Page 164 - Grade 10 C.Math

;dfwfg
oxfF, OBC = OCB

= 30° [;dlåafx' OBC sf cfwf/sf sf]0fx¿ ePsfn ]]

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k]ml/ BOC = 180° – (30° + 30°) = 120° [OBC df leqL sf0fx¿sf ofukmn 180° xg ePsfn] ]
Curriculum Development Centre
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To;n, x° = x 120° [Pp6} rfk BC df ag]sf kl/lwsf] sf]0f, s]Gb|Lo sf]0fsf] cfwf x'g] ePsfn] .]
  x = 60°
pbfx/0f 2 L
lrqdf QR //ST 5 . j[Ttsf] s]Gb| M 5 .
70°
olb QLT = 70° eP x° sf] dfg slt x'G5 < M
kTtf nufpg'xf];\ . Q R

x°
;dfwfg S T

oxfF QR // ST 5 . laGb'x¿ T / R hf]8\g'xf];\ .

To;}n] QS = RT [;dfgfGt/ /]vfx¿lar /x]sf rfkx¿ a/fa/ x'g] ePsfn] ]

k]ml/, QTR = 90° [cw{ j[Ttdf ag]sf] sf]0f ePsfn] ]

RQT = x° [ST || QR df PsfGt/ sf]0fx¿ ]
QLT = QRT = 70° [Pp6} rfk QT df ag]sf kl/lwsf sf]0fx¿ ePsfn] .]

To;}n], 70° + 90° + x° = 180° [QTR df leqL sf]0fx¿sf] of]ukmn 180° x'g] ePsfn] ]

cyjf, x° + 160° = 180°

cyjf x° = 180° – 160° = 20°

pbfx/0f 3
A
rj|mLo rt'e'{h ABCD sf e'hfx¿ AB / DC nfO{ a9fpFbf B
laGb' P df sfl6Psf 5g\ . olb DAB = CBP eP P

CP = BP x'G5 egL k|dfl0ft ug'{xf];\ .
C
D
;dfwfg
(1) PCB + BCD = 180° [l;wf /]vf PCD df ag]sf sf]0fx¿sf] of]ukmn 180° x'g] ePsfn] ]

(2) BCD + BAD = 180° [rj|mLo rt'e'{hsf ;Dd'v sf]0fx¿ kl/k"/s x'g] ePsfn] ]

161 ul0ft, sIff – 10