangle QRS ~ triangle OPS

If two triangles are similar, corresponding sides are in the same ratio:

QR / OP = QS / OS

You know the lengths of QR, OP, and QS....substitute those into the proportion and solve for the length of OS.....but when you do this, you'll find that the length of OS turns out to be less than 6. BIG problem, since OS is the hypotenuse of triangle OPS, and must be the longest side of that triangle.....however, PS is clearly AT LEAST 6 (since it is PQ + QS, and QS = 6)

Please re-check your problem...it seems that you've got an impossible situation here.

 

[Denis]

...I went nowhere with the hint. I see the concept of using the a^2+b^=c^ formula though.

I didn't look at the given info...now that I look at it, Mrspi is correct: your diagram makes no sense whatsoever!
Take a pencil and ruler and try to draw it with the dimensions you show...impossible.

 

[Loren]

Maybe I'm missing something, but I think we can develop several equations.
(6+PQ)^2 + 4^2 = (OS)^2
(RS)^2 + 5^2 = 6^2
(PQ)^2 + 4^2 = sqrt(4^2 + (OQ)^2)
(OQ)^2 = 4^2 + (PQ)^2
OR + RS = OS
4/(PQ) = (PQ+QS)/4

Out of that we should be able to get something???

 

[Aladdin]

Area of OQS is \(\displaystyle \frac{QR*OS}{2}\)

You need to find the length of OS :

OS^2=OP^2+PS^2

But you don't have the length of PS :

PS=PQ+QS

PQ= ??

Take right triangle ORQ: OR=OP=4

OQ^2=OR^2+RQ^2
OQ^2=16+25=41 ... OQ=\sqrt41

Now you can find the length of PQ & Substitute it :

Continue. . . .

 

[Deleted member 4993]

Area of OQS = 1/2 * (QS) * (OP) = 12 .... where is the problem???

 

[Mrspi]

Area of OQS = 1/2 * (QS) * (OP) = 12 .... where is the problem???

I don't think one can argue that triangles QRS and OPS are similar.

Using the proportion QR/OP = QS/OS, and substituting the given values, we find get

5/4 = 6/(OS) and OS = 4.8

So...we have a triangle OPS which is a RIGHT triangle....leg OP is 4, leg is PS (which is PQ + QS since Q is shown BETWEEN P and S), and the hypotenuse is 4.8. BUT....QS = 6 as given in the diagram. If we let x = PQ, then PS = x + 6

Using the Pythagorean theorem, we have

4[sup:3gcf5u35]2[/sup:3gcf5u35] + (x + 6)[sup:3gcf5u35]2[/sup:3gcf5u35] = 4.8[sup:3gcf5u35]2[/sup:3gcf5u35]
16 + (x + 6)[sup:3gcf5u35]2[/sup:3gcf5u35] = 23.04
(x + 6)[sup:3gcf5u35]2[/sup:3gcf5u35] = 7.04
x + 6 = + 2.65
x = -8.65 or x = -3.35

Don't you see a problem here? x was defined to represent the length of segment PQ. A length cannot be negative.

If you simply accept on faith that OS really is 4.8, then yes, the area of the triangle in question would compute to 12. But I stand my ground. The situation depicted in the diagram is impossible. Aren't we supposed to check to make sure the conditions in a problem make sense?

 

[Denis]

But I stand my ground. The situation depicted in the diagram is impossible.

AGREE once more. You ready to saddle Trigger and ride into the sunset with me, Mrspi :wink:

By the way, impossible to get an integer solution (QR = integer) with a primitve pythagorean triangle.
But, as example, triangle 9-12-15 has 2 integer solutions (QR = 3 and 6) fitting in triangles 3-4-5 and 6-8-10:
due to ye old similarity, of course.

 

[Mrspi]

AGREE once more. You ready to saddle Trigger and ride into the sunset with me, Mrspi :wink:

Yep!