FINDING AREA OF KITE

Area of kite ?

A kite is a quadrilateral which has two pairs of adjacent sides equal in length.

To find area of kite we need diagonals.

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

Find the area of each kite.

Problem 1 :

Solution :

Area of a kite = 1/2 d₁d₂

AC perpendicular BD

d₁ = 2 + 8 = 10

d₂ = 8 + 8 = 16

= 1/2 (10)(16)

= 80

So, area of a kite is 80 in².

Problem 2 :

Solution :

Area of a kite = 1/2 d₁d₂

AC perpendicular BD

d₁ = 3 + 3 = 6

d₂ = 2 + 4 = 6

= 1/2 (6)(6)

= 18

So, area of a kite is 18 m².

Problem 3 :

Solution :

Area of a kite = 1/2 d₁d₂

AC perpendicular BD

d₁ = 4 + 4 = 8

d₂ = 6

= 1/2 (8)(6)

= 24

So, area of a kite is 24 ft².

Problem 4 :

The area of a kite is 120 cm². The length of one diagonal is 20 cm. what is the length of the other diagonal?

a) 12cm     b) 20 cm       c) 24 cm     d) 48 cm

Solution :

Given, area of a kite is 120 cm²

Length of one diagonal d₁ = 20 cm.

Let length of the other diagonal d₂ be x.

Area of a kite = 1/2 d₁d₂

120 = 1/2 (20)(x)

120 = 10x

Divide both sides by 10.

120/10 = 10x/10

12 = x

So, length of the other diagonal is12 cm.

Problem 5 :

A kite has vertices at the points (2, 0) (3, 2) (4, 0) and (3, -3). Find the perimeter and area of the kite.

Solution :

Let the given points be A (2, 0) B (3, 2) C (4, 0) and D (3, -3)

Finding area :

Area of kite = (1/2) x AC x BD

AC = 2 units

BD = 5 units

= (1/2) x 4 x 5

= 10 square units.

Finding perimeter :

To find the perimeter of the kite, we need length of all sides. To figure out the length of the sides, we may use Pythagorean theorem.

Finding Perimeter :

Let AB = x

We know that diagonals of a kite bisect each other at right angles. Let O be the point of intersection of diagonals.

Using Pythagorean theorem, we get

Perimeter of kite ABCD = 2√5 + 2√10

= 2(2.23) + 2(3.16)

= 10.78 units.

So, perimeter of the kite is 10.78 units.

Problem 6 :

The area of a kite is 324 square inches. One diagonal is twice as long as the other diagonal. Find the length of each diagonal.

Solution :

Length of one diagonal = 2(length of other diagonal)

Let d₁ and d₂ be the lengths of diagonals.

Area of kite = (1/2) ⋅ d₁ ⋅ d₂

324 = (1/2) ⋅ d₁ ⋅ (2d₁)

d₁² = 324

d₁ = 18

d₂ = 2(18)

= 36 inches

So, the lengths of diagonals are 18 inches and 36 inches.

Problem 7 :

The vertices of the quadrilateral are A(2, 8), B (7, 9), C(11, 2) and D(3, 3). Show that ABCD is a kite and find the area of the kite.

Solution :

The diagonals will be perpendicular in kite.

Slope of AC = (y₂ - y₁) / (x₂ - x₁)

= (2 - 8) / (11 - 2)

= -6/9

= -2/3

Slope of BD = (9 - 3) / (7 - 3)

= 6/4

= 3/2

Slope of BD is the negative reciprocal of slope of AC.

Distance between AC = √(x₂ - x₁)² + (y₂ - y₁)²

= √(11 - 2)² + (2 - 8)²

= √9² + (-6)²

= √(81 + 36)

= √117

= √3 x 3 x 13

= 3√13

Distance between BD = √(3 - 7)² + (3 - 9)²

= √(-4)² + 6²

= √(16 + 36)

= √52

= √2 x 2 x 13

= 2√13

Area of kite ABCD = (1/2) x AC x BD

= (1/2) x 3√13 x 2√13

= 3√13√13

= 3(13)

= 39 square units.

Perimeter of kite ABCD :

Length of AD = √(3 - 2)² + (8 - 3)²

= √1² + 5²

= √(1 + 25)

= √26

Length of DB = √(11 - 3)² + (2 - 3)²

= √8² + (-1)²

= √(64 + 1)

 = √65

Perimeter of kite = 2(AD + BD)

= 2(√26 + √65)

= 2(5.09 + 8.06)

= 2(13.15)

= 26.3 units.

Problem 8 :

You and a friend are building a kite. You need a stick to place from X to W and a stick to place from W to Z to finish constructing the frame. You want the kite to have the geometric shape of a kite. How long does each stick need to be? Explain your reasoning

Solution :

Length of stick to be placed ZW = 18 inches

Length of stick to be placed WZ = 29 inches

Since it is kite, pair of sides will be equal.