Answer this question plz
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deepak70:
ur question is not appearing plz sent the it again..
Bhargav plz tell me section 3 12 and 13 questions plz
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This is the answer for your question in the give image hope you understand my solution.
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√25 is 5
Oh sorry I did mistake in hurry.. plz correct that
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Pythagorean theorem,
AB^2+AE^2=EB^2
16+9=EB^2
EB=5
ABE IS RIGHT ANGLE TRIANGLE
AREA=1/2*B*H=1/2*3*4=6
triangle EBD
AREA=1/2*EB*ED=1/2*12*5=30
BD^2=EB^2+ED^2
BD^2=144+25
BD=13
triangle BDC
AREA=1/2*BD*DC=1/2*24*13=156
TOTAL AREA=AREA OF(ABE+EBD+BDC)
156+30+5=191
AB^2+AE^2=EB^2
16+9=EB^2
EB=5
ABE IS RIGHT ANGLE TRIANGLE
AREA=1/2*B*H=1/2*3*4=6
triangle EBD
AREA=1/2*EB*ED=1/2*12*5=30
BD^2=EB^2+ED^2
BD^2=144+25
BD=13
triangle BDC
AREA=1/2*BD*DC=1/2*24*13=156
TOTAL AREA=AREA OF(ABE+EBD+BDC)
156+30+5=191
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