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 32.2 - Tangents giving Cyclic quadrilaterals

In the previous section we saw some tangents from an exterior point. In this section we will learn more details.
1. Consider the two tangents AP and BP in fig.32.15 below. 
• OA and OB are radii. 
    ♦ The tangents are drawn at the points of intersection of the radii with the circle

Fig.32.15

2. We know that ∠OAP = ∠OBP = 90^o
3. Now consider the quadrilateral OAPB. It is formed by the radii and the tangents
4. A and B are two opposite corners. The sum of  the angles at those opposite corners = 90+90 = 180^o
■ If the sum of opposite angles in any quadrilateral is 180^o, that quadrilateral will be a cyclic quadrilateral, 
• That is., all the four vertices of that quadrilateral will lie on a circle (We saw details here)
5. So we can draw a circle through all the four vertices of the quadrilateral OAPB. This circle is shown in green colour in fig.32.15(b) above

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We can write the above result in the form of a theorem:
Theorem 32.4
• If a quadrilateral has the following four vertices:
    ♦ First vertex at the centre of a circle
    ♦ Second vertex at the exterior point from which two tangents are drawn to the circle
    ♦ Third vertex at one tangent point
    ♦ Fourth vertex at the other tangent point
■ Then that quadrilateral is cyclic

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From the above, we get another useful result:
• In the fig.32.15(b) above, we considered the opposite vertices A and B. What about the other two opposite vertices O and P?
• Obviously, the sum of those two angles must also be 180^o (∵ the sum of interior angles of any quadrilaterals is 360^o)
This result is also useful for solving problems. We will write it as a theorem:
Theorem 32.5
1. In a circle, two radii are drawn 
2. Tangents are drawn at the end point of each of those radii
• These tangents meet at P
3. Then sum of the following two angles is 180^o
• Angle at O between the two radii
• Angle at P between the two tangents

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Now we will see an application of this theorem in the form of a solved example
Solved example 32.6
Draw an equilateral triangle in such a way that it's all three sides are tangential to a circle of radius 3 cm
Solution:
A rough sketch is shown in fig.32.16(a) below:

Fig.32.16

• The three sides of the triangle are: AB, BC and AC. They are all tangents to the circle of radius 3 cm
• The triangle is to be equilateral. So all it's angles are 60
• Based on this rough fig., we must obtain details to make the actual construction
We will write the steps:
1. From the center O, draw radial lines OP, OQ and OR. This is shown in fig.b
2. Based on theorem 32.5, we get:
• ∠A + ∠ROP = 180 ⟹ 60 + ∠ROP = 180 ⟹ ∠ROP = 120
• ∠B + ∠POQ = 180 ⟹ 60 + ∠POQ = 180 ⟹ ∠POQ = 120
• ∠C + ∠ROQ = 180 ⟹ 60 + ∠ROQ = 180 ⟹ ∠ROQ = 120
• So the three central angles are all 120^o. This is marked in fig.c
• We can now begin the actual construction
3. Step 1 is to draw a circle of radius 3 cm. Mark the center as 'O'. Then draw a radial line OP in any convenient direction. This is shown in fig.32.17(a) below:

Fig.32.17

4. Step 2 is to draw the other two radial lines:
• Draw OQ in such a way that it makes an angle of 120^o with OP 
• Draw OR in such a way that it makes an angle of 120^o with OQ 
• Then the third angle ∠POR will be naturally 120
This is shown in fig.32.17(b)
5. Step 3 is the final step. In this step we draw the tangents:
• Draw a line perpendicular to OP at P. This is the tangent at P
• Draw a line perpendicular to OQ at Q. This is the tangent at Q
• Draw a line perpendicular to OR at R. This is the tangent at R
■ The three tangents will meet at A, B and C, giving the required equilateral triangle. This is shown in fig.c

Solved example 32.7
Draw a triangle of angles 40^o, 60^o and 80^o in such a way that it's all three sides are tangential to a circle of radius 2.5 cm
Solution:
A rough sketch is shown in fig.32.18(a) below:

