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Area of a Kite

Area of a Kite

The area of a kite is half the product of the diagonals. Note: This formula works for the area of a rhombus as well, since a rhombus is a special kind of kite. Note that the diagonals of a kite are perpendicular.

 

Kite

d1 = long diagonal of kite
d2 = short diagonal of kite

Area = (½) d1d2


Kite shape with perpendicular diagonals labeled d₁ (long, vertical) and d₂ (short, horizontal), intersecting at a right angle.

Key Formula

A=12 d1 d2A = \frac{1}{2}\,d_1\,d_2
Where:
  • AA = Area of the kite
  • d1d_1 = Length of the longer diagonal
  • d2d_2 = Length of the shorter diagonal

Worked Example

Problem: A kite has diagonals of length 10 cm and 6 cm. Find its area.
Step 1: Identify the two diagonals.
d1=10 cm,d2=6 cmd_1 = 10 \text{ cm},\quad d_2 = 6 \text{ cm}
Step 2: Write the area formula for a kite.
A=12 d1 d2A = \frac{1}{2}\,d_1\,d_2
Step 3: Substitute the values and multiply.
A=12×10×6=602A = \frac{1}{2} \times 10 \times 6 = \frac{60}{2}
Step 4: Simplify to get the final area.
A=30 cm2A = 30 \text{ cm}^2
Answer: The area of the kite is 30 cm².

Another Example

This example differs from the first because the diagonals are not directly given. You must use the Pythagorean theorem with the perpendicular diagonals to find a missing diagonal before computing the area.

Problem: A kite has side lengths of 5 m and 13 m. The longer diagonal measures 24 m. Find the area of the kite.
Step 1: In a kite, one diagonal (the 'main' or symmetry diagonal) connects the two vertices where unequal sides meet. The other diagonal connects the vertices where equal sides meet. The longer diagonal splits the kite into two congruent triangles. Label half of the shorter diagonal as x. The longer diagonal (d₁ = 24 m) is split by the shorter diagonal into two segments; call them p and q.
d1=24 md_1 = 24 \text{ m}
Step 2: Because the diagonals are perpendicular, each side of the kite is the hypotenuse of a right triangle. Using the two pairs of sides (5 m and 13 m), set up equations. Let p be the portion of d₁ paired with the shorter sides, and q = 24 − p paired with the longer sides. Half the shorter diagonal is x.
p2+x2=52=25p^2 + x^2 = 5^2 = 25
Step 3: Write the second equation for the longer pair of sides.
q2+x2=132=169,q=24−pq^2 + x^2 = 13^2 = 169, \quad q = 24 - p
Step 4: From the first equation, x² = 25 − p². Try p = 3: x² = 25 − 9 = 16, so x = 4. Check: q = 24 − 3 = 21, and 21² + 16 = 441 + 16 = 457 ≠ 169. So instead, try p = 4 and q = 24 − 4 = 20: 4² + x² = 25 gives x² = 9, x = 3. Check: 20² + 9 = 400 + 9 = 409 ≠ 169. Reconsider: the shorter sides (5 m) should pair with shorter segments. Try p = 3, x = 4 again but with q = 24 − 3 = 21 — that's too large. The diagonal of 24 m must be the symmetry diagonal that is NOT bisected. Let the shorter diagonal d₂ = 2x be bisected by d₁. Then d₁ is split into segments p and 24 − p. With p = 3: x² = 25 − 9 = 16, x = 4. Then (24 − 3)² + 16 = 441 + 16 = 457 ≠ 169. Try p = 4, x = 3: (24 − 4)² + 9 = 409 ≠ 169. The sides 5 and 13 with a 24-diagonal suggest a different split. Actually, let's use side 13 with segment p: p² + x² = 169, and side 5 with segment (24 − p): (24 − p)² + x² = 25. Subtract: p² − (24 − p)² = 144. Expand: p² − 576 + 48p − p² = 144, so 48p = 720, p = 15. Then x² = 169 − 225 < 0 — impossible. Swap: side 5 with p, side 13 with (24 − p): p² + x² = 25 and (24 − p)² + x² = 169. Subtract: p² − (24 − p)² = 25 − 169 = −144. So 48p − 576 = −144, 48p = 432, p = 9. Then x² = 25 − 81 < 0, still impossible. The given numbers are inconsistent. Let me choose realistic numbers instead: sides 5 and 12, longer diagonal 13.
Step 4 (revised setup): Restart with clean numbers. A kite has side lengths 5 m and 5√2 m. The longer diagonal is 8 m. Actually, for clarity, let's use: sides 5 and 13, shorter diagonal unknown, and the longer diagonal splits into segments. Place the kite with the symmetry diagonal along the x-axis. Let the shorter diagonal be bisected into segments of length x on each side. Let the symmetry diagonal be split into segments of length p and q where p + q = d₁. Using sides 5 with segment p and sides 13 with segment q: p² + x² = 25 and q² + x² = 169. Then q² − p² = 144, so (q − p)(q + p) = 144. If d₁ = q + p = 24, then q − p = 6, giving q = 15, p = 3, and x² = 25 − 9 = 16, so x = 4. Check: 15² + 16 = 225 + 16 = 241 ≠ 169. This still fails because q goes with sides of length 13 means q² + x² = 13² only if the right triangle has legs q and x. Here q = 15 > 13 so it cannot work. The issue is the diagonal of 24 m is too long. Let d₁ = 18 instead: q − p = 144/18 = 8, q + p = 18, so q = 13, p = 5. Then x² = 25 − 25 = 0, degenerate. Try d₁ = 16: q − p = 9, q + p = 16, q = 12.5, p = 3.5. x² = 25 − 12.25 = 12.75 — messy. Let me just pick a totally clean example. Sides 10 and 17, find diagonals via right-triangle approach where the symmetry diagonal = 21 (split into 6 and 15), and half the other diagonal = 8. Check: 6² + 8² = 36 + 64 = 100 = 10². ✓ And 15² + 8² = 225 + 64 = 289 = 17². ✓ So d₁ = 21, d₂ = 16.
Step 4 (clean restart): New problem: A kite has sides of length 10 m and 17 m. Its longer diagonal (the axis of symmetry) measures 21 m. Find the shorter diagonal and then the area. The symmetry diagonal is split by the shorter diagonal into two segments p and q with p + q = 21. Each side forms a right triangle with half the shorter diagonal (call it x). Set up equations using the Pythagorean theorem.
p2+x2=102=100,q2+x2=172=289p^2 + x^2 = 10^2 = 100, \quad q^2 + x^2 = 17^2 = 289
Step 5: Subtract the first equation from the second to eliminate x².
q2−p2=189  ⟹  (q−p)(q+p)=189q^2 - p^2 = 189 \implies (q-p)(q+p) = 189
Step 6: Since q + p = 21, substitute and solve for q − p.
21(q−p)=189  ⟹  q−p=921(q - p) = 189 \implies q - p = 9
Step 7: Solve the system: q + p = 21 and q − p = 9 gives q = 15 and p = 6. Now find x from the first equation.
x2=100−36=64  ⟹  x=8x^2 = 100 - 36 = 64 \implies x = 8
Step 8: The shorter diagonal is d₂ = 2x = 16 m. Now apply the area formula.
A=12×21×16=168 m2A = \frac{1}{2} \times 21 \times 16 = 168 \text{ m}^2
Answer: The shorter diagonal is 16 m, and the area of the kite is 168 m².

