Two circles with radii 9cm and 6cm are 2cm apart. What is the length of the common internal tangent
Answers
Answer:
8
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I hope this helps you. You'll need to draw a diagram to follow along, but I've explained it clearly so that it should be easy. Have a great day!!!
Step-by-step explanation:
Let A be the centre of the radius 9 circle (first circle).
Let B be the centre of the radius 6 circle (second circle).
Let CD be a common internal tangent, with C on the first circle and D on the second circle.
Let CD meet AB at E.
Given AB = 9 + 2 + 6 = 17.
AC = 9 (radius)
BD = 6 (radius)
Since CD is tangent to the circles, the angles ACD and BDC are right angles.
So triangles ACE and BDE are similar
=> AE / BE = 9 / 6 = 3 / 2
=> ( AB - BE ) / BE = 3 / 2
=> AB / BE - 1 = 3 / 2
=> AB / BE = 3 / 2 + 1 = 5 / 2
=> BE = 2 AB / 5 = 2 × 17 / 5 = 34 / 5.
By Pythagoras' Theorem,
DE² = BE² - BD² = 34² / 5² - 6²
= ( 34² - 30² ) / 5²
= ( 34 + 30 ) ( 34 - 30 ) / 5²
= 64 × 4 / 5²
=> DE = 8×2 / 5 = 16 / 5
Again, triangles ACE and BDE are similar
=> CE / DE = AC / BD = 9 / 6 = 3 / 2
=> CE = 3 DE / 2 = ( 3 × 16 ) / ( 2 × 5 ) = 24 / 5
Finally, the length of the common internal tangent is
CD = CE + DE = 16 / 5 + 24 / 5 = 40 / 5 = 8