In the previous section we saw some solved examples on 30o, 60o and 45o right triangles. In this section, we will see a few more solved examples.
1. What is the ratio of the sides of a 30o, 60o, 90o triangle? If the length of the side opposite to the 30o angle is 10 cm, how much are the other sides?
Solution:
1. The ratio of the sides of a 30o, 60o, 90o triangle can be written as:
[smallest side : medium side : largest side] = [1 : √3 : 2]
2. Another way of writing this is:
• Medium side is √3 times the smallest side
• Largest side is 2 times the smallest side
Part 2:
1. In a 30o, 60o, 90o triangle, 30o is the smallest angle. So the side opposite 30o is the smallest side.
• This is given as 10 cm
2. So from (2) above, we get:
• Medium side = √3 × 10 = 10√3 cm
• Largest side = 2 × 10 = 20 cm
2. What is the ratio of the sides of a 45o, 45o, 90o triangle? If the length of the side opposite to the 45o angle is 6 cm, how much are the other sides?
Solution:
1. The ratio of the sides of a 45o, 45o, 90o triangle can be written as:
[smallest side : smallest side : largest side] = [1 : 1 : √2]
2. Another way of writing this is:
• Largest side is √2 times the smallest side
Part 2:
1. In a 45o, 45o, 90o triangle, 45o is the smallest angle. So the side opposite 45o is the smallest side.
• This is given as 6 cm
2. So from (2) above, we get:
• The other smallest side = 6 cm
• Largest side = √2 × 6 = 6√2 cm
3. If the height of an equilateral triangle is 12 cm, how much is it's sides?
Solution:
Fig (b) below shows rough sketch of the equilateral triangle and it's height
1. In the fig(b), BC is the height. It is given as 12 cm
2. Let the length of the side of the whole equilateral triangle ADC be 's' cm. We are required to find this 's'
3. AB will be equal to s⁄2 because BC is the perpendicular bisector of AD
4. Consider ⊿ABC on the left side. It is a right triangle. So we can find BC by applying the Pythagoras theorem:
BC2 = AC2 - AB2 ⟹ BC2 = s2 - (s⁄2)2 ⟹ BC2 = [(4s2 - s2) ⁄4] = [(3s2) ⁄4]
⟹ BC = √[(3s2) ⁄4] = (√3)s⁄2
5. This BC is given as 12 cm. So we can write:
(√3)s⁄2 = 12 ⟹ (√3)s = 24 ⟹ (√3)s = 8 × √3 × √3 ⟹ s = 8√3 cm
4. The diagonal of a square is 10 cm. What is the length of a side? What is the area of the square?
Solution:
Fig.(a) below shows a rough sketch.
1. We know that, the diagonal AC will bisect ∠DAB. So we get: ∠DAC = ∠BAC = 45o
2. Consider ⊿ABC. It is a 45o, 45o, 90o triangle
3. The ratio of the sides of a 45o, 45o, 90o triangle can be written as:
[smallest side : smallest side : largest side] = [1 : 1 : √2]
4. Another way of writing this is:
• Largest side is √2 times the smallest side
5. In a 45o, 45o, 90o triangle, 45o is the smallest angle. So the side opposite 45o is the smallest side. In the present problem it is BC
6. So from (4) above, we get:
• Largest side = side opposite 90o = 10 cm = √2 × BC
⟹ 5×√2×√2 = √2×BC ⟹ BC = 5√2 cm
7. So side of the square is 5√2 cm
Area = [5√2]2 = 25×2 = 50 cm2.
5. In each of the triangles shown in fig. below, the angles and one side are given. Find the lengths of the other two sides
Solution:
Fig(a):
1. ⊿ABC is a 45o, 45o, 90o triangle
2. The ratio of the sides of a 45o, 45o, 90o triangle can be written as:
[smallest side : smallest side : largest side] = [1 : 1 : √2]
3. Another way of writing this is:
• Largest side is √2 times the smallest side
• There are two smallest sides
4. In a 45o, 45o, 90o triangle, 45o is the smallest angle. So the side opposite 45o is the smallest side. In the present problem it is BC
• ∠ACB is also 45o. So AB is also the smallest side
• The two smallest sides are equal. So we get: AB = BC = 5 cm
5. So from (3) above, we get:
• Largest side = side opposite 90o = √2 × AB = 5√2 cm
Fig(b):
1. ⊿PQR is a 30o, 60o, 90o triangle
2. The ratio of the sides of a 30o, 60o, 90o triangle can be written as:
[smallest side : medium side : largest side] = [1 : √3 : 2]
3. Another way of writing this is:
• Medium side is √3 times the smallest side
• Largest side is 2 times the smallest side
4. In a 30o, 60o, 90o triangle, 30o is the smallest angle. So the side opposite 30o is the smallest side. In the present problem it is RQ which is 5 cm in length.
