Distance on the Coordinate Plane — Challenge
Draw the right triangle and the distance formula stops being something to memorise.
What these problems cover
These questions use 11 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.
- Perimeter — Distance all the way around the outside.
- Base — The number being multiplied. In 53 the base is 5.
- Work backwards — Start from the result and undo each step.
- Square root — For k ≥ 0, √k is the nonnegative number whose square is k.
- Leg — One of the two sides forming the right angle.
- Hypotenuse — The side opposite the right angle — always the longest.
- Midpoint — The point halfway between two others: average the x values and average the y values.
- Simplest radical form — No perfect square factors left under the root.
- Segment — Two endpoints and everything between them. Written AB.
- Diagonal — A segment joining two non-adjacent vertices.
- Height — The perpendicular distance from the base to the opposite corner.
Problem 1
What is the distance between (0,0) and (6,8)?
- 10
- 14
- 48
- 100
- 7
Worked answer
10
The horizontal gap is 6 and the vertical gap is 8, meeting at a right angle.
By Pythagoras the distance is √62+82 = √36+64 = √100 = 10.
Adding the gaps gives 14, the distance travelled along the axes rather than straight across.
Problem 2
In a right triangle, AB=6, AC=8, and the right angle is at A. A square has one corner at A, two sides along AB and AC, and its opposite corner on BC. What is its side length?
- 3
- 4
- 48/7
- 7/2
- 24/7
Worked answer
24/7
Place A=(0,0), B=(0,6) and C=(8,0). The hypotenuse is y=6-3x/4.
The opposite square corner is (s,s), so s=6-3s/4.
Thus 7s/4=6 and s=24/7.
Problem 3
A right triangle has hypotenuse 5 and legs whose lengths differ by 1. What is its shorter leg?
- 3
- 2
- 4
- 5/2
- √5
Worked answer
3
Let the legs be x and x+1. Then x2+(x+1)2=25.
This gives x2+x-12=(x+4)(x-3)=0.
A length is positive, so x=3.
Problem 4
From a rectangle of width 30 and height 10, two isosceles right triangles are cut from the upper corners. Each has legs of length 5 along the rectangle edges. What fraction of the rectangle is removed?
- 1/12
- 1/6
- 1/3
- 5/12
- 1/24
Worked answer
1/12
The removed area is 2·12·5·5=25.
The rectangle area is 30·10=300, so the fraction is 25/300=1/12.
Problem 5
What is the distance between (1,2) and (4,6)?
- 3
- 4
- 5
- 7
- 25
Worked answer
5
The horizontal gap is 4-1 = 3 and the vertical gap is 6-2 = 4.
The distance is √32+42 = √25 = 5.
Forgetting the square root leaves 25, which is the squared distance.
Problem 6
What is the distance between (-3,0) and (3,0)?
- 0
- 3
- 6
- 9
- 18
Worked answer
6
Both points lie on the x-axis, so the segment joining them is horizontal.
The distance is the gap in the x values: 3-(-3) = 6.
Subtracting a negative adds, which is why the answer is 6 rather than 0.
Problem 7
What is the distance between (2,3) and (2,8)?
- 2
- 3
- 5
- 8
- 11
Worked answer
5
The two points share the same x value, so the segment is vertical.
The distance is the gap in the y values: 8-3 = 5.
No square root is needed when the points differ in one direction only.
Problem 8
A triangle of area 4 is similar to a larger triangle. The ratio of a corresponding side of the larger triangle to the smaller is 5. What is the larger area?
- 20
- 25
- 100
- 80
- 125
Worked answer
100
Areas scale by the square of the side ratio.
The larger area is 4·52=100.