Math, asked by RockySince2007, 11 months ago

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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Answers

Answered by harshshrivastava554
1

Answer:x+y+z=193'

Step-by-step explanation:

Answered by Krishnakvinod
1

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

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