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
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(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 60o. Then draw a right triangle of the same area
Solution:
• 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:
• 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
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
In the next section we will see unchanging areas of parallelograms.
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 |
■ 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 |
• 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 60o. Then draw a right triangle of the same area
Solution:
 |
| Fig.14.17 |
• 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 |
• 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 |
• 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
In the next section we will see unchanging areas of parallelograms.
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