In the previous section we saw the details about √2. In this section, we will see √3. First, we will try to get a pictorial representation of √3. Consider the fig.16.7 below:
• ⊿ABC is a right triangle, right angled at A. The perpendicular sides AB and AC are 1 unit each. We have seen in the previous section that, the hypotenuse BC will be equal to √2. Now, we are going to add some extra geometrical figures, based on this ⊿ABC:
• ⊿BDC is drawn adjacent to ⊿ABC
• ⊿BDC is right angled at B
• One perpendicular side BC of ⊿BDC is the same hypotenuse of ⊿ABC. So BC = √2
• The other perpendicular side BD is equal to 1 unit
• Then, what is the hypotenuse CD of ⊿BDC?
• Applying Pythagoras theorem to ⊿BDC, we get:
CD2 = BD2 + BC2 ⇒ CD2 = 12 + (√2)2 ⇒ CD2 = 1 + 2 ⇒ CD2 = 3. ∴ CD = √3
• So CD is a pictorial representation of √3
• But there is more. With base CD, draw a square CDEF (coloured in green). All sides of this square will be equal to CD
• That is., all sides of the square CDEF will be equal to √3
• What is the area of a square with side √3?
• Obviously, the area will be equal to (√3×√3) = (√3)2 = 3
• So we can write: √3 is the side of a square whose area is 3 sq.units
Let us write a summary:
■ Length of the side of a square, whose area is 3 sq.units, is equal to √3
■ Hypotenuse of a right triangle, whose perpendicular sides are '1 unit' and '√2 unit', is equal to √3
■ Diagonal of a rectangle, whose adjacent sides are 1 unit and √2 unit, is equal to √3
Like √2, √3 is also an irrational number. The calculation of it's approximate value is shown in fig.16.8 below:
Based on the above, we can write:
• √3 ≈ 1.7 (correct to 1 decimal place)
• √3 ≈ 1.73 (correct to 2 decimal places)
• √3 ≈ 1.732 (correct to 3 decimal places) and so on...
The symbol '≈' is read as 'almost equal to'.
• In this way, we can calculate the approximate values of irrational numbers like √5, √7 etc.,
In the fig.16.9(a) below, the square on the hypotenuse of the top most right triangle is drawn.
Calculate the length of it's side, and it's area.
Solution:
• In the given fig.16.9(a), the lower most triangle has both it's perpendicular sides 1 m length.
• Also, bases of all the triangles are 1m, and all triangles are right angled.
• In this problem, it is better to give names (as shown in fig.b) to the triangles and squares. Then we will be able to write the steps easily.
1. Consider the lower most ⊿OAB. Applying Pythagoras theorem, we get:
2. OB2 = OA2 + AB2 ⇒ OB2 = 12 + 12 ⇒ OB2 = 1 + 1 ⇒ OB2 = 2. ∴ OB = √2
3. Consider the next upper ⊿OBC. Applying Pythagoras theorem, we get:
4. OC2 = OB2 + BC2 ⇒ OC2 = (√2)2 + 12 ⇒ OC2 = 2 + 1 ⇒ OC2 = 3. ∴ OC = √3
5. Consider the next upper ⊿OCD. Applying Pythagoras theorem, we get:
6. OD2 = OC2 + CD2 ⇒ OD2 = (√3)2 + 12 ⇒ OD2 = 3 + 1 ⇒ OD2 = 4. ∴ OD = √4
7. Consider the next upper ⊿ODE. Applying Pythagoras theorem, we get:
8. OE2 = OD2 + DE2 ⇒ OE2 = (√4)2 + 12 ⇒ OE2 = 4 + 1 ⇒ OE2 = 5. ∴ OE = √5
■ So the length of the side of the square EFGO is √5
■ ∴ Area = (√5×√5) = (√5)2 = 5 sq.m
Solved example 16.2
In the fig.16.10(a) below, a square is drawn on the altitude of an equilateral triangle. The side of the triangle is 2 m
(i) What is the altitude of the triangle?
(ii) What is the area of the square?
(iii) What are the lengths of the other two sides of the triangle shown in fig.16.10(b)?
