In figure below, circle with centre M touches the circle with centre N at point T. Radius RM touches

the smaller circl

2 thoughts on “In figure below, circle with centre M touches the circle with centre N at point T. Radius RM touches <br /><br />the smaller circl”

1. Answer:

   The figure is not visible….

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2. Answer:

   Hi,

   According to the question

   let r1 is the radius of bigger circle

   r1 = 9 cm

   so as to the line segment MR ,MR = 9 cm

   Now draw a perpendicular from R to T i.e. RT,RT is equal to radius of small circle,RT =

   2.5 cm

   Now look at the triangle ∆ MRT,here you know the base and perpendicular,have to calculate hypotenuse from Pythagoras theorem

   \begin{gathered} {(mt)}^{2} = \sqrt{( {MR)}^{2} + ( {RT)}^{2} } \\ {(MT)}^{2} = \sqrt{( {9)}^{2} + ( {2.5)}^{2} } \\ = \sqrt{81 + 6.25 } \\ = \sqrt{87.25} \\ MT= 9.34 \: m\end{gathered}

   (mt)

   2

   =

   (MR)

   2

   +(RT)

   2

   (MT)

   2

   =

   (9)

   2

   +(2.5)

   2

   =

   81+6.25

   =

   87.25

   MT=9.34m

   (2)

   length of segmennt MN= MT-TN

   = 9.34-2.5

   =6.84 cm

   3)

   angle MNS = 90°

   but I am not sure about it

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