OK. I did it. Now let's see if I can go through it without
getting too complicated.
I think the key to the whole thing is this fact:
A radius drawn perpendicular to a chord bisects the chord.
That tells us several things:
-- OM bisects AB.
'M' is the midpoint of AB.
AM is half of AB.
-- ON bisects AC.
'N' is the midpoint of AC.
AN is half of AC.
-- Since AC is half of AB,
AN is half of AM.
a = b/2
Now look at the right triangle inside the rectangle.
'r' is the hypotenuse, so
a² + b² = r²
But a = b/2, so (b/2)² + b² = r²
(b/2)² = b²/4 b²/4 + b² = r²
Multiply each side by 4: b² + 4b² = 4r²
- - - - - - - - - - -
0 + 5b² = 4r²
Repeat the
original equation: a² + b² = r²
Subtract the last
two equations: -a² + 4b² = 3r²
Add a² to each side: 4b² = a² + 3r² . <=== ! ! !
Answer:
A
Step-by-step explanation:
if you take 23 and put it in for p 18+23=41
Answer:
16 km due west
Step-by-step explanation:
The bearing of the school p from school q is 16 km due west.
To find the bearing of school q from school p, we have to find the direction that the school q is with respect to school p.
Since p is directly west of q, then it implies that q must be directly east of p.
We now know the direction.
Since the distance from q to p is exactly the same as the distance from p to q, then, the distance from p to q is 16 km.
Hence, the bearing of q from p is 16 km due west.
Answer:
the desired range is 0° < angle < 180°
Step-by-step explanation:
If 2x were just slightly less than b, then the angle opposite the base would be just less than 180 degrees.
The larger the value of x, the further the intersection of the two congruent sides is moved away from the base b. The angle between these two congruent sides would approach but never equal zero.
Thus, the desired range is 0° < angle < 180°
I think its either blue, black, or red
