In the previous section we saw theorem 27.10 and it's application. We also saw some solved examples. In this section we will see another application of the theorem.
Consider the two chords AB and CD in fig.27.55(a) below. They intersect at P.
1. Consider the chord AB. It is split into two pieces PA and PB.
2. Let us construct a rectangle with sides PA and PB. The longer piece PA will be the length and shorter piece PB will be the width. (In fact, the results will not be affected even if lengths and widths are interchanged)
3. Let us construct the rectangle over the circle itself. So the longer side of the rectangle will coincide with PA. This is shown in fig(b).
4. The width of the rectangle must be exactly equal to PB. For that, with P as centre and PB as radius, draw an arc (shown in yellow colour). Let this arc intersect the perpendicular from P at P'. Then PP' is the width of our rectangle. So The rectangle APP'A' can be easily completed.
5. Area of rectangle APP'A' = AP×PP' = PA×PB
8. Let us construct the rectangle over the circle itself. So the shorter side of the rectangle will coincide with PC. This is shown in fig(c).
9. The width of the rectangle must be exactly equal to PD. For that, with P as centre and PD as radius, draw an arc (shown in blue colour). Let this arc intersect the perpendicular from P at P''. Then PP'' is the length of our rectangle. So The rectangle CPP''C' can be easily completed.
10. Area of rectangle CPP''C' = CP×P''P = PC×PD
■ But applying theorem 27.10 , PA×PB = PC×PD. So the results in (5) and (10) are equal. Thus we find that, the areas of the two rectangles are equal.
Now let us put the above result for a practical application:
Consider the rectangle in fig.27.56(a) below.
• It has a length of 'a' and width of 'b'. We want to increase it's length by 'c'. So the new length would be '(a+c)'.
• But there is one condition: The area of the rectangle must remain the same.
• If the area is to remain the same, obviously, the width will have to decrease. Let the new width be 'x'. It is shown in fig.27.56(b). We want to find this 'x' graphically. Let us try:
• Consider fig27.56(c). Two chords AB and CD intersect at P. The lengths of the four pieces are:
PA = b, PB = a, PC = x, PD = (a+c)
• Imagine a rectangle with length a and width b
♦ It's area will be equal to ab = PA×PB
• Imagine another rectangle with length (a+c) and width x
♦ It's area will be equal to (a+c)x = PD×PC
■ Based on theorem 27.10 , the two areas will be equal.
• So our next aim is to construct a circle and the chords shown in fig.27.56(c).
• What we have is the given rectangle with length a and width b. We have to begin our work from that rectangle.
5. Thus we get the three points, A, B and D. Note that these are the same A, B and D in the fig.27.56(c) that we saw earlier.
6. In that fig., a circle passes through those three points.
• A circle passing through any three points is unique. That means there is one and only one circle which will pass through three points.
• We have learned about it earlier. See details here. Also we know how to draw the circle passing through any given three points.
7. So, once we have the three points A, B and D, we can draw the circle through them. This is shown in fig.27.57(b).
8. The circle will cut the vertical through P at C. PC will naturally be equal to 'x'. So we have drawn the circle and the chords in the earlier fig. 27.56(c). Now we can proceed to draw the new rectangle.
9. With P as centre and PC as radius, draw an arc (shown in blue colour in fig.27.57.c). It will intersect the horizontal through P at P'. So PP' = x.
10. So PP' is the width of the required rectangle. The length (=PD) is already available. Thus we can easily construct the new rectangle EDPP'
Solved example 27.20
Consider the two chords AB and CD in fig.27.55(a) below. They intersect at P.
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| Fig.27.55 |
2. Let us construct a rectangle with sides PA and PB. The longer piece PA will be the length and shorter piece PB will be the width. (In fact, the results will not be affected even if lengths and widths are interchanged)
3. Let us construct the rectangle over the circle itself. So the longer side of the rectangle will coincide with PA. This is shown in fig(b).
