Area and perimeter

(a) Count the squares: a 2 by 6 block (12) plus a 1 by 2 block (2), so $12 + 2 = 14$ cm².

(b) Go round the edge: $2 + 6 + 3 + 2 + 1 + 4 = 18$ cm.

(a) Area of a rectangle $=$ length $\times$ width $= 5 \times 4 = 20\text{ cm}^2$. Milo multiplied all four side lengths together, which uses each measurement twice.

(b) Area is measured in square units. An answer in cm is a length, so it should be in $\text{cm}^2$.

Missing sides: $10 - 4 = 6$ cm and $7 - 5 = 2$ cm.

Perimeter: $4 + 5 + 6 + 2 + 10 + 7 = 34$ cm

Split it into two rectangles.

Tall part: $10 \times 2 = 20$ cm²

Bottom right part: $(8 - 2) \times 2 = 12$ cm²

Total $= 20 + 12 = 32$ cm²

One side of a square is $40 \div 4 = 10$ cm.

The cross has 12 outside edges.

Perimeter $= 12 \times 10 = 120$ cm.

(a) $7 \times 3$ is the area of the rectangle. Perimeter means adding all four sides: $7 + 3 + 7 + 3 = 20$ cm.

(b) One side is $x$ cm. If $x = -2$, that side would be $-2$ cm long, and a length must be positive.

One rectangle stands upright (4 by 6), the other lies flat (6 by 4) beside it.

The step on top is $6 - 4 = 2$ cm.

Perimeter: $4 + 2 + 6 + 4 + 6 + 4 + 6 = 32$ cm

Or: $20 + 20 - 2 \times 4 = 32$ cm (the joined 4 cm edge is lost from both).

Triangle perimeter: $30 + 36 + 14 = 80$ cm

Rectangle perimeter: $80 \div 4 = 20$ cm

Two widths: $2 \times 4 = 8$ cm, so two lengths $= 12$ cm

Length $= 12 \div 2 = 6$ cm

(a) Area $= 10 \times 8 - 4 \times 3 = 80 - 12 = 68\text{ m}^2$

Paint: $3 \times 2.5 = 7.5$ litres, covering $7.5 \times 10 = 75\text{ m}^2$

$75 > 68$, so yes, she has enough.

(b) With 11 m² per litre it covers $82.5\text{ m}^2$, even more, so the answer does not change.

The shape is symmetrical, so the triangle's base $= 14 - 3.8 - 3.8 = 6.4$ cm.

Triangle area $= \frac{1}{2} \times 6.4 \times 6 = 19.2$ cm²

Rectangle area $= 3.5 \times 19.2 = 67.2$ cm²

$14w = 67.2$, so $w = 67.2 \div 14 = 4.8$

Area of a triangle is $\frac{1}{2} \times b \times h$, so $b \times h = 24$.

Choose base 6 cm and height 4 cm.

Put the top vertex directly above the middle of the base (3 cm along), so the two slanted sides are equal.

Along the length: $600 \div 30 = 20$ stones

Across the width: $120 \div 30 = 4$ stones

Number needed: $20 \times 4 = 80$

Cost: $80 \times £2.50 = £200$

$£200 < £220$, so he has enough money.

(a) Side $= \sqrt{81} = 9$ cm, so perimeter $= 4 \times 9 = 36$ cm.

(b) Triangle: $\frac{1}{2} \times 16 \times 9 = 72$ cm²

Parallelogram: $5 \times 72 = 360$ cm²

$30h = 360$, so $h = 12$

The parallel sides are 2 cm and 7 cm, and the height is 4 cm.

Area $= \frac{1}{2}(2 + 7) \times 4 = 18$ cm².

A triangle needs $\frac{1}{2} \times b \times h = 18$, so $b \times h = 36$, e.g. base 9 cm, height 4 cm.

Let each tile be $l$ long and $w$ wide.

Across the top: $2l + w = 11$. Down the side: $l + w = 7$.

Subtracting: $l = 4$, so $w = 3$.

One tile $= 4 \times 3 = 12$ cm², and there are 4 tiles: $4 \times 12 = 48$ cm².

Rectangle ABCO: $11 \times 7 = 77$

Rectangle DEFO: $7 \times 9 = 63$

Triangle CDO: $\frac12 \times 11 \times 9 = 49.5$

Quarter circle: $\frac14 \times \pi \times 7^2 = 38.48...$

Total $= 227.98...\text{ m}^2$, and $227.98 \div 14 = 16.28...$ so 17 bags.

$17 \times 10.95 = £186.15$

$QU = (3x + 5) - 2x = x + 5$ and $RV = 5$; the height is $4x$.

$A = \frac{1}{2}(QU + RV) \times 4x = \frac{1}{2}(x + 10) \times 4x$

$A = 2x(x + 10) = 2x^2 + 20x$

The big square has side $3 + 5 = 8$ cm, so its area is $64$ cm².

