Question

If a triangle and a parallelogram are on the same base and between the same parallels, then prove that the area of the triangles is equal to half the area of the parallelogram?

Answer 1

Hint: To prove the above statement we need to make some assumptions. We need to make a point outside the parallelogram and draw the line parallel to AB. Connect the point with A and B by drawing lines. Using the area of the triangle formulae we can prove the above given statement.

Complete step by step solution:
 For parallelogram ABCD

Properties of Parallelogram:
1. Opposite sides are congruent (AB=DC).
2. Opposite angles are congruent (D=C).
3. Consecutive angles are supplementary (A+D= 180 degrees).
4. If one angle is the right angle, then all angles are right angles.
5. The diagonals of a parallelogram bisect each other.
6. Each diagonal of a parallelogram separates it into two congruent triangles.
Since $DM\mathrm{\perp }AB$(DM is perpendicular to AB).
Height is DM, Base is AB
Area ABCD = Base $\times$Height
Area ABCD = AB$\times$DM … (1)
For $△PAB$
Since $PN\mathrm{\perp }AB$(PN is perpendicular to AB).
Height is PN, Base is AB
Area $△PAB$=${\displaystyle \frac{1}{2}}\times$Base$\times$Height
$\Rightarrow$Area $△PAB$=${\displaystyle \frac{1}{2}}\times$AB$\times$PN
Since opposite sides of parallelogram are parallel
So, $AB\parallel CD$
$\Rightarrow$Distances between parallel lines are equal
PN = DM (Based on the properties of Parallelogram).
$\Rightarrow$Area $△PAB$=${\displaystyle \frac{1}{2}}\times$AB$\times$DM … (2)
From (1) and (2)
Area $△PAB$=${\displaystyle \frac{1}{2}}\times$Area of ABCD.
Therefore, if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangles is equal to half the area of the parallelogram.

Note: We can also prove the above statement by drawing the diagonal and the perpendiculars meeting at some point. Based on the lines collinear and parallel lines we can say the perpendiculars drawn are equal. Therefore the area of the triangle is half the area of parallelogram when having the same height and same base. (Since diagonals of a parallelogram divide it into two congruent triangles).