Let AB be a chord of the given circle with centre and radius 13 cm.
Then, OA = 13 cm and ab = 10 cm
From O, draw OL⊥ AB
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
∴ AL = ½AB = (½ × 10)cm = 5 cm
From the right △OLA, we have
OA² = OL² + AL²
==> OL² = OA² – AL²
==> [(13)² – (5)²] cm² = 144cm²
==> OL = √144cm = 12 cm
Hence, the distance of the chord from the centre is 12 cm.
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First subtract the 7 from both sides and that would equal 2x+9=2x then subtract 2x and the answer is 9
Answer:
x= -2/3 and 6
Step-by-step explanation:
3x²-16x-12
3x²-18x+2x-12
(3x²-18x)+(2x-12)
3x(x-6)+2(x-6)
(3x+2)(x-6)
x= -2/3 and 6
Answer:
CE= 9cm, DE= 5cm, CD= 10.2956cm
Step-by-step explanation:
Let E be the point of the other triangle.
Using Pythagoras' Theorem,
CD= square root symbol (CE^2 + DE^2)
= square root symbol (9^2+5^2)
= 10.2956cm (6 sf)
Well according to the Inscribed Angle Theorem, the angle inscribed in a circle is equal to one-half of its intercepted arc. We can apply that theorem to this problem as well. So the measure of AED = 1/2( arc AD + arc BC). Plug in the numbers and simplify.
AED = 1/2( 62 + 28)
AED = 1/2( 90 )
AED = 45
So the measure of angle AED is B. 45 degrees.
Hope I helped!
