square problem?

Question

In the given figure , not drawn to scale , a semicircle is drawn on the hypotenuse of a right angled triangle ABC . It touches the remaining sides of the circle at P and Q . The two parts of the hypotenuse AC made by 'O' , the centre of the circle , i.e OA and OC , have lengths of 15 cm and 20 cm respectively . The radius of the semicircle :

(1) 12 cm
(2) 10 cm
(3) 16 cm
(4) 9 cm

Figure : http://img45.imageshack.us/my.php?image=semicirclejl9.jpg

how can i say OPBQ is a square ?

how can i prove PB=PO ?

because a square should have all the angles 90 degree AND all sides are equal , then we can call it a sqaure.

Answer 1

You know that OPBQ is a rectangle since you know three of the angles are right angles, so the fourth angle must be right as well.

You know that OP and OQ are equal, since they are the radius of the of the circle.

But then OP=BQ and OQ=BP because it is a rectangle. So all the sides are equal and all the angles are right. So it is a square.

Let the side of the square be x. Now, APO is similar to OQC. So
AO/OC = OP/QC
The left side is 15/20 = 3/4. The right side is x/QC. So
QC=(4/3)x.

Similarly, AP=(3/4)x.

So AB = (3/4)x + x = (7/4)x, and BC=(4/3)x + x = (7/3)x

And (AB)^2 + (BC)^2 = (AC)^2 = 35^2.

Dividing both sides by 7^2, we get:

So 5^2 = [(1/4)x]^2 + [(1/3)x]^2 = x^2 * ((1/4)^2 + (1/3)^2) =
x^2 * (25/144)

So x^2 = 144 and x=12.

Answer 2

Well all the angles do = 90 degrees since three are given = to 90 degrees so 4th angle must be 90 degrees.

Using Pythagorean theorem you can show AP = 9 and
CQ= 16
Now if OPBQ is to be a square PB = BQ =r
So let x = PB=BQ
Then (9+x)^2 + (16+x)^2 = (20+15)^2
If you solve above equation you will get x =12 = r
So OBPQ is a square.

Answer 3

OPBQ is a square because any oval inscribed within a rectangle, has the radius of the circle intersecting the sides at right angles. This problems looks at half a rectangle, (a triangle) and a semi-circle.
hence from the angle at BAC: sin(bac) = r/15
and sin(acb) = r/20; (note the 15 and 20 can be reversed without changing the answer, it is just notation).
but angle acb = 90-bac; then:
sin(acb) = sin(90-bac) = cos(bac) = r/20;

now have two equations and two unknowns, easiest to solve for the angle (bac)
sin(bac) = r/15
cos(bac)= r/20
or
tan(bac) = 20/15
bac = 53.13 degrees, then
sin(53.13) = r/12 or r = 15 * sin(53.13) = 15 * 0.8
r = 12 cm.

Answer 4

OPBQ is a square, because his opposite sides are parallel and all the internal angles are 90º,
PB = PO because they are parallel between parallels.
The radius of the semicircle is (1) 12 cm.

Answer 5

r=12cm
the proof is there
then AP=sqrt(15^2-12^2)=9cm
angle PAO=arcsin(12/15)=53.1
AB=35cos53.1=21
PB=21-9
=12cm
therefore
PO=PB