Fig.32.18

• The three sides of the triangle are: AB, BC and AC. They are all tangents to the circle of radius 2.5 cm
• The required angles are also marked in the fig.a
• Based on this rough fig., we must obtain details to make the actual construction
We will write the steps:
1. From the center O, draw radial lines OP, OQ and OR. This is shown in fig.b
2. Based on theorem 32.5, we get:
• ∠A + ∠ROP = 180 ⟹ 40 + ∠ROP = 180 ⟹ ∠ROP = 140
• ∠B + ∠POQ = 180 ⟹ 60 + ∠POQ = 180 ⟹ ∠POQ = 120
• ∠C + ∠ROQ = 180 ⟹ 80 + ∠ROQ = 180 ⟹ ∠ROQ = 100
• So the three central angles are 140^o, 120^o and 100^o. This is marked in fig.c
• We can now begin the actual construction
3. Step 1 is to draw a circle of radius 2.5 cm. Mark the center as 'O'. Then draw a radial line OP in any convenient direction. This is shown in fig.32.19(a) below:

Fig.32.19

4. Step 2 is to draw the other two radial lines:
• Draw OQ in such a way that it makes an angle of 120^o with OP 
• Draw OR in such a way that it makes an angle of 100^o with OQ 
• Then the third angle ∠POR will be naturally 140^o
This is shown in fig.32.19(b)
5. Step 3 is the final step. In this step we draw the tangents:
• Draw a line perpendicular to OP at P. This is the tangent at P
• Draw a line perpendicular to OQ at Q. This is the tangent at Q
• Draw a line perpendicular to OR at R. This is the tangent at R
■ The three tangents will meet at A, B and C, giving the required triangle. This is shown in fig.c

Solved example 32.8
In the fig.32.20(a), the blue triangle is equilateral. It's circumcircle is shown in green colour. All the sides of the red triangle are tangents to the green circle. The tangent points are the vertices of the blue triangle.

Fig.32.20

(i) Prove that the red triangle is also equilateral and it's sides are double that of the blue triangle
(ii) Draw the fig. with sides of the small triangle 3 cm
Solution:
Part 1:
Let ABC be the red triangle and PQR be the blue triangle. This is marked in fig.b
1. Mark the center O of the green circle. Draw radial lines OP, OQ and OR. They are shown in yellow colour
2. Given that the red lines are tangents at the vertices. So the yellow lines are perpendicular to the red lines. we can write:
AB is perpendicular to OP
BC is perpendicular to OQ
AC is perpendicular to OR
3. Each of the three blue lines PQ, QR and RP subtend an angle at the center O:
    ♦ PQ subtend ∠POQ
    ♦ QR subtend ∠QOR
    ♦ RP subtend ∠ROP
• But PQ = QR = RP (∵ they are sides of an equilateral triangle)
• So the angle subtended by them must be equal
    ♦ Thus we get: ∠POQ = ∠QOR = ∠ROP
• But the total angle at any point is 360^o 
    ♦ We can write: ∠POQ + ∠QOR + ∠ROP = 360^o
• So, since the three angles are equal, we get: ∠POQ = ∠QOR = ∠ROP = ³⁶⁰⁄₃ = 120^o
4. Based on theorem 32.5, we can write:
• ∠A + ∠ROP = 180 ⟹ ∠A+ 120 = 180 ⟹ ∠A = 60
• ∠B + ∠POQ = 180 ⟹ ∠B+ 120 = 180 ⟹ ∠B = 60
• ∠C + ∠ROQ = 180 ⟹ ∠C+ 120 = 180 ⟹ ∠C = 60
• So the three angles of the red triangle are all 60^o. This is marked in fig.b. Thus the red triangle ABC is an equilateral triangle. 
5. Consider fig.c: AB and AC are two tangents from an exterior point A
(i) P and R are the tangent points. Then by theorem 32.2 that we saw in the previous section, AP = AR
(ii) In the same way, we have: BP = BQ and CQ = CR
(iii) Let: AP = AR = a  
BP = BQ = b  
CQ = CR = c
(iv) Since the triangle is equilateral, we have: AB = BC = CA ⟹ (a+b) = (b+c) = (c+a)
From (a+b) = (b+c), we get: a = c
From (b+c) = (c+a), we get: a = b
• So we have: a = b = c
6. P is the midpoint of AB (∵ a = b)
R is the midpoint of AC (∵ a = c)
• So the line PR cuts the sides AB and AC at their midpoints
• Then by theorem 18.5, (Division of triangles by parallel lines) we get: 
    ♦ BC is parallel to PR
    ♦ BC is double the length of PR
In the same way: 
    ♦ AC is parallel to PQ
    ♦ AC is double the length of PQ
AND
    ♦ AB is parallel to RQ
    ♦ AB is double the length of RQ
■ (4) and (6) together gives the answer for part (i)
Part 2:
The steps for construction are given below:
1. Step 1 is to draw an equilateral triangle of side 3 cm. This is shown in fig.32.21(a) below