4. The width of the rectangle must be exactly equal to PB. For that, with P as centre and PB as radius, draw an arc (shown in yellow colour). Let this arc intersect the perpendicular from P at P'. Then PP' is the width of our rectangle. So The rectangle APP'A' can be easily completed.
5. Area of rectangle APP'A' = AP×PP' = PA×PB
6. Next, consider the chord CD. It is split into two pieces PC and PD.
7. Let us construct a rectangle with sides PC and PD. The longer piece PD will be the length and shorter piece PC will be the width. (In fact, the results will not be affected even if lengths and widths are interchanged) 8. Let us construct the rectangle over the circle itself. So the shorter side of the rectangle will coincide with PC. This is shown in fig(c).
9. The width of the rectangle must be exactly equal to PD. For that, with P as centre and PD as radius, draw an arc (shown in blue colour). Let this arc intersect the perpendicular from P at P''. Then PP'' is the length of our rectangle. So The rectangle CPP''C' can be easily completed.
10. Area of rectangle CPP''C' = CP×P''P = PC×PD
■ But applying theorem 27.10 , PA×PB = PC×PD. So the results in (5) and (10) are equal. Thus we find that, the areas of the two rectangles are equal.
Consider the rectangle in fig.27.56(a) below.
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| Fig.27.56 |
• But there is one condition: The area of the rectangle must remain the same.
• If the area is to remain the same, obviously, the width will have to decrease. Let the new width be 'x'. It is shown in fig.27.56(b). We want to find this 'x' graphically. Let us try:
• Consider fig27.56(c). Two chords AB and CD intersect at P. The lengths of the four pieces are:
PA = b, PB = a, PC = x, PD = (a+c)
• Imagine a rectangle with length a and width b
♦ It's area will be equal to ab = PA×PB
• Imagine another rectangle with length (a+c) and width x
♦ It's area will be equal to (a+c)x = PD×PC
■ Based on theorem 27.10 , the two areas will be equal.
• So our next aim is to construct a circle and the chords shown in fig.27.56(c).
• What we have is the given rectangle with length a and width b. We have to begin our work from that rectangle.
1. Consider fig.27.57(a) below. The base of the given rectangle is named as PB. So PB = a. For the ease of construction, the given rectangle should be placed in such a way that the side PB is exactly horizontal.
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| Fig.27.57 |
2. Next, PB is extended towards the right by a distance 'c', thus reaching the point B'. So PB' = (a+c)
3. With P as centre and PB' as radius, draw an arc (shown in red colour). It will intersect the vertical through P at D. So PD = (a+c).
• Note that, PB is horizontal and PD is vertical. So ∠BPD = 90o.
4. With P as centre and b as radius, draw an arc (shown in yellow colour). It will intersect the horizontal through P at A. So PA = b5. Thus we get the three points, A, B and D. Note that these are the same A, B and D in the fig.27.56(c) that we saw earlier.
6. In that fig., a circle passes through those three points.
• A circle passing through any three points is unique. That means there is one and only one circle which will pass through three points.
• We have learned about it earlier. See details here. Also we know how to draw the circle passing through any given three points.
7. So, once we have the three points A, B and D, we can draw the circle through them. This is shown in fig.27.57(b).
8. The circle will cut the vertical through P at C. PC will naturally be equal to 'x'. So we have drawn the circle and the chords in the earlier fig. 27.56(c). Now we can proceed to draw the new rectangle.
9. With P as centre and PC as radius, draw an arc (shown in blue colour in fig.27.57.c). It will intersect the horizontal through P at P'. So PP' = x.
10. So PP' is the width of the required rectangle. The length (=PD) is already available. Thus we can easily construct the new rectangle EDPP'
Draw a rectangle of Length 6 cm and width 4 cm. Draw a rectangle of the same area with length 7 cm.
Solution:
The required construction is shown in fig.27.58(a) below:
Let us see the steps:
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| Fig.27.58 |
1. Draw a rectangle of length 6 cm and width 4 cm. The length is named as PB. For the ease of construction, PB must be perfectly horizontal.
