Question

in the adjoining figure if ab+ad=5ad and ac_ad=8then the area of the rectangle abcd is

Answer 1

Let AB = x
AD = y
and AC = z

So, given

x + z = 5y

and

z - y = 8

So,

z = 8 + y


Now, x + z = 5y

=> x + 8 + y = 5y (since z = 8 + y)

=> x = 4y - 8

=> 4y = x + 8

=> y =
$\frac{x + 8}{4} \\ \\ = \frac{x}{4} + 2$

So, z = 8 + y

=> z = 8 + x/4 + 2

=> z = 10 + x/4


Now, by Pythagoras Theorem,

AC² = AB² + BC²

and BC = AD

=> BC² = AD²

=> AC² = AB² + AD²

=> z² = x² + y²

=>
$( 10 + \frac{x}{4} ) {}^{2} = {x}^{2} + ( \frac{x}{4} + 2) {}^{2}$

since, z = 10 + x/4 and y = x/4 + 2

=>
$100 + \frac{ {x}^{2} }{16} + 5x = {x}^{2} + \frac{ {x}^{2} }{16} + 4 + x$

cancel x²/16 from both sides.

=>
$100 + 5x = {x}^{2} + 4 + x \\ \\ = > {x}^{2} + x - 5x + 4 - 100 = 0 \\ \\ = > {x}^{2} - 4x - 96 = 0 \\ \\ = > {x}^{2} - 12x + 8x - 96 = 0$
(after splitting the middle terms)

=> x(x - 12) + 8(x - 12) = 0

=> (x + 8)(x - 12) = 0

=> (x - 12) = 0/(x + 8)

=> x - 12 = 0

=> x = 12

So, now y = x/4 + 2

=> y = 12/4 + 2

=> y = 3 + 2

=> y = 5

x = AB = 12

y = AD = 5

Area = AB × AD

= 12 × 5

= 60 sq. units.

Answer :- 60 option 3