Solution:
• We are given fig.16.10(a). It is better to give names (as shown in fig.c) to the triangle and square. Then we will be able to write the steps easily.
1. Given ABC is an equilateral triangle. So AB=BC=CA=2 m.
2. CD is the altitude. So AD=BD=1 m
3. Altitude CD is perpendicular to AB. So triangles ADC and BDC are right angled at D
4. Applying Pythagoras theorem to ⊿ADC, we get:
5. AC2 = AD2 + CD2 ⇒ CD2 = AC2 - AD2 ⇒ CD2 = 22 - 12 ⇒ CD2 = 4 - 1 ⇒ CD2 = 3. ∴ CD = √3
■ Thus we get: Altitude of the triangle = √3 m
6. Given that CDEF is a square. In (5), we calculated it’s one side CD as √3 m
■ So area of CDEF = (√3×√3) = (√3)2 = 3 sq.m
7. We are given a triangle PQR, with two angles and it's included side
8. Sum of the interior angles of any triangle is 180o
9. So we can write: ∠RPQ + ∠PQR + ∠PRQ = 180 ⇒ 60+30+∠PRQ =180 ⇒ 90+∠PRQ =180 ⇒ ∠PRQ = 180-90 =90
10. Thus we find that, PQR is a right angled triangle.
11. We are required to find the sides PR and QR
12. We cannot apply Pythagoras theorem because, only one side PQ is known. We must use some other method
13. Consider our first equilateral triangle ABC. It is shown separately in fig.d.
14. As it is an equilateral triangle, ∠CAB = ∠ABC = ∠BCA = 60o
15. Now consider ⊿ADC. The ∠CAD is same as ∠CAB.
16. But from (14), ∠CAB = 60o. So we get: ∠CAD = 60o
17. Now apply the sum of interior angle property to ⊿ADC
18. We can write: ∠CAD + ∠ADC + ∠DCA = 180 ⇒ 60+90+∠DCA =180 ⇒ 150 +∠DCA =180 ⇒ ∠DCA = 180-150 =30o
19. In fig.b, we have a triangle PQR, with two angles 30o and 60o, with an included side 2m
20. In fig.d, we have a triangle ADC, with two angles 30o and 60o, with an included side 2m
21. It is a case of ASA congruence. The two triangles are exactly the same. We can obtain the ‘missing details in ⊿PQR’ from ⊿ABC. For that, first we must write the correspondence:
22. From the figs.(b) and (d), it is obvious that P↔A, Q↔C and R↔D. Because,
♦ ∠P and ∠A have the same measure (60o)
♦ ∠Q and ∠C have the same measure (30o)
♦ ∠R and ∠D have the same measure (90o)
23. Now we write the corresponding of the sides.
♦ We have P↔A and Q↔C. So PQ↔AC
♦ We have Q↔C and R↔D. So QR↔CD
♦ We have P↔A and R↔D. So PR↔AD
■ Now we can write the missing details in ⊿PQR: We want the lengths PR and QR.
♦ From (23), we have PR↔AD. So PR = AD = 1 cm
♦ From (23), we have QR↔CD. So QR = CD = √3 cm [from (5)]
What we saw in the above problem, is a special triangle. It has one angle 30o , and another 60o. The third angle will then obviously be 90o. Let us randomly make a few such triangles. They are shown in the fig.16.11 below:
Fig.16.11.a shows the exact ⊿ADC which we saw in fig.16.10.c above. It has two angles 30o and 60o, and the included side 2 m. The other three are drawn randomly. But with the specified angles. How can we draw random triangles with the specified angles?
1. Draw a line XY of any convenient length, in any convenient direction. It is the first line of our triangle. It's length and direction does not matter for the construction.
2. But the second and third lines should be drawn in specified directions. That is., at the specified angles:
♦ Draw the second line from X, at 30o to XY. Here the length does not matter for the construction. But the angle does.
♦ Draw the third line at Y, at 60o to XY. Here also, the length does not matter for the construction, but the angle does.
3. The second and third lines will meet at Z. The triangle is complete. We can see that, in the construction of this triangle, we are not aware of the lengths of sides. We completed the triangle using angles only.
4. Once the triangle is complete, we must measure the sides and write them near the corresponding sides. We must also write the angles.
