Two triangles on the same base (or equal bases) and between the same parallels are equal in area . How do I prove this one ? :(

the formula for the area of a triangle is 1/2 base into height

if 2 triangles are on the same base then base will be equal for 2 of them.

we have learnt that distance between the 2 parallel lines are equal,so their height also become same.since both the triangles have same base and height their areas are equal

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ANOTHER METHOD IS GIVEN IN N.C.E.R.T. MATHS BOOK PAGE NO.160 (9TH)

  MY ANSWER  IS FOR STANDARD STUDENTS NOT FOR KIDS

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A  D

B

C

given- bc parallel to ad

to prove- ar[abc] = ar[ dcb]

proof - in triangle aob congruent to triangle cod

 ab = dc [between same parallels so height is same

angle bac = acd  [ad parallel to bc]

angle abd = bdc

triangle aob congruent to triangle dob [asa]

ar[aob] = ar[cod]

when equals are added to equals the wholes are equal

ar[aob] + ar[boc] = ar[cod] + ar[boc]

ar[abc] = ar[ dcb]

 

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