Frequently Asked Questions

Why does the area of a kite formula work?
The two perpendicular diagonals divide the kite into four right triangles. When you pair opposite triangles, they form two rectangles (each with dimensions equal to half of one diagonal and half of the other). The total area of these two rectangles simplifies to ½ d₁ d₂. This reasoning relies on the fact that the diagonals of a kite always intersect at right angles.
Is the area formula for a kite the same as for a rhombus?
Yes. A rhombus is a special case of a kite where all four sides are equal. Because both shapes have perpendicular diagonals, the formula A = ½ d₁ d₂ applies to both. The only difference is that a rhombus's diagonals bisect each other, while in a general kite only one diagonal is bisected by the other.
Does it matter which diagonal is d₁ and which is d₂?
No. Multiplication is commutative, so ½ d₁ d₂ gives the same result regardless of which diagonal you label as d₁ or d₂. The formula works whether you call the longer or shorter diagonal first.

Area of a Kite vs. Area of a Rhombus

Area of a KiteArea of a Rhombus
Shape definitionQuadrilateral with two pairs of consecutive congruent sidesQuadrilateral with all four sides congruent
FormulaA = ½ d₁ d₂A = ½ d₁ d₂
Diagonal propertiesPerpendicular; only the symmetry diagonal bisects the otherPerpendicular; both diagonals bisect each other
RelationshipGeneral caseSpecial case of a kite

Why It Matters

You encounter the area of a kite in geometry courses when studying quadrilaterals, and it frequently appears on standardized tests. The same formula applies to rhombuses, making it a versatile tool. Understanding why perpendicular diagonals lead to this formula also strengthens your grasp of how area formulas are derived from simpler shapes like triangles and rectangles.

Common Mistakes

Mistake: Multiplying the diagonals without dividing by 2.
Correction: The formula is A = ½ d₁ d₂, not d₁ d₂. The product d₁ d₂ gives the area of the rectangle that encloses the kite, which is exactly twice the kite's area. Always remember the factor of one-half.
Mistake: Using the side lengths instead of the diagonal lengths.
Correction: The formula requires the lengths of the two diagonals, not the sides. If you are given side lengths, you need additional information (like one diagonal or the angle between sides) and the Pythagorean theorem to find the diagonals first.

Related Terms

  • Kite — The shape whose area this formula computes
  • Area of a Rhombus — Uses the same diagonal-based area formula
  • Rhombus — A special kite with all sides equal
  • Diagonal of a Polygon — The two diagonals are the key measurements
  • Perpendicular — Kite diagonals always meet at right angles
  • Product — Area is half the product of the diagonals
  • Formula — General term for the area equation used