In the given figure, BC, BA and AC are tangents to the circle
touching the circle at D, E and F respectively. If BD = 30 cm,
CD = 7 cm and angle A = 90°, find the radius of the circle.
Answers
Step-by-step explanation:
Here, BD = 30 cm & CD = 7 cm
Theorems that can be implied :
1.) Length of two tangents are equal when drawn from same external point
2.) Pythagoras Theorem - which states that -
(Hypotenuse)^2 = (Perpendicular)^2 + (Base)^2(Hypotenuse)
2
=(Perpendicular)
2
+(Base)
2
Explanation
i) AB, BC, and AC are tangents to the circle at E, D, and F.
BD = 30 cm with DC = 7 cm further ∠BAC = 90°
As per the theorem,
BE = 30 cm = BD,
Also FC = 7 cm = DC
Let AE = x = AF…. (1)
Then AB = BE + AE = (30 + x)
AC = AF + FC = (7 + x)
BC = BD + DC = 30 + 7 = 37 cm
Consider right Δ ABC, by Pythagoras theorem we have
BC^2 = AB^2 + AC^2BC
2
=AB
2
+AC
2
⇒ (37)^2 = (30 + x)^2 + (7 + x)^2(37)
2
=(30+x)
2
+(7+x)
2
⇒ 1369 = 900 + 60x + x^2x
2
+ 49 + 14x + x^2x
2
⇒ 2x^22x
2
+ 74x + 949 – 1369 = 0
⇒ 2x^22x
2
+ 74x – 420 = 0
⇒ x^2x
2
+ 37x – 210 = 0
⇒ x^2x
2
+ 42x – 5x – 210 = 0
⇒ x (x + 42) – 5 (x + 42) = 0
⇒ (x – 5) (x + 42) = 0
⇒ (x – 5) = 0 or (x + 42) = 0
⇒ x = 5 or x = – 42
⇒ x = 5 [Since x cannot be negative]
∴ AF = 5 cm [From (1)]
Therefore AB =30 +x = 30 + 5 = 35 cm
ii)AC = 7 + x = 7 + 5 = 12 cm
Let ‘O’ be the centre of the circle and ‘r’ the radius of the circle.
Join two of the points O, F; points O, D and points O, E.
From the figure,
\frac {1}{2} \times AC \times AB = (\frac {1}{2} \times AB \times OE) + (\frac {1}{2} \times BC \times OD) + ( \frac {1}{2} \times AC \times OC)
2
1
×AC×AB=(
2
1
×AB×OE)+(
2
1
×BC×OD)+(
2
1
×AC×OC)
⇒ AC \times AB = AB \times OE + BC \times OD + AC \times OCAC×AB=AB×OE+BC×OD+AC×OC
⇒ 12 \times 35 = 35 \times r + 37 \times r + 12 \times r12×35=35×r+37×r+12×r
⇒ 420 = 84 r
∴ r = 5
Thus the radius of the circle is 5 cm..