In the previous section we saw the method to divide any line in any given ratio. We also saw two solved examples. In this section, we will see some more solved examples.
Solved example 18.3
.b)Draw another rectangle which have the same ratio
of sides, and with a perimeter 3 cm greater than that of ABCD
Solved example 18.3
In fig.18.15(a), ABCD is a rectangle. It's
sides are not given.
 |
| Fig.18.15 |
Solution:
First we will bring the width
BC 'in line' with the length AB. For that, the following steps (see fig.8.15.b) have
to be done:
1. Extend AB to a convenient
length AC'
2. With B as centre, and BC as radius, draw an arc
(shown in yellow colour) cutting AC' at C''
3. BC'' is equal to BC, and is 'in
line' with AB
4. So AC'' is 'half the
perimeter' of the original rectangle ABCD
5. Draw a line AP' in any
convenient direction towards the bottom side of AB
6. Bring AC'' on to this new
line. For that, with A as centre, and AC'' as radius, draw an arc
(shown in white colour) cutting AP' at P''.
7. So AP'' = AC'' = half of
original perimeter
• Now, the 'new perimeter' has to
be greater by 3 cm. So 'new half perimeter' is to be greater by 1.5 cm.
■ That is., new half perimeter =
original half perimeter + 1.5. The derivation is given below:
(i) Original half perimeter = original perimeter⁄2
(ii) New half perimeter = new perimeter⁄2
(iii) But new perimeter = original
perimeter + 3
(iv) Substituting (iii) in (ii) we get:
new half perimeter = (original perimeter + 3)⁄2
⇒ new half perimeter = original perimeter⁄2 + 3⁄2
⇒ new half perimeter = original half perimeter + 1.5
8. So we have to increase AP'' by
1.5 cm. For that, mark p''' at a distance 1.5 cm from P''. Thus,
AP''' is the 'new half perimeter'
new half perimeter = (original perimeter + 3)⁄2
⇒ new half perimeter = original perimeter⁄2 + 3⁄2
⇒ new half perimeter = original half perimeter + 1.5
9. We have to divide AP''' in the
ratio AB:BC. Let the point of division be P
10. Then AP:PP''' = AB:BC =
AB:BC'' (∵ BC = BC'')
11. That is., we have to divide
AP''' in the ratio AB:BC''
12. For that, first join C'' and P'''.
Then draw it's parallel through B. This parallel will intersect
AP''' at P. So we have two red parallel lines cutting through AP'' and AC'''. The ratios are equal. That is., AB:BC'' = AP:PP'''
13. Now draw a perpendicular PQ'
at P. With P as centre, and PP''' as radius, draw an arc, cutting PQ'
at Q
14. Draw QR parallel ro AP, and AR parallel to PQ. APQR is the required rectangle
14. Draw QR parallel ro AP, and AR parallel to PQ. APQR is the required rectangle
Solved example 18.4
Draw an 8 cm long line and
divide it in the ratio 2:3
Solution:
The construction is shown in
the fig.18.16 below:
 |
| Fig.18.16. Length of AB = 8 cm |
The steps can be written as
follows:
• AB is the 8 cm long line. We
have to divide it an in the ratio 2:3. Let E be the point of
division.
• So if AB is divided into 5 equal parts, AE will take up 2
such equal parts, and BE will take up 3 such equal parts.
• So AE = 2 × 8⁄5
= 16⁄5
= 31⁄5
and BE = 3 × 8⁄5
= 24⁄5
= 44⁄5
.
1. Draw a line AC of length 5
cm (∵ 2+3 = 5) in any convenient direction.
2. Mark a point D at a
distance of 2 cm from A
3. Join B and C. Draw a line
DE parallel to BC through D.
4. Then AE: BE = 2:3
Solved example 18.5
Draw a rectangle of perimeter
15 cm and sides in the ratio 3:4
Solution:
1. Draw a line AB (see fig.18.17 below) of length 15
cm, and divide it at E in such a way that AE:BE = 3:4
 |
| Fig.18.17 |
The steps for this division is
same as that for the previous problem. We will write them again:
• Draw a line AC of length
14 cm (∵ 2 times the sum of 3 and 4 is 14) in any convenient direction.
• Mark a point D at a
distance of 6 cm from A
• Join B and C. Draw a line
DE parallel to BC through D.
• Then AE : BE = 3:4
• Note that we took AC = 14
cm instead of 7. This is because, 7 is small compared to 15, and so,
we will have to draw very slanted lines, which may reduce accuracy.
However, the reader is advised to try that also.
