Chapter 32.11 - Solved examples on Incircles and Heron's formula

In the previous section we saw the Heron's formula. We also saw a solved example. In this section, we will see a few more solved examples.

Solved example 32.28
Draw an equilateral triangle and a semicircle touching it's two sides as shown in fig.32.70(a) below:

Fig.32.70

The diameter of the semicircle lies on the base of the triangle
Solution:
• A rough sketch is shown in fig.32.70(b)
(i) The triangle is named as ΔABC. For clarity, the full circle is drawn in the rough sketch.
(ii) Since the triangle is equilateral, the perpendicular dropped from C will pass through the midpoint of base AB
(iii) Also, the midpoint of Ab will be the center of the circle
(iv) The perpendicular OC will bisect the angle at C. So ∠ACO = ∠BCO = 30^o
(v) The radial line from O to the point of contact P will make an angle of 90^o with tangent BC. So in ΔOPC, ∠COP = 60^o.
Now we can begin the construction:
1. Draw a horizontal line shown in red colour in fig.32.71.a below:

Fig.32.71

2. Mark a point O on the line and through it, draw a vertical red dashed line
3. With O as center, draw the upper part of the circle with any convenient radius
4. Draw the green line at an angle of 60^o with the vertical dashed line. Let it meet the circle at P. This is shown in fig.b
5. Draw the red line perpendicular to OP at P. It intersect the vertical dashed line at C and the horizontal red line at B
6. Through C, draw a line inclined at 30^o with the vertical dashed line. It will intersect the horizontal red line at A. This is shown in fig.c
7. Thus ΔABC is completed. Sides AC and BC will be tangential to the circle

Solved example 32.29
Prove that the radius of the incircle of an equilateral triangle is half the radius of it's circumcircle
Solution:
We have to consider the two quantities:
(i) Radius of the incircle of an equilateral triangle
(ii) Radius of the circumcircle of that same equilateral triangle
We have to prove that (i) is half of (ii)
Let us write the steps:
1. In fig.32.72 below, ABC is the equilateral triangle

Fig.32.72

The circumcircle is shown in yellow colour
The incircle is shown in blue colour
Since the triangle is equilateral, both the circles will have the same center O
2. The sides of any triangle will be tangential to the incircle. 
So side AB is tangential to the blue circle
Thus, the radial line OP will be perpendicular to the side AB
3. The center O will lie on the angle bisector of the angle at A
So O will lie on a line which is inclined at 30^o with base AB (∵ ⁶⁰⁄₂ = 30)
4. Thus we get a 30 60 right triangle OAP. 
In such a triangle, the hypotenuse will be 2 times the smallest side. (Details here)
So OA is two times OP
In other words, OP is half of OA
But OP is the radius of the incircle and OA is the radius of the circumcircle. Hence proved

Solved example 32.30
Prove that if the hypotenuse of a right triangle is h and the radius of it's incircle is r, then it's area is r(h+r)
Solution:
• Fig.32.73(a) shows the given right triangle

Fig.32.73

• Let us name it as ABC and split the sides. This is shown in fig.b
1. Consider the radial lines OP and OQ:
• The radial line OP will be parallel to side AB (∵ BC is perpendicular to AB)
• The radial line OQ will be parallel to side BC (∵ AB is perpendicular to BC)
• The lengths of all radial lines is r 
2. So in quadrilateral OQBP, we have:
• Two adjacent sides OP and OQ equal in length
• 90^o at all the four vertices
• So OQBP is a square
• Thus we get: OP = OQ = r = y
3. Perimeter of ΔABC = [AB+BC+AC] = [(x+y)+(y+z)+(x+z)] = [(x+r)+(r+z)+h] 
= [(x+z)+2r+h] = [h+2r+h] = [2r+2h] = 2[h+r]
4. So half of perimeter = s = [h+r]
• We know that Area of the triangle = A = rs (Details here)
• Thus area of our ΔABC = rs = r[h+r]

Solved example 32.31
Calculate the area of a triangle of sides 13, 14 and 15 cm
Solution:
1. Using Heron's formula:
• s = ^(a+b+c)⁄₂ = ^(13+14+15)⁄₂ = ⁴²⁄₂ = 21 cm
    ♦ (s-a) = (21-13) = 8 cm
    ♦ (s-b) = (21-14) = 7 cm
    ♦ (s-c) = (21-15) = 6 cm
• A = √[s(s-a)(s-b)(s-c)] = √[21×8×7×6] = √[7056] = 84 cm².

Solved example 32.32
In fig.32.74(a), PQ is a diameter of the circle with center O. AB and CD are two tangents drawn at p and Q respectively. 

