In the given figure ABCD is a trapezium ECB is a triange and AECD is a parallelogram. Area of triangle BCE=210cm square EB= 6cm s CD= 24cm. Find the area of trapezium.
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area of triangleBCE= 1/2*b*h
210=1/2*6*h
210=1/2*6h
210=3h
210/3= h
70= h
therefore height of triangle is70 cm
area of trapezium = 1/2*sum of its parallel sides * distance between them
=1/2*(6+24+24)*70
=1/2*54*70
= 27*70
= 180 cm sq
hope it'll help u:-)
210=1/2*6*h
210=1/2*6h
210=3h
210/3= h
70= h
therefore height of triangle is70 cm
area of trapezium = 1/2*sum of its parallel sides * distance between them
=1/2*(6+24+24)*70
=1/2*54*70
= 27*70
= 180 cm sq
hope it'll help u:-)
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Answered by
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At first
find the h from the
Area of triangle =1/2*base *height
1/2 * 3 * h = 210
Then, h = 70
Area of trap. = 1/2*(24+30)*70
" ". " =1/2*54*70
" ". " = 189 cm*cm
Area of trap. = 189cm*cm
Samajh me aaya, bachu
find the h from the
Area of triangle =1/2*base *height
1/2 * 3 * h = 210
Then, h = 70
Area of trap. = 1/2*(24+30)*70
" ". " =1/2*54*70
" ". " = 189 cm*cm
Area of trap. = 189cm*cm
Samajh me aaya, bachu
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