Calculate the QM and the radius

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chijioke

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An isosceles triangle PQR has it vertices on the circumference of a circle. If |PQ| = |QR| = 17 cm, |PR| = 16 cm and m is the midpoint of |PR|. Calculate
a. |QM|
b. calculate the radius of the circle to the nearest whole number.
waec mat gce pic jpj.jpg
 
chijioke said:
b. calculate the radius of the circle to the nearest whole number.
Click to expand...
Have you heard of the formula: Circumradius formula.

\(\displaystyle r = \frac{abc}{4A}\)

where \(\displaystyle a,b,c\) are the side lengths of the triangle and \(\displaystyle A\) is the triangle area.

Or

\(\displaystyle r = \frac{a}{2\sin A}\), where \(\displaystyle a = \overline{PR}\) and \(\displaystyle A\) the angle opposite to it.

The formula gives the radius of the circle for any triangle inscribed inside when all vertices of the triangle touch the circumference of the circle.

If you are not allowed to use this formula and you want to solve the problem using pure geometry, you can construct the perpendicular bisectors.
 
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I'd use these facts:
  1. QO = OR = r
  2. QO + OM = QM = 15
  3. QM^2 + MR^2 = QM^2 + 64 = r^2
but this is not the only approach.
 
Beer drenched reaction follows.
chijioke said:
An isosceles triangle PQR has it vertices on the circumference of a circle. If |PQ| = |QR| = 17 cm, |PR| = 16 cm and m is the midpoint of |PR|. Calculate
a. |QM|
b. calculate the radius of the circle to the nearest whole number.
View attachment 39607
Click to expand...

You're clearly suffering from some kind amnesia.

How do I find the radius of the circle? (An isosceles triangle ABC has its vertices on a circle....)

i. Let the height \overline{BM}~\text{be x} x=\sqrt{13^2-5^2}=12 cm \therefore the height of the triangle \overline{BM}=12cm ii. So how can I obtain the radius of the circle?
www.freemathhelp.com
 
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jonah2.0 said:
You're clearly suffering from some kind amnesia.
Click to expand...
Relax man! It is very normal that people forget how to solve the same problem twice after some time passes. I myself who is considered the greatest Lord of mathematics in the history of this forum, I forget how to solve the same problem after one week from the solution. Now imagine this @chijioke who did it last time two years ago!

😒😒
 
Beer drenched non sequitur ramblings follow.
logistic_guy said:
...
If I mind only my own biz, where is the fun in that? ...
Click to expand...

Cockroach community dynamics is complicated. They eat their dead so that nothing goes to waste. It's best to stay out of their biz.
 
logistic_guy said:
Have you heard of the formula: Circumradius formula.

\(\displaystyle r = \frac{abc}{4A}\)

where \(\displaystyle a,b,c\) are the side lengths of the triangle and \(\displaystyle A\) is the triangle area.

Or

\(\displaystyle r = \frac{a}{2\sin A}\), where \(\displaystyle a = \overline{PR}\) and \(\displaystyle A\) the angle opposite to it.

The formula gives the radius of the circle for any triangle inscribed inside when all vertices of the triangle touch the circumference of the circle.

If you are not allowed to use this formula and you want to solve the problem using pure geometry, you can construct the perpendicular bisectors.
Click to expand...
1753937003388.jpeg
 
Since you got a different answer, what you did is simply wrong. The question is, why?

First, this work accomplished nothing:

1754050648370.png​

Then, the following is just wrong; there's a major sign error:

1754050699700.png​

In fact, [imath]r^2-(15-r)^2=r^2-(225-30r+r^2)=30r-225[/imath]. So the next line should be

[imath]30r-225=64\\30r=289\\r=289/30\approx9.63[/imath]​
 
I believe you've made a mistake here: the circled equation does not look right to me:
eq4.jpg
 
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