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# Properties of Trapezoids and Kites: Problem-Solving

Aug 16, 2023

## 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.

Proof:

Draw perpendicular lines AE and BF between the parallel sides of the trapezoid.

In ΔAED and ΔBFC,

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, 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:

∠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 a trapezoid ABCD, AB∥CD, and X is the midpoint of AD, and Y is the midpoint of BC.

To prove: XY = 1/2 x (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

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: XZꞱWY

Proof: Draw diagonals XZ and WY. Let the diagonals intersect at O.

In ΔWXY and ΔWZY,

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,

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

1. Find the value of k if STUV is a trapezoid.

2. If EFGH is an isosceles trapezoid, find the value of p.

3. Find the length of PQ if LMQP is a trapezoid O is the midpoint of LP and N is the midpoint of MQ.

4. What must be the value of m if JKLM is a kite?

5. If WXYZ is a kite, find the length of diagonal XZ.

#### 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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