5. So from (3) above, we get:
• Medium side = √3 × 5 = 5√3 cm
• Largest side = 2 × 5 = 10 cm
Fig(c):
1. ⊿XYZ is a 30o, 60o, 90o triangle
2. The ratio of the sides of a 30o, 60o, 90o triangle can be written as:
[smallest side : medium side : largest side] = [1 : √3 : 2]
3. Another way of writing this is:
• Medium side is √3 times the smallest side
• Largest side is 2 times the smallest side
4. In a 30o, 60o, 90o triangle, 30o is the smallest angle. So the side opposite 30o is the smallest side. In the present problem it is YZ
5. So from (3) above, we get:
• Largest side = 10√3 = 2 × XY ⟹ XY = 5√3
• Medium side = √3 × XY = √3 × 5√3 = 5 × 3 = 15 cm
6. Calculate the perimeter of the triangle ABC shown in fig.(a) below. It's altitude is given as 6 cm.
Solution:
1. The two base angles are both 60o. So it is an equilateral triangle
2. Let the altitude meet the base at D. Then CD = 6 cm see fig.(b)
3. Since the triangle is equilateral, the altitude will be perpendicular to base. So ADC will be equal to 90o
4. Consider ⊿ADC. We have: ∠A = 60o and ∠D = 90o. So ∠C = 30o
5. It is a 30, 60, 90 triangle. AB is the shortest side and CD is the medium side.
6. In a 30, 60, 90 triangle, the medium side is √3 times the shortest side. So we can write:
CD = √3 × AD ⟹ 6 = √3 × AD ⟹ 2 × √3 × √3 = √3 × AD ⟹ AD = 2√3
7. AB = 2 × AD ⟹ AB = 2 × 2√3 ⟹ AB = 4√3
8. So one side is 4√3. Total perimeter = 3 × 4√3 = 12√3
7. Calculate the perimeter of the quadrilateral ABCD shown in fig.(a) below
Solution:
1. Consider ΔBCD. Two of it's angles are 45o. So it is an isosceles right triangle. The third angle at C will be 90o. This is shown in fig.(b)
2. It is a 45, 45, 90 triangle.
• BC and CD are the shortest sides. BC = CD = 10 cm
• BD is the hypotenuse
3. In a 45, 45, 90 triangle, hypotenuse is √2 times the shortest side. So we can write:
BD = 10√2
4. Consider ⊿ABD. It is a right triangle.
∠A = 90o and ∠D = 30o. So ∠B = 60o
• It is a 30, 60, 90 triangle. This is shown in fig.(b). AB is the shortest side, AD is the medium side and BD is the hypotenuse
5. In a 30, 60, 90 triangle, the hypotenuse is twice the shortest side. So we can write:
BD = 2 × AB ⟹10√2 = 2 × AB ⟹ AB = 5√2
6. In a 30, 60, 90 triangle, the medium side is √3 times the shortest side. So we can write:
AD = √3 × AB ⟹AD = √3 × 5√2 = 5√6
7. So perimeter = AB + BC + CD + DA = 5√2 + 10 + 10 + 5√6 = [5(√2+√6) + 20] cm
8. Calculate the perimeter of the quadrilateral ABCD shown in fig.(a) below:
Solution:
1. Three interior angles of the quadrilateral ABCD is given as 90o. So the fourth angle will also be 90o. It is a rectangle.
2. Consider the ⊿ADE in fig.(b). It's angle at E will be 60o. So ⊿ADE is a 30, 60, 90 triangle.
• AE is it's largest side and DE is the smallest side.
3. In a 30, 60, 90 triangle, the largest side will be 2 times the smallest side. So we can write:
AE = 2 × DE ⟹10 = 2 × DE ⟹ DE = 5 cm
4. Also in a 30, 60, 90 triangle, medium side is √3 times the smallest side. So we can write:
AD = √3 × DE ⟹ AD = √3 × 5 = 5 √3
• So we have calculated the width of the rectangle. Now we have to calculate the length
5. Consider ∠DAB. We can write:
∠DAB = ∠DAE + ∠EAC + ∠CAB ⟹ 90 = 30 + ∠EAC + 30 ⟹ 90 = 60 +∠ EAC ⟹ ∠EAC = 30o
6. Now consider ΔEAC. We can write:
∠EAC + ∠ACE + ∠CEA = 180o (∵ sum of interior angles in any triangle is 180)
⟹ 30 + ∠ACE + 120 = 180 ⟹150 + ∠ACE = 180 ⟹ ∠ACE = 30o
7. So the base angles in ΔEAC are both 30o. It is an isosceles triangle. The sides opposite the equal angles will be of the same length
• Thus we get: AE = CE = 10 cm
8. So length of the rectangle = ED + EC = 5 + 10 = 15 cm
9. Thus perimeter = 2(l+b) = 2 × (5√3 + 15) = (30 + 10√3) cm
1. What is the ratio of the sides of a 30o, 60o, 90o triangle? If the length of the side opposite to the 30o angle is 10 cm, how much are the other sides?
