Area of Rectangles, Squares & Triangles

A Counting tiles — the birth of area

Imagine covering a floor with square tiles of the same size. The area of a shape is simply how many tiles fit inside it. Let's start by counting — and then discover a shortcut.

A1 Count the tiles in a 3 × 4 rectangle Starter ▼

Each small square below is one tile. How many tiles fill the rectangle?

Hint: Count every tile one by one. Then look at how many rows and columns there are.

Show solution

Counting one by one: 1, 2, 3 … 12 tiles.

There are 4 rows and 3 columns.

$4 \times 3 = 12$

12 tiles → Area = 12 square units

A2 Count the tiles in a 5 × 4 rectangle Starter ▼

Count every tile. Then verify your count by multiplying rows × columns.

Hint: Count row by row — each row has the same number of tiles.

Show solution

5 columns, 4 rows.

Each row has 5 tiles. 4 rows × 5 = 20.

20 tiles → Area = 20 square units

A3 A big floor: 12 × 10 tiles — count or multiply? Easy ▼

A room floor is covered with 12 columns and 10 rows of square tiles. How many tiles in total? Try counting — then try multiplying. Which is faster?

Hint: Counting all 120 tiles is tedious! Notice each row has exactly 12 tiles, and there are 10 rows. So you could just compute $12 \times 10$.

Show solution

12 columns × 10 rows

$12 \times 10 = 120$

120 tiles → Area = 120 square units

Key insight: Counting is slow. Multiplying is instant. This is why Area of a rectangle = length × width.

A4 A square floor: 6 × 6 tiles Starter ▼

How many tiles? What is special about a square compared to a rectangle?

Hint: A square has equal length and width. How does that change the multiplication?

Show solution

6 columns × 6 rows

$6 \times 6 = 36$

36 tiles → Area = 36 square units

For a square with side $s$: Area $= s \times s = s^2$. That's why we say "6 squared"!

A5 Odd one out: 7 × 3 rectangle Easy ▼

A rectangular garden is 7 tiles wide and 3 tiles tall. Without counting tile by tile, can you find the total area?

Hint: Use multiplication — you have everything you need.

Show solution

$7 \times 3 = 21$

Area = 21 square units

B Area of rectangles and squares

Formula: Area of a rectangle $= \text{length} \times \text{width}$ (sometimes written $\ell \times w$)
Formula: Area of a square $= \text{side}^2 = s^2$

What does "square unit" mean?

If tiles are 1 cm × 1 cm, the area is measured in square centimetres (cm²). If tiles are 1 m × 1 m, the area is in m². The unit always gets squared.

B1 Rectangle: length = 9 cm, width = 4 cm Starter ▼

Find the area of a rectangle with length 9 cm and width 4 cm.

Show solution

Area $= \ell \times w = 9 \times 4$

Area = 36 cm²

B2 Square: side = 7 m Starter ▼

Find the area of a square with side length 7 m.

Show solution

Area $= s^2 = 7^2 = 7 \times 7$

Area = 49 m²

B3 Rectangle: length = 13 cm, width = 5 cm Easy ▼

Find the area of a rectangle 13 cm long and 5 cm wide.

Show solution

Area $= 13 \times 5 = 65$

Area = 65 cm²

B4 Rectangle: length = $\frac{3}{2}$ m, width = $\frac{4}{3}$ m Medium ▼

Find the area of a rectangle with length $\dfrac{3}{2}$ m and width $\dfrac{4}{3}$ m.

Hint: Multiply the fractions: $\dfrac{3}{2} \times \dfrac{4}{3}$. Simplify before multiplying.

Show solution

Area $= \dfrac{3}{2} \times \dfrac{4}{3} = \dfrac{3 \times 4}{2 \times 3} = \dfrac{12}{6} = 2$

Area = 2 m²

B5 Find the missing side: area = 48 cm², length = 8 cm Medium ▼

A rectangle has area 48 cm² and length 8 cm. What is the width?