2. Extend PB towards the right by 1 cm upto B'. So PB' = 7 cm
3. With P as centre and PB' as radius, draw an arc (shown in red colour). It will intersect the vertical through P at D. So PD = 7 cm.
5. Thus we get the three points, A, B and D.
6. Once we have the three points A, B and D, we can draw the circle through them.
7. The circle will cut the vertical through P at C. PC will naturally be equal to 'x'.
8. With P as centre and PC as radius, draw an arc (shown in blue colour). It will intersect the horizontal through P at P'. So PP' = x.
9. So PP' is the width of the required rectangle. The length (=PD) is already available. Thus we can easily construct the new rectangle EDPP'
Solved example 27.21
Draw a rectangle of Length 7 cm and width 3 cm. Draw a rectangle of the same area with length 5 cm.
2. Extend PB towards the right by 1 cm upto B'. So PB' = 7 cm
3. With P as centre and PB' as radius, draw an arc (shown in red colour). It will intersect the vertical through P at D. So PD = 7 cm.
• Note that, PB is horizontal and PD is vertical. So ∠BPD = 90o.
4. With P as centre and 4 cm as radius, draw an arc (shown in yellow colour). It will intersect the horizontal through P at A. So PA = 4 cm 5. Thus we get the three points, A, B and D.
6. Once we have the three points A, B and D, we can draw the circle through them.
7. The circle will cut the vertical through P at C. PC will naturally be equal to 'x'.
8. With P as centre and PC as radius, draw an arc (shown in blue colour). It will intersect the horizontal through P at P'. So PP' = x.
9. So PP' is the width of the required rectangle. The length (=PD) is already available. Thus we can easily construct the new rectangle EDPP'
Solved example 27.21
Draw a rectangle of Length 7 cm and width 3 cm. Draw a rectangle of the same area with length 5 cm.
Solution:
The required construction is shown in fig.27.58(b) above.
Let us see the steps:
1. Draw a rectangle of length 7 cm and width 3 cm. The length is named as PB. For the ease of construction, PB must be perfectly horizontal.
2. In this problem, length of the new rectangle is less. So there is no need for extension. We can mark B' within PB. We mark B' in such a way that PB' = 5 cm
3. With P as centre and PB' as radius, draw an arc (shown in red colour). It will intersect the vertical through P at D. So PD = 5 cm.
5. Thus we get the three points, A, B and D.
6. Once we have the three points A, B and D, we can draw the circle through them.
7. The circle will cut the vertical through P at C. PC will naturally be equal to 'x'.
8. With P as centre and PC as radius, draw an arc (shown in blue colour). It will intersect the horizontal through P at P'. So PP' = x.
9. So PP' is the width of the required rectangle. The length (=PD) is already available. Thus we can easily construct the new rectangle EDPP'
2. In this problem, length of the new rectangle is less. So there is no need for extension. We can mark B' within PB. We mark B' in such a way that PB' = 5 cm
3. With P as centre and PB' as radius, draw an arc (shown in red colour). It will intersect the vertical through P at D. So PD = 5 cm.
• Note that, PB is horizontal and PD is vertical. So ∠BPD = 90o.
4. With P as centre and 3 cm as radius, draw an arc (shown in yellow colour). It will intersect the horizontal through P at A. So PA = 4 cm 5. Thus we get the three points, A, B and D.
6. Once we have the three points A, B and D, we can draw the circle through them.
7. The circle will cut the vertical through P at C. PC will naturally be equal to 'x'.
8. With P as centre and PC as radius, draw an arc (shown in blue colour). It will intersect the horizontal through P at P'. So PP' = x.
9. So PP' is the width of the required rectangle. The length (=PD) is already available. Thus we can easily construct the new rectangle EDPP'
So we saw an application of theorem 27.10. In the next section, we will see a special case of this application.
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