Let a, b and c be the sides of a triangle and c be the longest side.
If a, b and c are the sides of a right triangle, then by Pythagorean theorem,
c2 = a2 + b2
If c2 ≠ a2 + b2, then the triangle is not a right triangle.
In each case, determine if the side lengths lengths form a right triangle.
Example 1 :
20, 21, 29
Solution :
Let a = 20, b = 21 and c = 29.
c2 = 292
= 841 ----(1)
a2 + b2 = 202 + 212
= 400 + 441
= 841 ----(2)
From (1) and (2),
c2 = a2 + b2
The side lengths 20, 21 and 29 form a right triangle.
Example 2 :
8, 10, 12
Solution :
Let a = 8, b = 10 and c = 12.
c2 = 122
= 144 ----(1)
a2 + b2 = 82 + 102
= 64 + 100
= 164 ----(2)
From (1) and (2),
c2 ≠ a2 + b2
The side lengths 8, 10 and 12 do not form a right triangle.
Example 3 :
30, 40, 50
Solution :
Let a = 30, b = 40 and c = 50.
c2 = 502
= 2500 ----(1)
a2 + b2 = 302 + 402
= 900 + 1600
= 2500 ----(2)
From (1) and (2),
c2 = a2 + b2
The side lengths 30, 40 and 50 form a right triangle.
Example 4 :
6, 12, 18
Solution :
Let a = 6, b = 12 and c = 18.
c2 = 182
= 324 ----(1)
a2 + b2 = 62 + 122
= 36 + 144
= 180 ----(2)
From (1) and (2),
c2 ≠ a2 + b2
The side lengths 6, 12 and 18 do not form a right triangle.
Example 5 :
24, 30, 36
Solution :
Let a = 24, b = 30 and c = 36.
c2 = 362
= 1296 ----(1)
a2 + b2 = 242 + 302
= 576 + 900
= 1476 ----(2)
From (1) and (2),
c2 ≠ a2 + b2
The side lengths 24, 30 and 36 do not form a right triangle.
Example 6 :
A triangle can have _________ right angle.
Solution :
A triangle can have one right angle.
Example 7 :
Each angle of an acute angled triangle is ________than 900
Solution :
Each angle of an acute angled triangle is lesser than 900
Example 8 :
The sum of the two acute angles in a right angled triangle is __
Solution :
The sum of the two acute angles in a right angled triangle is 900
Example 9 :
The triangle DEF is shown below. Find the length of EF to 1 decimal place.
Solution :
Since the triangle DEF is a right triangle, it must satisfy the Pythagorean theorem.
c2 = a2 + b2
DF2 = DE2 + EF2
(8.9)2 = 7.92 + EF2
79.21 = 62.41 + EF2
EF2 = 79.21 - 62.41
EF2 = 16.8
EF = √16.8
EF = 4.09 cm
Approximately 4.1 cm
Example 10 :
To wash a window that is 8 meters off the ground, Ben leans a 10 meter ladder against the side of the building. To reach the window, how far from the building should Ben place the base of the ladder?
Solution :
Let x be the distance between base of the building to the foot of the ladder.
102 = 82 + x2
100 = 64 + x2
x2 = 100 - 64
x2 = 36
x = √36
x = 6 m
So, the distance between base of the building to the foot of the ladder is 6 m.
Example 11 :
A rectangular swimming pool is 21 meters wide and 50 metres long. Calculate the length of the diagonal to 1 decimal place.
Solution :
AC2 = AB2 + BC2
AC2 = 502 + 212
= 2500 + 441
= 2941
AC = √2941
AC = 54.23
So, the length of diagonal is 54.23 m
Example 12 :
Miss Barker is teaching a 5th grade class. She is standing 12 feet in front of Jim. Francisco is sitting 5 feet to Jim’s right. How far apart are Miss Barker and Francisco?
Solution :
Let x be the distance between Miss Barker and Francisco.
x2 = 122 + 52
x2 = 144 + 25
x2 = 169
x = √169
x = 13 ft
So, the distance between Miss Barker and Francisco is 13 ft.
Example 13 :
A triangle has sides with lengths of 10 metres, 16 metres and 20 metres. Is it a right angled triangle? Explain your reasoning.
Solution :
Let a = 10 m, b = 16 m and c = 20 m
c2 = a2 + b2
202 = 102 + 162
400 = 100 + 256
400 ≠ 356
So, the given sides are not measures of the right triangle.
Example 14 :
a) One side of a right angled triangle is 10 cm. The other two are both of length x. Calculate x to 2 decimal places.
b) Find the perimeter of the triangle in part a)
Solution :
a) 102 = x2 + x2
100 = 2x2
x2 = 100/2
x2 = 50
x = √50
x = 7.07
b) Perimeter of the triangle = x + x + 10
= 7.07 + 7.07 + 10
= 24.14 cm
So, the required length of diagonal is 24.14 cm
Example 15 :
Find the length of the diagonal of a square of side 4 cm to 2 decimal places.
Solution :
In square all sides will be equal. Let x be the length of diagonal.
x2 = 42 + 42
x2 = 16 + 16
x2 = 32
x = √32
x = 5.65
So, the required length of diagonal is 5.65 cm.
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