In the previous section we saw an application of theorem 27.10 . We also saw some solved examples. In this section we will see a special case of that application.
■ In the previous section we saw this:
• A rectangle was given to us
♦ We made a new rectangle of the same area but different dimensions
■ In this section we will see this:
• A rectangle will be given to us
♦ We will make a square of the same area.
■ We will learn the method by analysing an actual example:
Solved example 27.22
■ In the previous section we saw this:
• A rectangle was given to us
♦ We made a new rectangle of the same area but different dimensions
■ In this section we will see this:
• A rectangle will be given to us
♦ We will make a square of the same area.
■ We will learn the method by analysing an actual example:
Solved example 27.22
Draw a rectangle of Length 4 cm and width 2 cm. Draw a square of the same area.
Solution:
Consider the rectangle in fig.27.59(a) below:
• It has a length of 4 cm and width of 2 cm. We want to change it into a square.
• But there is one condition: The area of the rectangle and the new square must be the same.
• Let the side of the new square be 'x'. It is shown in fig.27.59(b). We want to find this 'x' graphically. Let us try:
• Consider fig27.59(c). Two chords AB and CD intersect at P.
• Out of the two chords, AB is a diameter. Because it is passing through the centre 'O'
• This diameter intersects the other chord CD in a perpendicular direction. So CD is bisected.
♦ That means, PC = PD = x. See theorem 17.1.
• The lengths of the four pieces are:
PA = 2, PB = 4, PC = x, PD = x
• Imagine a rectangle with length 4 and width 2
♦ It's area will be equal to 4×2 = PA×PB
• Imagine another rectangle with length x and width x
♦ It's area will be equal to x×x = PD×PC
■ Based on theorem 27.10 , the two areas will be equal.
• A rectangle with length x and width also x is a square.
• So our next aim is to construct a circle and the chords shown in fig.27.59(c).
• What we have, is the given rectangle with length 4 cm and width 2 cm. We have to begin our work from that rectangle.
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| Fig.27.59 |
• But there is one condition: The area of the rectangle and the new square must be the same.
• Let the side of the new square be 'x'. It is shown in fig.27.59(b). We want to find this 'x' graphically. Let us try:
• Consider fig27.59(c). Two chords AB and CD intersect at P.
• Out of the two chords, AB is a diameter. Because it is passing through the centre 'O'
• This diameter intersects the other chord CD in a perpendicular direction. So CD is bisected.
♦ That means, PC = PD = x. See theorem 17.1.
• The lengths of the four pieces are:
PA = 2, PB = 4, PC = x, PD = x
• Imagine a rectangle with length 4 and width 2
♦ It's area will be equal to 4×2 = PA×PB
• Imagine another rectangle with length x and width x
♦ It's area will be equal to x×x = PD×PC
■ Based on theorem 27.10 , the two areas will be equal.
• A rectangle with length x and width also x is a square.
• So our next aim is to construct a circle and the chords shown in fig.27.59(c).
• What we have, is the given rectangle with length 4 cm and width 2 cm. We have to begin our work from that rectangle.
1. Consider fig.27.60(a) below. The base of the given rectangle is named as PB. So PB = 4. For the ease of construction, the given rectangle should be placed in such a way that the side PB is exactly horizontal.
2. With P as centre and PA as radius, draw an arc (shown in yellow colour) which will cut the horizontal through P at A. So PA = 2 cm
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| Fig.27.60 |
3. With AB as diameter, draw a circle. This is shown in fig(b).
• For drawing the circle, first draw the perpendicular bisector of AB. It will cut AB at centre 'O'.
• With centre 'O' and OA as radius, draw the circle. This step is not shown in the fig.
4. Draw a vertical through P. It will intersect the circle at C and D.
• AB is a diameter of the circle. CD is a chord
• Since PB is horizontal and CD is vertical, ∠PBD = 90o
• So diameter AB is perpendicular to CD. Then by theorem 17.1, AB is the perpendicular bisector of CD. Thus we get PC = PD
5. Once we get any one side of a square, we can easily construct it. Here, Both PC and PD are equal to the side of the required square. We can use any one of them.
• We will use the lower PD So that the square will be distinct from the given rectangle
• Thus in the fig(c), the square PP'ED is constructed
• With centre 'O' and OA as radius, draw the circle. This step is not shown in the fig.
4. Draw a perpendicular to AB through P. This perpendicular will meet the semi-circle at D
5. PD is the required side. The square PP'ED can then be easily drawn.
■ In this method, we do not even have to draw the original rectangle
■ Also note that, the semi-circle can be drawn on the upper side of AB
Solved example 27.23
• For drawing the circle, first draw the perpendicular bisector of AB. It will cut AB at centre 'O'.
