In the previous section we saw the details about excircles. We saw that any triangle will have three excircles. For our discussion in this section, we will need one of those excircles.
1. Consider the cyan excircle in fig.32.65(a) below:
P, Q and R are the points at which the excircle touches the sides of the ΔABC
2. Lengths of the sides of ΔABC are: a cm, b cm and c cm (a opposite vertex A, b opposite vertex B and c opposite vertex C)
3. We know that the lengths BP and BR will be equal (∵ they are tangents from the same point B).
• But we want to find their actual lengths
4. We have:
• Length of tangent BP = BA + AP = c + AP (∵ BA = c)
• Length of tangent BR = BC + CR = a + CR (∵ BC = a)
5. Now we change AP and CR.
• What other value can be put in the place of AP?
Ans: We can put AQ in the place of AP because AP = AQ (Both are tangents from the same point A)
♦ This is indicated by the two yellow lines in fig.b
• What other value can be put in the place of CR?
Ans: We can put CQ in the place of CR because CQ = CR (Both are tangents from the same point C)
♦ This is indicated by the two magenta lines in fig.b
6. So (4) will become:
• Length of tangent BP = c + AQ
• Length of tangent BR = a + CQ
7. Let us add the two lengths:
(Length of tangent BP + Length of tangent BR) = [(c+AQ) + (a+CQ)] = [(a+b) + (AQ+CQ)]
8. But (AQ+CQ) = AC = b
• So (7) becomes:
(Length of tangent BP + Length of tangent BR) = [(a+b+c)]
9. But length of tangent BP = length of tangent BR
• Let us put it as l. So l is the length of each of the two tangents of an excircle from the remote vertex of the triangle.
• Also we know that (a+b+c) = perimeter p of the triangle
• Thus we get:
2l = (a+b+c) = p ⟹ l =p⁄2 = s
♦ Where s is the semi perimeter
10. So we can write:
■ 'Length of each of the two tangents of an excircle from the remote vertex of the triangle' is equal to the 'semi perimeter of the triangle'.
Now we will see an application of the result in (10) above
1. In fig.32.66(a) below, both the excircle and incircle are shown.
Note: The line joinig B and O is passing through O' also. Why is that so?
Ans: • Lines BA and BC are tangential to both the circles
• So centers of both the circle will lie on the bisector of the angle between BA and BC
• Line BO' is indeed that angle bisector
2. Two known lengths are already marked:
(i) The length BP will be equal to the semi perimeter s
This we get from (10) above
(ii) The length BT will be equal to (s-b)
This we get from the pattern that we saw in the previous section
3. The radius of the excircle is marked as r1. The radius of the incircle is marked as r
• We have a larger outer triangle: ΔO'PB
• We have a smaller inner triangle: ΔOTB
4. We will prove that, these two triangles are similar:
• The two triangles are shown separately in fig.b
• The angle at B will obviously be the same in both triangles. They are shown in white colour.
• Angle at P in ΔO'PB = Angle at T in ΔOTB (Both being 90o)
• Two angles are same. So the third angle will also be the same. So we get:
• Angle at O' in O'PB = Angle at O in OTB. They are shown in magenta colour.
Thus the two triangles are indeed similar
5. Now we apply a special property that is applicable to any two similar triangles (Details here):
= side opposite P in ΔO'PB⁄side opposite T in ΔOTB
= side opposite B in ΔO'PB⁄side opposite B in ΔOTB
8. So we get: BP⁄BT = O'B⁄OB = OP⁄OT
This is same as: s⁄(s-b) = O'B⁄OB = r1⁄r
9. Take the second and the third ratios. We get:
In the next section, we will see a few more solved examples.
1. Consider the cyan excircle in fig.32.65(a) below:
 |
| Fig.32.65 |
2. Lengths of the sides of ΔABC are: a cm, b cm and c cm (a opposite vertex A, b opposite vertex B and c opposite vertex C)
3. We know that the lengths BP and BR will be equal (∵ they are tangents from the same point B).
• But we want to find their actual lengths
4. We have:
• Length of tangent BP = BA + AP = c + AP (∵ BA = c)
• Length of tangent BR = BC + CR = a + CR (∵ BC = a)
5. Now we change AP and CR.
• What other value can be put in the place of AP?
