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or, 1600  80 xx^2 h^2 1369
or, 1600  80 x 169 1396 [with the help of (1)]
or, 1600  169  1396 80 x; or, 80 x 400 ;? x 5
Now putting the value of x in equation (1), we get
52 h^2 163 or, h 169  25 144 ; ?h 12

Area of ABCD = (ABCD)˜h
2

1

= ( 91 51 ) 12
2

1

 u square cm.

= 852 square cm.
The required area is 852 square cm.
16 ⋅3 Area of regular polygon
The lengths of all sides of a regular polygon are equal. Again, the angles are also
equal. Regular polygon with n sides produces n isocsceles triangles by adding centre
to the vertices.
? Area of the regular polygon = nu area of one triangular
region.
Let ABCDEFbe a regular polygon whose centre is O.
? It has n sides and the length of each side is a. We join O, A ;
O, B.
? Let in 'AOB height OM h and ‘OAB θ
? The angle produced at each of the vertices of regular polygon = 2 θ.
? Angle produced by n number of vertices in the polygon = θ˜n

θ

2
Angle produced in the polygon at the centre = 4 right angles.
The sum of angles of n number of triangles = 2 θ(n 4 ) right angles.
?Sum of 3 angles of 'OAB = 2 right angles.
? The wise, summation of the angles of n numbers of triangles = n˜ 2 right angles
? 2 θ(n 4 ) right angles = n˜ 2 right angles
or, 2 θ˜n ( 2 n 4 ) right angles

or,
n

n

2

2  4

θ right angles

or, ̧
¹

̈ ·

©

§

n

θ 1 2 right angles

or, 90 $
2

(^1) ̧u
¹
·
̈
©
§ 
n
θ = 90q
180 q
n
Now,
a
h
a
h 2
2
tanθ ;? 2 tanθ
h a
M