In the previous section we completed the discussion on the division of triangles. We also saw some solved examples. In this section we will see some more solved examples that demonstrate the topics that we have discussed in this chapter as a whole.
Solved example 14.20
ABCD is a trapezium. It’s diagonals AC and BD meet at O. Prove that the magenta coloured ΔAOD and the red coloured ΔBOC have the same area.
Solution:
1. In any trapezium, two opposite sides are parallel. In the fig., the parallel sides are AB and CD
2. In the fig., ΔABD and ΔABC have the same area. [since they are triangles between the same parallels, and they have the same base AB]
3. So we can write: ar (ABD) = ar (ABC)
4. But [ar (ABD) = ar (ABO) + ar (AOD)] and [ar (ABC) = ar (ABO) + ar (BOC)]
5. Substituting these values in 3, we get:
[ar (ABO) + ar (AOD)] = [ar (ABO) + ar (BOC)] That is., the yellow ΔABO is common to both
6. ar (ABO) present on both sides will cancel out. So we get:
ar (AOD) = ar (BOC)
Solved example 14.21
In the previous example, what is the total area of the trapezium ABCD, if the area of the blue triangle is 4 cm2 and yellow triangle is 9 cm2
Solution:
1. In the previous example, we have already proved that ar (AOD) = ar (BOC). Let each be equal to x. That is.,
2. Let ar (AOD) = ar (BOC) = x cm2
3. From fig.14.33(b) we get • ar (AOD) = 1⁄2 × AO × h1 and • ar (COD) = 1⁄2 × CO × h1
4. Also we get • ar (AOB) = 1⁄2 × AO × h2 and • ar (COB) = 1⁄2 × CO × h2
5. From (3) and (4) we get
6. So we get, total area of the trapezium = 9 + 4 + x + x = 9 + 4 + 6 + 6 = 25 cm2
Solved example 14.22
In fig.14.34(a), CD is a median of the ABC. This median CD is divided at E in such a way that CE : DE = 2:1. Prove that area of each triangle in fig.13.34(b) is one third of the whole area of ΔABC
Solution:
1. Consider ΔADC. It is split into two triangles: ΔADE and ΔACE.
2. Given that CE : DE = 2 :1. So ar (ACE) : ar (ADE) = 2 :1
3. That means ar (ACE) = 2 × ar (ADE)
4. In a similar way, ar (BCE) = 2 × ar (BDE)
5. Consider ΔABE. It is split into two triangles: ΔADE and ΔBDE
6. CD is a median. So AD : BD = 1 : 1. So ar (ADE) = ar (BDE)
7. So we can put ar (ADE) in the place of ar (BDE) in (4)
8. We get ar (BCE) = 2 × ar (ADE)
9. Compare (3) and (8). The right sides are the same. So left sides also must be equal
10. We get ar (ACE) = ar (BCE)
11. Take the sum of two triangles: ΔADE and ΔBDE:
12. We get ar (ADE) + ar (BDE) = ar (ABE). But from (6), ar (BDE) = ar (ADE)
13. ∴ 2 × ar (ADE) = ar (ABE) - - -(6)
14. Comparing the above with (3) we get ar (ACE) = ar (ABE)
15. Comparing (14) and (10) we get ar (ACE) = ar (ABE) = ar (BCE)
16. The sum of the 3 triangles in (15) is the total area ar (ABC). Each of the three are equal. That means each triangle is equal to one third of the total area
Solved example 14.23
In fig.14.35(a), ABCD is a parallelogram. AB is extended to any point P. Line AQ is drawn through A, parallel to PC. AQ meets CB produced at Q. Parallelogram BQRP is completed by drawing QR and RP. Prove that ar (ABCD) = ar (BQRP). [Hint: Draw the diagonals of the parallelograms]
Solution:
1. The diagonals AC and PQ are added to the fig. in 14.25(a). The modified fig. is shown in (b)
2. Given that PC is parallel to AQ
3. So AQC and AQP are two triangles with the same base, and between same parallels.
4. Thus, they have the same area. That is., ar (AQC) = ar (AQP)
5. Let us split the above two areas:
• ar (AQC) = ar (ABC) + ar (AQB)
• ar (AQP) = ar (BQP) + ar (AQB)
6. Let us equate as in (4): [ar (ABC) + ar (AQB)] = [ar (BQP) + ar (AQB)]
7. ar (AQB) is common. It will cancel out.