Ans: We can put AQ in the place of AP because AP = AQ (Both are tangents from the same point A)
♦ This is indicated by the two yellow lines in fig.b
• What other value can be put in the place of CR?
Ans: We can put CQ in the place of CR because CQ = CR (Both are tangents from the same point C)
♦ This is indicated by the two magenta lines in fig.b
6. So (4) will become:
• Length of tangent BP = c + AQ
• Length of tangent BR = a + CQ
7. Let us add the two lengths:
(Length of tangent BP + Length of tangent BR) = [(c+AQ) + (a+CQ)] = [(a+b) + (AQ+CQ)]
8. But (AQ+CQ) = AC = b
• So (7) becomes:
(Length of tangent BP + Length of tangent BR) = [(a+b+c)]
9. But length of tangent BP = length of tangent BR
• Let us put it as l. So l is the length of each of the two tangents of an excircle from the remote vertex of the triangle.
• Also we know that (a+b+c) = perimeter p of the triangle
• Thus we get:
2l = (a+b+c) = p ⟹ l =p⁄2 = s
♦ Where s is the semi perimeter
10. So we can write:
■ 'Length of each of the two tangents of an excircle from the remote vertex of the triangle' is equal to the 'semi perimeter of the triangle'.
Now we will see an application of the result in (10) above
1. In fig.32.66(a) below, both the excircle and incircle are shown.
 |
| Fig.32.66 |
Ans: • Lines BA and BC are tangential to both the circles
• So centers of both the circle will lie on the bisector of the angle between BA and BC
• Line BO' is indeed that angle bisector
2. Two known lengths are already marked:
(i) The length BP will be equal to the semi perimeter s
This we get from (10) above
(ii) The length BT will be equal to (s-b)
This we get from the pattern that we saw in the previous section
3. The radius of the excircle is marked as r1. The radius of the incircle is marked as r
• We have a larger outer triangle: ΔO'PB
• We have a smaller inner triangle: ΔOTB
4. We will prove that, these two triangles are similar:
• The two triangles are shown separately in fig.b
• The angle at B will obviously be the same in both triangles. They are shown in white colour.
• Angle at P in ΔO'PB = Angle at T in ΔOTB (Both being 90o)
• Two angles are same. So the third angle will also be the same. So we get:
• Angle at O' in O'PB = Angle at O in OTB. They are shown in magenta colour.
Thus the two triangles are indeed similar
5. Now we apply a special property that is applicable to any two similar triangles (Details here):
side opposite smallest angle in ΔO'PB⁄side opposite smallest angle in ΔOTB
= side opposite medium angle in ΔO'PB⁄side opposite medium angle in ΔOTB
= side opposite largest angle in ΔO'PB⁄side opposite largest angle in ΔOTB
6. But angles in the two triangles are the same. That is.,
• Smallest angle in ΔO'PB = Smallest angle in ΔOTB
• Medium angle in ΔO'PB = Medium angle in ΔOTB
• Largest angle in ΔO'PB = Largest angle in ΔOTB
7. So we can write this:
7. So we can write this:
• Ratio of the sides opposite equal angles in the two similar triangles are the same. That is.,
side opposite O' in ΔO'PB⁄side opposite O in ΔOTB= side opposite P in ΔO'PB⁄side opposite T in ΔOTB
= side opposite B in ΔO'PB⁄side opposite B in ΔOTB
8. So we get: BP⁄BT = O'B⁄OB = OP⁄OT
This is same as: s⁄(s-b) = O'B⁄OB = r1⁄r
9. Take the second and the third ratios. We get:
s⁄(s-b) = r1⁄r ⟹ (s-b)⁄s = r⁄r1
• We will need this result after a few more steps
10. In the fig.32.67(a) below, the angle bisector O'B is avoided. Instead two new lines are drawn:
O'A and OA
11. Two known lengths are already marked:
(i) The length AT will be equal to (s-a)
This we get from the pattern that we saw in the previous section
(ii) The length AP will be equal to (s-c)
• This is calculated as follows:
AP = (BP-AB) = (s-c)
