Area from Coordinates

Worked example

1. 1

   $A(-2,1),B(4,1),C(1,6)$ — The question. A and B both have y = 1, so AB lies flat along a grid line. Use it as the base.

2. 2

   $\text{base}=4-(-2)$ — For a flat side, take the smaller x from the bigger x.

3. 3

   $=4+2=6$ — Taking away a negative is adding: −(−2) = +2. So the base is 6.

4. 4

   $\text{height}=6-1=5$ — The height goes straight up from the base (y = 1) to C (y = 6). So take the y's: height 5.

5. 5

   $\text{area}=\frac{1}{2}\times 6\times 5$ — Area of a triangle is half base times height.

6. 6

   $=\frac{1}{2}\times 30=15$ — 6 × 5 = 30, and half of 30 is 15. On a centimetre grid the answer is 15 cm².

Worked example

1. 1

   $P(1,1),Q(6,3),R(3,6)$ — The question. No two corners share an x or a y, so no side lies along a grid line.

2. 2

   $\text{width}=6-1=5$ — The box runs from the smallest x (1) to the biggest x (6).

3. 3

   $\text{height}=6-1=5$ — It runs from the smallest y (1) to the biggest y (6).

4. 4

   $\text{box}=5\times 5=25$ — The area of the whole box. The triangle is inside it.

5. 5

   $\frac{1}{2}\times 5\times 2=5$ — Corner under PQ: across 6 − 1 = 5, up 3 − 1 = 2. It has a right angle, so half of across times up.

6. 6

   $\frac{1}{2}\times 3\times 3=4.5$ — Corner by QR: across 6 − 3 = 3, up 6 − 3 = 3.

7. 7

   $\frac{1}{2}\times 2\times 5=5$ — Corner by RP: across 3 − 1 = 2, up 6 − 1 = 5.

8. 8

   $5+4.5+5=14.5$ — Add the three corners. They are the parts of the box that are not the triangle.

9. 9

   $25-14.5=10.5$ — Take the corners away from the box. What is left is the triangle: 10.5.

Worked example

1. 1

   $A(-3,-1),B(4,-1),C(2,3),D(-1,3)$ — The question. AB is flat (y = −1) and DC is flat (y = 3), so the shape is a trapezium.

2. 2

   $\text{rectangle}:2-(-1)=3\text{wide}$ — Drop straight lines down from D and C. The middle piece is a rectangle as wide as DC: 2 + 1 = 3.

3. 3

   $3-(-1)=4\text{tall}$ — Its height runs from y = −1 up to y = 3. Taking away −1 adds 1, so 4.

4. 4

   $3\times 4=12$ — Area of the middle rectangle.

5. 5

   $\frac{1}{2}\times 2\times 4=4$ — Left triangle: across from A to below D is −1 − (−3) = 2, and it is 4 tall.

6. 6

   $\frac{1}{2}\times 2\times 4=4$ — Right triangle: across from below C to B is 4 − 2 = 2, and it is 4 tall.

7. 7

   $12+4+4=20$ — Add the three pieces. The area is 20 squares, so 20 cm² on a centimetre grid.

Worked example

1. 1

   $A(1,1),B(5,4),C(-1,12)$ — Three corners of rectangle ABCD. You need two sides: AB and BC.

2. 2

   $AB:\text{across}5-1=4,\text{up}4-1=3$ — Subtract the x's, then the y's. These are the two short sides of a right-angled triangle under AB.

3. 3

   $A{B}^2=4^2+3^2=16+9=25$ — Pythagoras: the sloping side squared equals across squared plus up squared.

4. 4

   $AB=\sqrt{25}=5$ — Square root to undo the square: AB = 5.

5. 5

   $BC:\text{across}-1-5=-6$ — C is to the left of B, so the difference is negative. The minus only shows the direction.

6. 6

   $(-6{)}^2=36$ — Squaring removes the minus, because minus times minus is plus. Keep the brackets so the minus is squared too.

7. 7

   $\text{up}12-4=8,8^2=64$ — Subtract the y's, then square.

8. 8

   $BC=\sqrt{36+64}=\sqrt{100}=10$ — Pythagoras again: BC = 10. A length can never be negative.

9. 9

   $2(5+10)=30$ — A rectangle has two sides of each length, so add one of each and double it.