O is the centre of the circle having radius 5cm . Triangle ABC is inscribed in the circle such that AB = AC = 6cm . Find the length of side BC.
Answers
Triangle ABC is inscribed in the circle of radius 5 cm such that AB = AC = 6 cm, then the length of side BC is 9.6 cm .
Step-by-step explanation:
It is given that,
A circle with centre O and radius 5 cm.
∆ ABC is inscribed inside the circle such that AB = AC = 6 cm
Step 1:
We know that the angle bisector of an angle between two equal chords of a circle passes through the centre of the circle.
Here, AB and AC are given as two equal chords of a circle, therefore, the centre of the circle O lies on the bisector of ∠BAC.
⇒ OA is the bisector of ∠BAC
Let’s join the points B & C intersecting OA at P.
Again we know that the internal bisector of an angle divides the opposite sides in the ratio of the sides containing the angle i.e., bisector OA will divide BC in the ratio of AB : AC.
∴ The ratio in which P divides BC = 6 : 6 = 1 : 1
i.e., P is mid-point of BC ∴ CP = BP …… (i)
Since we know that if a line from the centre to chord, divides the chord into two equal parts, then the line joining the chord will be perpendicular to it. i.e., OP ⊥ BC
Step 2:
Now,
In right-angled triangle ΔABP, by applying the Pythagoras theorem, we get
AB² = AP² + BP²
⇒ BP² = AB² – AP²
⇒ BP² = 6² - AP² ............. (ii)
And,
In right-angled triangle OBP, by applying the Pythagoras theorem, we get
OB² = OP² + BP²
⇒ OB² = (AO - AP)² + BP²
⇒ 5² = (5 - AP)² + BP²
⇒ BP² = 25 - (5 - AP)² ........... (iii)
Equating (ii) and (iii), we get
62 - AP² = 25 - (5 - AP)²
⇒ 36 – AP² = 25 – (25 – 10AP + AP²)
⇒ 11 – AP² = - 25 + 10AP – AP²
⇒ 36 = 10AP
⇒ AP = 3.6 cm
Substituting the value of AP in (ii), we get
BP² = 6² - (3.6)² = 23.04
⇒ BP = 4.8 cm ….. (iv)
Therefore, from (i) and (iv), we get
BC = 2 * BP = 2 * 4.8 = 9.6 cm
Thus, the length of side BC is 9.6 cm.
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