12. There are two triangles: ΔO'PA and ΔOTA.
We have to prove them they are similar
13. We have:
∠O'AP = ∠O'AC = 1⁄2 × ∠PAC [∵ O'A is the bisector of ∠PAC]
14. But ∠PAC = (180-∠BAC) [∵ ∠PAC and ∠BAC form a linear pair]
• So (13) becomes:
∠O'AP = ∠O'AC = 1⁄2 × (180-∠BAC) = [90- (1⁄2 × ∠BAC)]
• Thus the angle O'AP in ΔO'PA is [90- (1⁄2 × ∠BAC)]
♦ This is shown in white colour in fig.32.67(b)
• The same angle is present in ΔOTA also. Which is it?
15. We have:
∠O'AT = ∠OAC = 1⁄2 × ∠BAC [∵ OA is the bisector of ∠BAC]
• So in the right triangle OAT,
♦ Angle at T is 90
♦ Angle at A is 1⁄2 × ∠BAC
• Then clearly, the third angle is: [90- (1⁄2 × ∠BAC)]
• So, the angle AOT in ΔOAT is [90- (1⁄2 × ∠BAC)]
♦ This is shown in white colour in fig.32.67(b)
16. So two angles are the same in both the triangles: The 90o angle and the one marked in white colour
• Then the third angle will also be the same. This is shown in magenta colour
• Thus the two triangles are indeed similar
17. Now we apply a special property that is applicable to any two similar triangles (Details here):
= side opposite A in ΔO'PA⁄side opposite O in ΔOTA
= side opposite P in ΔO'PA⁄side opposite T in ΔOTA
20. So we get: PA⁄OT = O'P⁄AT = O'A⁄OA
This is same as: (s-c)⁄r = r1⁄(s-a) = O'A⁄OA
21. Take the first and the second ratios. We get:
• We will need this result after a few more steps
10. In the fig.32.67(a) below, the angle bisector O'B is avoided. Instead two new lines are drawn:
O'A and OA
 |
| Fig.32.67 |
(i) The length AT will be equal to (s-a)
This we get from the pattern that we saw in the previous section
(ii) The length AP will be equal to (s-c)
• This is calculated as follows:
AP = (BP-AB) = (s-c)
12. There are two triangles: ΔO'PA and ΔOTA.
We have to prove them they are similar
13. We have:
∠O'AP = ∠O'AC = 1⁄2 × ∠PAC [∵ O'A is the bisector of ∠PAC]
14. But ∠PAC = (180-∠BAC) [∵ ∠PAC and ∠BAC form a linear pair]
• So (13) becomes:
∠O'AP = ∠O'AC = 1⁄2 × (180-∠BAC) = [90- (1⁄2 × ∠BAC)]
• Thus the angle O'AP in ΔO'PA is [90- (1⁄2 × ∠BAC)]
♦ This is shown in white colour in fig.32.67(b)
• The same angle is present in ΔOTA also. Which is it?
15. We have:
∠O'AT = ∠OAC = 1⁄2 × ∠BAC [∵ OA is the bisector of ∠BAC]
• So in the right triangle OAT,
♦ Angle at T is 90
♦ Angle at A is 1⁄2 × ∠BAC
• Then clearly, the third angle is: [90- (1⁄2 × ∠BAC)]
• So, the angle AOT in ΔOAT is [90- (1⁄2 × ∠BAC)]
♦ This is shown in white colour in fig.32.67(b)
16. So two angles are the same in both the triangles: The 90o angle and the one marked in white colour
• Then the third angle will also be the same. This is shown in magenta colour
• Thus the two triangles are indeed similar
17. Now we apply a special property that is applicable to any two similar triangles (Details here):
side opposite smallest angle in ΔO'PA⁄side opposite smallest angle in ΔOTA
= side opposite medium angle in ΔO'PA⁄side opposite medium angle in ΔOTA
= side opposite largest angle in ΔO'PA⁄side opposite largest angle in ΔOTA
18. But angles in the two triangles are the same. That is.,
• Smallest angle in ΔO'PA = Smallest angle in ΔOTA
• Medium angle in ΔO'PA = Medium angle in ΔOTA
• Largest angle in ΔO'PA = Largest angle in ΔOTA
19. So we can write this:
19. So we can write this:
• Ratio of the sides opposite equal angles in the two similar triangles are the same. That is.,
side opposite O' in ΔO'PA⁄side opposite A in ΔOTA= side opposite A in ΔO'PA⁄side opposite O in ΔOTA
= side opposite P in ΔO'PA⁄side opposite T in ΔOTA
20. So we get: PA⁄OT = O'P⁄AT = O'A⁄OA
This is same as: (s-c)⁄r = r1⁄(s-a) = O'A⁄OA
21. Take the first and the second ratios. We get:
(s-c)⁄r = r1⁄(s-a) =
• Cross multiplying, we get:
rr1 = (s-a)(s-c)