• With centre 'O' and OA as radius, draw the circle. This step is not shown in the fig.
4. Draw a vertical through P. It will intersect the circle at C and D.
• AB is a diameter of the circle. CD is a chord
• Since PB is horizontal and CD is vertical, ∠PBD = 90o
• So diameter AB is perpendicular to CD. Then by theorem 17.1, AB is the perpendicular bisector of CD. Thus we get PC = PD
5. Once we get any one side of a square, we can easily construct it. Here, Both PC and PD are equal to the side of the required square. We can use any one of them.
• We will use the lower PD So that the square will be distinct from the given rectangle
• Thus in the fig(c), the square PP'ED is constructed
Now, there is an easier method to obtain the required square:
Consider fig.27.60(c) above. Our real aim is to obtain the side PD. If we can obtain this side directly, a lot of work can be saved. Let us see the method for doing it:
Consider fig.27.61(a) below:
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| Fig.27.61 |
1. Draw a line AB, 6 cm in length
2. Mark a point P such that PA = 2 cm and PB = 4 cm
3. Draw a semi-circle with AB as diameter
• For drawing the circle, first draw the perpendicular bisector of AB. It will cut AB at centre 'O'. • With centre 'O' and OA as radius, draw the circle. This step is not shown in the fig.
4. Draw a perpendicular to AB through P. This perpendicular will meet the semi-circle at D
5. PD is the required side. The square PP'ED can then be easily drawn.
■ In this method, we do not even have to draw the original rectangle
■ Also note that, the semi-circle can be drawn on the upper side of AB
Let a rectangle be of length 6 cm and width 4 cm. Draw a square of the same area.
Solution:
5. PD is the required side. The square PP'ED can then be easily drawn.
(i) In the solved example 27.22, we got a square whose area is same as a rectangle of length 4 cm and width 2 cm
• So the area of the newly formed square is 8 cm2.
• If the area of a square is 8 cm2, then obviously, it's side would be √8 cm
• Thus the length of PD = √8 cm
(ii) In the solved example 27.23, we got a square whose area is same as a rectangle of length 6 cm and width 4 cm
• So the area of the newly formed square is 24 cm2.
• If the area of a square is 24 cm2, then obviously, it's side would be √24 cm
• Thus the length of PD = √24 cm
■ So this is an excellent method to find the values of irrational numbers graphically.
• Note that in an earlier chapter, we had learned another method for doing this. Details here.
Another example:
Consider fig.27.62(a) below:
1. Out of the two chords, chord AB is a diameter. Chord CD is drawn perpendicular to AB.
• So AB will be the perpendicular bisector of CD. Thus PC = PD
2. Now let us apply theorem 27.10:
Multiplying opposite pieces of the same chord, we get:
PA×PB = PC×PD ⇒ PA×PB = PC2 (∵ PC = PD)
3. This is a very useful result.
• We no longer need the piece PD. So we need not consider the lower part of the circle.
• That means, all our further calculations will be related to the portion above the diameter AB.
• That is., we will be dealing with a semi-circle only.
4. In fig.27.62(b) above, the diameter AB (= 8 cm) is split into two parts at 'P'.
PA is 6 cm and PB is 2 cm.
5. A perpendicular PC is erected at P in such a way that C lies on the semi-circle with AB as diameter.
6. Using the equation in (2) above, we get:
PA×PB = PC2. That is., 6×2 = PC2 ⇒ 12 = PC2 ⇒ PC = √12
■ Let us write a summary of what we have done above:
• We drew a semi-circle with diameter AB = 8 cm
• We split the diameter at P such that PA = 6 cm and PB = 2 cm
• Finally we erected a perpendicular at P in such a way that it meets the circle at C
• We find that PC = √12 cm
• We know that √12 is an irrational number. We cannot obtain √12 on a scale.
• But using the above method, we are able to draw a line of length √12 cm with out using a scale.
• Once we obtain a line of length √12 cm, if required, we can construct a square of area 12 cm2.
• Because area of a square of side √12 cm = √12 ×√12 = 12 cm2
■ The procedure for constructing the square is shown in fig(c)
1. First draw a horizontal line through C
2. Then draw an arc with C as centre and CP as radius. This arc will cut the horizontal line at a point. Name this point as D
3. Through D, drop a perpendicular to AB. Name the foot of this perpendicular as F
4. Then FPCD is a square of area 12 cm2
Now let us see some solved examples:
Solution:
Case (a): x2 = 8×18 ⇒ x2 = 144 ⇒ x = √144 ⇒ x = 12
Case (b): x2 = 9×4 ⇒ x2 = 36 ⇒ x = √36 ⇒ x = 6
Case (c): x2 = 2×5 ⇒ x2 = 10 ⇒ x = √10
1. Draw a line AB, (6+4) = 10 cm in length. See fig.27.61(b) above.
2. Mark a point P such that PA = 4 cm and PB = 6 cm
3. Draw a semi-circle with AB as diameter
4. Draw a perpendicular to AB through P. This perpendicular will meet the semi-circle at D5. PD is the required side. The square PP'ED can then be easily drawn.
