Congruent Triangles G Amritsar

GUIDED BY MR .HARISH KUMAR GOVT. GIRLS. SEC. SCHOOL. PUTLIGHAR, AMRITSAR

Take a line GH=x cm

By taking G as centre and radius=Ycm draw an arc after that taking H as centre and radius Zcm draw in arc which meet first arc at I

Join IG and IH . We get a triangle GHI where GH=x cm, IG=y cm, IH=z cm

Take a line JK=x cm

By taking J as centre and radius=Ycm draw an arc after that taking K as centre and radius Zcm draw in arc which meet first arc at L

Join LJ and LK . We get a triangle JKL where JK=x cm, LJ=y cm, LK=z cm

We see that GH = JK, IG = LJ and IH = LK. So triangle “GHI” is congruent to triangle “JKL”

Draw a line segment EF = m cm.

Draw an angle XEF= 90 at E.

Draw an arc by taking F as centre and radius= M cm . Which cut ray EX at D.

Join D and F . We get required triangle DEF.

Draw a line segments UV =M cm.

Draw an angle XUV=90 degree at U.

Draw an arc by taking V as centre and radius = M cm . Which cut ray UX at W.

Join w and v .we get required triangle WUV.

We see that EF=UV.

Draw an angle ABC=a degree Take AB=x c.m and BC=y c.m

Join A and C .We get a triangle ABC

Draw an angle DEF=a degree Take DE=x cm and EF=y cm

Join D and F We get a triangle DEF

We see that AB=DE, BC=EF and angle “ABC” = angle “DEF” so, Triangle ABC is congruent to Triangle DEF.

Draw a line PQ=x c.m

Draw angles XPQ=b degree at P and YQP=c degree at Q.Take a point R where PX and YQ meet each other. We get a trianglePQR.

Draw a line ST =x cm

Draw angles XST =b degree at s YTS =c degree at T.Take a point U where SX and YT meet each other. We get a triangle STU.

We see that angle “RPQ” = angle “UST” , PQ=ST and angle “RQP” = angle “UTS”. So, Triangle PQR is congruent to Triangle STU.