• When we use 14 cm instead of
7, AD should be 6 instead of 3, and also, DC should be 8 instead of
4
2. Draw EP’ perpendicular to
AB
3. With E as centre, and EA as
radius, draw an arc, cutting EP’ at P
4. Draw PQ parallel to AB and
BQ parallel to EP
5. BEPQ is the required
rectangle
[Note that from the
examination point of view, it may be sufficient to show the
construction alone. The steps need to be written only if they are
specifically required]
Solved example 18.6
Draw triangles specified
below, each of perimeter 10 cm
(i) Equilateral triangle (ii)
Sides in the ratio 3:4:5 (iii) Sides in the ratio 2:3:4
Solution:
(i) The perimeter is given as
10 cm. So draw a line AB 10 cm long. See fig.18.18 below:
1. Divide the line AB in the
ratio 1:1:1. This ratio is to be used because, we want 3 equal sides for an
equilateral triangle. The steps for this division are as follows:
 |
| Fig.18.18. Length of AB = 10 cm |
• Draw a line AC of length 9 cm (∵ 9 is comparable with 10, and can be easily divided into 3 equal parts) in any convenient direction.
• Mark a point E at a distance of 3 cm from A. Also another point D at a distance of 3 cm from E
• Join B and C. Draw a lines DG and EF parallel to BC.
• Then AF : FG : BG = 3:3:3 = 1:1:1
2. We will fix FG as the base and fold the segments AF and BG downwards to form the triangle. See fig.b
3. With F as centre, and AF as radius, draw an arc (shown in green colour)
4. With G as centre, and BG as radius, draw another arc (shown in white colour)
5. The two arcs will intersect at H
6. HGF is the required triangle
(ii) The perimeter is given as 10 cm. So draw a line AB 10 cm long. See fig.18.19 below:
1. Divide the line AB in the ratio 3:4:5. The steps for this division are as follows:
2. We will fix FG as the base and fold the segments AF and BG downwards to form the triangle. See fig.b
3. With F as centre, and AF as radius, draw an arc (shown in green colour)
4. With G as centre, and BG as radius, draw another arc (shown in white colour)
5. The two arcs will intersect at H
6. HGF is the required triangle
(ii) The perimeter is given as 10 cm. So draw a line AB 10 cm long. See fig.18.19 below:
 |
| Fig.18.19. Length of AB = 10 cm |
• Draw a line AC of length 12 cm (∵ 3+4+5 =12) in any convenient direction.
• Mark a point E at a distance of 3 cm from A. Also another point D at a distance of 4 cm from E
• Join B and C. Draw a lines DG and EF parallel to BC.
• Then AF : FG : BG = 3:4:5
2. We will fix FG as the base and fold the segments AF and BG downwards to form the triangle. See fig.b
3. With F as centre, and AF as radius, draw an arc (shown in green colour)
4. With G as centre, and BG as radius, draw another arc (shown in white colour)
5. The two arcs will intersect at H
6. HGF is the required triangle
(iii) The perimeter is given as 10 cm. So draw a line AB 10 cm long. See fig.18.20 below:
1. Divide the line AB in the ratio 2:3:4. The steps for this division are as follows:
2. We will fix FG as the base and fold the segments AF and BG downwards to form the triangle. See fig.b
3. With F as centre, and AF as radius, draw an arc (shown in green colour)
4. With G as centre, and BG as radius, draw another arc (shown in white colour)
5. The two arcs will intersect at H
6. HGF is the required triangle
(iii) The perimeter is given as 10 cm. So draw a line AB 10 cm long. See fig.18.20 below:
 |
| Fig.18.20. Length of AB = 10 cm |
• Draw a line AC of length 9 cm (∵ 2+3+4 =12) in any convenient direction.
• Mark a point E at a distance of 2 cm from A. Also another point D at a distance of 3 cm from E
• Join B and C. Draw a lines DG and EF parallel to BC.
• Then AF : FG : BG = 2:3:4
2. We will fix FG as the base and fold the segments AF and BG downwards to form the triangle. See fig.b
3. With F as centre, and AF as radius, draw an arc (shown in green colour)
4. With G as centre, and BG as radius, draw another arc (shown in white colour)
5. The two arcs will intersect at H
6. HGF is the required triangle
Solved example 18.7
2. We will fix FG as the base and fold the segments AF and BG downwards to form the triangle. See fig.b
3. With F as centre, and AF as radius, draw an arc (shown in green colour)
4. With G as centre, and BG as radius, draw another arc (shown in white colour)
5. The two arcs will intersect at H
6. HGF is the required triangle
Solved example 18.7
The fig.18.21(a) shows a
non-isosceles trapezium.
 |
| Fig.18.21 |
It's diagonals intersect at P. Prove that PA
× PD = PB × PC
Solution:
■ The fig.a shows a
non-isosceles trapezium. Whether it is an isosceles trapezium, or a
non-isosceles trapezium, the top and bottom sides will be parallel. So AB and DC are parallel
1. Through P, draw a red line
parallel to AB. This is shown in fig.b
2. Now there are 3 parallel
lines: AB, 'the red line' and DC
3. These three parallel lines cut
through the diagonals AC and BD
4. The distances cut will be in
the same ratio. So we get: PA:PC = PB:PD
5. ⇒ PA⁄PC
= PB⁄PD ⇒ PA × PD = PB × PC
• It may be noted that, the above result is applicable to the two diagonals of any trapezium
One more solved example is given as a presentation video here:
It is a problem involving shadows cast by objects
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