Connected ideas

Rectangle, expansion and factorising

Split and combine areas with a common height.

Predict, test and explain

Step 1 of 3

Predict

What is 3(2x + 5) after expansion?

3(2x + 1)

Scroll the diagram sideways to see every label.

x=4

\text{total area}=27

Both regions have the same height. Multiply that height by each width to expand the product. Physical widths and heights stay positive.

Check the working

Quantity	Value
First region area	24
Second region area	3
Whole rectangle area	27
Product and sum	$3(2x + 1)=6x + 3$

Explore and compare

Can you break the rule?

Two rectangles with the same area have the same perimeter.

Always, sometimes or never? Test it in the simulation.

Hint

Use the height and total width. Compare a two-by-six rectangle with a three-by-four rectangle.

Check the reasoning

Sometimes. Equal area can give different perimeters. Rectangles with the same dimensions do have the same perimeter, so the claim is sometimes true.

$2\times6=3\times4=12,\quad 2(2+6)=16\ne14=2(3+4)$

One counterexample disproves “always”. Examples alone do not prove it.

Build two differently shaped rectangles with the same area. Save the first, then change the dimensions.

Predict

Try it in the simulation. Change one thing at a time.

Hint

Use total width: a times x plus b. Try height two and width six, then height three and width four.

Explain

Notes and saved values stay in this tab. They disappear when you reload.

Hints, self-check and connections

Try a self-check

Fixed values, separate from the diagram controls.

A rectangle has height 3 cm and width (2x + 5) cm, where x is positive. Expand its area expression, then find the area at x = 2.

$A=3(2x+5)\,\text{cm}^2$

Hint

Multiply the height by both parts of the width.

Reveal answer

The expanded area is (6x + 15) cm². At x = 2, the area is 27 cm².

$A=(6x+15)\,\text{cm}^2,\qquad A(2)=27\,\text{cm}^2$

Build an explanation

Hint 1

For a rectangle of height h and width ax + b, split the width into ax and b. Use positive h, a, b and x so both pieces have positive dimensions.

Hint 2

Add the two areas: h(ax + b) = hax + hb. This is expansion through a shared height.

Hint 3

Reverse the split by taking out the common height: hax + hb = h(ax + b). This is factorising.

Worked example

This example uses fixed values, separate from the diagram controls.

A rectangle is 4 cm high and (3x + 2) cm wide, where x > 0. Split it at width 3x, expand its area, factorise again, and find its area when x = 5.

The two widths are 3x cm and 2 cm, and both pieces have height 4 cm.
Their areas are 4 × 3x = 12x cm² and 4 × 2 = 8 cm².
Add them to expand the area: 4(3x + 2) = 12x + 8.
Both terms have common factor 4, so reverse the expansion: 12x + 8 = 4(3x + 2).
For x = 5, the width is 3 × 5 + 2 = 17 cm and the area is 4 × 17 = 68 cm².

Answer: 4(3x + 2) = 12x + 8; area at x = 5 is 68 cm².

Connect this idea

Watch out: The height multiplies both width parts: 4(3x + 2) = 12x + 8, not 12x + 2.