Hint: Area $= \ell \times w$, so $w = \text{Area} \div \ell$.

Show solution

$48 = 8 \times w$

$w = 48 \div 8 = 6$

Width = 6 cm

B6 Find the side: square area = 144 m² Medium ▼

A square has area 144 m². What is the length of one side?

Hint: Area $= s^2$, so $s = \sqrt{\text{Area}}$.

Show solution

$s^2 = 144$

$s = \sqrt{144} = 12$

Side = 12 m

C A right triangle is half a rectangle

Here is the key idea: take any rectangle and draw a diagonal line from one corner to the opposite corner. You get two identical right triangles. So the area of each triangle is exactly half the area of the rectangle.

Area of a right triangle

$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ where base and height are the two sides that meet at the right angle.

C1 Verify: right triangle with base = 6, height = 4 Starter ▼

A rectangle is 6 cm wide and 4 cm tall. If you draw a diagonal, what is the area of the right triangle you get?

The rectangle has area $6 \times 4 = 24$ cm².
The triangle is exactly half of that.

Show solution

Rectangle area $= 6 \times 4 = 24$ cm²

Triangle area $= \dfrac{1}{2} \times 24 = 12$ cm²

Or directly: $\dfrac{1}{2} \times 6 \times 4 = 12$ cm²

Area = 12 cm²

C2 Right triangle: base = 10, height = 6 Easy ▼

Find the area of a right triangle with base 10 cm and height 6 cm.

Hint: What rectangle would this triangle be half of?

Show solution

Rectangle area $= 10 \times 6 = 60$ cm²

Triangle area $= \dfrac{60}{2} = 30$ cm²

Area = 30 cm²

C3 Right isosceles triangle: base = height = 8 Easy ▼

A right triangle has base = 8 cm and height = 8 cm (both legs equal). Find the area.

Show solution

Area $= \dfrac{1}{2} \times 8 \times 8 = \dfrac{64}{2} = 32$ cm²

Area = 32 cm²

D Practice: right triangle area

Use the formula $\text{Area} = \dfrac{1}{2} \times b \times h$ for all of these. Some questions ask you to find a missing side given the area.

D1 base = 5, height = 8 Starter ▼

Find the area of a right triangle with base 5 cm and height 8 cm.

Show solution

Area $= \dfrac{1}{2} \times 5 \times 8 = \dfrac{40}{2} = 20$

Area = 20 cm²

D2 base = 12, height = 9 Easy ▼

Find the area of a right triangle with base 12 cm and height 9 cm.

Show solution

Area $= \dfrac{1}{2} \times 12 \times 9 = \dfrac{108}{2} = 54$

Area = 54 cm²

D3 base = 7, height = 4 Easy ▼

Find the area of a right triangle with base 7 cm and height 4 cm.

Show solution

Area $= \dfrac{1}{2} \times 7 \times 4 = \dfrac{28}{2} = 14$

Area = 14 cm²

D4 base = 11, height = 6 Easy ▼

Find the area of a right triangle with base 11 m and height 6 m.

Show solution

Area $= \dfrac{1}{2} \times 11 \times 6 = \dfrac{66}{2} = 33$

Area = 33 m²

D5 Fraction: base = $\frac{5}{2}$, height = 4 Medium ▼

Find the area of a right triangle with base $\dfrac{5}{2}$ cm and height 4 cm.

Hint: $\dfrac{1}{2} \times \dfrac{5}{2} \times 4$ — multiply the fractions step by step.

Show solution

Area $= \dfrac{1}{2} \times \dfrac{5}{2} \times 4 = \dfrac{1}{2} \times 10 = 5$

Area = 5 cm²

D6 Find the missing base: area = 30, height = 12 Medium ▼

A right triangle has area 30 cm² and height 12 cm. Find the base.

Hint: Start from $30 = \dfrac{1}{2} \times b \times 12$ and solve for $b$.