Irrational numbers
Let us analyse the above two solved examples.(i) In the solved example 27.22, we got a square whose area is same as a rectangle of length 4 cm and width 2 cm
• So the area of the newly formed square is 8 cm2.
• If the area of a square is 8 cm2, then obviously, it's side would be √8 cm
• Thus the length of PD = √8 cm
(ii) In the solved example 27.23, we got a square whose area is same as a rectangle of length 6 cm and width 4 cm
• So the area of the newly formed square is 24 cm2.
• If the area of a square is 24 cm2, then obviously, it's side would be √24 cm
• Thus the length of PD = √24 cm
■ So this is an excellent method to find the values of irrational numbers graphically.
• Note that in an earlier chapter, we had learned another method for doing this. Details here.
Another example:
Consider fig.27.62(a) below:
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| Fig.27.62 |
• So AB will be the perpendicular bisector of CD. Thus PC = PD
2. Now let us apply theorem 27.10:
Multiplying opposite pieces of the same chord, we get:
PA×PB = PC×PD ⇒ PA×PB = PC2 (∵ PC = PD)
3. This is a very useful result.
• We no longer need the piece PD. So we need not consider the lower part of the circle.
• That means, all our further calculations will be related to the portion above the diameter AB.
• That is., we will be dealing with a semi-circle only.
4. In fig.27.62(b) above, the diameter AB (= 8 cm) is split into two parts at 'P'.
PA is 6 cm and PB is 2 cm.
5. A perpendicular PC is erected at P in such a way that C lies on the semi-circle with AB as diameter.
6. Using the equation in (2) above, we get:
PA×PB = PC2. That is., 6×2 = PC2 ⇒ 12 = PC2 ⇒ PC = √12
■ Let us write a summary of what we have done above:
• We drew a semi-circle with diameter AB = 8 cm
• We split the diameter at P such that PA = 6 cm and PB = 2 cm
• Finally we erected a perpendicular at P in such a way that it meets the circle at C
• We find that PC = √12 cm
• But using the above method, we are able to draw a line of length √12 cm with out using a scale.
• Once we obtain a line of length √12 cm, if required, we can construct a square of area 12 cm2.
• Because area of a square of side √12 cm = √12 ×√12 = 12 cm2
■ The procedure for constructing the square is shown in fig(c)
1. First draw a horizontal line through C
2. Then draw an arc with C as centre and CP as radius. This arc will cut the horizontal line at a point. Name this point as D
3. Through D, drop a perpendicular to AB. Name the foot of this perpendicular as F
4. Then FPCD is a square of area 12 cm2
Solved example 27.24
Find the value of 'x' in the figs.27.63(a), (b) and(c) below: |
| Fig.27.63 |
Case (a): x2 = 8×18 ⇒ x2 = 144 ⇒ x = √144 ⇒ x = 12
Case (b): x2 = 9×4 ⇒ x2 = 36 ⇒ x = √36 ⇒ x = 6
Case (c): x2 = 2×5 ⇒ x2 = 10 ⇒ x = √10
Solved example 27.25
In the fig.27.63(d) above, AD = 10 cm, BD = 6 cm, CD = 2 cm. Find the value of CP, CQ and PQ.
Solution:
• CQ2 = AC×CD = (AD-CD)×CD = (10-2)×2 = 8×2 = 16 cm
⇒ CQ = √16 = 4 cm
• CP2 = BC×CD = (BD-CD)×CD = (6-2)×2 = 4×2 = 8 cm
⇒ CP = √8 = 2√2 cm
• PQ = (CQ - CP) = (4 - 2√2) = 2(2-√2) cm
• CQ2 = AC×CD = (AD-CD)×CD = (10-2)×2 = 8×2 = 16 cm
⇒ CQ = √16 = 4 cm
• CP2 = BC×CD = (BD-CD)×CD = (6-2)×2 = 4×2 = 8 cm
⇒ CP = √8 = 2√2 cm
• PQ = (CQ - CP) = (4 - 2√2) = 2(2-√2) cm
In the next section, we will see a few more solved examples.
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