find the area of shaded region
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where is the region
jaysharma5:
region is in photo
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where is the shaded region
and if you are saying about the figure AEHFG
which is something like a Pentagon
first a fall we Will find the length of AE in the triangle ADE with the help of Pythagoras theorem
so
(hyp)^2=(base)^2+(perp)^2
=(14)^2+(10)^2
=196+100
=296cm
=17.20cm
Now we will find the value of GF with same theorem thta is used above.
(hyp)^2=(base)^2+(perp)^2
=(2)^2+(6)^2
=4+36
=40cm
=6.32cm
Since it is a rectangle
so opposite sides are equal
then AD=BC
10cm=BF+FC
10=2+FC
10-2=FC
8cm=FC
So in figure FHEC
opposite sides are equal
HE=FC=8cm
EC=HF=8cm
perimeter=sum of all sides
=17.20+8+8+6.32+16
=55.52cm
and if you are saying about the figure AEHFG
which is something like a Pentagon
first a fall we Will find the length of AE in the triangle ADE with the help of Pythagoras theorem
so
(hyp)^2=(base)^2+(perp)^2
=(14)^2+(10)^2
=196+100
=296cm
=17.20cm
Now we will find the value of GF with same theorem thta is used above.
(hyp)^2=(base)^2+(perp)^2
=(2)^2+(6)^2
=4+36
=40cm
=6.32cm
Since it is a rectangle
so opposite sides are equal
then AD=BC
10cm=BF+FC
10=2+FC
10-2=FC
8cm=FC
So in figure FHEC
opposite sides are equal
HE=FC=8cm
EC=HF=8cm
perimeter=sum of all sides
=17.20+8+8+6.32+16
=55.52cm
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