Question

Af
a triangle and a parallelogram are
on the same base and between the same
farallels, then prove that the area of the
triangle is equal to half the area of the
Parallelogram
19
С C
boyove?
А​

Answer 1

Answer:

Therefore, area of ∆ABD = 1/2 area of parallelogram ABCD

                                  = 1/2 (AB × AE);

                                      [Since, DE is the altitude of parallelogram ABCD]

Here, AB is the base and AE is the height of ∆ABD.

Notes:

1. If a triangle and parallelogram are on the same base and have the same altitude, the area of the triangle will be half that of the parallelogram.

If they have same altitude, they will lie between the same parallels. Hence the area of the triangle will be equal to half that of the parallelogram.

2. If a triangle and a rectangle be on the same base and between the same parallels, the area of the triangle will be half that of the rectangle.

3. Area of a triangle = 1/2 × base × altitude.

∆ ABC and rectangle BCDE are on the same base BC and between the same parallels BC and ED.



Therefore, ∆ ABC = 1/2 rectangle BCDE = 1/2 BC ∙ CD

                        = 1/2 BC ∙ AP [Since APCD is a rectangle]

Solved example for the triangle and parallelogram on same base and between same parallels:

1. ∆ ABD and parallelogram ABCD are on the same base AB. If base and altitude of the parallelogram are 15 cm and 10 cm, find the area of the triangle.

Solution:            

Base of parallelogram = 15 cm

Altitude of parallelogram = 10 cm



Therefore Area of parallelogram = 15 × 10 cm2

                                            = 150 cm2

∆ ABD and parallelogram ABCD are on the same base AB.

Therefore of ∆ ABD = 1/2 the area of parallelogram ABCD

                           = 1/2 × 150 cm2

                           = 75 cm2

Answer 2

Answer:

we can prove by diagnose of the parllelogram which divides the parallelogram into 2 equal triangles