Question

$Question:$

A tile is in the shape of a rhombus whose diagonals are (x +5) units and (x-8)units . Then the number of such tiles required to tile on the floor of area (x² + x-20)sq.units is _______

Explain you answer briefly :)

Answer 1

Answer:

Ans) 2(x-4)/(x-8)

Step-by-step explanation:

Given :

Diagonal 1=(x+5)

Diagonal 2=(x-8)

Area of Rhombus =1/2*diagonal 1*diagonal 2

=1/2*(x+5)*(x-8)

Now,

Given Area of the floor

=x²+x-20 Sq units

=(x+5)(x-4) Sq units

Number of tiles

=2(x+5)(x-4)/(x+5)(x-8)

=2(x-4)/(x-8)

Answer 2

Solution:

Area of diagonal$= \frac{1}{2} \times D_{2} \times D_{2}$

$= \frac{1}{2} \times (x + 5)(x - 8) \\ = \frac{ {x}^{2} - 8x + 5x - 40}{2} \\ = \frac{ {x}^{2} - 3x - 40}{2}$

Area of floor=$( {x}^{2} + x - 20)$

No. Of titles required=$\frac{ {x}^{2} + x - 20 }{ \frac{ {x}^{2} - 3x - 40}{2} }$

$= \frac{ {2x}^{2} + 2x - 40 }{ {x}^{2} - 3x - 40 } \\ = \frac{ {2x}^{2} + 10x - 8x - 40}{ {x}^{2} - 8x + 5x - 40 } \\ = \frac{( 2x +( x +5)}{(x + 5)(x - 8)} \\ = \frac{2(x - 4)}{x - 8}$