Question

There are 2 questions
Plz solve for me​

Answer 1

Let l, b and h are length, width and height of a rectangular box .

Area of the base = 48 cm² (given)

=> lb = 48 ---(1)

=> b = 48/l ----(2)

h = 3 cm and

diagonal (d) = 13 cm

=> d² = 13²

=> l² + b² + h² = 169

=> l² + b² + 3² = 169

=> l²+b² = 160

=> l² + b² + 2lb = 160 + 2lb

=> ( l + b )² = 160 + 2 ×48 [From (1) ]

=> ( l + b )² = 160 + 96

=> ( l + b )² = 256

=> ( l + b )² = 16²

=> l + b = 16 ---(3)

$\implies l + \frac{48}{l} = 16$

$\implies l^{2} + 48 = 16l$

$\implies l^{2} - 16l + 48 = 0$

$\implies l^{2} - 12l - 4l + 48 = 0$

$\implies l( l -12) - 4( l - 12 ) = 0$

$\implies (l-12)(l-4) = 0$

$\implies l - 12 = 0 \:Or \: l - 4 = 0$

$\implies l = 12 \:Or \: l = 4$

/* Put value of l in equation (3), we get */

$b = 4 \: Or \: b = 12$

Therefore.,

$Values \: of \: (l,b) = ( 12,4) \:or \: ( 4,12)$

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