show that angle BAE = 90 degrees

B

bumblebee123

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question:

the diagram shows two circles. A,B,D and E are points on one circle. B,C and D are points on the second circle. BC is a diameter of the second circle. ABC and CDE are straight lines

a) show that angle BAE = 90 degrees You must give a reason for each stage in your working

AB = 7cm, BC = 5cm, DE = 10cm and CD= x cm

b) (i) show that x^2 + 10x - 60 = 0

(ii) calculate the length of CE. give your answer to 3.s.f.

11873

I have no idea where to start- any help would be really appreciated!
 
Draw in BD. Use triangle BDC to find angle BDC, and therefore angle BDE. Then think about implications within the larger circle. (Theorems about angles inscribed in a circle will be of use.)
 
does angle BDC make a right angle? BC is the diameter, line BD must also = x

If it does make a right angle, wouldn't x = 5/2
 
Looking at the little circle BC is a diameter (given information). What do you know about the angle formed in a semicircle?
(BD is not necessarily the same as x)
 
bumblebee123 said:
does angle BDC make a right angle? BC is the diameter, line BD must also = x

If it does make a right angle, wouldn't x = 5/2
Click to expand...
Yes, it's a right angle because it is inscribed in a semicircle.

But, no, not all right triangles are right isosceles triangles, which seems to be what you are assuming.

To find x, you will need information from both circles. For now, focus on part a, which involves no lengths.

Do what I suggested in the larger circle now. What is angle BDE?
 
Dr.Peterson said:
Yes, it's a right angle because it is inscribed in a semicircle.

But, no, not all right triangles are right isosceles triangles, which seems to be what you are assuming.

To find x, you will need information from both circles. For now, focus on part a, which involves no lengths.

Do what I suggested in the larger circle now. What is angle BDE?
Click to expand...

yep- I assumed they were both the same length, I don't know why I thought that!!!
 
BAED forms a cyclic quadrilateral. opposite angles in a cyclic quadrilateral add up to 180 degrees. Hence, angle BAE = 180 - BDE. so BAE = 180 -90 = 90 degrees
 
b) (i)

x * ( 10 + x ) = 12 * 5

10x + x^2 = 60

x^2 + 10x - 60 = 0

( ii )
by putting the equation into the quadratic formula, I found that x = 4.2 ( to one decimal place )

CE = ED + x
CE = 10 + 4.2
CE = 14.2cm ( 3 s.f.) which is correct!

have I used the right methods? :)
 
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