Problem: A ladder 5 m long is leaning against a wall so that its foot is 3 m from the wall. To what height h does the ladder reach on the wall?
The ladder forms the hypotenuse of a right triangle; the distance from the foot to the wall and the height on the wall form the legs.
h2=52−32=25−9=16
h=16=4 m
Answer: The ladder reaches a height of 4 m.
The diagonal of a rectangle
Problem: A rectangle has dimensions a=6 cm and b=8 cm. Compute the length of its diagonal u.
The diagonal of a rectangle divides the rectangle into two right triangles. The sides of the rectangle are the legs, and the diagonal is the hypotenuse.
u2=a2+b2=62+82=36+64=100
u=100=10 cm
Answer: The diagonal of the rectangle has length 10 cm.
The diagonal of a square
Problem: A square has side a=5 cm. What is the length of its diagonal?
In a square, both legs of the right triangle are equal:
u2=a2+a2=2a2=2⋅25=50
u=50≈7.07 cm
Answer: The diagonal of the square has length approximately 7.07 cm.
💡 Formula for the diagonal of a square:u=a2
Distance between two points
Problem: City A is 30 km north and 40 km east of city B. What is the straight-line distance between the cities?
Imagine a right triangle: one leg is 30 km (north), the other is 40 km (east), and the hypotenuse is the straight-line distance.
Problem: An isosceles triangle has base z=12 cm and legs r=10 cm. Compute its height v.
The height of an isosceles triangle bisects the base. This creates a right triangle where the leg of the isosceles triangle is the hypotenuse, half the base is one leg, and the height is the other leg.
v2=r2−(2z)2=102−62=100−36=64
v=64=8 cm
Answer: The height of the triangle is 8 cm.
The shortest route
Problem: A park has the shape of a rectangle with dimensions 80 m and 60 m. A path runs around the perimeter of the park, but you can also walk diagonally across the park. By how many meters is the diagonal path shorter than the path along two adjacent sides?
Step 1: The diagonal of the park.