Show solution

$30 = \dfrac{1}{2} \times b \times 12 = 6b$

$b = 30 \div 6 = 5$

Base = 5 cm

D7 Find the missing height: area = 42, base = 7 Medium ▼

A right triangle has area 42 m² and base 7 m. Find the height.

Show solution

$42 = \dfrac{1}{2} \times 7 \times h = \dfrac{7h}{2}$

$h = \dfrac{42 \times 2}{7} = \dfrac{84}{7} = 12$

Height = 12 m

D8 Two equal right triangles form a rectangle — find the rectangle's dimensions Hard ▼

Two identical right triangles are placed together to form a rectangle. Each triangle has area 40 cm². The height of each triangle is 8 cm. What are the length and width of the rectangle?

Hint: First find the base of one triangle, then think about how two triangles form the rectangle.

Show solution

Triangle area $= 40$ cm², height $= 8$ cm

$40 = \dfrac{1}{2} \times b \times 8 = 4b \;\Rightarrow\; b = 10$ cm

Rectangle = base × height $= 10 \times 8$

Rectangle is 10 cm × 8 cm (area = 80 cm²)

E Deriving the general triangle area formula

So far we used the right-angle idea. Now let's show that $\text{Area} = \tfrac{1}{2} \times \text{base} \times \text{height}$ works for any triangle — not just right triangles.

E1 Why does ½ × base × height work for any triangle? Medium ▼

Take any triangle with base $b$ and drop a perpendicular from the opposite vertex to the base. Call its length $h$ (the height). Why is the area always $\dfrac{1}{2} b h$?

The altitude $h$ splits the triangle into two right triangles. Each is half of a small rectangle. The left right-triangle has area $\tfrac{1}{2} b_1 h$ and the right has $\tfrac{1}{2} b_2 h$. Together: $\tfrac{1}{2}(b_1 + b_2)h = \tfrac{1}{2} b h$.

Show formal derivation

Drop altitude $h$ from top vertex to base, splitting base into lengths $b_1$ and $b_2$ with $b_1 + b_2 = b$.

Left right-triangle: Area $= \dfrac{1}{2} b_1 h$

Right right-triangle: Area $= \dfrac{1}{2} b_2 h$

Total: $\dfrac{1}{2} b_1 h + \dfrac{1}{2} b_2 h = \dfrac{1}{2}(b_1 + b_2)h = \dfrac{1}{2} b h$

$$\boxed{\text{Area} = \frac{1}{2} \times b \times h}$$

E2 Does the choice of base matter? Medium ▼

A triangle has three sides: 10 cm, 8 cm, and 6 cm. You measure the height to the 10 cm side and get $h_1 = 4.8$ cm. Verify the area, then confirm that choosing the 6 cm side as base gives the same area if $h_2 = 8$ cm.

Hint: Compute $\tfrac{1}{2} \times 10 \times 4.8$ and also $\tfrac{1}{2} \times 6 \times 8$.

Show solution

Using base 10: Area $= \dfrac{1}{2} \times 10 \times 4.8 = 24$ cm²

Using base 6: Area $= \dfrac{1}{2} \times 6 \times 8 = 24$ cm²

Both give 24 cm² ✓ — the formula works regardless of which side you call the base, as long as you use the perpendicular height to that base.

F Area of acute, obtuse, and equilateral triangles

The formula $\tfrac{1}{2} b h$ works for all triangles. The only challenge is finding $h$ — the perpendicular height, not the slant side.

F1 Acute triangle: base = 12 cm, height = 8 cm Easy ▼

An acute triangle (all angles less than 90°) has a base of 12 cm. The perpendicular height from the opposite vertex to the base is 8 cm. Find the area.

Show solution

Area $= \dfrac{1}{2} \times 12 \times 8 = 48$

Area = 48 cm²

F2 Obtuse triangle: base = 10 cm, height = 6 cm Medium ▼

An obtuse triangle (one angle greater than 90°) has base 10 cm. The height is measured outside the base line and equals 6 cm. Find the area.

