1. Applying Pythagorean Theorem

2. Baseball Problem A baseball “diamond” is really a square.
You can use the
Pythagorean
theorem to find
how far does a
catcher have to
throw the ball from
home plate to second
base?

3. The legs are from home to first and from first to second.
Baseball Problem
The distance between
consecutive bases is 90
feet. How far does a
catcher have to throw
the ball from home
plate to second base?
Use the Pythagorean theorem to solve for x. a2 + b2 = c2
Which side is the hypotenuse?
Which sides are the legs?
The hypotenuse is the distance from home to second, or side x in the picture.
The legs are from home to first and from first to second.
c2 = a2 + b2
x2 =
x2 =
x2 = 16,200
Square Root Each Side
x =
The distance from home plate to 2nd base is ft

4. Ladder Problem
A ladder leans against a second-story window of a house. If the ladder is 25 meters long, and the base of the ladder is 7 meters from the house, how high is the window?
Distance from house: 7 meters
First draw a diagram that shows the sides of the right triangle.
Label the sides: Ladder is 25 m
Distance from house is 7 m
Use a2 + b2 = c2 to solve for the missing side.
72 + b2 = 252
49 + b2 = 625
b2 = 576
b = 24
The height of the window is 24 meters.

5. Application of Pythagoras’ Theorem
1.A car travels 16 km from east to west. Then it turns left and travels a further 12 km. Find the displacement between the starting point and the destination point of the car. N
The hypotenuse is the distance unknown or x in the picture.
The legs are the directions 16 km & 12 km
c2 = a2 + b2
x2 =
x2 =
x2 = 400
Square Root Each Side
x = 20
16km
12km x
The car displacement is 20 km SW.

6. The height of a tree is 5 m. The distance between
the top of it and the tip of its shadow is 13 m.
5 m
13 m
Find the length of the shadow.
Solution: (Pythagoras’ Theorem)
c2 = a2 + b2
132 = 52 + x2
x2 =
x2 =
x2 = 144
x = 12 x
13 m
5 m
The length of the shadow is 12 m. x

7. Classwork
Page 4ab,5ab,6,7,11,13,16,18