Abs one of the direct common tangent of two circles of radii 12 centimetre and 4 cm respectively touching each other find the area of the region enclosed by circles and the tangent
Answers
Answer:
18.53 Square cm (Appr)
Step-by-step explanation:
Let P and Q be the centers of the circles with radii 12 and 4 cm respectively.
PA = 12, QB = 4 PQ = 12 + 4 = 16
PR = PA – RA = 12 – 4 = 8
In ∆PQR, PQ^2 = PR^2 + QR^2
QR^2 = 16*16 – 8*8 = 192
QR = 8√3
AB = 8√3
In quadrilateral ABQP, PA is parallel to QB.
ABQP is a trapezium.
Area of ABQP = (½)*AB(PA + QB)
= ½*8√3*(12 + 4) = 63√3 s.cm.
In right ∆PQR,
Tan(P) = QR / PR = 8√3/8 = √3
So Angle P = 60 degrees.
Angle Q = 180 – 60 = 120 degrees.
Area of region enclosed by circles and common tangent = Area of the shaded region
= Area of PABQ – Area(Sector PAS) – Area(Sector QSB)
= 64√3 – ((60/360)*π*12*12) – ((120/360)* π*4*4
= 64√3 – 24 π – (16 π/3)
= 18.53 Square cm (Appr)
Answer:
Consider two circles of radii 12cm and 4cm with centers B and A respectively.
The length of direct common tangent to both the circles is 15cm.
CD=15 cm
CD=BE=15 cm
AE=AC−CE
AE=12−4=8 cm
AB=x cm
In the right triangle BEA,
Using Pythagoras theorem,
BE
2
+EA
2
=BA
2
x
2
=8
2
+15
2
x
2
=289
x=17 cm