Hint: Even though the height falls outside the triangle, the formula $\tfrac{1}{2} b h$ still works. Use base = 10 and height = 6.

Show solution

Area $= \dfrac{1}{2} \times 10 \times 6 = 30$

Area = 30 cm²

F3 Equilateral triangle with side = 10 cm Hard ▼

An equilateral triangle has all three sides equal to 10 cm. First find the height using the Pythagorean theorem, then find the area.

Hint: The altitude bisects the base into two halves of 5 cm each. In the right triangle formed: $h^2 + 5^2 = 10^2$.

Show solution

Using Pythagoras: $h^2 = 10^2 - 5^2 = 100 - 25 = 75$

$h = \sqrt{75} = 5\sqrt{3}$ cm

Area $= \dfrac{1}{2} \times 10 \times 5\sqrt{3} = 25\sqrt{3}$

Area $= 25\sqrt{3} \approx 43.3$ cm²

General formula for equilateral triangle with side $s$: Area $= \dfrac{\sqrt{3}}{4} s^2$

F4 Scalene triangle: base = 15, height = 7 Easy ▼

A scalene triangle (all sides different) has base 15 m and perpendicular height 7 m. Find the area.

Show solution

Area $= \dfrac{1}{2} \times 15 \times 7 = \dfrac{105}{2} = 52.5$

Area = 52.5 m²

F5 Isosceles triangle: base = 8, equal sides = 5 Hard ▼

An isosceles triangle has a base of 8 cm and two equal sides of 5 cm. Find the height, then the area.

Hint: The altitude from the apex bisects the base into two segments of 4 cm each. Use Pythagoras: $h^2 + 4^2 = 5^2$.

Show solution

$h^2 = 5^2 - 4^2 = 25 - 16 = 9$

$h = 3$ cm

Area $= \dfrac{1}{2} \times 8 \times 3 = 12$

Area = 12 cm²

G Combined shapes — rectangles and triangles together

Real-world shapes are often made by combining simpler shapes. Split the shape into rectangles and triangles, find each area, then add them up.

G1 House shape: rectangle + triangle roof Easy ▼

A house shape consists of a rectangle (base 10 m, height 6 m) topped by an isosceles triangular roof (base 10 m, height 4 m). Find the total area.

Show solution

Rectangle area $= 10 \times 6 = 60$ m²

Triangle area $= \dfrac{1}{2} \times 10 \times 4 = 20$ m²

Total $= 60 + 20 = 80$ m²

Total area = 80 m²

G2 Right trapezoid: rectangle + right triangle Medium ▼

A right trapezoid has a rectangular part (8 m × 5 m) with a right triangle attached to the right side (base 6 m, height 5 m). Find the total area.

Show solution

Rectangle area $= 8 \times 5 = 40$ m²

Triangle area $= \dfrac{1}{2} \times 6 \times 5 = 15$ m²

Total $= 40 + 15 = 55$ m²

Total area = 55 m²

G3 L-shape: two rectangles Medium ▼

An L-shaped room consists of two rectangles: a large one (10 m × 8 m) and a smaller one (4 m × 3 m) cut from one corner. Find the area of the L-shape.

Hint: Compute the area of the full large rectangle, then subtract the cut corner. (Or split the L into two smaller rectangles and add.)

Show solution

Full rectangle area $= 10 \times 8 = 80$ m²

Cut corner area $= 4 \times 3 = 12$ m² (but we'll do this properly in Section H!)

Alternatively: split L into two pieces:
Top rectangle (10 m × 3 m cut from 8 m tall → bottom piece 6 m × 10 m = 60 m²? Let's use direct subtraction.)
L area $= 80 - 12 = 68$ m²

Area = 68 m²

G4 Pentagon: rectangle + triangle on top Medium ▼

A pentagon is formed by a rectangle (12 cm × 7 cm) with an isosceles triangle (base 12 cm, height 5 cm) sitting on top. Find the total area.

