KITES IN GEOMETRY

A kite is a quadrilateral that has two pairs of consecutive congruent sides, but opposite sides are not congruent.  

Theorems on Kites

Theorem 1 : 

If a quadrilateral is a kite, then its diagonals are perpendicular. 

It has been illustrated in the diagram shown below. 

Theorem 2 : 

If a quadrilateral is a kite, then exactly one pair of opposite angles are congruent. 

It has been illustrated in the diagram shown below. 

Kites in Geometry - Practice Problems

Problem 1 : 

In the kite WXYZ shown below, find the length of each side. 

Solution : 

Because WXYZ is a kite, the diagonals are perpendicular. We can use Pythagorean theorem to find the side lengths. 

In the kite WXYZ shown above, let us consider the triangle part WUZ. 

Because the diagonals WY and XZ are perpendicular and they intersect at U, angle U is a right angle. 

So, WUZ is a right triangle. 

By Pythagoren theorem, we have

WZ²  =  WU² + UZ²

Take radical on both sides. 

√WZ²  =  √(WU² + UZ²)

WZ  =  √(20² + 12²)

WZ  =  √(400 + 144)

WZ  =  √544

WZ  ≈  23.32

Similarly, in the right triangle triangle YUZ, we have

YZ²  =  YU² + UZ²

Take radical on both sides. 

√YZ²  =  √(YU² + UZ²)

YZ  =  √(12² + 12²)

YZ  =  √(144 + 144)

YZ  =  √288

YZ  ≈  16.97

We know that a kite is  a quadrilateral that has two pairs of consecutive congruent sides, but opposite sides are not congruent. 

So, in the kite WXYZ shown above, we have

WX  ≅  WZ

YX  ≅  YZ

Hence, we have

WX  =  WZ  ≈  23.32 

YX  =  YZ  ≈  16.97

Problem 2 : 

Find m∠G and m∠J in the diagram shown below. 

Solution : 

The quadrilateral GHJK shown above has two pairs of consecutive congruent sides, but opposite sides are not congruent. 

So, the quadrilateral GHJK is a kite. 

By theorem 2 above, exactly one pair of opposite angles of a kite are congruent. 

But, in the diagram shown above, the pair of m∠H and m∠K are not congruent.

Then, the pair of m∠G and m∠J must be congruent.

That is, 

m∠G  ≅  m∠J

Let , m∠G  =  m∠J  =  x°

We know that the four angles of a quadrilateral add up to 360°. 

So, we have

m∠G + m∠H + m∠J + m∠K  =  360°

x° + 132° + x° + 60°  =  360°

Simplify.

2x° + 192°  =  360°

Subtract 192° from both sides. 

2x° + 192°  =  360°

2x°  =  168°

Divide both sides by 2. 

x°  =  84°

Hence, we have

m∠G  =  m∠J  =  84°

Problem 3 :

ABCD is a kite. Find ∠A, ∠C and ∠D.

Solution :

∠A = x + 30,  ∠C = 125,  ∠B = 125 and ∠D = x

∠A + ∠B +  ∠C +  ∠D = 360

x + 30 + 125 + 125 + x = 360

2x + 280 = 360

2x = 360 - 280

2x = 80

x = 80/2

x = 40

Applying the value of x, we get

∠A = 40 + 30 ==> 70

∠C = 125

∠D = 40

Problem 4 :

Find the value of the missing variable(s). Figure is a kite

Solution :

x = 114

x + 88 + 114 + y = 360

114 + 88 + 114 + y = 360

316 + y = 360

y = 360 - 316

y = 44

Problem 5 :

In kite PQRS, SQ is perpendicular to PR and PO = OR. IF PQ = 5x - 10 and QR = 4x + 5. Find x, PO and QR.

Solution :

PQ = QR

5x - 10 = 4x + 5

5x - 4x = 5 + 10

x = 15

PQ = 5(15) - 10

= 75 - 10

= 65

QR = 65

Problem 6 :

In kite ABCD, DB is perpendicular to AC, if AC = 10 and BD = 7. Find the area of the kite ABCD.

Solution :

Area of kite = 1/2 x diagonal 1 x diagonal 2

= (1/2) x 10 x 7

= 70/2

= 35 square units.

Problem 7 :

Solve for x.

Solution :

2x + 4 = 10

2x = 10 - 4

2x = 6

x = 6/2

x = 3

So, the value of x is 3.

Problem 8 :

Find the perimeter of the kite.

Solution :

In the right triangle at the top left and right,

let x be the unknown side which is hypotenuse.

x^2 = 3^2 + 4^2

x^2 = 9 + 16

x^2 = 25

x = 5

Let y be the length of the longer leg.

y^2 = 4² + 5^2

= 16 + 25

= 41

y = √41

Perimeter of the kite = 2(5) + 2√41

= 10 + 2(6.4)

= 10 + 12.8

= 22.8

So, the required perimeter of the kite is 22.8 units.

Problem 9 :

Find the value of x.

Solution :

In the kite, the diagonals will be perpendicular to each other.

6x + 6 = 90

6x = 90 - 6

6x = 84

x = 84/6

= 14

So, the value of x is 14.

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