radius of a circle with Centre O is 25 cm find the distance of a chord from the centre if length of the chord is 48 CM
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Answered by
6
Answer:
48 cm
Step-by-step explanation:
First, sketch it.
You find a right-angle triangle with hypotenuse 25 (the radius) and one side 7 (the altitude) and the unknown 3rd side
Pythagoras:
7² + x² = 25²
x² = 625 - 49 =576
x = 24
The chord is 48cm
rosnaaugustine460:
i am sorry
answer is wrong
with sketch
7 cm is answer
Answered by
0
Answer:
Given,
seg OP ⊥ chord CD and l(CD) = 48 cm
Radius of circle = 25 cm, so OD = 25 cm
Now,
l(PD) = ½ l(CD)
[Perpendicular drawn from the centre of a circle to its chord bisects the chord]
l(PD) = ½ x 48
⇒ l(PD) = 24 cm … (i)
In ∆OPD, we have
m ∠OPD = 90°
So, by Pythagoras theorem
[l(OD)]² = [l(OP)]² + [l(PD)]²
(25)² = [l(OP)]² + (24)² [From (i)]
(25)² – (24)² = [l(OP)]²
(25 + 24) (25 – 24) = [l(OP)]² [Since, a² – b² = (a + b) (a – b)]
49 x 1 = [l(OP)]²
[l(OP)]² = 49
∴l(OP) = √49 = 7 cm [Taking square root of both sides]
Thus, the distance of the chord from the centre of the circle is 7 cm.
Step-by-step explanation:
Given,
seg OP ⊥ chord CD and l(CD) = 48 cm
Radius of circle = 25 cm, so OD = 25 cm
Now,
l(PD) = ½ l(CD)
[Perpendicular drawn from the centre of a circle to its chord bisects the chord]
l(PD) = ½ x 48
⇒ l(PD) = 24 cm … (i)
In ∆OPD, we have
m ∠OPD = 90°
So, by Pythagoras theorem
[l(OD)]² = [l(OP)]² + [l(PD)]²
(25)² = [l(OP)]² + (24)² [From (i)]
(25)² – (24)² = [l(OP)]²
(25 + 24) (25 – 24) = [l(OP)]² [Since, a² – b² = (a + b) (a – b)]
49 x 1 = [l(OP)]²
[l(OP)]² = 49
∴l(OP) = √49 = 7 cm [Taking square root of both sides]
Thus, the distance of the chord from the centre of the circle is 7 cm.
Step-by-step explanation:
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