Mensuration (2D)

high importance~2 Q in Tier 126 formulas⚡ 15 shortcuts5 subtopics

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Areas of quadrilaterals

⏱ 4 min read🧩 5 question types🎯 13 practice Q

The idea in one minute

A rectangle, square, parallelogram, rhombus or trapezium each has a two-part area formula. Most questions give two facts and ask for a third, so the fast route is an identity, not a quadratic. Any quadrilateral splits into two triangles along a diagonal.

The family card

Figure	Area	Extras
Rectangle $l\times b$	$lb$	$P=2(l+b)$ , diagonal $\sqrt{l^2+b^2}$
Square, side $a$	$a^2$	diagonal $a\sqrt2$
Parallelogram	$bh$ or $ab\sin\theta$	adjacent angles add to $180^\circ$
Rhombus	$\dfrac{1}{2}d_1d_2$	diagonals cross at $90^\circ$
Trapezium	$\dfrac{1}{2}(a+b)h$	$a$ , $b$ parallel

Rule: Any quadrilateral: $K=\dfrac{1}{2}\times\text{diagonal}\times(\text{sum of the two perpendiculars on it})$ . Every row above is a special case.

Rectangle from two facts

Given $P=34$ and diagonal $13$ : $l+b=17$ and $l^2+b^2=169$ . Use the identity

(l+b)^2=l^2+b^2+2lb \Rightarrow 289=169+2lb \Rightarrow lb=60

Area $60$ without ever finding $l$ and $b$ . Check: $l+b=17$ with $lb=60$ gives sides $12$ and $5$ , and $5$ - $12$ - $13$ confirms the diagonal.

A square of side $a$ has diagonal $a\sqrt2$ . Given the diagonal $8\sqrt2$ , the side is $8$ and the area is $64$ .

Tip: Two facts about a rectangle almost always combine through $(l+b)^2$ or $(l-b)^2$ . Avoid solving a quadratic.

Rhombus through half-diagonals

The diagonals bisect each other at right angles, so the halves and one side form a right triangle.

Diagonals in ratio $5:12$ with area $120$ : $\dfrac{1}{2}(5k)(12k)=120$ gives $k=2$ , so diagonals $10$ and $24$ . Halves $5$ and $12$ give side $13$ and perimeter $52$ . All four sides are equal, so the perimeter is always four times one side.

Watch: Halves of rhombus diagonals are triplets in disguise: (5, 12, 13), (8, 6, 10), (12, 35, 37).

Parallelogram and trapezium

Parallelogram sides $10$ and $8$ with a $30^\circ$ angle between them: $K=10\times8\times\sin30^\circ=80\times\dfrac{1}{2}=40$ .

Trapezium: parallel sides differ by $8$ , height $6$ , area $96$ . Then $\dfrac{1}{2}(a+b)\times6=96$ gives $a+b=32$ ; with $a-b=8$ , the sides are $20$ and $12$ . The height on the base $10$ above is $8\sin30^\circ=4$ cm, and the midsegment between the slanted sides is $\dfrac{20+12}{2}=16$ .

Remember: The height of a parallelogram is perpendicular to the base, never the slant side.

Split along a diagonal

A general quadrilateral with diagonal $12$ and perpendiculars $5$ and $9$ dropping onto it from the other two corners:

K=\frac{1}{2}\times12\times(5+9)=84

The diagonal makes two triangles; add their two areas to finish.

Tip: A square with diagonal $8\sqrt2$ has side $8$ , area $64$ . Match the diagonal formula $a\sqrt2$ backwards to the side.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Rectangle from two of {area, perimeter, diagonal}

How to spot it:

Two rectangle facts given, a third one asked.

Method

Turn the perimeter into $l+b$ .
Write the second fact as $l^2+b^2$ or $lb$ .
Bridge with $(l+b)^2=l^2+b^2+2lb$ .
Read off the asked quantity.

Why it works:

Identities move between the three facts faster than solving for the sides.

Try this

The perimeter of a rectangle is 34 cm and its diagonal is 13 cm. Its area is:

Show solution

$l+b=17$ and $l^2+b^2=169$ .
$2lb=289-169=120$ .
$lb=60$ sq cm.

Answer

60 sq cm

Type 2very common2 practice Q

Rhombus: diagonals, area and perimeter

How to spot it:

Rhombus with diagonals in a ratio, or a mix of area and side asked.

Method

Set the diagonals as $d_1$ and $d_2$ from the ratio.
Use $K=\dfrac{1}{2}d_1d_2$ to fix them.
Halve both diagonals.
Side $=\sqrt{(\text{half}_1)^2+(\text{half}_2)^2}$ , perimeter is $4$ sides.

Why it works:

Ratio plus area pins both diagonals; the half-diagonal right triangle does the rest.

Try this

The diagonals of a rhombus are in the ratio 5 : 12 and its area is 120 sq cm. Its perimeter is:

Show solution

$\dfrac12\times5k\times12k=120 \Rightarrow k=2$ : diagonals 10, 24.
Halves 5, 12 give side 13.
$P=4\times13=52$ cm.

Answer

52 cm

Type 3common2 practice Q

Parallelogram area (base-height or ab sin theta)

How to spot it:

Parallelogram with two sides and the angle between them, or base and height.

Method

If a height is given, use $K=bh$ .
With two sides and the angle: $K=ab\sin\theta$ .
Recall $\sin30^\circ=\dfrac12$ , $\sin90^\circ=1$ .