Solution:
1. The ratio of the sides of a 30o, 60o, 90o triangle can be written as:
[smallest side : medium side : largest side] = [1 : √3 : 2]
2. Another way of writing this is:
• Medium side is √3 times the smallest side
• Largest side is 2 times the smallest side
Part 2:
1. In a 30o, 60o, 90o triangle, 30o is the smallest angle. So the side opposite 30o is the smallest side.
• This is given as 10 cm
2. So from (2) above, we get:
• Medium side = √3 × 10 = 10√3 cm
• Largest side = 2 × 10 = 20 cm
2. What is the ratio of the sides of a 45o, 45o, 90o triangle? If the length of the side opposite to the 45o angle is 6 cm, how much are the other sides?
Solution:
1. The ratio of the sides of a 45o, 45o, 90o triangle can be written as:
[smallest side : smallest side : largest side] = [1 : 1 : √2]
2. Another way of writing this is:
• Largest side is √2 times the smallest side
Part 2:
1. In a 45o, 45o, 90o triangle, 45o is the smallest angle. So the side opposite 45o is the smallest side.
• This is given as 6 cm
2. So from (2) above, we get:
• The other smallest side = 6 cm
• Largest side = √2 × 6 = 6√2 cm
3. If the height of an equilateral triangle is 12 cm, how much is it's sides?
Solution:
Fig (b) below shows rough sketch of the equilateral triangle and it's height
2. Let the length of the side of the whole equilateral triangle ADC be 's' cm. We are required to find this 's'
3. AB will be equal to s⁄2 because BC is the perpendicular bisector of AD
4. Consider ⊿ABC on the left side. It is a right triangle. So we can find BC by applying the Pythagoras theorem:
BC2 = AC2 - AB2 ⟹ BC2 = s2 - (s⁄2)2 ⟹ BC2 = [(4s2 - s2) ⁄4] = [(3s2) ⁄4]
⟹ BC = √[(3s2) ⁄4] = (√3)s⁄2
5. This BC is given as 12 cm. So we can write:
(√3)s⁄2 = 12 ⟹ (√3)s = 24 ⟹ (√3)s = 8 × √3 × √3 ⟹ s = 8√3 cm
4. The diagonal of a square is 10 cm. What is the length of a side? What is the area of the square?
Solution:
Fig.(a) below shows a rough sketch.
2. Consider ⊿ABC. It is a 45o, 45o, 90o triangle
3. The ratio of the sides of a 45o, 45o, 90o triangle can be written as:
[smallest side : smallest side : largest side] = [1 : 1 : √2]
4. Another way of writing this is:
• Largest side is √2 times the smallest side
5. In a 45o, 45o, 90o triangle, 45o is the smallest angle. So the side opposite 45o is the smallest side. In the present problem it is BC
6. So from (4) above, we get:
• Largest side = side opposite 90o = 10 cm = √2 × BC
⟹ 5×√2×√2 = √2×BC ⟹ BC = 5√2 cm
7. So side of the square is 5√2 cm
Area = [5√2]2 = 25×2 = 50 cm2.
5. In each of the triangles shown in fig. below, the angles and one side are given. Find the lengths of the other two sides
Fig(a):
1. ⊿ABC is a 45o, 45o, 90o triangle
2. The ratio of the sides of a 45o, 45o, 90o triangle can be written as:
[smallest side : smallest side : largest side] = [1 : 1 : √2]
3. Another way of writing this is:
• Largest side is √2 times the smallest side
• There are two smallest sides
4. In a 45o, 45o, 90o triangle, 45o is the smallest angle. So the side opposite 45o is the smallest side. In the present problem it is BC
• ∠ACB is also 45o. So AB is also the smallest side
• The two smallest sides are equal. So we get: AB = BC = 5 cm
5. So from (3) above, we get:
• Largest side = side opposite 90o = √2 × AB = 5√2 cm
Fig(b):
1. ⊿PQR is a 30o, 60o, 90o triangle
2. The ratio of the sides of a 30o, 60o, 90o triangle can be written as:
[smallest side : medium side : largest side] = [1 : √3 : 2]
3. Another way of writing this is:
• Medium side is √3 times the smallest side
• Largest side is 2 times the smallest side
4. In a 30o, 60o, 90o triangle, 30o is the smallest angle. So the side opposite 30o is the smallest side. In the present problem it is RQ which is 5 cm in length.
