In the previous section we completed the discussion on the multiplication of irrational numbers. We also saw a practical case. In this section, we will see some solved examples.
Solved example 16.12
Fig.16.28 below shows four
equilateral triangles. Yellow, magenta, red and cyan. The yellow and
magenta are cut into two equal halves, thus giving 4 right triangles.
Red and cyan are kept uncut.
 |
| Fig.16.28 |
The four right triangles and the two
uncut triangles are put together to form a rectangle. If the side of
each of the original equilateral triangles is 1 m, what is the area
and perimeter of the rectangle?
Solution:
1. When the equilateral
triangle of side 1 m is cut in half, two right triangles are obtained
2. The base of each right
triangle is 1⁄2 m
3. The hypotenuse of each
right triangle is the same 1 m of the original equilateral triangle
4. So we have the hypotenuse
and the base. Applying Pythagoras theorem we get:
Altitude = √[12 –
(1⁄2)2] = √[1 – (1⁄4)]
= √[3⁄4] = √3⁄2
5. Now, the length of the
final rectangle is 2 times the altitude = 2 × √3⁄2
= √3 m
6. Width of the final
rectangle is the same base of the original equilateral triangle = 1 m
7. So area = length × width =
√3 × 1 = √3 sq.m = 1.73 sq.m (correct to 1 cm)
8. Perimeter = 2 × (length +
width) = 2(√3 + 1) = 2 × (1.73 +1) = 2 × 2.73 = 5.46 m (correct to 1 cm)
Solved example 16.13
A square and an equilateral
triangle of side twice the side of the square, are cut, and the
pieces are rearranged to form a trapezium, as shown in fig.16.29 below:
 |
| Fig.16.29 |
If the side of the square is 2 cm, what is the area and perimeter of the trapezium?
Solution:
1. When the square is cut
along the diagonal, it gives two right angled triangles.
2. The side of the square is
given as 2 cm. Applying Pythagoras theorem, the hypotenuse of the
triangles = √[22 + 22] = √[4 + 4] = √[8] = √[4×2] = √4 × √2 = 2√2
3. This hypotenuse becomes the
sloping side of the trapezium. The side of the square becomes height
of the trapezium
4. Now consider the
equilateral triangle. It's side is given as twice that of the square.
So side of the equilateral triangle = 2 × 2 = 4 cm
5. When it is cut into halves,
we get two right triangles. Hypotenuse of this right triangle = 4 cm.
Base = 2 cm. Applying Pythagoras theorem we get: Altitude = √[42 - 22] = √[16 - 4] = √[12] = √[4×3] = √4 × √3 = 2√3
6. This altitude becomes the
length of the 'middle rectangle' of the trapezium.
7. So now we have all the
required dimensions of the trapezium:
♦ Top parallel side = 2√3
♦ Bottom parallel side = 2 +
2 + 2√3 = 4 + 2√3
♦ Height = 2
8. Area of a trapezium = [(Top
parallel side + Bottom parallel side)⁄2] × height
♦ Top parallel side + Bottom
parallel side = 2√3 + 4 + 2√3 = 4 + 4√3 = 4(1 + √3)
∴ Area = [4(1+√3) ⁄2]
× 2 = 4(1 + √3) = 4 × (1 + 1.73) = 4 × 2.73 = 10.92 sq.cm
2. Through the vertex B, a perpendicular line BD is drawn to AC. So ∠ADB is 90o
3. Consider ⊿ABD. We know two of it's angles: ∠DAB = 60o and ∠ADB = 90o
4. So ∠ABD = 30o (∵ sum of interior angles of any triangle is 180o)
5. Thus we find that ⊿ABD is a 30, 60, 90 triangle. We have seen the details here. In such a triangle, if we know any one side, the other two sides can be easily calculated.
