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 cmÂ², mÂ², 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 âˆ†ABC and âˆ†BCD are on the same base BC. Find the area of âˆ†BCD.

**Solution:**

In right angled âˆ† ABC, AC = 15 cm and BC = 10 cm. using Pythagoras theorem, we get

ACÂ² = ABÂ² + BCÂ²

15Â² = xÂ² + 10Â²

xÂ² = 15Â² â€“ 10Â²

xÂ² = 225 â€“ 100

xÂ² = 125

x = âˆš125

x = 11.18 cm

Now, since âˆ† ABC and âˆ†BCD are on the same base BC.

Therefore, area of âˆ† ABC = Area of âˆ†BCD

1/2 Ã— base Ã— height = Area of âˆ†BCD

1/2 Ã— 10 Ã— 11.18 = Area of âˆ†BCD

Area of âˆ†BCD = 5 x 11.18 = 55.9 cmÂ²

Therefore, area of âˆ†BCD is 55.9 cmÂ².

**Example 2.**

AD is the median of âˆ†ABC. E is any point on AD. Show that area of âˆ†ABE = area of âˆ†ACE

**Solution:**

The common base for âˆ†ABE, âˆ†ACE is AE.

ABC is a triangle with AD as the median.

i.e BD = CD

E is any point on AD.

In âˆ†ABC,

Given that,

D is the midpoint of BC.

âˆ´ AD is median

Since median divides the triangle into two triangles of equal area.

So, Area of âˆ†ABD = Area of âˆ†ABC â€”â€“ (i)

In âˆ†EBC

Given that, D is the midpoint of BC.

âˆ´ ED is median.

Since median divides the triangle into two triangles of equal area.

So, Area of âˆ†EBD = Area of âˆ†EDC â€”â€” (ii)

Subtracting equation (1) from equation (2)

Area of âˆ†ABD â€“ Area of âˆ†EBD = Area of âˆ†ADC â€“ Area of âˆ†EDC

Area of âˆ†ABE = Area of âˆ†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 cmÂ² 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 cmÂ²

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.