5. So from (3) above, we get:
• Medium side = √3 × 5 = 5√3 cm
• Largest side = 2 × 5 = 10 cm
Fig(c):
1. ⊿XYZ is a 30o, 60o, 90o triangle
2. The ratio of the sides of a 30o, 60o, 90o triangle can be written as:
[smallest side : medium side : largest side] = [1 : √3 : 2]
3. Another way of writing this is:
• Medium side is √3 times the smallest side
• Largest side is 2 times the smallest side
4. In a 30o, 60o, 90o triangle, 30o is the smallest angle. So the side opposite 30o is the smallest side. In the present problem it is YZ
5. So from (3) above, we get:
• Largest side = 10√3 = 2 × XY ⟹ XY = 5√3
• Medium side = √3 × XY = √3 × 5√3 = 5 × 3 = 15 cm
6. Calculate the perimeter of the triangle ABC shown in fig.(a) below. It's altitude is given as 6 cm.
1. The two base angles are both 60o. So it is an equilateral triangle
2. Let the altitude meet the base at D. Then CD = 6 cm see fig.(b)
3. Since the triangle is equilateral, the altitude will be perpendicular to base. So ADC will be equal to 90o
4. Consider ⊿ADC. We have: ∠A = 60o and ∠D = 90o. So ∠C = 30o
5. It is a 30, 60, 90 triangle. AB is the shortest side and CD is the medium side.
6. In a 30, 60, 90 triangle, the medium side is √3 times the shortest side. So we can write:
CD = √3 × AD ⟹ 6 = √3 × AD ⟹ 2 × √3 × √3 = √3 × AD ⟹ AD = 2√3
7. AB = 2 × AD ⟹ AB = 2 × 2√3 ⟹ AB = 4√3
8. So one side is 4√3. Total perimeter = 3 × 4√3 = 12√3
7. Calculate the perimeter of the quadrilateral ABCD shown in fig.(a) below
1. Consider ΔBCD. Two of it's angles are 45o. So it is an isosceles right triangle. The third angle at C will be 90o. This is shown in fig.(b)
2. It is a 45, 45, 90 triangle.
• BC and CD are the shortest sides. BC = CD = 10 cm
• BD is the hypotenuse
3. In a 45, 45, 90 triangle, hypotenuse is √2 times the shortest side. So we can write:
BD = 10√2
4. Consider ⊿ABD. It is a right triangle.
∠A = 90o and ∠D = 30o. So ∠B = 60o
• It is a 30, 60, 90 triangle. This is shown in fig.(b). AB is the shortest side, AD is the medium side and BD is the hypotenuse
5. In a 30, 60, 90 triangle, the hypotenuse is twice the shortest side. So we can write:
BD = 2 × AB ⟹10√2 = 2 × AB ⟹ AB = 5√2
6. In a 30, 60, 90 triangle, the medium side is √3 times the shortest side. So we can write:
AD = √3 × AB ⟹AD = √3 × 5√2 = 5√6
7. So perimeter = AB + BC + CD + DA = 5√2 + 10 + 10 + 5√6 = [5(√2+√6) + 20] cm
8. Calculate the perimeter of the quadrilateral ABCD shown in fig.(a) below:
1. Three interior angles of the quadrilateral ABCD is given as 90o. So the fourth angle will also be 90o. It is a rectangle.
2. Consider the ⊿ADE in fig.(b). It's angle at E will be 60o. So ⊿ADE is a 30, 60, 90 triangle.
• AE is it's largest side and DE is the smallest side.
3. In a 30, 60, 90 triangle, the largest side will be 2 times the smallest side. So we can write:
AE = 2 × DE ⟹10 = 2 × DE ⟹ DE = 5 cm
4. Also in a 30, 60, 90 triangle, medium side is √3 times the smallest side. So we can write:
AD = √3 × DE ⟹ AD = √3 × 5 = 5 √3
• So we have calculated the width of the rectangle. Now we have to calculate the length
5. Consider ∠DAB. We can write:
∠DAB = ∠DAE + ∠EAC + ∠CAB ⟹ 90 = 30 + ∠EAC + 30 ⟹ 90 = 60 +∠ EAC ⟹ ∠EAC = 30o
6. Now consider ΔEAC. We can write:
∠EAC + ∠ACE + ∠CEA = 180o (∵ sum of interior angles in any triangle is 180)
⟹ 30 + ∠ACE + 120 = 180 ⟹150 + ∠ACE = 180 ⟹ ∠ACE = 30o
7. So the base angles in ΔEAC are both 30o. It is an isosceles triangle. The sides opposite the equal angles will be of the same length
• Thus we get: AE = CE = 10 cm
8. So length of the rectangle = ED + EC = 5 + 10 = 15 cm
9. Thus perimeter = 2(l+b) = 2 × (5√3 + 15) = (30 + 10√3) cm
In the next section we will see a few more solved examples.
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