Chapter 14.3 - Polygon to Triangle - Solved examples

In the previous section we saw how a quadrilateral, or even a polygon, can be transformed into a triangle having the same area. In this section, we will see some solved examples.

Solved example 14.8
Draw the two quadrilaterals shown in fig.14.15(a) and 14.16(a) below. For each of them, draw a triangle of same area, and calculate the area. Lengths required for the area calculations may be measured

Fig.14.15

Solution:
■ Fig.14.15(a) is a rough sketch. But it contains all the details required to construct the actual quadrilateral. The construction of ABCD is done in fig.b
• Once the construction is done, we can proceed to do the transformation:
• Draw diagonal BD 
• Draw MN parallel to BD through C
• On MN, mark E, in line with AB
• Join B and D to E. This gives the transformed shape: ΔAED
• Now we have to find the area of ΔAED
• From the actual construction, the required lengths can be measured. Base AE is already drawn. To obtain height, draw a perpendicular DF to AB
• By measurement, base AE = 10.6 cm and Height DF = 2.9 cm
• So Area = ¹⁄₂ × b × h = ¹⁄₂ × 10.6 × 2.9 = 15.66 cm²

Fig.14.16

■ Fig. 14.16(a) is a rough sketch. But it contains all the details required to construct the actual quadrilateral. The construction of ABCD is done in fig.b
• Once the construction is done, we can proceed to do the transformation:
• Draw diagonal BD 
• Draw MN parallel to BD through C
• On MN, mark E, in line with AB
• Join B and D to E. This gives the transformed shape: ΔAED
• Now we have to find the area of AED
• From the actual construction, the required lengths can be measured. Base AE is already drawn. To obtain height, draw a perpendicular DF to AB
• By measurement, base AE = 12.6 cm and Height DF = 5.2 cm
• So Area = ¹⁄₂ × b × h = ¹⁄₂ × 12.6 × 5.2 = 32.76 cm² 

Solved example 14.9
Draw a rhombus of sides 6 cm, and one angle 60^o. Then draw a right triangle of the same area
Solution:

Fig.14.17

• We have learned to construct a rhombus in lower classes. In fig.14.17, ABCD is the required rhombus. It's construction details are not shown here.
• Once the construction is done, we can proceed to do the transformation:
• Draw diagonal BD 
• Draw MN parallel to BD through C
• On MN, mark E, in line with AB
• Join B and D to E. This gives the transformed shape: ΔAED
• But we want a right triangle of the same area as ABCD
• So we need to transform ΔAED into a right triangle
• For that, extend CD to the left. Draw a perpendicular AF to this extended line
• Join A and E to F. ΔAEF is the required right triangle

Solved example 14.10
Draw a regular pentagon, and then a triangle of the same area. Calculate the area of the triangle.
Solution:

Fig.14.18

• We have learned to construct a regular pentagon in lower classes. In this problem, we can draw a regular pentagon of any side. In fig.14.18, ABCDE is the required regular pentagon with sides 4 cm. It's construction details are not shown here.
• Once the construction is done, we can proceed to do the transformation:
• Draw diagonal BD 
• Draw MN parallel to BD through C
• On MN, mark F, in line with AB
Join B and D to F. The pentagon ABCDE has now become a quadrilateral AFDE. Now work on the left side:
• Draw diagonal AD 
• Draw M'N' parallel to AD through E
• On M'N', mark G, in line with AB
Join A and D to G. GFD is the required triangle. Now we want it's area
• From the actual construction, the required lengths can be measured. Base GF is already drawn. To obtain height, draw a perpendicular DH to AB
• By measurement, base AE = 8 cm and Height DF = 6.5 cm
• So Area = ¹⁄₂ × b × h = ¹⁄₂ × 8 × 6.5 = 26 cm²

Solved example 14.11
Fig.14.19(a) shows a rectangle ABCD. It is split into an yellow part and a green part. The lines EF and FG does this separation. Instead of the two lines, draw a single line to make the separation in such a way that the areas of the two parts remain the same. Calculate the area of the two parts