In this way, three random triangles ⊿XYZ, ⊿MNO, and ⊿UVW are constructed in the fig.16.11. All of them are right angled triangles
5.All of them have:
• One smallest side. This is the side opposite to the smallest angle which is the 30o angle. We will call it 'small side'
• One medium side. This is the side opposite to the intermediate angle which is the 60o angle. We will call it 'medium side'
• One largest side. This is the side opposite to the largest angle which is the 90o angle. We will call it 'large side'
6. Now, in each triangle, consider the ratio:
small side : medium side : large side
7. In fig.a we will get 1:√3:2 [remember that √3 ≈ 1.73, which lies in between 1 and 2. So √3 is indeed the medium side]
8. In ⊿XYZ of fig.b, we will get 1.81 : 3.14 : 3.62.
• Is there a common factor? Let us divide by the smallest side:
• 1.81/1.81 = 1, 3.14/1.81 = 1.73, 3.62/1.81 = 2
• Consider the second value: 3.14/1.81 = 1.73. But this 1.73 is an approximate value of √3. So we can write:
■ small side : medium side : large side
= 1.81 : 3.14 : 3.62 ⇒ 1 : 1.73 : 2 ⇒ 1 : √3 : 2
9. In ⊿MNO of fig.c, we will get 1.45 : 2.51 : 2.90
• Is there a common factor? Let us divide by the smallest side:
• 1.45/1.45 = 1, 2.51/1.45 = 1.73, 2.90/1.45 = 2
• Consider the second value: 2.51/1.45 = 1.73. But this 1.73 is an approximate value of √3. So we can write:
■ small side : medium side : large side
= 1.45 : 2.51 : 2.90 ⇒ 1 : 1.73 : 2 ⇒ 1 : √3 : 2
10. In ⊿UVW of fig.d, we will get 2.19 : 3.79 : 4.38
• Is there a common factor? Let us divide by the smallest side:
• 2.19/2.19 = 1, 3.79/2.19 = 1.73, 4.38/2.19 = 2
• Consider the second value: 3.79/2.19 = 1.73. But this 1.73 is an approximate value of √3. So we can write:
■ small side : medium side : large side
= 2.19 : 3.79 : 4.38 ⇒ 1 : 1.73 : 2 ⇒ 1 : √3 : 2
So we saw 3 random 30o, 60o , 90o triangles. In all of them we got the same ratio. In fact, in any such triangle that we consider, we will get the same ratio. So we can write:
■ If the angles of a triangle are 30o, 60o and 90o, then the ratio
small side : medium side : large side = 1 : √3 : 2
Another way of saying this is:
■ If the angles of a triangle are 30o, 60o and 90o, then:
• The medium side will always be √3 times the smallest side
• The largest side will always be 2 times the smallest side
Both sides are approximately equal.
• Let us repeat the steps from (3) using a more exact value of √3:
So, when we increase the number of decimal places, the accuracy increases, and we get exactly equal quantities on both sides.
In the next section, we will see more solved examples.
 |
| Fig.16.7 |
• ⊿BDC is drawn adjacent to ⊿ABC
• ⊿BDC is right angled at B
• One perpendicular side BC of ⊿BDC is the same hypotenuse of ⊿ABC. So BC = √2
• The other perpendicular side BD is equal to 1 unit
• Then, what is the hypotenuse CD of ⊿BDC?
• Applying Pythagoras theorem to ⊿BDC, we get:
CD2 = BD2 + BC2 ⇒ CD2 = 12 + (√2)2 ⇒ CD2 = 1 + 2 ⇒ CD2 = 3. ∴ CD = √3
• So CD is a pictorial representation of √3
• But there is more. With base CD, draw a square CDEF (coloured in green). All sides of this square will be equal to CD
• That is., all sides of the square CDEF will be equal to √3
• What is the area of a square with side √3?