Fig.32.74

Another tangent intersects AB and CD at M and N. Prove that ∠MON is 90^o.
Solution:
1. AB and CD are tangents at the ends of a diameter. So AB and CD are parallel. (See Solved example 32.5)
2. MN is a transversal cutting two parallel lines. So we have: ∠MNC = ∠BMN
3. Two tangents are drawn from N. Then ON is an angle bisector. 
• So we have: ∠ONQ = ∠ONR = (¹⁄₂ ×  ∠MNC)  
• Let us denote this as θ. In fig.b, ∠ONR is marked as θ.
4. From (3) we get ∠MNC = 2∠ONR
• So from (2) we get: ∠BMN = 2∠ONR = 2θ 
5. But ∠BMN and ∠PMN forms a linear pair. So we get:
∠PMN = (180-∠BMN) = (180-2θ)
6. Two tangents are drawn from M. Then OM is an angle bisector. 
• So we have: ∠OMP = ∠OMR = (¹⁄₂ ×  ∠PMN) 
7. But from (5), ∠PMN = (180-2θ)
• So we get: ∠OMP = ∠OMR = (¹⁄₂ ×  ∠PMN) = [¹⁄₂ × (180-2θ)] = (90-θ)
• Thus, ∠OMR is marked as (90-θ) in fig.b
8. Now consider ⊿ONR. It is a right triangle.
• One of it's angle is θ. So the third angle will be (90-θ)
• Thus ∠RON is marked as (90-θ) in fig.c
9. Now consider ⊿OMR. It is a right triangle.
• One of it's angle is (90-θ). So the third angle will be θ
• Thus ∠ROM is marked as θ in fig.c
10. Now we get: ∠MON = [θ+(90-θ)] = 90^o

Solved example 32.33
The radii of two concentric circles are 13 cm and 8 cm. AB is the diameter of the larger circle. A chord BE of the larger circle touches the smaller circle at D. Find the distance AD
Solution:
1. The rough sketch based on given data is shown in fig.32.75(a) below:

Fig.32.75

• Join O and D. Then OD will be perpendicular to BC. (since BC is a tangent to the inner circle at D). This is shown in fig.b
2. So we have a right triangle: ⊿OBD
• Applying Pythagoras theorem, we get:
BD² = OB² - OD² ⟹ BD² = 13² - 8² ⟹ BD² = 169 - 64 ⟹ BD² = 105 ⟹ BD = √105
3. BC is a chord of the larger circle and OD is a perpendicular from the center. So OD will bisect the chord BC.
• Thus we get: BC = 2BD = 2√105
4. Since BCA is a semi circle, ∠BCA will be 90^o
• So we have a right triangle: ⊿BCA
• Applying Pythagoras theorem, we get:
AC² = AB² - BC² ⟹ AC² = 26² - (2√105)² ⟹ AC² = [676 - (4×105)] 
⟹ AC² = [676 - (420)] = [256] ⟹ AC = √256 = 16 cm
5. Consider the right triangle: ⊿ACD
• Applying Pythagoras theorem, we get:
AD² = AC² + DC² ⟹ AD² = 16² + (√105)² ⟹ AD² = [256 + (105)] = [361] 
⟹ AC = √361 = 19 cm

Solved example 32.34
In the fig.32.76(a) below, tangents are drawn at P and R. The tangent at P is AT and tangent at R is BT. They intersect at T such that ∠ATB = 30^o. 

Fig.32.76

If PQ is parallel to BT, calculate ∠PRQ
Solution:
1. We know that tangents drawn from a point have the same length. So TP = TR
2. Thus the ΔPTR is an isosceles triangle. The base angles will be equal. So we get:
∠TPR = ∠TRP = [¹⁄₂ ×(180-30)] = [¹⁄₂ ×(150)] = 75^o. This is shown in fig.b
3. The angles at the extremities of chord PR on the 'tangent side' is 75^o
• The angle made by chord PR at any point on the circle on the 'non-tangent side' will also be 75^o. (Theorem 32.7)
• Thus we get: PQR = 75^o
4. Given that PQ is parallel to BT. So we get: ∠QPR = ∠PRT = 75^o
• Thus we get two angles in the PQR: ∠PQR and ∠QPR
• The third angle ∠PRQ = [180-(75+75)] = [180-150] = 30^o

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In the next section, we will see Solids.

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Chapter 32.10 - The Heron's Formula

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:

Fig.32.65

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'.

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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

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 90^o)
• 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:

• 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

Fig.32.67

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 90^o 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:

• 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.

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Let us see a quick example:
A polygon ABCD has an irregular shape as shown in fig.32.68 below:

Fig.32.68

• 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.

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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 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

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In the next section, we will see a few more solved examples.

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