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:
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 = 30o
(v) The radial line from O to the point of contact P will make an angle of 90o with tangent BC. So in ΔOPC, ∠COP = 60o.
Now we can begin the construction:
1. Draw a horizontal line shown in red colour in fig.32.71.a below:
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 60o 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 30o 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
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 30o 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
• 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
• 90o 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.
Another tangent intersects AB and CD at M and N. Prove that ∠MON is 90o.
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)] = 90o
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:
• 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 90o
• 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 = 30o.
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)] = 75o. This is shown in fig.b
3. The angles at the extremities of chord PR on the 'tangent side' is 75o
• The angle made by chord PR at any point on the circle on the 'non-tangent side' will also be 75o. (Theorem 32.7)
• Thus we get: PQR = 75o
4. Given that PQ is parallel to BT. So we get: ∠QPR = ∠PRT = 75o
• Thus we get two angles in the PQR: ∠PQR and ∠QPR
• The third angle ∠PRQ = [180-(75+75)] = [180-150] = 30o
In the next section, we will see Solids.
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 |
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 = 30o
(v) The radial line from O to the point of contact P will make an angle of 90o with tangent BC. So in ΔOPC, ∠COP = 60o.
Now we can begin the construction:
1. Draw a horizontal line shown in red colour in fig.32.71.a below:
 |
| Fig.32.71 |
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 60o 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 30o 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 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 30o 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 |
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
• 90o 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 |
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)] = 90o
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 |
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 90o
• 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 = 30o.
 |
| Fig.32.76 |
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)] = 75o. This is shown in fig.b
3. The angles at the extremities of chord PR on the 'tangent side' is 75o
• The angle made by chord PR at any point on the circle on the 'non-tangent side' will also be 75o. (Theorem 32.7)
• Thus we get: PQR = 75o
4. Given that PQ is parallel to BT. So we get: ∠QPR = ∠PRT = 75o
• Thus we get two angles in the PQR: ∠PQR and ∠QPR
• The third angle ∠PRQ = [180-(75+75)] = [180-150] = 30o
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