• Obviously, the area will be equal to (√3×√3) = (√3)2 = 3
• So we can write: √3 is the side of a square whose area is 3 sq.units
Let us write a summary:
■ Length of the side of a square, whose area is 3 sq.units, is equal to √3
■ Hypotenuse of a right triangle, whose perpendicular sides are '1 unit' and '√2 unit', is equal to √3
■ Diagonal of a rectangle, whose adjacent sides are 1 unit and √2 unit, is equal to √3
 |
| Fig.16.8 |
• √3 ≈ 1.7 (correct to 1 decimal place)
• √3 ≈ 1.73 (correct to 2 decimal places)
• √3 ≈ 1.732 (correct to 3 decimal places) and so on...
The symbol '≈' is read as 'almost equal to'.
• In this way, we can calculate the approximate values of irrational numbers like √5, √7 etc.,
We will now see some solved examples
Solved example 16.1In the fig.16.9(a) below, the square on the hypotenuse of the top most right triangle is drawn.
 |
| Fig.16.9 |
Solution:
• In the given fig.16.9(a), the lower most triangle has both it's perpendicular sides 1 m length.
• Also, bases of all the triangles are 1m, and all triangles are right angled.
• In this problem, it is better to give names (as shown in fig.b) to the triangles and squares. Then we will be able to write the steps easily.
1. Consider the lower most ⊿OAB. Applying Pythagoras theorem, we get:
2. OB2 = OA2 + AB2 ⇒ OB2 = 12 + 12 ⇒ OB2 = 1 + 1 ⇒ OB2 = 2. ∴ OB = √2
3. Consider the next upper ⊿OBC. Applying Pythagoras theorem, we get:
4. OC2 = OB2 + BC2 ⇒ OC2 = (√2)2 + 12 ⇒ OC2 = 2 + 1 ⇒ OC2 = 3. ∴ OC = √3
5. Consider the next upper ⊿OCD. Applying Pythagoras theorem, we get:
6. OD2 = OC2 + CD2 ⇒ OD2 = (√3)2 + 12 ⇒ OD2 = 3 + 1 ⇒ OD2 = 4. ∴ OD = √4
7. Consider the next upper ⊿ODE. Applying Pythagoras theorem, we get:
8. OE2 = OD2 + DE2 ⇒ OE2 = (√4)2 + 12 ⇒ OE2 = 4 + 1 ⇒ OE2 = 5. ∴ OE = √5
■ So the length of the side of the square EFGO is √5
■ ∴ Area = (√5×√5) = (√5)2 = 5 sq.m
Solved example 16.2
In the fig.16.10(a) below, a square is drawn on the altitude of an equilateral triangle. The side of the triangle is 2 m
 |
| Fig.16.10 |
(ii) What is the area of the square?
(iii) What are the lengths of the other two sides of the triangle shown in fig.16.10(b)?
Solution:
• We are given fig.16.10(a). It is better to give names (as shown in fig.c) to the triangle and square. Then we will be able to write the steps easily.
