Question

A tile 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 (x2 + x – 20) sq. units is
(1)
2( 6)
( 2)
x
x
 
(2)
4
2
x
x
 
(3)
2( 4)
( 8)
x
x
 
(4)
8
2

Answer 1

Refer the attachment for complete question.
Hey there!

Given,
A tile is in the shape of a rhombus .
Length of diagonals = ( x + 5 ) , ( x - 8 )

Area of the rhombus shaped tile.
= 1/2 ( x + 5 ) ( x - 8 )
= 1/2 (x² - 8x + 5x - 40 )
= 1/2 ( x² - 3x - 40 )

Given area to be tiled = x² + x - 20

Factorising to get the answer easily.

Area to be tiled
= x² + 5x - 4x - 20
= x ( x + 5 ) - 4 ( x + 5 )
= ( x - 4 ) ( x + 5 )

Number of tiles required = $\frac{ \textbf{ Area to be tiled} } { \textbf{Area of each tile }}$

$= \frac{ (x - 4 )( x + 5) }{ 1/2 (x-8)(x+5)} \\ \\ \\ = \frac{2(x-4)}{ (x - 8)} = \frac{ 2x -8}{x-8}$

Final answer : $\frac{2( x - 4 )}{ x - 8 }$

The number of such tiles required to tile the area of ( x² +x - 20 ) is $\frac{2( x - 4 )}{ x - 8 }$

Therefore, Answer is Option 3

Answer 2

$\underline{\bold{Given:-}}$

The tile is in the shape of rhombus.

Length of diagonals of rhombus = (x+5) and (x-8).

Area of floor to be tiled
$= ( {x}^{2} + x - 20 ) {unit}^{2}$

$\underline{\bold{To\:find:-}}$

Number of tiles required to tile the floor.

$\underline{\bold{Solution:-}}$

Area of tile
$= \frac{1}{2} \times \: diagonal \: 1 \times diagonal \: 2 \\ \\ = \frac{1}{2} \times (x + 5) \times (x - 8)$

Number of tiles required to tile the floor

$= \frac{area \: of \: floor}{area \: of \: one \: tile} \\ \\ = \frac{ {x}^{2} + x - 20 }{ \frac{1}{2}(x + 5) (x - 8)} \\ \\ = 2( \frac{ {x}^{2} +5x - 4x - 20 }{(x + 5)(x - 8)} ) \\ \\ = 2 \times \frac{(x + 5)(x - 4)}{(x + 5)(x - 8)} \\ \\ = 2 \times \frac{(x - 4)}{(x - 8)}$

So, $2 \times \frac{(x - 4)}{(x - 8)}$ tiles will be required to tile the floor.