Fig.14.19

Solution:
• We are given the fig.14.19(a). It has all the required details for us to draw it in our note book:
• The rectangle has length 5 cm, and width 3 cm. So the rectangle can be easily drawn
• Mark a point G on CD, 1 cm from D. Mark a point E on AB 2 cm from A. So we get G and E. The only remaining point is F. It's details are also given
• Point F satisfies two conditions: It is 3 cm from side AD and 2 cm from side AB
• So draw a line UV (see fig.b), parallel to AD, at a distance of 3 cm from AD
• Draw line XY, parallel to AB, at a distance of 2 cm from AB
• The point of intersection of UV and XY is the point F
• Now we can proceed to solve the actual problem
• AEFGD is a pentagon. Two of it's sides, EF and FG is separating the rectangle into two parts.
• We want the two lines to become one. Then the pentagon will become a quadrilateral.
• The area of the new quadrilateral should be same as the original pentagon
• For that, draw the diagonal EG. Draw MN parallel to EG through F
• MN will intersect AB at H. Join H to G. AHGD is the required quadrilateral.
• The final two areas are shown in fig.(d). We can now proceed to find the areas:
• AHGD is a quadrilateral. Not just a quadrilateral, it is a non isosceles trapezium. We have seen how to find it's area
• We have, Area = ¹⁄₂ × (a + b)d.  In this problem, a = 1, b = 4 (by measurement), and d = 3 cm
• Area = ¹⁄₂ × (1 + 4) × 3 = 7.5 cm²
• Consider the green portion HBCG. It is an inverted shape of the yellow portion (∵ it has the parallel sides equal to 1 cm and 4 cm, and the height 3 cm)
• So the area of the green portion is also 7.5

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In the next section we will see unchanging areas of parallelograms.

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Chapter 14.2 - Triangle with the same area as a Quadrilateral

In the previous section we completed the discussion on 'unchanging areas of triangles between parallel lines'. In this section, we will see the relation between the areas of quadrilaterals and triangles.

• Fig.14.11(a) below, shows a quadrilateral ABCD

Fig.14.11

• In the fig.b, a diagonal BD is drawn. This diagonal splits the quadrilateral into two triangles: ΔABD and ΔCBD. Through C, draw a line MN, parallel to the diagonal BD. This is shown in fig.b. 
• Extend the side AB towards the right. It will meet MN at E. This is shown in the fig.c
• Join B and D to E. We will get a new ΔEBD as shown in fig.d
• What is the peculiarity of this new ΔEBD ?
ΔEBD and ΔCBD has the same base BD. Their third vertices E and C lies on the same line MN parallel to the base BD, and passing through C. So ΔEBD has the same area as ΔCBD
• Now ignore the original ΔCBD. Because, we have it's area in another shape, which is ΔEBD.
We split the original quadrilateral into two triangles, and now, one of them has transformed into another shape with the same area. So the total area of the quadrilateral remains unchanged
• Besides 'keeping the area the same', we achieve one more thing in this transformation:
Look at the final shape in fig.e: AED is a triangle. This was achieved by marking 'E'  in line with the side AB. When one vertex comes in line with another two vertices, one side will disappear. This is shown in the animation in fig.14.12 below:

Fig.14.12

In fig.b, the vertex 'C' comes in line with side AB. So two sides: AB and BC becomes one side AE
• So we transformed a '4 sided quadrilateral' into a '3 sided triangle' with the same area.

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Figs. (a) to (e) in 14.11 were drawn to show each step of the transformation. In an actual construction, We can do all the steps in a single fig. This is shown in the fig. 14.13 below:

Fig.14.13

The fig.14.13 was drawn quickly by following the steps below:
• Draw diagonal BD 
• Draw MN parallel to BD through C
• On MN, mark E, in line with AB
• Join B and D to E. This gives the transformed shape: ΔAED

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So now we know how to transform a quadrilateral into a triangle of same area. The same method can be used to transform any polygon into a triangle of same area. But we may have to do the above process many number of times. The 'number of times' will depend on the number of sides of the polygon. For example, if we are given a pentagon, we will apply the method once to get a quadrilateral. So the five sides of the pentagon will become four sides of a quadrilateral. Then we apply the method a second time to change the quadrilateral into a triangle. Like wise, if we are given a hexagon, we will have to apply the method 3 times. Fig.14.14 below shows the transformation of a pentagon.

• Fig.14.14(a) shows the original pentagon
• In fig.(b), the pentagon is split into a quadrilateral and a triangle, by drawing the diagonal BD
• In fig(c), one side of the pentagon is reduced, by bringing the vertex C to F. Thus we get a quadrilateral AFDE. This quadrilateral has the same area as the original pentagon ABCDE
• In fig(d), the method is applied again. This time, on the left side of the quadrilateral. (This is because, if we continue working on the right side, the resulting triangle will be too elongated towards the right. The reader may try that too). In fig.(d), the quadrilateral is split into two triangles.   
• In fig(e), the vertex E is brought down to G. We get the final ΔGFD

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■ While doing the actual construction, the above steps can be done in a single fig. This is shown in fig(f). The steps involved are:
• Draw the diagonal BD. Draw MN through C, parallel to BD
• Extend AB towards the right, to meet MN at F. Join B and D to F
• Now work on the left side of the pentagon. Draw diagonal AD. Draw M'N' through E, parallel to AD
• Extend AB towards the left, to meet M'N' at G. Join A and D to G. ΔGFD is the required triangle

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In the next section we will see some solved examples.

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