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