Fig.32.22

2. Step 2 is to draw the circumcircle of this triangle. The method can be seen here. In this step, we get the position of the circumcentre  'O' also. This is shown in fig.b
3. Step 3 is to join all the three vertices to the centre O. These are the yellow lines in fig.c
Now draw lines perpendicular to the yellow lines through the vertices. These lines (shown in red colour in fig.c) will give the outer triangle

Solved example 32.9
Fig.32.23(a) shows the two tangents (red lines) drawn from an external point. And also the radii (green lines) through the points of contact.

Fig.32.23

(i) Prove that the tangents have the same length
(ii) In fig(b), the external point is joined to the center of the circle (cyan line). Prove that this cyan line bisects the angle between the radii
(iii) Prove that the cyan line bisects the angle between the tangents also
(iv) In fig(c), a chord is drawn between the points of contact (magenta line). prove that the cyan line is the perpendicular bisector of the magenta chord
Solution:
Part (i): The tangents in fig32.23 will be of equal length.We proved it earlier and wrote Theorem 32.2.
Part (ii): We have to prove that the white and grey colored angles in fig.32.24(b) below are equal

Fig.32.24

We have proved it already and wrote the first part of Theorem 32.3
Part (iii): We have to prove that the green and yellow colored angles in fig.32.24(b) above are equal
We have proved it already and wrote the second part of Theorem 32.3 
Part (iv): We have to prove that the cyan line is perpendicular to the magenta chord. 
    ♦ And also that it divides the magenta chord into two equal parts as shown in fig.32.24(c) above.
• We proved it in solved example 32.4. See fig.32.13. It is shown again below:

Fig.32.13

• We proved that TO is the perpendicular bisector of PQ

Solved example 32.10
(i) In the fig.32.25(a) below, the two magenta chords are perpendicular to each other. 

Fig.32.25

Red colored tangents are drawn at the ends of those chords. Those red tangents meet at four points to give a quadrilateral. Prove that this quadrilateral is cyclic
(ii) What sort of quadrilateral will it be, if one chord is a diameter?
(iii) What sort of quadrilateral will it be, if both chords are diameters?
Solution:
Part (i):
1. Let the magenta chords be PQ and RS. See fig.32.25(b)
2. Red tangents are drawn at P, Q, R and S
• Those tangents meet at four points to give a quadrilateral ABCD
3. The tangent points P, Q, R and S are joined to the centre O by green radial lines
• So, if ∠POR = x, we will get: ∠PAR = (180-x)
• Similarly, if ∠SOQ = y, we will get: ∠SCQ = (180-y)
4. In the quadrilateral ABCD, ∠PAR and ∠SCQ are opposite angles
• Their sum is: [(180-x) + (180-y)] = [360-(x+y)]
5. Consider the arc RP. It is drawn from the ends of perpendicular chords
• Similarly, the arc QS is drawn from the ends of the same perpendicular chords
• If we align the two arcs together, they will form a semi-circle. We proved this as a solved example in a video presentation)
6. The central angles of the arcs are x^o and y^o. Since they form a semi -circle, we get: (x+y) = 180^o 
7. So the sum in (4) becomes: [360-(x+y)] = [360-180] = 180
8. That is., sum of opposite angles in the quadrilateral ABCD is 180^o. So it is a cyclic quadrilateral. This is shown in fig.32.25(c)
Part (ii):
1. Let the magenta chords be PQ and RS. See fig.32.26(a)
• This time, RS is a diameter

Fig.32.26

2. Red tangents are drawn at P, Q, R and S
• Those tangents meet at four points to give a quadrilateral ABCD
    ♦ We have to find which type of quadrilateral is ABCD

