please solve it and step by step solutions because i need for subjective exam
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go through the solution step by step
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mysticd:
Thank u rishi.
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ur most welcome
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look ABC is this triangle right angle
so, use Pythagoras theorem
AC=5cm
now again ACD is right angle triangle
also given angle ADC =30
use sin30=AC/DC
1/2=5/DC
DC=10cm
a/c DC=DE
and CDE is also right angle triangle and two equal side so, two angle 45degree
now use
sin45=CD/CE
CE=10root2
now
look CE =EF and CEF is right angle triangle so, sin45=10root2/CF
CF=20 cm=FG=CG
now P, Q and R is the midpoint of CF , GF, CG
area of PQR=1/4area of CFG
1/4 x root3/4 x (20)^2=25root3
so, use Pythagoras theorem
AC=5cm
now again ACD is right angle triangle
also given angle ADC =30
use sin30=AC/DC
1/2=5/DC
DC=10cm
a/c DC=DE
and CDE is also right angle triangle and two equal side so, two angle 45degree
now use
sin45=CD/CE
CE=10root2
now
look CE =EF and CEF is right angle triangle so, sin45=10root2/CF
CF=20 cm=FG=CG
now P, Q and R is the midpoint of CF , GF, CG
area of PQR=1/4area of CFG
1/4 x root3/4 x (20)^2=25root3
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