SME
Grade 10 Math · Problem type
Kite
Use this for “A kite’s diagonals cross at O; BD = 12, OA = 8 and OC = 4.5. How long are its sides?”
Diagram
A kite: two pairs of equal sides next to each other; the diagonals meet at right angles.
BD = 12
This is an example. Type your own values to start a new problem.
AB
=
BD
2
²
+
OA
2
CB
=
BD
2
²
+
OC
2
Formulas
AB
=
√(
(
BD
2
)
2
+
OA²)
10
=
√(
(
12
2
)
2
+
8
2
)
CB
=
√(
(
BD
2
)
2
+
OC²)
7.5
=
√(
(
12
2
)
2
+
4.5
2
)
Assumptions
•
A kite has two pairs of equal sides next to each other: AB = AD and CB = CD.
•
Its diagonals cross at right angles, and diagonal AC cuts BD in half.
•
So each side is the hypotenuse of a right triangle with legs half of BD and OA or OC.
•
Lengths are in any one unit.
Step-by-step
WE KNOW
BD = 12, OA = 8, OC = 4.5
FIND
side AB = AD (AB), side CB = CD (CB)
Step 1 · Find side AB = AD (AB)
Use AB = √((BD ÷ 2)² + OA²)
AB is the hypotenuse of the right triangle AOB.
AB
=
√(
(
12
2
)
2
+
8
2
)
AB
=
√
6
2
+
8
2
AB
=
√
36 + 64
AB
=
√
100
AB = 10
Step 2 · Find side CB = CD (CB)
Use CB = √((BD ÷ 2)² + OC²)
CB is the hypotenuse of the right triangle COB.
CB
=
√(
(
12
2
)
2
+
4.5
2
)
CB
=
√
6
2
+
4.5
2
CB
=
√
36 + 20.25
CB
=
√
56.25
CB = 7.5
Check
Put the numbers back in. Both sides should match.
10
=
√(
(
12
2
)
2
+
8
2
)
✓
7.5
=
√(
(
12
2
)
2
+
4.5
2
)
✓
Topic check
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