A surveyor wants to find the height of a building. at a point 109.0 ft from the base of the building he sights to the top of the
building and finds the distance to be 194.0 ft. how high is the building (to the nearest tenth)?
1 answer:
Draw an illustration
We could draw an illustration of the problem by drawing a right triangle.
109 ft will be the base of the triangle, 194 ft will be the hypotenuse of the triangle, and the height of the building will be the height of the triangle.
See image attached.
Solve the problem with phytagorean theorem
For an instance, h represents the height of the building. We could find the value of h by using pythagorean theorem
a² + b² = c²
a and b act as the base and the height, c acts as the hypotenuse
in this case, we could write it as below
h² + 109² = 194²
Solve the equation
h² + 109² = 194²
h² = 194² - 109²
h² = 37,636 - 11,881
h² = 25,755
h = √25,755
h = 160.4836
To the nearest tenth
h = 160.5
The building is 160.5 ft high
We could draw an illustration of the problem by drawing a right triangle.
109 ft will be the base of the triangle, 194 ft will be the hypotenuse of the triangle, and the height of the building will be the height of the triangle.
See image attached.
Solve the problem with phytagorean theorem
For an instance, h represents the height of the building. We could find the value of h by using pythagorean theorem
a² + b² = c²
a and b act as the base and the height, c acts as the hypotenuse
in this case, we could write it as below
h² + 109² = 194²
Solve the equation
h² + 109² = 194²
h² = 194² - 109²
h² = 37,636 - 11,881
h² = 25,755
h = √25,755
h = 160.4836
To the nearest tenth
h = 160.5
The building is 160.5 ft high
You might be interested in