Q.3. (b) in the figure given above, DE ‖ BC. Prove that:
(i) area of ∆ ACD = area of ∆ ABE
(ii) area of ∆ OBD = area of ∆ OCE
Dear Student,
Please find below the solution to the asked query:
We know : Area of triangle = Base Height
Let , Height between parallel lines ( DE | | BC ) = h
So,
Area of BDE = DE h --- ( 1 ) ( Here base = DE and Height = h as that is lies between DE | | BC )
And
Area of CDE = DE h --- ( 2 ) ( Here base = DE and Height = h as that is lies between DE | | BC )
From equation 1 and 2 we get :
Area of CDE = Area of BDE --- ( 3 )
i ) Now we add " Area of ADE " on both hand side of equation 3 and get :
Area of CDE + Area of ADE = Area of BDE + Area of ADE
Area of ACD = Area of ABE ( Hence proved )
ii ) Now we subtract " Area of ODE " on both hand side of equation 3 and get :
Area of CDE - Area of ODE = Area of BDE - Area of ODE
Area of OCE = Area of OBD ( Hence proved )
Hope this information will clear your doubts about topic.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards
Please find below the solution to the asked query:
We know : Area of triangle = Base Height
Let , Height between parallel lines ( DE | | BC ) = h
So,
Area of BDE = DE h --- ( 1 ) ( Here base = DE and Height = h as that is lies between DE | | BC )
And
Area of CDE = DE h --- ( 2 ) ( Here base = DE and Height = h as that is lies between DE | | BC )
From equation 1 and 2 we get :
Area of CDE = Area of BDE --- ( 3 )
i ) Now we add " Area of ADE " on both hand side of equation 3 and get :
Area of CDE + Area of ADE = Area of BDE + Area of ADE
Area of ACD = Area of ABE ( Hence proved )
ii ) Now we subtract " Area of ODE " on both hand side of equation 3 and get :
Area of CDE - Area of ODE = Area of BDE - Area of ODE
Area of OCE = Area of OBD ( Hence proved )
Hope this information will clear your doubts about topic.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards