Question

A lite is in the shape of a rhombus whose diagonals are $(x+5)$ units and $(x-8)$ units. The number of such tiles required to tile on the floor of area $(x^2+x-20)sq.units$ id

• A. $\sqrt{2}(x+1)=t\frac{2(x+6)}{(x+2)}$
• B. $\frac{x+4}{x-2}$
• C. $\sqrt{2}(x+1)=t\frac{2(x-4)}{(x-8)}$
• D. $\frac{x-8}{x+2}$

Answer 1

The correct option is A $\sqrt{2}(x+1)=t\frac{2(x+6)}{(x+2)}$

The tile is in the shape of rhombus whose diagonals are ( x + 5 ) and ( x - 8 ) units respectively.

We know, area of rhombus $=\frac{1}{2}\times \left(\text{product of diagonals}\right)$

Hence, area of one tile $ = \dfrac{1}{2} (x+5) (x-8)

Now the area of the floor is $x^2+x-20$

$\therefore$ tiles needed $=\frac{x^2+x-20}{\frac{1}{2}(x+5)(x-8)}=\frac{2(x^2+5x-4x-20)}{(x+5)(x-8)}=\frac{2(x+5)(x-4)}{(x+5)(x-8)}=\frac{2(x-4)}{(x-8)}$