Co ord Geom of Circles

[mattgad]

Question:

The line AB is a chord of circle centre (2, -1), where A(3,7) and B(-5, 3). AC is the diameter of the circle. Find the area of triangle ABC

Answer: 60

My working so far:

\(\displaystyle (x - 2)^2 + (y + 1)^2 = r^2\)

Written as a circle.

Subbing in A(3, 7) to give

\(\displaystyle (x - 2)^2 + (y + 1)^2 = 65\)

I then decided as to get from A(3, 7) to Centre(2, -1), that to get to C, i need to go back 1 x value and down 8 y values. So C must be (1, -9)

I know this is probably the worst way to work it out, is there any other way of doing it?

Length of AB is
\(\displaystyle \sqrt {(3 - - 5)^2 + (7 - 3)^2 }\) = \(\displaystyle \sqrt {80}\)

Length of BC

\(\displaystyle \sqrt {( - 5 - 1)^2 + (3 - - 9)^2 } = \sqrt {180}\)

So area,

1/2bh

\(\displaystyle 1/2(\sqrt {80} * \sqrt {180)} \cr
\sqrt {20} *\sqrt {45} \cr
2\sqrt 5 *3\sqrt 5 \cr\)

= 30

Where have I gone wrong? Am I even going in the right direction?

Thanks

 

[pka]

You MULTIPLY \(\displaystyle \left( {.5} \right)\sqrt {80} \sqrt {180} = (.5)(2\sqrt {20} )(2\sqrt {45} ) = 60\) !

 

[kangaroo]

Hi. Actually, the answer of 60 is wrong. One way to find the area of a triangle is one half times the base times the height: A = bh/2 -- but this assumes that the base is PERPENDICULAR to the height. In the problem given, AB is not perpendicular to BC, so you can't use the formula.

However, you can fit a rectangle to the triangles from D(-5,-1) to A(3,7) and subtract off the areas of the three right triangles that get cut out. First let E(3,-1) and F(-5,7).

Area of the rectangle AEDF = (3- -5)(7- -1) = 64

Area of triangle AEC = bh / 2 = (1)(8)/2 = 4

Area of triangle BDC = bh / 2 = (7)(4)/2 = 14

Area of triangle AFB = bh / 2 = (8)(4)/2 = 16

So the area of triangle ABC = 64 - (4+14+16) = 30

 

[mattgad]

Thanks both for your replies.

However, when I was doing 1/2(b x h), I realised that is 1/2b1/2h, when I only want 1/2bh, so using the correct method, I come to the same answer as pka and the answer in my textbook.

 

[kangaroo]

I'm afraid your text book is mistaken!!! What text book is it? What page and what number problem?

 

[mattgad]

Core Mathematics 2 (Heinemann Modular Mathematics for London AS & A-level S.)
Keith Pledger

ISBN 0435510983

pg 67 question 17

BUT, I still believe the answer of 60 to be correct.

 

[pka]

Kangaroo, you are absolutely wrong yourself.
First, because AC is a diameter then the angle ABC is inscribed in a semicircle.
That means that it is a right angle. If |AB| is the length of line segment AB the then area of ΔABC is (.5)(|AB|)(|BC|).
PLEASE CORRECT YOUR MISTAKE.

 

[kangaroo]

I stand corrected!!

(I thought C was the center of the circle, and didn't realize that AC was a diameter. Sorry, I misread the question!!!)

 

[Denis]

I stand corrected!!
(I thought C was the center of the circle, and didn't realize that AC was a diameter. Sorry, I misread the question!!!)

Well don't feel bad; not bad for a kangaroo

 

[soroban]

Hello, mattgad!

It's much easier than you think . . .

The line AB is a chord of circle centre (2, -1), where A(3,7) and B(-5, 3).
AC is the diameter of the circle. Find the area of triangle ABC.

Did you make a sketch?

If \(\displaystyle AC\)is a diameter, then \(\displaystyle \angle B\,=\,90^o\)
\(\displaystyle \;\;\;\)We have a right triangle!

\(\displaystyle AB\:=\:\sqrt{(-5-3)^2\,+\,(3-7)^2}\:=\:\sqrt{80}
:=\:4\sqrt{5}\)

\(\displaystyle BC\:=\:\sqrt{(1-[-5])^2\,+\,(-9-3)^2}\:=\:\sqrt{180}\:=\:6\sqrt{5}\)

\(\displaystyle \text{Area }\,=\;\frac{1}{2}(AB)(BC)\;=\;\frac{1}{2}(4\sqrt{5})(6\sqrt{5})\;=\;\frac{1}{2}(24\cdot5) \;=\;60\)