SPECIAL RIGHT TRIANGLES

Find the value of x in each triangle

Problem 1 :

Solution :

It comes under 45-45-90 right triangle. 

Hypotenuse (y) = 2(Smallest sides)

y = 2(5)

y = 10

x = 5

Problem 2 :

Solution :

It comes under 30-60-90 right triangle. 

Hypotenuse = b

Smallest leg = 12

Longer leg (a) = √3(shorter leg)

= 12√3

Hypotenuse = 2(shorter leg)

= 2(12)

= 24

Problem 3 :

Solution :

Hypotenuse = 8

Shorter leg = c

c = d

Hypotenuse = 2(Shorter leg)

8 = 2c

c = 4 and d = 4

Problem 4 :

Solution :

Shorter leg = c, hypotenuse = 10, longer leg = d

Problem 5 :

Solution :

Shorter leg = r = 16, hypotenuse = q

Hypotenuse = 2(Shorter leg)

q = 2(16)

q = 32

Problem 6 :

Solution :

Hypotenuse  = m

Shorter leg = 6 and 

longer leg = p

Hypotenuse = 2(Shorter leg)

m = 2(6)

m = 12

Longer leg = √3 (shorter leg)

P = 6√3

Problem 7 :

Solution :

Hypotenuse  = h

Shorter leg = f and 

longer leg = 8

Longer leg = √3 (shorter leg)

8 = √3(f)

f = 8/√3

f = (8/√3) ⋅ (√3/√3)

f = 8√3/3

Hypotenuse (h) = 2(shorter leg)

= 2(8√3/3)

= 16√3/3

Problem 8 :

Solution :

Hypotenuse = 6√2

Shorter leg = n

Hypotenuse = 2(Shorter leg)

6√2 = 2n

n = 3√2

Problem 9 :

The side length of an equilateral triangle is 5 cm. Find the length of an altitude of the triangle.

Solution :

Shorter leg = CD = 2.5

Longer leg = AC

Hypotenuse = AD = 5

Longer leg = √3(Shorter leg)

= √3(2.5)

= 2.5√3

Problem 10 :

The perimeter of the square is 36 inches. Find the length of the diagonal.

Solution :

Let x be the side length of square.

Perimeter of the square = 36

4x = 36

x = 9

Using Pythagorean theorem :

x² + x² = (Length of diagonal)²

9² + 9² = (Length of diagonal)²

81 + 81 = (Length of diagonal)²

Length of diagonal = √162

= 9√2

Problem 11 :

The road sign is shaped like an equilateral triangle. Estimate the area of the sign by finding the area of the equilateral triangle.

Solution :

First fi nd the height h of the triangle by dividing it into two 30°- 60°- 90° triangles. The length of the longer leg of one of these triangles is h. The length of the shorter leg is 18 inches.

h = 18 ⋅ √3

= 18 √3

Area = 1/2 bh

= 1/2 (36) (18√3)

≈ 561.18

The area of the sign is about 561 square inches.

Problem 12 :

A tipping platform is a ramp used to unload trucks. How high is the end of an 80-foot ramp when the tipping angle is 30°? 45°?

When the tipping angle is 30°, the height h of the ramp is the length of the shorter leg of a 30°- 60°- 90° triangle. The length of the hypotenuse is 80 feet.

80 = 2h

40 = h

When the tipping angle is 45°, the height h of the ramp is the length of a leg of a 45°- 45°- 90° triangle. The length of the hypotenuse is 80 feet.

80 = h ⋅ √2

80/√2 = h

56.6 ≈  56 feet 7 inches

When the tipping angle is 30°, the ramp height is 40 feet. When the tipping angle is 45°, the ramp height is about 56 feet 7 inches.

Problem 13 :

The body of a dump truck is raised to empty a load of sand. How high is the 14-foot-long body from the frame when it is tipped upward by a 60° angle?

Solution :

In the 30-60-90 right triangle above,

The side which is opposite to 60 degree is √3 (smaller side)

Let half of the base = x (smaller side)

2x = 14

x = 7

Height =  √3 (7)

= 7√3

= 7(1.732)

= 12.124 foot

So, the required height is 12.124 foot.

Problem 14 :

A nut is shaped like a regular hexagon with side lengths of 1 centimeter. Find the value of x. (Hint: A regular hexagon can be divided into six congruent triangles.)

In a hexagon, we have six identitcal triangles. So, each angle measure at the center will be 360/6, that is 60.

Half of the triangle is 30-60-90 right triangle. Half of the base = 0.5 and half of the height is x/2.

Smaller side = 0.5, side which is opposite to 60 degree = x/2

x/2 = √3(smaller side)

x/2 = √3(0.5)

x = 2√3(0.5)

x = √3

x = 1.732 cm

So, the value of x is 1.732 cm.