Show solution

Rectangle: $12 \times 7 = 84$ cm²

Triangle: $\dfrac{1}{2} \times 12 \times 5 = 30$ cm²

Total: $84 + 30 = 114$ cm²

Area = 114 cm²

G5 Arrow / chevron: rectangle + two triangles Hard ▼

An arrow shape is formed from a rectangle (14 cm × 4 cm) with two right triangles attached on the left side, each with base 4 cm and height 2 cm. The two triangles form a point. Find the total area.

Hint: The two triangles together form one rectangle (4 cm × 2 cm). Or add them individually: each triangle is $\tfrac{1}{2} \times 4 \times 2$.

Show solution

Rectangle: $14 \times 4 = 56$ cm²

Two triangles: $2 \times \dfrac{1}{2} \times 4 \times 2 = 8$ cm²

Total: $56 + 8 = 64$ cm²

Area = 64 cm²

H Finding area by subtraction

Sometimes it is easier to compute the area of a large, simple shape and then subtract the parts you don't want. Area of shape = Area of outer shape − Area of removed part(s).

H1 Picture frame: big rectangle minus small rectangle Easy ▼

A rectangular picture frame has outer dimensions 20 cm × 15 cm. The inner opening is 14 cm × 9 cm. Find the area of the frame itself.

Show solution

Outer area $= 20 \times 15 = 300$ cm²

Inner area $= 14 \times 9 = 126$ cm²

Frame area $= 300 - 126 = 174$ cm²

Frame area = 174 cm²

H2 Square minus a corner triangle Easy ▼

A square has side 10 cm. A right triangle with legs 4 cm and 6 cm is cut from one corner. Find the area of the remaining shape.

Show solution

Square area $= 10 \times 10 = 100$ cm²

Triangle area $= \dfrac{1}{2} \times 4 \times 6 = 12$ cm²

Remaining area $= 100 - 12 = 88$ cm²

Remaining area = 88 cm²

H3 Rectangle minus two triangles = parallelogram Medium ▼

A rectangle is 16 cm × 8 cm. Two identical right triangles (each with base 3 cm and height 8 cm) are cut from the left and right ends to form a parallelogram. Find the parallelogram's area.

Show solution

Rectangle area $= 16 \times 8 = 128$ cm²

Each triangle area $= \dfrac{1}{2} \times 3 \times 8 = 12$ cm²

Two triangles $= 2 \times 12 = 24$ cm²

Parallelogram area $= 128 - 24 = 104$ cm²

Parallelogram area = 104 cm²

H4 Square minus 4 corner triangles — rotated inner square Hard ▼

A square has side 10 cm. From each corner, a right isosceles triangle with legs 3 cm is cut. The remaining shape is a regular octagon (almost). Find its area.

Hint: Each corner triangle has area $\dfrac{1}{2} \times 3 \times 3$. Subtract all four.

Show solution

Square area $= 10^2 = 100$ cm²

Each triangle area $= \dfrac{1}{2} \times 3 \times 3 = \dfrac{9}{2} = 4.5$ cm²

Four triangles $= 4 \times 4.5 = 18$ cm²

Octagon area $= 100 - 18 = 82$ cm²

Octagon area = 82 cm²

H5 Challenge: large triangle minus small triangle inside Challenge ▼

A large right triangle has base 20 cm and height 15 cm. Inside it, a smaller similar triangle (half the linear dimensions) is removed from the corner. Find the area of the remaining region.

Hint: If linear dimensions are halved, the area scales by $\left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}$.

Show solution

Large triangle area $= \dfrac{1}{2} \times 20 \times 15 = 150$ cm²

Small triangle dimensions: base $= 10$ cm, height $= 7.5$ cm

Small triangle area $= \dfrac{1}{2} \times 10 \times 7.5 = 37.5$ cm²

Also: $\dfrac{1}{4} \times 150 = 37.5$ cm² ✓ (similarity check)

Remaining area $= 150 - 37.5 = 112.5$ cm²

Remaining area = 112.5 cm²

Area of Shapes