In the given figure,
ABCD is a rhombus FHI is an equilateral
triangle and CEFG is a trapezium with
EC parallel to FG DCFI BCG and GFH
are straight lines. Find x + y + z.
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In the given figure,
ABCD is a rhombus FHI is an equilateral
triangle and CEFG is a trapezium with
EC parallel to FG DCFI BCG and GFH
are straight lines. Find x + y + z.
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Answer:x+y+z=193'
Step-by-step explanation:
x+y+z=193
Step-by-step explanation:
ABCD is a rhombus
<B=<C=108(opposite angles)
therefore <A=<C=180--108=72
<GCF=<C=72 (opposite angles)
FHI is a equilateral triangle all angles are 60
therefore <GFC=<HFI=60
<G=180--72+60=180--132=48=x
<GHF (outer angle)=180--60=120
therefore <GHF=<CFE=120=z
y=180--120+35=180--155=25
therefore x+y+z=48+25+120=193