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Year 8 Maths Area and Perimeter

Area of Kites and Rhombuses

20 practice questions 2 video lessons Theory + worked examples
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Practice questions

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Theory

The area of a kite or a rhombus is half the product of its diagonals: \(A = \dfrac{1}{2}xy\), where \(x\) and \(y\) are the full diagonals. A kite with diagonals \(6\) cm and \(9\) cm has area \(\dfrac{1}{2} \times 6 \times 9 = 27~\text{cm}^2\). This Year 8 maths page shows how.

A diagonal joins two opposite corners. For a kite or a rhombus with diagonals \(x\) and \(y\): \(A = \dfrac{1}{2}xy\). The side lengths are never used.

Why it works. The shape fills exactly half of the rectangle built on its diagonals.

The diagonals cross at right angles. In a rhombus each diagonal cuts the other in half. In a kite the axis of symmetry cuts the other diagonal in half, but is not cut in half itself. A square is a special rhombus.

Given only a half? In a rhombus every centre-to-corner length is half a diagonal, so double it. In a kite only the diagonal cut by the axis has equal halves.

Missing diagonal: substitute, double both sides to remove the half, then divide. If one diagonal is a multiple of the other, call the shorter one \(x\); you get an \(x^2\), so square root.

Units. Change both diagonals to one unit first; \(1\) ha \(= 10\,000~\text{m}^2\). To count tiles, divide the total area by one tile’s area.

Kite inside a rectangle A kite with its vertical diagonal x and horizontal diagonal y dashed, drawn inside a dashed rectangle x by y. The kite fills exactly half of the rectangle. x y
The kite is half of the rectangle built on its diagonals, so \(A = \dfrac{1}{2}xy\).
Rhombus with half diagonals A rhombus with its diagonals crossing at the centre. The half diagonal a runs from the centre to the right corner and b from the centre to the top, so the full diagonals are 2a and 2b. a b x = 2a, y = 2b
The centre-to-corner lengths \(a\) and \(b\) are only half diagonals: double each one before substituting.

Kites and rhombuses:

\[ A = \dfrac{1}{2}xy \]
A =1/2 x y
A side is not a diagonal. Use only the two diagonals, never a slanting side.
Same unit first. Diagonals of \(70\) cm and \(0.4\) m: \(\dfrac{1}{2} \times 0.7 \times 0.4 = 0.14~\text{m}^2\).

How to find the area

  1. Find the full lengths of both diagonals, in one unit (double any halves).
  2. Multiply them and halve.
  3. Give square units, then answer the question asked.
Example 1 — A kite
Find the area of a kite with diagonals of \(6\) cm and \(9\) cm.
Solution
\(A\)\(=\)\(\dfrac{1}{2} \times 6 \times 9 = 27\)
27 cm

\(27~\text{cm}^2\)

Example 2 — A rhombus
Find the area of a rhombus with diagonals of \(8\) cm and \(14\) cm.
Solution
\(A\)\(=\)\(\dfrac{1}{2} \times 8 \times 14 = 56\)
56 cm

\(56~\text{cm}^2\)

Example 3 — Larger diagonals
A rhombus has diagonals of \(16\) cm and \(11\) cm. Find its area.
Solution
\(A\)\(=\)\(\dfrac{1}{2} \times 16 \times 11\)
\(\)\(=\)\(8 \times 11 = 88\)
88 cm

\(88~\text{cm}^2\)

Example 4 — Only half a diagonal
A kite has an axis of symmetry \(18\) cm long. The other diagonal is marked only from the axis to one corner: \(5\) cm. Find the area.
Solution

The axis cuts the other diagonal in half, so double the \(5\) cm: \(10\) cm.

\(A\)\(=\)\(\dfrac{1}{2} \times 10 \times 18 = 90\)
90 cm

\(90~\text{cm}^2\)

Example 5 — Missing diagonal, mixed units
A kite-shaped sign has an area of \(2.4~\text{m}^2\). One diagonal is \(160\) cm. Find the other diagonal in metres.
Solution
\(2.4\)\(=\)\(\dfrac{1}{2} \times 1.6 \times y\)
\(2.4\)\(=\)\(0.8y\)
\(y\)\(=\)\(3\)
3 m

\(160\) cm \(= 1.6\) m, so the other diagonal is \(3\) m. Check: \(\dfrac{1}{2} \times 1.6 \times 3 = 2.4\).

Example 6 — A paddock in hectares
A kite-shaped paddock has diagonals of \(240\) m and \(150\) m. Find its area in hectares.
Solution
\(A\)\(=\)\(\dfrac{1}{2} \times 240 \times 150\)
\(\)\(=\)\(18\,000~\text{m}^2\)
\(18\,000 \div 10\,000\)\(=\)\(1.8\)
1.8 ha

\(1.8\) ha

Example 7 — Diagonals in a ratio
A rhombus has an area of \(50~\text{cm}^2\). One diagonal is \(4\) times as long as the other. Find both diagonals.
Solution
\(\dfrac{1}{2} \times x \times 4x\)\(=\)\(50\)
\(2x^2\)\(=\)\(50\)
\(x^2\)\(=\)\(25\)
\(x\)\(=\)\(5\)
5 cm and 20 cm

A length is positive, so \(x = 5\): the diagonals are \(5\) cm and \(20\) cm.

Common pitfalls

Using half diagonals as full ones. A centre-to-corner length in a rhombus is only half a diagonal.
Forgetting the half. \(A = \dfrac{1}{2}xy\), not \(xy\).
Halving the wrong kite diagonal. The axis of symmetry is not cut in half; only the other diagonal is.
Using the sides. The formula needs the diagonals.

Frequently asked questions

What is the area of a rhombus?

Half the product of its diagonals: \(A = \dfrac{1}{2}xy\).

What is the area of a kite?

Also \(A = \dfrac{1}{2}xy\), using its two diagonals.

Why does the formula have a half?

The shape fills exactly half of the rectangle built on its diagonals.

How do I find a missing diagonal?

Double the area, then divide by the known diagonal: \(y = 2A \div x\).

Can I use this formula for a square?

Yes. A square is a special rhombus, so half the product of its equal diagonals gives its area.