Why it works:

Two entry points cover every parallelogram question; the angle version needs no height.

Try this

Two adjacent sides of a parallelogram are 10 cm and 8 cm and the angle between them is $30^\circ$ . Its area is

Show solution

$K=10\times8\times\sin30^\circ$ .
$=80\times\dfrac12=40$ sq cm.

Answer

40 sq cm

Type 4common2 practice Q

Square diagonal and trapezium area

How to spot it:

A square given by its diagonal, or a trapezium with a side difference.

Method

Square: side $=\dfrac{\text{diagonal}}{\sqrt2}$ , then $K=a^2$ .
Trapezium: write $a+b$ from $\dfrac{2K}{h}$ .
Combine with $a-b$ if a difference is given.

Why it works:

Both shapes resolve with one substitution instead of full side-by-side solving.

Try this

The parallel sides of a trapezium differ by 8 cm and its height is 6 cm. If its area is 96 sq cm, the longer parallel side is:

Show solution

$a+b=\dfrac{2\times96}{6}=32$ and $a-b=8$ .
$2a=40 \Rightarrow a=20$ cm.

Answer

20 cm

Type 5common

Any quadrilateral split by a diagonal

How to spot it:

A quadrilateral that is not a named shape, with a diagonal and two heights.

Method

Draw the given diagonal; it makes two triangles.
Collect both perpendicular heights onto it.
$K=\dfrac{1}{2}\times d\times(h_1+h_2)$ .

Why it works:

One formula covers every irregular quadrilateral, and it is just two triangle areas added.

Try this

A quadrilateral has a 12 cm diagonal; the perpendiculars to it from the other two corners are 5 cm and 9 cm. Its area is:

Show solution

$K=\dfrac12\times12\times(5+9)$ .
$=6\times14=84$ sq cm.

Answer

84 sq cm

Formula sheet

Rectangle

K=lb,\quad P=2(l+b),\quad d=\sqrt{l^2+b^2}

Three linked facts; any two fix the third.

Square

K=a^2,\quad d=a\sqrt2

Diagonal is root-two times the side.

Parallelogram

K=bh=ab\sin\theta

theta = angle between the two given sides.

Rhombus

K=\frac{1}{2}d_1d_2

Diagonals cross at right angles and halve each other.

Trapezium

K=\frac{1}{2}(a+b)h

a, b = the two parallel sides.

Any quadrilateral

K=\frac{1}{2}d(h_1+h_2)

d = a diagonal; h1, h2 = perpendiculars onto it.

Shortcuts that save time

⚡ Rectangle identities, not quadratics

With perimeter and diagonal, jump straight to $(l+b)^2=l^2+b^2+2lb$ and read off the area.

Example

The perimeter of a rectangle is 34 cm and its diagonal is 13 cm. Its area is:

Show solution

$l+b=17$ , so $(l+b)^2=289$ .
$l^2+b^2=13^2=169$ .
$2lb=289-169=120$ , so $lb=60$ .

Answer

60 sq cm

⚡ Rhombus ratio to perimeter

Ratio plus area fixes both diagonals. Halve them, spot the triplet, read the side.

Example

The diagonals of a rhombus are in the ratio 5 : 12 and its area is 120 sq cm. Its perimeter is:

Show solution

$\dfrac{1}{2}\times5k\times12k=120$ gives $k=2$ .
Diagonals $10$ , $24$ ; halves $5$ , $12$ ; side $13$ .
Perimeter $=4\times13=52$ cm.

Answer

52 cm

⚡ Diagonal splits any quadrilateral

No special shape? Draw one diagonal. The area is half the diagonal times the sum of the two heights.

Example

A quadrilateral has a diagonal of 12 cm, and the perpendiculars to it from the other corners are 5 cm and 9 cm. Find the area.

Show solution

$K=\dfrac{1}{2}\times12\times(5+9)$ .
$=6\times14$ .
$=84$ sq cm.

Answer

84 sq cm

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Treating the slant side of a parallelogram as the height.

Height is perpendicular. Sides 10 and 8 at $30^\circ$ give area $10\times8\times\dfrac12=40$ .

Mistake 02

Using $d_1d_2$ for the rhombus area.

The formula has the half: $\dfrac{1}{2}\times10\times24=120$ , not 240.

Mistake 03

Halving only one diagonal of a rhombus.

Both diagonals bisect each other, so both become halves: 5 and 12.

Mistake 04

Solving a quadratic for $l$ and $b$ when only $lb$ is asked.

Use $(l+b)^2-(l^2+b^2)=2lb$ and stop at the product.

Mistake 05

Adding the parallel sides after multiplying by the height.

Trapezium area is $\dfrac12(a+b)h$ ; add first, then halve and multiply.

Quick revision

Read this the night before the exam.

Rectangle: $lb$ , $2(l+b)$ , $\sqrt{l^2+b^2}$ ; two facts combine through an identity.
Square: diagonal $a\sqrt2$ ; backwards from the diagonal, divide by $\sqrt2$ .
Parallelogram: $bh=ab\sin\theta$ ; the height is perpendicular.
Rhombus: $\dfrac12 d_1d_2$ ; half-diagonals plus a side form a right triangle.
Trapezium: $\dfrac12(a+b)h$ with $a$ , $b$ the parallel sides.
Any quadrilateral: $\dfrac12 \times$ diagonal $\times$ (sum of perpendiculars).

Practice: 13 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 13 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.

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