Two geometric shapes are said to be on the same base and between the same parallels, if they have a common side called the base and vertices opposite to the common base on the line parallel to the base. In the below sections, you can learn about the How to Identify figure on the same base and between the same parallels. We know the measure of the plane region enclosed by a closed figure called the area.
The area is measured in cm², m², and other square units. We also know how to calculate the area of different figures using various formulas. Here you will use the formulas to study the relationship between the area of figures when they lie on the same base and between the same parallels.
Some Geometric Figures on the Same Base and between Same Parallels
Below-mentioned is the geometric figures for which you can change the same base and between the same parallels.
The common base for triangles ABC, BCD is BC. And both triangles lie on the same base.
Parallelograms ABCD and triangle CDE lie on the same base CD.
Parallelograms ABCD and EFCD are on the same base DC.
Trapezium ABCD and parallelogram EFCD have a common side DC. We say that trapezium ABCD and parallelogram EFCD are on the same base DC.
Example Questions on Figures that Lie on the Same Base and Between Same Parallels
Example 1.
ABC is right-angled at B with BC = 10 cm and AC = 15 cm. also ∆ABC and ∆BCD are on the same base BC. Find the area of ∆BCD.
Solution:
In right angled ∆ ABC, AC = 15 cm and BC = 10 cm. using Pythagoras theorem, we get
AC² = AB² + BC²
15² = x² + 10²
x² = 15² – 10²
x² = 225 – 100
x² = 125
x = √125
x = 11.18 cm
Now, since ∆ ABC and ∆BCD are on the same base BC.
Therefore, area of ∆ ABC = Area of ∆BCD
1/2 × base × height = Area of ∆BCD
1/2 × 10 × 11.18 = Area of ∆BCD
Area of ∆BCD = 5 x 11.18 = 55.9 cm²
Therefore, area of ∆BCD is 55.9 cm².
Example 2.
AD is the median of ∆ABC. E is any point on AD. Show that area of ∆ABE = area of ∆ACE
Solution:
The common base for ∆ABE, ∆ACE is AE.
ABC is a triangle with AD as the median.
i.e BD = CD
E is any point on AD.
In ∆ABC,
Given that,
D is the midpoint of BC.
∴ AD is median
Since median divides the triangle into two triangles of equal area.
So, Area of ∆ABD = Area of ∆ABC —– (i)
In ∆EBC
Given that, D is the midpoint of BC.
∴ ED is median.
Since median divides the triangle into two triangles of equal area.
So, Area of ∆EBD = Area of ∆EDC —— (ii)
Subtracting equation (1) from equation (2)
Area of ∆ABD – Area of ∆EBD = Area of ∆ADC – Area of ∆EDC
Area of ∆ABE = Area of ∆ACE
Hence, proved.
Example 3.
Parallelogram PQRS and PQTU are on the same base PQ and between the same parallels PQ and UR. Area of parallelogram PQRS = 56 cm² and the altitude of the parallelogram PQTU = 7 cm. Find the length of the common side of two parallelograms.
Solution:
Given that,
Area of parallelogram PQRS = 56 cm²
The altitude of the parallelogram PQTU = 7 cm
Area of parallelogram PQRS = base x height
56 = PQ x 7
PQ = 56/7
PQ = 8 cm
Therefore, the length of the common side of the two parallelograms is 8 cm.