Key Concepts
- Define a trapezoid.
- Explain the properties of a trapezoid.
- Solve problems based on properties of a kite.
Trapezoid
A quadrilateral having only one pair of parallel sides is called a trapezoid.
- A trapezium in which the non-parallel sides are equal is an isosceles trapezium.
Property of a trapezoid related to base angles
Theorem 1:
In an isosceles trapezoid, each pair of base angles is congruent.
Given: ABCD is a trapezoid where AB∥CD.
To prove: ∠ADC = ∠BCD and ∠BAD = ∠ABC
Proof:
Draw perpendicular lines AE and BF between the parallel sides of the trapezoid.
In ΔAED and ΔBFC,
AD = BC [Isosceles trapezoid]
AE = BF [Distance between parallel lines will always be equal]
∠AEB = ∠BFC=90° [AEꞱCD and BFꞱCD]
If two right-angled triangles have their hypotenuses equal in length and a pair of shorter sides are equal in length, then the triangles are congruent.
∴ ΔAED ≌ ΔBFC [RHS congruence rule]
We know that the corresponding parts of congruent triangles are equal.
Hence, ∠ADC = ∠BCD
And ∠EAD = ∠FBC
Now, ∠BAD = ∠BAE + ∠EAD
∠BAD = 90° + ∠EAD
∠BAD = 90° + ∠EAD
∠BAD = ∠ABC
Hence, each pair of base angles of an isosceles trapezoid is congruent.
Property of trapezoid related to the length of diagonals
Theorem 2:
The diagonals of an isosceles trapezoid are congruent.
Given: In trapezoid ABCD, AB∥CD, and AD = BC
To prove: AC = BD
Proof:
In ΔADC and ΔBCD,
AD = BC [Isosceles trapezoid]
∠ADC = ∠BCD [Base angles of isosceles trapezoid]
CD = CD [Common]
Therefore, ΔAED ≌ ΔBFC [SAS congruence rule]
We know that the corresponding parts of congruent triangles are equal.
So, AC = BD
Hence, the diagonals of an isosceles trapezoid are congruent.
Property of trapezoid related to the length of diagonals
Theorem 3:
In a trapezoid, the midsegment is parallel to the bases, and the length of the midsegment is half the sum of the lengths of the bases.
Given: In trapezoid ABCD, AB∥CD, and X is the midpoint of AD, Y is the midpoint of BC.
To prove: XY = 1/2 × (AB+CD)
Proof: Construct BD such that the midpoint of BD passes through XY.
In ΔADB, X is the midpoint of AD, and M is the midpoint of DB.
So, XM is the midsegment of ΔADB.
We know that a line segment joining the midpoints of two sides of the triangle is parallel to the third side and has a length equal to half the length of the third side. [Midsegment theorem]
∴ XM ∥ AB and XM = 1/2 × AB …(1)
In ΔBCD, Y is the midpoint of BC and M is the midpoint of BD.
So, MY is the midsegment of ΔBCD.
∴ MY ∥ CD and MY = 1/2 × CD …(2) [Midsegment theorem]
Since XM ∥ AB and MY ∥ CD, so, XY
Now, XY=XM+MY
= 1/2 × AB + 1/2 × CD
XY = 1/2 × (AB+CD)
Kite
A kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other.
(or)
A parallelogram also has two pairs of equal-length sides, but they are opposite each other in a kite.
- Only one diagonal of a kite bisects the other diagonal.
Property of kite related to the angle between the diagonals
Theorem:
The diagonals of a kite are perpendicular.
Given: In kite WXYZ, XY=YZ, WX=ZW
To prove: XZ Ʇ WY
Proof: Draw diagonals XZ and WY. Let the diagonals intersect at O.
In ΔWXY and ΔWZY,
WX=WZ [Adjacent sides of kite]
XY=ZY [Adjacent sides of kite]
WY=YW [Reflexive property]
∴ ΔWXY ≌ ΔWZY [SSS congruence rule]
We know that the corresponding parts of congruent triangles are equal.
So, ∠XYW= ∠ZYW …(1)
In ΔOXY and ΔOZY,
XY=ZY [Adjacent sides of kite]
OY=YO [Reflexive property]
∠XYW= ∠ZYW [from (1)]
∴ ΔOXY ≌ ΔOZY [SAS congruence rule]
So, ∠YOX = ∠YOZ [CPCT]
But ∠YOX+∠YOZ = 180°
2∠YOX = 180°
∠YOX = 90°
Hence, the diagonals of a kite are perpendicular.
Exercise
- Find the value of k if STUV is a trapezoid.
- If EFGH is an isosceles trapezoid, find the value of p.
- Find the length of PQ if LMQP is a trapezoid and O is the midpoint of LP and N is the midpoint of MQ.
- What must be the value of m if JKLM is a kite?
- If WXYZ is a kite, find the length of diagonal XZ.
Concept Map
What we have learned
- A quadrilateral having only one pair of parallel sides is called a trapezoid.
- A kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other.
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