Question

In figure 3.86, circle with centre M touches the circle with centre N at point T. Radius RM touches the smaller circle at S. Radii of circles are 9 cm and 2.5 cm. Find the answers to the following questions hence find the ratio MS:SR.
(1) Find the length of segment MT
(2) Find the length of seg MN
(3) Find the measure of ∠NSM.

Answer 1

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
${(mt)}^{2} = \sqrt{( {MR)}^{2} + ( {RT)}^{2} } \\ {(MT)}^{2} = \sqrt{( {9)}^{2} + ( {2.5)}^{2} } \\ = \sqrt{81 + 6.25 } \\ = \sqrt{87.25} \\ MT= 9.34 \: m$
(2)
length of segmennt MN= MT-TN

= 9.34-2.5
=6.84 cm
3)
angle MNS = 90°

Answer 2

Answer:

Hope it helps u

Step-by-step explanation:

1. length of segment MT is 9cm
2. length of segmemt MN is 6.5 cm
3. angle NSM=90°and MS:SR=2:1