Question

ʜʟᴡ

ᴜʀ Qᴜᴇꜱᴛɪᴏɴ ʜᴇʀᴇ
ᴘʟᴢ ᴛᴇʟʟ ᴄᴏʀʀᴇᴄᴛ ᴀɴꜱᴡᴇʀ

ᴅᴏɴ'ᴛ ꜱᴘᴀᴍ


Mujha attachment nhi chahiya iska solution chahiya.....​

Answer 1

Answer:

Construct a circle with radius =6cm and centre be C

Locate a point A which is 10 cm from C

Then find the perpendicular bisector of line joining points C and A. lets say the mid point will be M.

Now draw an 'another circle with centre at M and radius =6cm(

2

CA

) and lets say this circle cuts the previous circle at

points N and Q then draw the lines AB and AN which are our external tangents.

Sinces △CSA is right angle triangle at L.S=90

o

Therefore using pythagoras theorem

(CQ)

2

+(SN)

2

=(CA)

2

(10)

2

−(6)

2

=(QN)

2

QN=8cm is the length of the tangent.

hope it helps...

Answer 2

Step-by-step explanation:

A pair of targets to the given cricle can be constraced as follows :

Step 1}. Taking any point O of the given plane as center, draw a cricle of 6 cm radius. Locate a point P, 10 cm away from O. Join OP.

Step 2}. Bisect OP. Let M be the midpoint of PO.

Step 3}. Taking M as center and MO as a radius, draw a cricle.

Step 4}. Let the cricle intersect the previous circle at point Q and R.

Step 5}. Join PQ and PR are the required targets. The length of targets PQ and PR are 8 cm each

I hope it may help to you...☺️☺️