Each corner triangle has area $\frac{1}{2} \times 3 \times 5 = 7.5$ cm².

Shaded area $= 64 - 4 \times 7.5 = 34$ cm².

(Or: side$^2 = 3^2 + 5^2 = 34$.)

$CB^2 = 9^2 - 6^2 = 81 - 36 = 45$, so $CB = \sqrt{45}$

Quarter circle area $= \frac{1}{4} \pi r^2 = \frac{1}{4} \times \pi \times 45 = 35.34\ldots$

$= 35.3\text{ m}^2$ (3 s.f.)

Area of PQRS: $PQ \times 10 = 45$, so $PQ = 4.5$ cm, so $BC = 4.5$ cm.

Perimeter of ABCD: $2 \times AB + 2 \times 4.5 = 26$

$2 \times AB = 17$, so $AB = 8.5$ cm

Area of A $= \frac{1}{4} \times \pi \times 15^2 = 56.25\pi$

Area of B $= 56.25\pi \div 9 = 6.25\pi$

$\pi r^2 = 6.25\pi$, so $r^2 = 6.25$ and $r = \sqrt{6.25} = 2.5$ cm

Area $= \frac{1}{2}(10 + 16) \times 7 = 91$ m²

Paint $= 91 \div 2 = 45.5$ litres

Tins $= 45.5 \div 5 = 9.1$, so 10 tins

Cost $= 10 \times 16.99 = £169.90 > £160$, so No.

Let each tile be $l$ cm long and $w$ cm wide.

Across the top: $2l + w = 11$. Down the left side: $l + w = 7$.

Subtracting gives $l = 4$, so $w = 3$.

There are 4 tiles, each $4 \times 3 = 12$ cm², so the total area is $4 \times 12 = 48$ cm².

Rectangle: $19 \times 16 = 304$ cm²

Circle radius 8: half-circle area $= \frac{1}{2} \pi \times 8^2 = 100.53$ cm²

Shaded: $304 - 100.53 = 203.47$ cm²

$\frac{203.47}{304} \times 100 = 66.9\%$

Rectangle: $19 \times 16 = 304$ cm²

Radius $= 8$, semicircle $= \frac{1}{2} \pi \times 8^2 = 100.53$ cm²

Shaded $= 304 - 100.53 = 203.47$ cm²

$\frac{203.47}{304} \times 100 = 66.9\%$

The rectangle is 5 by 3, so its perimeter is 16.

(i) Cut a triangle off a corner: the slanted cut is shorter than the two edges it replaces, so the perimeter drops.

(ii) Cut a 1 by 1 square from a corner: two edges of 1 are removed and two new edges of 1 appear, so the perimeter stays 16.

(iii) Cut a 1 by 1 notch from the middle of a side: 1 unit of edge goes but 3 units appear, so the perimeter becomes 18.

(a) Wall area $= \frac{1.8 + 2.7}{2} \times 0.8 = 1.8\text{ m}^2 = 18000\text{ cm}^2$

Tile area $= 15 \times 7.5 = 112.5\text{ cm}^2$

Tiles: $18000 \div 112.5 = 160$. Packs: $160 \div 9 = 17.8$, round up to 18.

(b) $160 \times 1.1 = 176$ tiles, $176 \div 9 = 19.6$, so 20 packs.

That is 2 more packs, but 10% of 18 is only 1.8, so the assumption is not correct.

(a) Area of wall $= \frac{1.8 + 2.7}{2} \times 0.8 = 1.8\text{ m}^2 = 18\,000\text{ cm}^2$

Area of a tile $= 15 \times 7.5 = 112.5\text{ cm}^2$

Tiles $= 18\,000 \div 112.5 = 160$; packs $= 160 \div 9 = 17.8$, so 18 packs.

(b) $160 \times 1.1 = 176$ tiles; $176 \div 9 = 19.6$, so 20 packs.

10% more packs would be $18 \times 1.1 = 19.8$. She actually needs 2 more packs, which is about 11%, so her assumption is not correct.

Each octagon angle is 135°, so where two octagons meet at a point the gap is $360 - 2 \times 135 = 90°$.

Each point of the star is a right-angled triangle with legs $a$: area $\frac{1}{2}a^2$, hypotenuse $a\sqrt{2}$.

Each side of the central square is an octagon side, a triangle base and another octagon side: $a + a\sqrt{2} + a = (2 + \sqrt{2})a$.

Area $= (2+\sqrt{2})^2a^2 + 4 \times \frac{1}{2}a^2 = (6 + 4\sqrt{2})a^2 + 2a^2 = (8 + 4\sqrt{2})a^2 = 4(2+\sqrt{2})a^2$