6. We know the side AB as 4 m. This is the longest side, the hypotenuse, which is opposite the largest angle 90o
7. The longest side is twice the smallest side. Here, the smallest side is AD, as it is opposite the smallest angle 30o
8. So we get: AB = 2AD ⇒ AD = AB/2 = 4/2 = 2 cm
9. The medium side is √3 times the smallest side. The medium side is BD, as it is opposite the intermediate angle 60o
10. So we get: BD = √3 × AD = √3 × 2 = 2√3
11. Thus we get all the sides of ⊿ABD
12. Now consider ⊿BCD. In this triangle, we will get ∠DBC = 75 - 30 = 45o
13. We will also get ∠DCB = 180 -(60 + 75) = 180 - 135 = 45o
14. So, in ⊿BCD, ∠B = ∠C = 45o. Base angles are equal. So it is an isosceles triangle
15. The sides opposite to the base angles will be equal. That is., BD = CD
16. But in (10), we obtained BD as 2√3 cm. So, CD is also equal to 2√3 cm
17. Now we have the two components of the side AC of the original triangle ABC:
18. AC = {AD + CD} = {2 [from 8] + 2√3 [from 16]} = 2 + 2√3
19. Now come back to ⊿BCD. It is a right triangle, and it's two legs are now known:
BD = 2√3 cm [from 10], and, CD = 2√3 cm [from 16]
20. Applying Pythagoras theorem, we get: BC = √[(2√3)2 + (2√3)2] = √[12 + 12] = √24 = √[4 × 6] = √4 × √6 = 2√6
21. Thus we get all the sides of the original triangle ABC
• AB = 4 cm [given] • BC = 2√6 [from 20] • AC = (2 + 2√3) [from 18]
22. Thus perimeter = 4 + 2√6 + 2 + 2√3 = 6 + 2√3 + 2√6 cm
23. Area = 1/2 × Base × Altitude = 1/2 × AC × BD = 1/2 × (2 + 2√3) × 2√3
⇒ Area = (2 + 2√3) × √3 ⇒ Area = [(2 × √3) + (2√3 × √3)] = [2√3 + (2 × 3)] = 2√3 + 6 sq.cm
Solved example 16.14
Calculate the area and
perimeter of the triangle shown in the fig.16.30(a) below
 |
| Fig.16.30 |
Solution:
1. Let us name the triangle as ABC. This is shown in fig.b2. Through the vertex B, a perpendicular line BD is drawn to AC. So ∠ADB is 90o
3. Consider ⊿ABD. We know two of it's angles: ∠DAB = 60o and ∠ADB = 90o
4. So ∠ABD = 30o (∵ sum of interior angles of any triangle is 180o)
5. Thus we find that ⊿ABD is a 30, 60, 90 triangle. We have seen the details here. In such a triangle, if we know any one side, the other two sides can be easily calculated.
6. We know the side AB as 4 m. This is the longest side, the hypotenuse, which is opposite the largest angle 90o
7. The longest side is twice the smallest side. Here, the smallest side is AD, as it is opposite the smallest angle 30o
8. So we get: AB = 2AD ⇒ AD = AB/2 = 4/2 = 2 cm
9. The medium side is √3 times the smallest side. The medium side is BD, as it is opposite the intermediate angle 60o
10. So we get: BD = √3 × AD = √3 × 2 = 2√3
11. Thus we get all the sides of ⊿ABD
12. Now consider ⊿BCD. In this triangle, we will get ∠DBC = 75 - 30 = 45o
13. We will also get ∠DCB = 180 -(60 + 75) = 180 - 135 = 45o
14. So, in ⊿BCD, ∠B = ∠C = 45o. Base angles are equal. So it is an isosceles triangle
15. The sides opposite to the base angles will be equal. That is., BD = CD
16. But in (10), we obtained BD as 2√3 cm. So, CD is also equal to 2√3 cm
17. Now we have the two components of the side AC of the original triangle ABC:
18. AC = {AD + CD} = {2 [from 8] + 2√3 [from 16]} = 2 + 2√3
19. Now come back to ⊿BCD. It is a right triangle, and it's two legs are now known:
BD = 2√3 cm [from 10], and, CD = 2√3 cm [from 16]
20. Applying Pythagoras theorem, we get: BC = √[(2√3)2 + (2√3)2] = √[12 + 12] = √24 = √[4 × 6] = √4 × √6 = 2√6
21. Thus we get all the sides of the original triangle ABC
• AB = 4 cm [given] • BC = 2√6 [from 20] • AC = (2 + 2√3) [from 18]
22. Thus perimeter = 4 + 2√6 + 2 + 2√3 = 6 + 2√3 + 2√6 cm
23. Area = 1/2 × Base × Altitude = 1/2 × AC × BD = 1/2 × (2 + 2√3) × 2√3
⇒ Area = (2 + 2√3) × √3 ⇒ Area = [(2 × √3) + (2√3 × √3)] = [2√3 + (2 × 3)] = 2√3 + 6 sq.cm
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