ppt

2. TRIANGLE

3. RECTANGLE

4. AREA
AREA OF TRIANGLE IN A
RECTANGLE WITH SAME BASE

5. The area of the rectangle =
AB×AD
D
A
C
B
P
Q

6. The area of the triangle
=½bh
=½AB×PQ
=½AB×AD
=½×area of
rectangle

7. So where ever be the point P on
CD, the area of the triangle APB is
half the area of the rectangle ABCD

8. Problem
To get an isosceles triangle of
area half that of the rectangle,
where must be the point P ?

9. P should be the mid point of CD
such that AP=BP
D C
A B
P

10. Problem
If ABP is to be right , where must be the
position of P ?
P must be in the position of C
D
A B
C
P