To find diagonal, we have the following ways.
(i) From the given area and one diagonal, find the other diagonal.
(ii) Using Pythagorean theorem, find length of diagonal.
The diagonals of a kite are perpendicular to each other. The longer diagonal of the kite bisects the shorter diagonal.
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
Problem 1 :
The area of this shape is 48 ft^{2}. Solve for x.
Solution :
By observing the figure, length of one diagonal is given.
Area of a kite = 1/2 d_{1}d_{2}
Let x be the another diagonal.
48 = 1/2 (8)(x)
48 = 4x
Divide both sides by 4.
48/4 = 4x/4
12 = x
So, the value of x is 12
Problem 2 :
The area of this shape is 32 in^{2}. Solve for x.
Solution :
This is a rhombus.
Area of a rhombus = 1/2 d_{1}d_{2}
d_{1 }= 8 + 8 = 16
d_{2} = x + x = 2x
32 = 1/2 (16)(2x)
32 = 8(2x)
x = 32/16
x = 2
Problem 3 :
Find the area of the kite given below,
Solution :
In kite, the diagonal will bisect each other at right angles.
To figure out OC,
Use Pythagorean Theorem :
(BC)^{2} = (CO)^{2} + (BO)^{2}
(13)^{2} = (CO)^{2} + (12)^{2}
169 = (CO)^{2} + 144
Subtract 144 from both sides.
25 = (CO)^{2}
CO = 5, then CA = 2(5) ==> 10
Area of a kite = 1/2 d_{1}d_{2}
= 1/2
(10)(20)
= 1/2 (200)
= 100
So, area of a kite is 100.
Find the length of missing diagonal in each kite.
Problem 4 :
Find TR if QS = 24 ft
Solution :
Area of kite = (1/2) x TR x SQ
Here area = 384 and SQ = 24 ft
Applying these values, we get
384 = (1/2) x TR x 24
TR = 384/12
TR = 32
Problem 5 :
Find VX if WU = 28 in.
Solution :
Area of kite = (1/2) x WU x VX
Here area = 154 and WU = 28 in
Applying these values, we get
154 = (1/2) x 28 x VX
154 = 14 VX
VX = 154/14
VX = 11 in
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