8. So we get ar (ABC) = ar (BQP)
9. In (8) above, each is half of the corresponding parallelogram. (∵ AC and PQ are diagonals)
10. Doubling each will give the corresponding parallelogram. So we get ar (ABCD) = ar (BQRP)
Solved example 14.24
Prove that the perpendiculars drawn from any point on the angle bisector to the sides are equal
Solution:
1. The diagram for this example is given in fig.14.36(a) below:
We have:
• an ∠BCA, • it's bisector BG, • 'any point' D on the bisector, • perpendiculars DE and DF from D, to the sides
2. We have to prove that DF and DE are equal
3. We have proved the above equality when we discussed Theorem 14.7
4. In fact, the fig. 14.36(a) given above is the same fig.14.28(b) that we saw when we discussed the theorem
5. There we proved that DF and DE are equal, and so, indicated each of them as 'h'. The same steps can be followed in this example also
Solved example 14.25
In the fig.14.36(b), ABCD is a rectangle of length 14 cm, and width 6 cm. E is the midpoint of BC. F is the mid point of AE. Find the areas of ΔABF and ΔEBF
Solution:
1. Consider ΔABE. Base AB = 14 cm. Height BE = 6/2 = 3 cm (∵ E is the midpoint of BC)
2. So ar (ABE) = 1⁄2 × 14 × 3 = 21 cm2
3. F is the midpoint of AE. So BF is a median of ΔABE
4. So ar (ABF) = ar (EBF) [By Theorem 14.5]
5. If the two areas are equal, each must be exact half of the total ar (ABE)
6. But ar (ABE) = 21 cm2
7. So we get ar (ABF) = ar (EBF) = 21/2 = 10.5 cm2
This completes the discussion on Triangles. In the next section we will see 'Pairs of Equations'.
Solved example 14.20
ABCD is a trapezium. It’s diagonals AC and BD meet at O. Prove that the magenta coloured ΔAOD and the red coloured ΔBOC have the same area.
 |
| Fig.14.33 |
1. In any trapezium, two opposite sides are parallel. In the fig., the parallel sides are AB and CD
2. In the fig., ΔABD and ΔABC have the same area. [since they are triangles between the same parallels, and they have the same base AB]
3. So we can write: ar (ABD) = ar (ABC)
4. But [ar (ABD) = ar (ABO) + ar (AOD)] and [ar (ABC) = ar (ABO) + ar (BOC)]
5. Substituting these values in 3, we get:
[ar (ABO) + ar (AOD)] = [ar (ABO) + ar (BOC)] That is., the yellow ΔABO is common to both
6. ar (ABO) present on both sides will cancel out. So we get:
ar (AOD) = ar (BOC)
Solved example 14.21
In the previous example, what is the total area of the trapezium ABCD, if the area of the blue triangle is 4 cm2 and yellow triangle is 9 cm2
Solution:
1. In the previous example, we have already proved that ar (AOD) = ar (BOC). Let each be equal to x. That is.,
2. Let ar (AOD) = ar (BOC) = x cm2
3. From fig.14.33(b) we get • ar (AOD) = 1⁄2 × AO × h1 and • ar (COD) = 1⁄2 × CO × h1
4. Also we get • ar (AOB) = 1⁄2 × AO × h2 and • ar (COB) = 1⁄2 × CO × h2
5. From (3) and (4) we get
Solved example 14.22
In fig.14.34(a), CD is a median of the ABC. This median CD is divided at E in such a way that CE : DE = 2:1. Prove that area of each triangle in fig.13.34(b) is one third of the whole area of ΔABC
