Area of a Parallelogram and Trapezium

Worked example

1. 1

   $b=9,h=5,\text{sloping side }6$ — The question gives three lengths. Only two go in the formula: the base and the height at a right angle to it.

2. 2

   $A=9\times 5$ — Put the base and the height into the formula. The sloping 6 is left out, because it is not at a right angle to the base.

3. 3

   $A=45{\text{ cm}}^2$ — 9 × 5 = 45. An area is in square units, so the answer is in cm².

Worked example

1. 1

   $a=7,b=12,h=5$ — Find the two parallel sides first. The height is the gap between them, measured at a right angle.

2. 2

   $\frac{1}{2}(7+12)\times 5$ — Put the numbers into the formula. Writing this line earns a mark, even if you slip later.

3. 3

   $=\frac{1}{2}\times 19\times 5$ — Work out the bracket first: 7 + 12 = 19.

4. 4

   $=9.5\times 5$ — Half of 19 is 9.5. It is the average of the two parallel sides.

5. 5

   $=47.5{\text{ cm}}^2$ — 9.5 × 5 = 47.5. An area is in square units, so cm².

Worked example

1. 1

   $\frac{1}{2}(5+y)\times 8=76$ — Put the known numbers into the formula. The unknown side $y$ goes in its place, and the whole thing equals the area.

2. 2

   $4(5+y)=76$ — Half of 8 is 4, so the half and the 8 join up as one number.

3. 3

   $5+y=19$ — Divide both sides by 4: 76 ÷ 4 = 19. This is the two parallel sides added.

4. 4

   $y=14$ — Take 5 from both sides: 19 − 5 = 14. So y = 14 cm.

Worked example

1. 1

   $9\times h=63$ — Base × height = area, with the height as the unknown.

2. 2

   $h=63÷9=7$ — Divide both sides by the base, 9. So the height is 7 cm.

Worked example

1. 1

   $a=3,b=8$ — The two upright sides both stand at right angles to the bottom, so they never meet. They are the parallel pair.

2. 2

   $h=12$ — The height is the distance between the parallel sides. The bottom side meets both at a right angle, so it is the height.

3. 3

   $\frac{1}{2}(3+8)\times 12$ — Put them in the formula. The sloping 13 cm is not used: it is not at a right angle to the parallel sides.

4. 4

   $=\frac{1}{2}\times 11\times 12$ — Work out the bracket: 3 + 8 = 11.

5. 5

   $=66{\text{ cm}}^2$ — Half of 11 is 5.5, and 5.5 × 12 = 66.

Worked example

1. 1

   $8^2+d^2={10}^2$ — Draw a line across at 3 cm high. Above it is a right-angled triangle: 8 cm across, d cm up, and the 10 cm slope as its longest side.

2. 2

   $d^2=100-64=36$ — Square the lengths: 10² = 100 and 8² = 64. Then take 64 from both sides.

3. 3

   $d=6$ — Square root: √36 = 6. Ignore −6, because a length cannot be negative.

4. 4

   $3+6=9$ — The right upright side is the 3 cm under the line plus the 6 cm above it: 9 cm.

5. 5

   $\frac{1}{2}(3+9)\times 8$ — Now use the trapezium formula. The parallel sides are 3 and 9, and the 8 cm base is the height.

6. 6

   $=6\times 8=48{\text{ cm}}^2$ — 3 + 9 = 12, half of 12 is 6, and 6 × 8 = 48.

Worked example

1. 1

   $\frac{1}{2}(4+7)\times 3$ — Find the area of the wall first, with the trapezium formula.

2. 2

   $=\frac{1}{2}\times 11\times 3$ — Work out the bracket: 4 + 7 = 11.

3. 3

   $=16.5{\text{ m}}^2$ — Half of 11 is 5.5, and 5.5 × 3 = 16.5. This is one coat.

4. 4

   $2\times 16.5=33{\text{ m}}^2$ — Two coats means painting the wall twice, so double the area.

5. 5

   $33÷10=3.3$ — Each tin covers 10 m², so divide by 10 to find how many tins the paint fills.

6. 6

   $4\text{ tins}$ — Round up, not down: 3 tins cover only 30 m², which leaves 3 m² bare.

Worked example

1. 1

   $\frac{1}{2}(120+280)\times 150$ — Area first, with the trapezium formula.

2. 2

   $=\frac{1}{2}\times 400\times 150$ — Work out the bracket: 120 + 280 = 400.

3. 3

   $=30000{\text{ m}}^2$ — Half of 400 is 200, and 200 × 150 = 30 000.

4. 4

   $30000÷10000=3$ — Change to hectares: every 10 000 m² is one hectare, so the field is 3 hectares.

5. 5

   $27600÷3=9200$ — Per hectare means the price of one hectare, so share the price between the 3 hectares: 9 200 dollars.

Worked example

1. 1

   $\frac{1}{2}(x+x+4)\times x=48$ — Put the letters into the trapezium formula and set it equal to the area, 48.

2. 2

   $\frac{1}{2}(2x+4)\times x=48$ — Collect like terms in the bracket: x + x = 2x.

3. 3

   $(x+2)\times x=48$ — Halve the bracket: half of 2x is x, and half of 4 is 2.

4. 4

   $x^2+2x=48$ — Multiply out: x × x = x², and 2 × x = 2x.

5. 5

   $x^2+2x-48=0$ — Take 48 from both sides, so the quadratic equals 0 and can be factorised.

6. 6

   $(x+8)(x-6)=0$ — Factorise: +8 and −6 multiply to −48 and add to +2.

7. 7

   $x=-8\text{or}x=6$ — A product is 0 only when one bracket is 0, so there are two answers.

8. 8

   $x=6$ — Reject −8, because a length cannot be negative. So x = 6.