22. Multiplying both sides by r⁄r1, we get: r⁄r1 × rr1 = r⁄r1 × (s-a)(s-c) ⟹ r2 = r⁄r1 × (s-a)(s-c)
23. But from (9), we have: r⁄r1 = (s-b)⁄s.
• Substituting this in (22) we get:
r2 = (s-b)⁄s × (s-a)(s-c) = [(s-a)(s-b)(s-c)]⁄s
24. From theorem 32.10 that we saw in the previous section, we have: r = A⁄s
♦ Where A is the area of the triangle
• Substituting this in (23), we get: (A⁄s)2 = [(s-a)(s-b)(s-c)]⁄s ⟹ A2 = [s(s-a)(s-b)(s-c)]
■ From this we get:
A = √[s(s-a)(s-b)(s-c)]
• This is known as the Heron's formula. It is named after Heron of Alexandria, a Greek Engineer and Mathematician.
Let us see a quick example:
A polygon ABCD has an irregular shape as shown in fig.32.68 below:
• It's area is calculated using a cad program. The above fig.32.62 is it's screen shot.
• The area is obtained as. It can be seen at the bottom left corner.
• Let us check using Heron's formula:
■ Applying Heron's formula for ΔABC:
• s = (a+b+c)⁄2 = (9.1+12.0+19.4)⁄2 = 40.5⁄2 = 20.25 cm
♦ (s-a) = (20.25-9.1) = 11.15 cm
♦ (s-b) = (20.25-12.0) = 8.25 cm
♦ (s-c) = (20.25-19.4) = 0.85 cm
• A = √[s(s-a)(s-b)(s-c)] = √[20.25×11.15×8.25×0.85] = √[1583.33] = 39.79 cm2.
■ Applying Heron's formula for ΔACD:
• s = (a+b+c)⁄2 = (15+15.2+19.4)⁄2 = 49.6⁄2 = 24.8 cm
♦ (s-a) = (24.8-15) = 9.8 cm
♦ (s-b) = (24.8-15.2) = 9.6 cm
♦ (s-c) = (24.8-19.4) = 5.4 cm
• A = √[s(s-a)(s-b)(s-c)] = √[24.8×9.8×9.6×5.4] = √[12599.19] = 112.25 cm2.
■ Total area = 39.79+112.25 = 152.04 cm2.
• The cad program gives an area of 151.13 cm2. They are approximately equal. If we increase the accuracy by increasing the number of decimal places, both results will become equal.
Now we will see a solved example:
Solved example 32.27
Draw a triangle of sides 4, 5 and 6 cm. (i) Draw it's incircle. (ii) Calculate the radius of that incircle.
Solution:
• Let us name the triangle as ABC.
• Let AB = 6 cm, AC = 5 cm and BC = 4 cm
Part (i):
Step 1: Draw ΔABC as usual
Step 2: Draw the angle bisectors of any two angles. We will choose the angles A and B.