 |
| Fig.14.34 |
1. Consider ΔADC. It is split into two triangles: ΔADE and ΔACE.
2. Given that CE : DE = 2 :1. So ar (ACE) : ar (ADE) = 2 :1
3. That means ar (ACE) = 2 × ar (ADE)
4. In a similar way, ar (BCE) = 2 × ar (BDE)
5. Consider ΔABE. It is split into two triangles: ΔADE and ΔBDE
6. CD is a median. So AD : BD = 1 : 1. So ar (ADE) = ar (BDE)
7. So we can put ar (ADE) in the place of ar (BDE) in (4)
8. We get ar (BCE) = 2 × ar (ADE)
9. Compare (3) and (8). The right sides are the same. So left sides also must be equal
10. We get ar (ACE) = ar (BCE)
11. Take the sum of two triangles: ΔADE and ΔBDE:
12. We get ar (ADE) + ar (BDE) = ar (ABE). But from (6), ar (BDE) = ar (ADE)
13. ∴ 2 × ar (ADE) = ar (ABE) - - -(6)
14. Comparing the above with (3) we get ar (ACE) = ar (ABE)
15. Comparing (14) and (10) we get ar (ACE) = ar (ABE) = ar (BCE)
16. The sum of the 3 triangles in (15) is the total area ar (ABC). Each of the three are equal. That means each triangle is equal to one third of the total area
Solved example 14.23
In fig.14.35(a), ABCD is a parallelogram. AB is extended to any point P. Line AQ is drawn through A, parallel to PC. AQ meets CB produced at Q. Parallelogram BQRP is completed by drawing QR and RP. Prove that ar (ABCD) = ar (BQRP). [Hint: Draw the diagonals of the parallelograms]
 |
| Fig.14.35 |
1. The diagonals AC and PQ are added to the fig. in 14.25(a). The modified fig. is shown in (b)
2. Given that PC is parallel to AQ
3. So AQC and AQP are two triangles with the same base, and between same parallels.
4. Thus, they have the same area. That is., ar (AQC) = ar (AQP)
5. Let us split the above two areas:
• ar (AQC) = ar (ABC) + ar (AQB)
• ar (AQP) = ar (BQP) + ar (AQB)
6. Let us equate as in (4): [ar (ABC) + ar (AQB)] = [ar (BQP) + ar (AQB)]
7. ar (AQB) is common. It will cancel out.
8. So we get ar (ABC) = ar (BQP)
9. In (8) above, each is half of the corresponding parallelogram. (∵ AC and PQ are diagonals)
10. Doubling each will give the corresponding parallelogram. So we get ar (ABCD) = ar (BQRP)
Solved example 14.24
Prove that the perpendiculars drawn from any point on the angle bisector to the sides are equal
Solution:
1. The diagram for this example is given in fig.14.36(a) below:
 |
| Fig.14.36 |
• an ∠BCA, • it's bisector BG, • 'any point' D on the bisector, • perpendiculars DE and DF from D, to the sides
2. We have to prove that DF and DE are equal
3. We have proved the above equality when we discussed Theorem 14.7
4. In fact, the fig. 14.36(a) given above is the same fig.14.28(b) that we saw when we discussed the theorem
5. There we proved that DF and DE are equal, and so, indicated each of them as 'h'. The same steps can be followed in this example also
Solved example 14.25
In the fig.14.36(b), ABCD is a rectangle of length 14 cm, and width 6 cm. E is the midpoint of BC. F is the mid point of AE. Find the areas of ΔABF and ΔEBF
Solution:
1. Consider ΔABE. Base AB = 14 cm. Height BE = 6/2 = 3 cm (∵ E is the midpoint of BC)
2. So ar (ABE) = 1⁄2 × 14 × 3 = 21 cm2
3. F is the midpoint of AE. So BF is a median of ΔABE
4. So ar (ABF) = ar (EBF) [By Theorem 14.5]
5. If the two areas are equal, each must be exact half of the total ar (ABE)
6. But ar (ABE) = 21 cm2
7. So we get ar (ABF) = ar (EBF) = 21/2 = 10.5 cm2
This completes the discussion on Triangles. In the next section we will see 'Pairs of Equations'.
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