Chapter 14 - Triangles between Parallel lines

In the previous section we completed the discussion on the area of Quadrilaterals. In this chapter, we will discuss some advanced features about areas of Triangles.

Unchanging Areas of Triangles

Fig.14.1(a) below shows a triangle of base 8 cm and height 4 cm. 
We know that, it's area =  ¹⁄₂ × b × h = ¹⁄₂ × 8 × 4 = 16cm². 
We need only the length of base and height of a triangle to find it's area.

The same height 4 cm can be at different positions as shown in fig.(b) and (c). The base remains the same. The height also remains the same. The top vertex is at different positions. But in all the three cases, the area is 16cm² because, base and height are the same.

The fig.14.2(a) below shows two triangles: ΔABC (shown in yellow colour) and ΔABP (shown in green colour) drawn between two parallel lines.

Fig.14.2

The following conditions are satisfied in the fig.(a):
• The blue lines are parallel
• The two triangles have the same base AB
• The third vertex C of ΔABC and the third vertex P of ΔABP lies on the same line
When the above conditions are satisfied, we can see a peculiarity about their areas:
Area of ΔABC = ¹⁄₂ × b × h = ¹⁄₂ × AB × h₁
Area of ΔABP = ¹⁄₂ × b × h = ¹⁄₂ × AB × h₁
■ Both the areas are same.

In the above discussion, we have chosen AB as the base. There are two more sides BC and CA. The result is applicable to those sides also, as shown in figs 14.1 (b) and (c)
• In fig.(b), ΔABC and ΔACQ are drawn between parallel sides. The side AC is chosen as the base
• One line is drawn through this base AC
• The parallel line is drawn through the third vertex B
• The third vertex Q of the other ΔACQ also lies on this parallel line
Area of ΔABC = ¹⁄₂ × b × h = ¹⁄₂ × AC × h₂
Area of ΔACQ = ¹⁄₂ × b × h = ¹⁄₂ × AC × h₂
■ Both areas are the same

• In fig.(c), ΔABC and ΔBCR are drawn between parallel sides. The side BC is chosen as the base
• One line is drawn through this base BC
• The parallel line is drawn through the third vertex A
• The third vertex R of the other ΔBCR also lies on this parallel line
Area of ΔABC = ¹⁄₂ × b × h = ¹⁄₂ × BC × h₂
Area of ΔACQ = ¹⁄₂ × b × h = ¹⁄₂ × BC × h₂
■ Both areas are the same

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In all such cases, it is important to measure the distance between parallel lines in a perpendicular direction. An example is shown in the fig.14.3 below:

Fig.14.3

However, when we say ‘distance between parallel lines’, it is implied that, the measurement is taken in the perpendicular direction.

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We can write the above results in the form of a theorem:
Theorem 14.1:
■ Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
The converse of this theorem can also be written:
Theorem 14.2:
■ Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.

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We will now see some solved examples
Solved example 14.1
Draw a triangle ABC of sides 4, 5 and 6 cm. Draw an isosceles triangle of the same area and same base.
Solution:
All the three sides of ΔABC are given. So we can easily draw it. It is shown in fig.14.4(a) below. 

Fig.14.4

Now we want another triangle which satisfies two conditions:
■ It must have the same area and base as ΔABC
■ It must be an isosceles triangle
• AB is the base. Through the third vertex C, draw a line MN, parallel to AB (shown in fig.b)
• If the new triangle has the same base AB, it's third vertex must lie on the line MN. This is to ensure that it will have the same area as the original ΔABC
• But having the same area is not enough. It must be isosceles too. We can indeed draw an isosceles triangle with base AB. And also with it's third vertex on MN
• For that, draw the perpendicular bisector a'b' of the base AB. This will intersect MN at P
• Join A and B to P. ΔABP is the required triangle (∵ in an isosceles triangle, the third vertex will lie on the perpendicular bisector of the base)

Solved example 14.2
Draw a new triangle having the same area of ΔABC in fig.14.4(a). The new triangle should have one side 7 cm
Solution:
We have already constructed ABC. We can use it again for this problem. 

Fig.14.5

Here we want another triangle which satisfies two conditions:
■ It must have the same area as ABC
■ One side must be 7 cm in length
• Take AB as the base. Through the third vertex C, draw a line MN, parallel to AB (shown in fig.14.5.b)
• If the new triangle has the same base AB, it's third vertex must lie on the line MN. This is to ensure that it will have the same area as the original ΔABC
• But having the same area is not enough. One side must be 7 cm.
• With A as center, draw an arc (shown in magenta colour) of radius 7 cm, cutting MN at Q
• Join A and B to Q. ΔABQ is the required triangle

Solved example 14.3
Draw a new triangle having the same area of ABC in fig.14.4(a). The new triangle should be an isosceles one with base equal to 5 cm
Solution:
We have already constructed ABC. We can use it again for this problem.

Fig.14.6

 Here we want another triangle which satisfies two conditions:
■ It must have the same area as ΔABC
■ It must be an isosceles triangle with base 5 cm
• Take AC as the base (∵ it has a length equal to 5 cm). Through the third vertex B, draw a line MN, parallel to AC. This is shown in fig.b
• If the new triangle has the same base AC, it's third vertex must lie on the line MN. This is to ensure that it will have the same area as the original ΔABC
• But having the same area and base 5 cm is not enough. It must be isosceles too. We can indeed draw an isosceles triangle with base AC. And also with it's third vertex on MN
• For that, draw the perpendicular bisector a'c' of the base AC. This will intersect MN at R
• Join A and C to R. ΔACR is the required triangle.(∵ in an isosceles triangle, the third vertex will lie on the perpendicular bisector of the base)

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In the next section we will see more solved examples.

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