DETERMINE IF THREE SIDES FORM A RIGHT TRIANGLE

By using Pythagorean theorem, we can determine whether the three sides will create a right triangle.

a² = b² + c²

Here the longest side can be considered as hypotenuse. We have to check, if the square of the longest side is equal to the some of squares of the remaining sides.

• If it is true, then we can decide the given sides will be sides of right triangle.
• If it is not true, then we can decide the given sides will not create right triangle.

The following figures are not drawn to scale. Which of the triangles are right angled?

Problem 1 :

Solution :

Using Pythagorean theorem.

 7² = 4² + 5²

49 = 16 + 25

49 ≠ 41

So, it is not a right triangle.

The following figures are not drawn to scale. Which of the triangles are right angled?

Problem 2 :

Solution :

The measure of longest side = 15

a = 15, b = 12 and c = 9

Using Pythagorean theorem.

15² = 12² + 9²

225 = 144 + 81

225 = 225

So, it is a right triangle.

Problem 3 :

Solution :

The measure of longest side = 9 cm

a = 9, b = 8 and c = 5

Using Pythagorean theorem.

9² = 8² + 5²

81 = 64 + 25

81 ≠ 89

So, it is not a right triangle.

Problem 4 :

Solution :

The measure of longest side = √12

a = √12, b = √7 and c = 3

Using Pythagorean theorem.

(√12)² = (√7)² + 3²

12 = 7 + 9

12 ≠ 16

So, it is not a right triangle.

Problem 5 :

Solution :

The measure of longest side = √75

a = √75, b = √48 and c = √27

Using Pythagorean theorem.

 (√75)² = (√48)² + (√27)²

75 = 48 + 27

75 = 75

So, it is a right triangle.

Problem 6 :

Solution :

The measure of longest side = √75

a = √75, b = √48 and c = √27

Using Pythagorean theorem.

17² = 15² + 8²

289 = 225 + 64

289 = 289

So, it is a right triangle.

Problem 7 :

Determine the area and perimeter of the triangle shown on the right.

Solution :

To find area of the triangle, we need the height of the triangle.

AB² = AC² + BC²

13² = 12² + BC²

BC² = 169 - 144

BC² = 25

BC = 5 cm

Perimeter of the triangle = AB + BC + CA

= 13 + 5 + 12

= 30 cm

Area of triangle = (1/2) x AB x BC

= (1/2) x 13 x 5

= 65/2

= 32.5 cm²

Problem 8 :

A rectangular park has a straight path from one corner to the opposite corner. If the park has dimensions of 2 km by 1.4 km, determine the length of the path.

Solution :

DB² = AB² + AD²

DB² = 2² + 1.4²

= 4 + 1.96

= 5.96

DB = 2.44 km

So, the length of the path is 2.44 km.

Problem 9 :

A 12-foot ladder is leaned against the side of a building. If the bottom of the ladder rests 3 feet from the wall, determine the height at which the top of the ladder rests on the wall.

Solution :

Height of the top of ladder rests on the wall :

AC² = AB² + BC²

12² = AB² + 3²

144 - 9 = AB²

 AB² = 135

AB =  √135

= √(5 x 3 x 3 x 3)

= 3 √15

So, the required height is 3 √15 ft

Problem 10 :

When a triangle is inscribed in a semicircle, a right angle is always formed at the point on the circular arc. Calculate the area of the shaded region in the diagram below.

Solution :

Angle in a semi circle is a right angle.

Diameter of the circle = base of the triangle

Diameter² = 37.8² + 27.1²

Diameter² = 1428.84 + 734.41

Diameter² = 2163.25

Diameter = √2163.25

= 46.5 m

Radius = 23.25 ft

Area of semicircle = (1/2) πr²

= (1/2) ⋅ 3.14 ⋅ (23.25)²

= 1.57 ⋅ 540.5625

= 848.68 square feet

Area of triangle = (1/2) ⋅ base ⋅ height

= (1/2) ⋅ 46.5 ⋅ 27.1

= 23.25 ⋅ 27.1

= 630.075 square feet

Area of the shaded region = area of semicircle - area of triangle

= 848.68 - 630.075

= 218.605 square ft

Problem 11 :

Determine the perimeter of the triangle shown on the right.

Solution :

Let x be the measure of equal sides.

28.3² = x² + x²

2x² = 800.39

x² = 800.39/2

x² = 400.445

x = √ 400.445

x = 20 (approximately)

Perimeter of the triangle = 2x + 28.3

= 2(20) + 28.3

= 40 + 28.3

= 68.3 mm

Problem 12 :

A television screen has a width-to-height ratio of 16:9. If the diagonal of the screen is 65 inches, determine its width and height.

Solution :

Width = 16x 

height = 9x

65² = (16x)² + (9x)²

4225 = 256x² + 81x²

4225 = 337x²

x² = 4225/337

x² = 12.53

x = √12.53

= 3.54

Width = 16 x ==> 16(3.54) ==> 56.64 inches

height = 9x ==> 9(3.54) ==> 31.86 inches

So, width and height are 57 inches and 32 inches respectively.