Question

Classify each of them on the basis of the following.

Simple curve (b) Simple closed curve (c) Polygon

(d) Convex polygon (e) Concave polygon
ʙᴏʜᴏᴛ ᴋᴀʀᴇᴇʙɪ ɪɴsᴀɴ ʜᴀɪ ᴍᴇʀɪ..
ᴀᴜʀ ᴠɪɴᴇᴇᴛ ᴊɪᴊᴀ ᴊɪ..xᴅ ɪᴛɴᴇ ᴊᴀʟᴅɪ ʙʜᴜʟ ɢʏᴇ​

Answer 1

Answer:

ʜᴜʜ.. ᴀʙ ᴍᴇʀᴀ ᴅɪᴍᴀɢ ᴋʜᴀʀᴀʙ ʜᴏ ᴄʜᴜᴋᴀ ʜᴀɪ..

sᴀʙᴋᴇ ʟɪʏᴇ ᴛɪᴍᴇ ʜᴀɪ ᴍᴇʀᴇ ʟɪʏᴇ ᴛʜᴏᴅᴀ ʙʜɪ ɴʜɪ ɪᴛɴᴀ ᴋʏᴀ ʙᴜsʏ ʜᴏ..

(￣へ ￣)

ɪᴛ's ᴇɴᴏᴜɢʜ ʏʀʀ..

ᴍᴀɪ ʏʜᴀ ᴘᴀɢʟᴏ ᴋɪ ᴛᴀʀᴀʜ ᴡᴀɪᴛ ᴋʀ ʀʜᴀ ʜᴜ ᴍᴀᴅᴀᴍ ᴋᴏ ᴋᴏɪ ғᴀʀǫ ɪᴄʜ ɴʜɪ ᴘᴅ ʀᴇʟᴀ ʜᴀɪ..

ʟɪᴛᴇʀᴀʟʟʏ ᴛᴜᴍɴᴇ ᴍᴏᴏᴅ ᴋᴀ ғᴀʟᴜᴅᴀ ᴋʀ ʀᴀᴋʜᴀ ʜᴀɪ..

ʙᴀᴋɪ ᴋᴇ ʙᴀɴᴅɪʏᴏ ᴋᴏ ᴅᴇᴋʜᴏ ɪᴛɴᴀ ᴘʏᴀʀ ᴀᴜʀ ᴍᴇʀɪ ʙᴀɴᴅɪ ᴋᴏ ᴅᴇᴋʜᴏ ᴘʏᴀʀ ᴋᴀ ᴘ ʙʜɪ ɴʜɪ ᴋʀɴᴀ.. ɪɢɴᴏʀᴇ ᴋʀɴᴀ ʜᴀɪ ʙᴀs ᴋɪᴛɴɪ sᴀᴅɪ ʜᴜɪ ᴋɪsᴍᴀᴛ ʜᴀɪ..

Let us take another quadrilateral ABCD which is not convex .

Join BC, Such that it divides ABCD into two triangles ΔABC and ΔBCD. In ΔABC,

∠1 + ∠2 + ∠3 = 180° (angle sum property of triangle)

In ΔBCD,

∠4 + ∠5 + ∠6 = 180° (angle sum property of triangle)

∴, ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 180° + 180°

⇒ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 360°

⇒ ∠A + ∠B + ∠C + ∠D = 360°

Thus, this property hold if the quadrilateral is not convex.