■ Since PQ and RS are perpendicular chords, the quadrilateral ABCD will be cyclic. This we have proved in part (i)
3. Since RS is a diameter, the tangents at R and S will be parallel. 
• That is., the opposite sides AD and BC are parallel
    ♦ This we proved in solved example 32.5 in the previous section
4. If the opposite are parallel in a quadrilateral, it will be a trapezium  
■ If a trapezium is cyclic, it will be an isosceles trapezium.
The proof is given in this video presentation.
Another method:
1. The tangent points P, Q, R and S are joined to the centre O by green radial lines.  See fig.32.26(b)
Note that, The radial lines for R and S will be OR and OS themselves. Because RS is a diameter.
• So, if ∠POR = x, we will get: ∠PAR = (180-x)

• Similarly, if ∠SOQ = y, we will get: ∠SCQ = (180-y)
2. Since RS is a diameter, the tangents at R and S will be parallel. 
• That is., the opposite sides AD and BC are parallel
    ♦ This we proved in solved example 32.5 in the previous section 
3. So we have two parallel lines cut by a transversal CD
• If the angle at C is (180-y), angle at D will be y (∵ AD and BC are parallel)
See proof in video presentation.
4. But from (6) in part (i), we have: (x+y) = 180
• So we get: y = (180-x) 
5. But from (1) above, we have: angle at A = (180-x)
• So angles at A and D are equal
■ If AD and BC are parallel and angles at A and D are equal, it will be an isosceles trapezium.
Part (iii):
1. Let the magenta chords be PQ and RS. See fig.32.27(a)

Fig.32.27

• This time, both PQ and RS are diameters
2. So opposite sides AB and CD will be parallel
The opposite sides AD and BC will also be parallel
3. Since the opposite sides are parallel to the perpendicular lines PQ and RS, we get:
∠A = ∠B = ∠C = ∠D = 90^o
• This is shown in fig.32.27(b) 
4. Since the corner angles are all 90^o, we get:
• Lengths AB = BC = CD = AD
5. From (2), (3) and (4), it is clear that, ABCD is a square.

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In the next section, we will see more details about Tangents.

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Chapter 27.12 - Chords inside a circle - Solved examples

In the previous section we saw how to draw a square whose area is same as that of a given rectangle. We also saw a method to represent irrational numbers. In this section we will see a few more solved examples.

Solved example 27.26

Draw a triangle of 4, 5, 6 cm and draw a square of equal area

Solution:

1. Draw the triangle ABC as shown in the fig.27.64(a) shown below

We have AB = 6 cm, BC = 5 cm and AC = 4 cm.  We will not write the construction steps for this triangle. However, it may be noted that, intersection of two arcs shown in green colour gives the vertex C

2. We want a square whose area is same as that of △ ABC. So first we must know the area of this triangle

3. Draw a line CD perpendicular to the side AB of the triangle. This is shown in the above fig.27.64(a)

4. Now we have AB as the base and CD as the altitude. So area of △ ABC =

¹⁄₂ × AB × CD ⇒ Area of △ABC = AB × (¹⁄₂ × CD)

5. So, if we get half of the altitude CD, we can multiply it with AB, to get the required area.

6. To get half of CD, draw a perpendicular bisector of CD. This is shown in red colour in fig.(b). Note that, the procedure for drawing the perpendicular bisector is not shown here.

7. Let the perpendicular bisector intersect CD at E. So we get:

 CE = ED = ¹⁄₂×CD

So we have to multiply ED with AB

8. Extend AB towards the right. It is shown by the red dotted line.

9. With ED as radius, and B as center, draw an arc, cutting the extension at F. 

So we have BF = ED

10. Now we want a semi circle with AF as the diameter. For that, draw the perpendicular bisector of AF. This is shown in magenta colour in fig.(b)

11. Let the perpendicular bisector intersect AF at O. Then O is the center of the required semi circle

12. With O as center and OA as radius, draw a semi circle. This is shown in green colour in fig. (c)

13. Erect a perpendicular at B. let it meet the semi circle at G. BG is shown in blue colour in fig(c).

14. By the properties of chords, we have:

AB × BF = BG × BG

15. But AB × BF is the area of △ABC.

• BG × BG is the area of a square whose side is BG

16. So, if we draw a square with side BG, the area of that square will be equal to the area of △ABC

That means, a square with side BG is our required square.   