1. Given ABC is an equilateral triangle. So AB=BC=CA=2 m.
2. CD is the altitude. So AD=BD=1 m
3. Altitude CD is perpendicular to AB. So triangles ADC and BDC are right angled at D
4. Applying Pythagoras theorem to ⊿ADC, we get:
5. AC2 = AD2 + CD2 ⇒ CD2 = AC2 - AD2 ⇒ CD2 = 22 - 12 ⇒ CD2 = 4 - 1 ⇒ CD2 = 3. ∴ CD = √3
■ Thus we get: Altitude of the triangle = √3 m
6. Given that CDEF is a square. In (5), we calculated it’s one side CD as √3 m
■ So area of CDEF = (√3×√3) = (√3)2 = 3 sq.m
7. We are given a triangle PQR, with two angles and it's included side
8. Sum of the interior angles of any triangle is 180o
9. So we can write: ∠RPQ + ∠PQR + ∠PRQ = 180 ⇒ 60+30+∠PRQ =180 ⇒ 90+∠PRQ =180 ⇒ ∠PRQ = 180-90 =90
10. Thus we find that, PQR is a right angled triangle.
11. We are required to find the sides PR and QR
12. We cannot apply Pythagoras theorem because, only one side PQ is known. We must use some other method
13. Consider our first equilateral triangle ABC. It is shown separately in fig.d.
14. As it is an equilateral triangle, ∠CAB = ∠ABC = ∠BCA = 60o
15. Now consider ⊿ADC. The ∠CAD is same as ∠CAB.
16. But from (14), ∠CAB = 60o. So we get: ∠CAD = 60o
17. Now apply the sum of interior angle property to ⊿ADC
18. We can write: ∠CAD + ∠ADC + ∠DCA = 180 ⇒ 60+90+∠DCA =180 ⇒ 150 +∠DCA =180 ⇒ ∠DCA = 180-150 =30o
19. In fig.b, we have a triangle PQR, with two angles 30o and 60o, with an included side 2m
20. In fig.d, we have a triangle ADC, with two angles 30o and 60o, with an included side 2m
21. It is a case of ASA congruence. The two triangles are exactly the same. We can obtain the ‘missing details in ⊿PQR’ from ⊿ABC. For that, first we must write the correspondence:
22. From the figs.(b) and (d), it is obvious that P↔A, Q↔C and R↔D. Because,
♦ ∠P and ∠A have the same measure (60o)
♦ ∠Q and ∠C have the same measure (30o)
♦ ∠R and ∠D have the same measure (90o)
23. Now we write the corresponding of the sides.
♦ We have P↔A and Q↔C. So PQ↔AC
♦ We have Q↔C and R↔D. So QR↔CD
♦ We have P↔A and R↔D. So PR↔AD
■ Now we can write the missing details in ⊿PQR: We want the lengths PR and QR.
♦ From (23), we have PR↔AD. So PR = AD = 1 cm
♦ From (23), we have QR↔CD. So QR = CD = √3 cm [from (5)]
 |
| Fig.16.11 |
There is an easy method:
Consider for example ⊿XYZ in fig.16.11.b. It can be drawn by the following procedure:1. Draw a line XY of any convenient length, in any convenient direction. It is the first line of our triangle. It's length and direction does not matter for the construction.
2. But the second and third lines should be drawn in specified directions. That is., at the specified angles:
♦ Draw the second line from X, at 30o to XY. Here the length does not matter for the construction. But the angle does.
♦ Draw the third line at Y, at 60o to XY. Here also, the length does not matter for the construction, but the angle does.
3. The second and third lines will meet at Z. The triangle is complete. We can see that, in the construction of this triangle, we are not aware of the lengths of sides. We completed the triangle using angles only.
4. Once the triangle is complete, we must measure the sides and write them near the corresponding sides. We must also write the angles.
In this way, three random triangles ⊿XYZ, ⊿MNO, and ⊿UVW are constructed in the fig.16.11. All of them are right angled triangles
5.All of them have:
• One smallest side. This is the side opposite to the smallest angle which is the 30o angle. We will call it 'small side'
• One medium side. This is the side opposite to the intermediate angle which is the 60o angle. We will call it 'medium side'
• One largest side. This is the side opposite to the largest angle which is the 90o angle. We will call it 'large side'
6. Now, in each triangle, consider the ratio:
small side : medium side : large side
7. In fig.a we will get 1:√3:2 [remember that √3 ≈ 1.73, which lies in between 1 and 2. So √3 is indeed the medium side]
8. In ⊿XYZ of fig.b, we will get 1.81 : 3.14 : 3.62.
• Is there a common factor? Let us divide by the smallest side:
• 1.81/1.81 = 1, 3.14/1.81 = 1.73, 3.62/1.81 = 2
• Consider the second value: 3.14/1.81 = 1.73. But this 1.73 is an approximate value of √3. So we can write:
■ small side : medium side : large side
= 1.81 : 3.14 : 3.62 ⇒ 1 : 1.73 : 2 ⇒ 1 : √3 : 2
9. In ⊿MNO of fig.c, we will get 1.45 : 2.51 : 2.90
• Is there a common factor? Let us divide by the smallest side:
• 1.45/1.45 = 1, 2.51/1.45 = 1.73, 2.90/1.45 = 2
• Consider the second value: 2.51/1.45 = 1.73. But this 1.73 is an approximate value of √3. So we can write:
■ small side : medium side : large side
= 1.45 : 2.51 : 2.90 ⇒ 1 : 1.73 : 2 ⇒ 1 : √3 : 2
10. In ⊿UVW of fig.d, we will get 2.19 : 3.79 : 4.38
• Is there a common factor? Let us divide by the smallest side:
• 2.19/2.19 = 1, 3.79/2.19 = 1.73, 4.38/2.19 = 2
• Consider the second value: 3.79/2.19 = 1.73. But this 1.73 is an approximate value of √3. So we can write:
■ small side : medium side : large side
= 2.19 : 3.79 : 4.38 ⇒ 1 : 1.73 : 2 ⇒ 1 : √3 : 2
So we saw 3 random 30o, 60o , 90o triangles. In all of them we got the same ratio. In fact, in any such triangle that we consider, we will get the same ratio. So we can write:
■ If the angles of a triangle are 30o, 60o and 90o, then the ratio
small side : medium side : large side = 1 : √3 : 2
Another way of saying this is:
■ If the angles of a triangle are 30o, 60o and 90o, then:
• The medium side will always be √3 times the smallest side
• The largest side will always be 2 times the smallest side
• 30o, 60o, 90o triangles are encountered many times in science and engineering topics.
