Question

(b)
In the given figure BC is parallel to DE. Prove that:
area AABE = area AACD​

Answer 1

Step-by-step explanation:

Given

AB ‖ DC

To prove that : (i) area(∆ACD) = area(∆ABE)

(ii) area(∆OCE) = area(∆OBD)

(i)

Here in the given figure Consider

BDE and

ECD,

we find that they have same base DE and lie between two parallel lines BC and DE

According to the theorem: triangles on the same base and between same parallel lines have equal

areas.

Area of

BDE = Area of

ECD

Now,

Area of

ACD = Area of

ECD + Area of

ADE ---1

Area of

ABE = Area of

BDE + Area of

ADE ---2

From 1 and 2

We can conclude that area(∆AOD) = area(∆BOC) (Since Area of

ADE is common)

Hence proved

(ii)

Here in the given figure Consider

BCD and

BCE,

we find that they have same base BC and lie between two parallel lines BC and DE

According to the theorem : triangles on the same base and between same parallel lines have equal

areas.

Area of

BCD = Area of

BCE

Now,

Area of

OBD = Area of

BCD - Area of

BOC ---1

Area of

OCE = Area of

BCE - Area of

BOC ---2

From 1 and 2

We can conclude that area(∆OCE) = area(∆OBD) (Since Area of

BOC is common)

Hence proved