• Mark the point of intersection as O. This is shown in fig.32.69(a) below:
Step 3: Drop a perpendicular from O onto the side AB. This is shown in fig.b
Step 4: With O as center and the perpendicular distance as radius, draw the incircle. This is shown in fig.c
Part (ii):
1. Area of ΔABC using Heron's formula:
• s = (a+b+c)⁄2 = (4+5+6)⁄2 = 15⁄2 = 7.5 cm
♦ (s-a) = (7.5-4) = 3.5 cm
♦ (s-b) = (7.5-5) = 2.5 cm
♦ (s-c) = (7.5-6) = 1.5 cm
• A = √[s(s-a)(s-b)(s-c)] = √[7.5×3.5×2.5×1.5] = √[98.44] = 9.9 cm2
2. From theorem 32.10, we have: r = A⁄s = 9.9⁄7.5 = 1.32 cm
• Cross multiplying, we get:
rr1 = (s-a)(s-c)
22. Multiplying both sides by r⁄r1, we get: r⁄r1 × rr1 = r⁄r1 × (s-a)(s-c) ⟹ r2 = r⁄r1 × (s-a)(s-c)
23. But from (9), we have: r⁄r1 = (s-b)⁄s.
• Substituting this in (22) we get:
r2 = (s-b)⁄s × (s-a)(s-c) = [(s-a)(s-b)(s-c)]⁄s
24. From theorem 32.10 that we saw in the previous section, we have: r = A⁄s
♦ Where A is the area of the triangle
• Substituting this in (23), we get: (A⁄s)2 = [(s-a)(s-b)(s-c)]⁄s ⟹ A2 = [s(s-a)(s-b)(s-c)]
■ From this we get:
A = √[s(s-a)(s-b)(s-c)]
• This is known as the Heron's formula. It is named after Heron of Alexandria, a Greek Engineer and Mathematician.
Let us see a quick example:
A polygon ABCD has an irregular shape as shown in fig.32.68 below:
 |
| Fig.32.68 |
• The area is obtained as. It can be seen at the bottom left corner.
• Let us check using Heron's formula:
■ Applying Heron's formula for ΔABC:
• s = (a+b+c)⁄2 = (9.1+12.0+19.4)⁄2 = 40.5⁄2 = 20.25 cm
♦ (s-a) = (20.25-9.1) = 11.15 cm
♦ (s-b) = (20.25-12.0) = 8.25 cm
♦ (s-c) = (20.25-19.4) = 0.85 cm
• A = √[s(s-a)(s-b)(s-c)] = √[20.25×11.15×8.25×0.85] = √[1583.33] = 39.79 cm2.
■ Applying Heron's formula for ΔACD:
• s = (a+b+c)⁄2 = (15+15.2+19.4)⁄2 = 49.6⁄2 = 24.8 cm
♦ (s-a) = (24.8-15) = 9.8 cm
♦ (s-b) = (24.8-15.2) = 9.6 cm
♦ (s-c) = (24.8-19.4) = 5.4 cm
• A = √[s(s-a)(s-b)(s-c)] = √[24.8×9.8×9.6×5.4] = √[12599.19] = 112.25 cm2.
■ Total area = 39.79+112.25 = 152.04 cm2.
• The cad program gives an area of 151.13 cm2. They are approximately equal. If we increase the accuracy by increasing the number of decimal places, both results will become equal.
Now we will see a solved example:
Solved example 32.27
Draw a triangle of sides 4, 5 and 6 cm. (i) Draw it's incircle. (ii) Calculate the radius of that incircle.
Solution:
• Let us name the triangle as ABC.
• Let AB = 6 cm, AC = 5 cm and BC = 4 cm
Part (i):
Step 1: Draw ΔABC as usual
Step 2: Draw the angle bisectors of any two angles. We will choose the angles A and B.
• Mark the point of intersection as O. This is shown in fig.32.69(a) below:
 |
| Fig.32.69 |
Step 4: With O as center and the perpendicular distance as radius, draw the incircle. This is shown in fig.c
Part (ii):
1. Area of ΔABC using Heron's formula:
• s = (a+b+c)⁄2 = (4+5+6)⁄2 = 15⁄2 = 7.5 cm
♦ (s-a) = (7.5-4) = 3.5 cm
♦ (s-b) = (7.5-5) = 2.5 cm
♦ (s-c) = (7.5-6) = 1.5 cm
• A = √[s(s-a)(s-b)(s-c)] = √[7.5×3.5×2.5×1.5] = √[98.44] = 9.9 cm2
2. From theorem 32.10, we have: r = A⁄s = 9.9⁄7.5 = 1.32 cm
