Ninth Grade Mathematics: Parallelograms on Same Base & Between Parallels
One of the most profound discoveries in classical geometry is that the area of a parallelogram does not depend on the slant of its sides or the length of its perimeter; it depends exclusively on the length of its base and its perpendicular altitude. In this chapter, you will master the theorem proving that parallelograms on the same base and between the same parallels are equal in area.
Statement and Significance of the Theorem
The central theorem governing parallelogram areas states:
The Parallelogram Area Theorem:
Parallelograms on the same base (or equal bases) and between the same parallels
are equal in area.
If Parallelogram ABCD and Parallelogram EBCF share base BC
and lie between parallel lines BC and AF:
ar(ABCD) = ar(EBCF)
A E D F (Line AF || Base BC)
*---------*---------*---------*
/ / / /
/ / / /
/ / / /
/ / / /
*---------*---------*---------*
B C
Base BC is shared by both Parallelogram ABCD and EBCF!
This theorem has an astounding implication: you can shear or tilt a parallelogram sideways as far as you wish—making its perimeter arbitrarily large—without changing its enclosed area by even a fraction of a square millimeter, as long as its base and parallel ceiling remain fixed.
Deductive Proof of the Theorem
We prove this theorem using triangle congruence and the Area Addition Axiom:
Given: Parallelogram ABCD and Parallelogram EBCF on the common base BC,
lying between parallel lines BC and line AF.
To Prove: ar(ABCD) = ar(EBCF).
Proof:
Step 1: Consider Triangle ADE and Triangle DCF (or Triangle AEB and Triangle DFC):
Examine Triangle ABF and the two end triangles Triangle AEB and Triangle DFC.
1. In Parallelogram ABCD: side AB = DC and AB || DC.
Since AB || DC and line AF acts as a transversal line:
Angle EAB = Angle FDC (Corresponding Angles)
2. In Parallelogram EBCF: side EB = FC and EB || FC.
Since EB || FC and line AF acts as a transversal line:
Angle AEB = Angle DFC (Corresponding Angles)
3. Since two pairs of angles are equal, the third pair must also be equal:
Angle ABE = Angle DCF (Angle Sum Property)
Step 2: In Triangle AEB and Triangle DFC:
- Angle EAB = Angle FDC (Proven)
- Side AB = DC (Opposite sides of parallelogram ABCD)
- Angle ABE = Angle DCF (Proven)
Step 3: By the ASA (or AAS) Congruence Criterion:
Triangle AEB =~ Triangle DFC.
Step 4: By the Congruent Area Axiom:
ar(Triangle AEB) = ar(Triangle DFC).
Step 5: Now express the areas using the Area Addition Axiom:
ar(ABCD) = ar(Trapezoid EBCD) + ar(Triangle AEB)
ar(EBCF) = ar(Trapezoid EBCD) + ar(Triangle DFC)
Step 6: Since ar(Triangle AEB) = ar(Triangle DFC), adding the common region
ar(Trapezoid EBCD) to both sides yields:
ar(ABCD) = ar(EBCF).
The proof is complete. By cutting off triangle AEB from the left and pasting the congruent triangle DFC onto the right, parallelogram ABCD transforms into parallelogram EBCF without losing or gaining any area.
Derivation of the Formula: Area = Base * Altitude
If one of the parallelograms between the parallels is a rectangle (where the sides are perpendicular to the base at 90 degrees), its area is simply Base * Height.
D C
*---------------------* (Line CD || Base AB)
/| |\
/ | | \
/ | h | \
*---+---------------------+---*
A P Q B
Area = Base * Altitude = b * h
Because any slanted parallelogram on base AB between the same parallels has the exact same area as the perpendicular rectangle on that base:
Area of any Parallelogram = Base * Altitude = b * h
Chapter Exercises
Exercise 1: Parallelogram ABCD has base AB = 15 cm and altitude corresponding to AB equal to 8 cm. If the length of the adjacent side AD is 10 cm, find the area of parallelogram ABCD, and find the altitude corresponding to base AD.
Exercise 2: In the standard figure, parallelogram ABCD and rectangle ABEF share the common base AB and lie between the same parallel lines AB and CF. If AB = 12 cm and the perimeter of rectangle ABEF is 40 cm, calculate the area of parallelogram ABCD.
Exercise 3: Prove that the perimeter of a slanted parallelogram is strictly greater than the perimeter of a rectangle standing on the same base and between the same parallels.
Solutions and Explanations
Solution 1: The area of a parallelogram is the product of its base and corresponding altitude: Area = Base * Altitude = AB * h_1 = 15 * 8 = 120 square cm. Since the area of the parallelogram remains invariant regardless of which side is treated as the base, we can express the same area using base AD: Area = AD * h_2. Substitute Area = 120 and AD = 10: 120 = 10 * h_2, which yields h_2 = 120 / 10 = 12 cm. The area is 120 square cm, and the altitude corresponding to side AD is 12 cm.
Solution 2: In rectangle ABEF, the perimeter is given as 40 cm. The perimeter of a rectangle is 2 * (Base + Height): 2 * (AB + BE) = 40. Given AB = 12: 2 * (12 + BE) = 40, so 12 + BE = 20, which gives height BE = 8 cm. The area of rectangle ABEF is Base * Height = 12 * 8 = 96 square cm. By the Parallelogram Area Theorem, since parallelogram ABCD and rectangle ABEF share the common base AB and lie between the same parallels AB and CF, their areas are equal: ar(ABCD) = ar(ABEF) = 96 square cm.
Solution 3: Let rectangle ABEF and slanted parallelogram ABCD share the common base AB and lie between parallel lines AB and CD. In rectangle ABEF, the sides AF and BE are perpendicular to AB, so AF = BE = h (the altitude). In right triangle AFD, hypotenuse AD is opposite the 90-degree angle AFD. In any right triangle, the hypotenuse is strictly longer than either leg, so AD > AF. The perimeter of the rectangle is 2(AB + AF) = 2(AB + h). The perimeter of the parallelogram is 2(AB + AD). Since AD > AF, it follows that 2(AB + AD) > 2*(AB + AF). Therefore, the perimeter of the slanted parallelogram is strictly greater than the perimeter of the rectangle, even though their areas are identical.