contestada

A rectangle's base is 8 in shorter than three times its height. The rectangle's area is 16 in?. Find this rectangle's dimensions.

The rectangle's height is..
The rectangle's base is..

Respuesta :

Let's height be x

Base=3x-8

ATQ

[tex]\\ \sf\longmapsto x(3x-8)=16[/tex]

[tex]\\ \sf\longmapsto 3x^2-8x=16[/tex]

[tex]\\ \sf\longmapsto 3x^2-8x-16=0[/tex]

[tex]\\ \sf\longmapsto 3x^2-12x+4x-16=0[/tex]

[tex]\\ \sf\longmapsto 3x(x-4)+4(x-4)=0[/tex]

[tex]\\ \sf\longmapsto (3x+4)(x-4)=0[/tex]

  • Dimensions should be whole nos

[tex]\\ \sf\longmapsto x=4[/tex]

[tex]\\ \sf\longmapsto 3x-8=3(4)-8=12-8=4[/tex]

  • Base=Height=4in
DᴀʀᴋPᴀʀᴀᴅᴏx

[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]

Let the height of rectangle be = x,

  • height = x

  • base = 3x - 8

Area of rectangle = base × height

  • [tex](3x - 8) \times (x) = 16[/tex]

  • [tex]3 {x}^{2} - 8x = 16[/tex]

  • [tex]3 {x}^{2} - 8x - 16 = 0[/tex]

  • [tex]3 {x}^{2} - 12x + 4x - 16 = 0[/tex]

  • [tex]3x(x - 4) + 4(x - 4) = 0[/tex]

  • [tex](3x + 4)(x - 4) = 0[/tex]

so, there are two cases now,

Case - 1 :

  • [tex]3x + 4= 0[/tex]

  • [tex]3x = - 4[/tex]

  • [tex]x = \dfrac{ - 4}{3} [/tex]

here, value of x (length) is negative, measure of length can't be negative, so length is not equal to -4/3

Case - 2 :

  • [tex]x - 4 = 0[/tex]

  • [tex]x = 4[/tex]

here, value of x = 4, and satisfys the condition so

  • length = x = 4 in

  • base = 3x - 8 = (3 × 4) - 8 = 12 - 8 = 4 in

Otras preguntas