17. With B as center and BG as radius, draw an arc cutting the red dotted line at H. 

Then BH = BG

18. With G as center and BG as radius, draw an arc

With H as center and BG as radius, draw an arc

These two arcs, which are shown in red colour in fig. (d), will intersect at I

19. So BHIG is the required square

Solved example 27.27
Draw an equilateral triangle of height 3 cm
Solution:
• If we are given the three sides of a triangle, we can easily construct it. 
    ♦ If it is an equilateral triangle, we will need only one side. 
• But in this problem, we are given the height. Consider fig27.65(a) below. An equilateral triangle of side 's' is drawn. Let it's height be 'h'. 

• This height will be perpendicular to the base. So we get two right triangles in side the given equilateral triangle. 
• Consider any one of those right triangles. Applying Pythagoras theorem, we get:  
h² =  s² - (^s⁄₂)² ⇒ h² =  s² - ^s2⁄₄ ⇒ h² =  ^3s2⁄₄ ⇒ h = ^√3s⁄₂.
• In the present problem we have h = 3 cm. So we can write:
3 = ^√3s⁄₂ ⇒ √3 × √3 = ^√3s⁄₂ 
⇒ √3 = ^s⁄₂ ⇒ s = 2 × √3 ⇒ s = √4 × √3 ⇒ s = √12  
• Thus, the side of an equilateral triangle whose height is 3 cm is √12 cm. Now we can do the construction:
1. Draw AB with length 4 cm and BC with length 3 cm. This is shown in fig.27.65(b)
2. Draw the perpendicular bisector of AC. It is shown in magenta colour in fig(b)
3. Let the perpendicular bisector intersect AC at O
4. Draw a semi circle with O as center and  AO as radius
5. Erect a perpendicular BD at B. Length of BD will be equal to √12 cm
6. With B as center and BD as radius, draw an arc. This is shown in fig(c)
With D as center and BD as radius, draw another arc
7. The two arcs will intersect at a point. Name this point as E
8. Join DE and BE. Then △BED is the required triangle.

Solved example 27.28
Draw an isosceles right triangle whose hypotenuse is 4 cm
Solution:
We have to draw an isosceles triangle which is right angled. We know that, in an isosceles triangle, two sides will be equal
• If we are given the length of equal sides and the base angle, we can easily draw it. 
• But in this problem, we are given the hypotenuse. Consider fig27.66(a) below. An isosceles right triangle is drawn

• Let the equal sides be 's'. Then we get:
4² =  s² + s² ⇒ 4² =  2s² ⇒ 16 = 2s² ⇒ s² = 8 ⇒ s = √8
• So the equal sides of an isosceles right triangle, whose hypotenuse is 4 cm is √8 cm. Now we can do the construction
1. Draw AB with length 4 cm and BC with length 2 cm. This is shown in fig.27.66(b)
2. Draw the perpendicular bisector of AC. It is shown in magenta colour in fig(b)
3. Let the perpendicular bisector intersect AC at O
4. Draw a semi circle with O as center and  AO as radius
5. Erect a perpendicular BD at B. Length of BD will be equal to √8 cm
6. Extend BC towards the right. This is shown as red dashed line in fig(c)
7. With B as centre and BD as radius, draw an arc, cutting the extension at E
8. Join DE. Then BED is the required triangle  

Solved example 27.28
Draw a line of length √12 cm. Construct a square with this length as the side. Can you construct a line of length √48 cm in the same figure
Solution:
1. Draw AB with length 4 cm and BC with length 3 cm. This is shown in fig.27.67

2. Draw the perpendicular bisector of AC. It is shown in magenta colour
3. Let the perpendicular bisector intersect AC at O₁
4. Draw a semi circle with O₁ as center and  AO₁ as radius
5. Erect a perpendicular BD at B. Length of BD will be equal to √12 cm
6. Once BD is obtained, the required square BEFD can be easily constructed
7. Now we want to draw √48. We will get 48 as 4 × 12
8.  We already have AB = 4 cm. So extend AC towards the left in such a way that, AG = 12 cm  
9. Draw the perpendicular bisector of BG. It is shown in magenta colour
10. Let the perpendicular bisector intersect AC at O₂
12. Draw a semi circle with O₂ as center and  GO₂ as radius
13. Erect a perpendicular AH at H. Length of AH will be equal to √48 cm

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In the next section, we will see chord which intersect at a point outside the circle.

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