Question

18.ABCD is a rectangle and O is the point of intersection of the diagonals AC
and 'BD. If OA = (6x+ 1) cm, OB = (8x-7) cm, and BC = 48 cm, then find
(a) Value of x (b) Length of AB (C) Perimeter of ABCD (d) Area of ABCD
​

Answer 1

[Note : refer the figure as shown in the attachment ]

${\textbf{\large{Step-by-step explanation:}}}$

⬛ABCD is a rectangle ,

As from the properties of rectangle

Diagonal AC = Diagonal BD

∴ AO = BO

⟹ (6x+ 1) = (8x-7)

⟹ 6x - 8x = - 7 - 1

⟹ - 2x = - 8

⟹ x = - 8 / - 2

∴ The value of x is

$\boxed{\textbf{\large{x = 4}}}$

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Diagonal AC = 2(AO )

AO = ((6( 4 )+ 1))

= ( 6 (4) + 1)

= ( 24 + 1)

= 25

diagonal AC = diagonal BD =

= 25 x 2 = 50

let us suppose A triangle ABC

/_ ABC = 90°

therefor, by Pythagoras theorem

( AC )^2 = ( AB) ^2 + ( BC) ^2

(50)^2 = ( AB )^2 + ( 48 )^2

2500 = ( AB) ^2 + ( 2304 )

2500 - 2304 = ( AB) ^2

AB = √ 196

$\boxed{\textbf{\large{AB = 14 cm }}}$

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Now, find the perimeter of ⬛ABCD

As we know, the formula to find the perimeter of rectangle

★perimeter of rectangle = 2 ( l + b)

where length is BC and breadth is AB

therefor ,

perimeter of rectangle ABCD

= 2 ( ( BC ) + ( AB))

= 2 ((48 + 1 4 )

= 2 ( 62 )

$\boxed{\textbf{\large{124 cm }}}$

Therefor, perimeter of rectangle ABCD is 124 cm

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Now,

★ Area of rectangle = l x b

Area of rectangle ABCD

= ((BC) x (AB))

= ( 48 x 14 )

$\boxed{\textbf{\large{672 cmSquare}}}$

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