• One of the 'Set squares' in the geometry box have 30o, 60o, 90o angles.
we will now see a solved example.
Solved example 16.3
Fig.16.12 below shows a triangle ABC. Find the missing lengths
Solution:
In the problem, AB is given as 2.47. We have to find AC and BC
1. AB is the side opposite to the intermediate angle 60o. So AB is the medium side. Thus AC is the smallest side (opposite to 30o), and BC is the largest side (opposite to 90o)
2. We know that, the medium side is √3 times the smallest side.
3. So AB = √3 × AC ⇒ AC = AB/√3 ⇒ AC = 2.47/1.73 = 1.43
4. Thus we get: Smallest side = AC = 1.43
5. We know that, the largest side is 2 times the smallest side.
6. So BC = 2 × AC = 2 × 1.43 = 2.86
7. Thus we get: largest side = BC = 2.86
8. Check: Applying Pythagoras theorem, we get: BC2 = AB2 + AC2
⇒ 2.862 = 2.472 + 1.432 ⇒ 8.18 = 6.1 + 2.045 ⇒ 8.18 = 8.145• One of the 'Set squares' in the geometry box have 30o, 60o, 90o angles.
we will now see a solved example.
Solved example 16.3
Fig.16.12 below shows a triangle ABC. Find the missing lengths
 |
| Fig.16.12 |
In the problem, AB is given as 2.47. We have to find AC and BC
1. AB is the side opposite to the intermediate angle 60o. So AB is the medium side. Thus AC is the smallest side (opposite to 30o), and BC is the largest side (opposite to 90o)
2. We know that, the medium side is √3 times the smallest side.
3. So AB = √3 × AC ⇒ AC = AB/√3 ⇒ AC = 2.47/1.73 = 1.43
4. Thus we get: Smallest side = AC = 1.43
5. We know that, the largest side is 2 times the smallest side.
6. So BC = 2 × AC = 2 × 1.43 = 2.86
7. Thus we get: largest side = BC = 2.86
8. Check: Applying Pythagoras theorem, we get: BC2 = AB2 + AC2
Both sides are approximately equal.
• Let us repeat the steps from (3) using a more exact value of √3:
3. AB = √3 × AC ⇒ AC = AB/√3 ⇒ AC = 2.47/1.7320508 = 1.426055
4. Thus we get: Smallest side = AC = 1.426055
5. We know that, the largest side is 2 times the smallest side.
6. So BC = 2 × AC = 2 × 1.426055 = 2.85211
7. Thus we get: largest side = BC = 2.85211
8. Check: Applying Pythagoras theorem, we get: BC2 = AB2 + AC2
⇒ 2.852112 = 2.472 + 1.4260552 ⇒ 8.1345 = 6.1009 + 2.0336 ⇒ 8.1345 = 8.13454. Thus we get: Smallest side = AC = 1.426055
5. We know that, the largest side is 2 times the smallest side.
6. So BC = 2 × AC = 2 × 1.426055 = 2.85211
7. Thus we get: largest side = BC = 2.85211
8. Check: Applying Pythagoras theorem, we get: BC2 = AB2 + AC2
So, when we increase the number of decimal places, the accuracy increases, and we get exactly equal quantities on both sides.
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