Question

Plzzz I need help with this question I tried to solve it many times but I can't

Let x represents a length in cm where x > 1.5.
AB = 2x – 3, BC = 4x – 6,
AD = 2x + 3 and BE = 4x +1
• ABC is right triangle of area Sj.
• CBFG is a rectangle of area S2.
. ADEB is a right trapezoid of area S3.
1) Express S1, S2 and S3 in terms of x.
2) Let S = Si + S2 - Sz.
a) Show that S = (2x - 3)(5x - 9).
b) For what values of x, the areas of AFGC and ADEB are equal?
c) Expand and reduce S.
d) What is the value of x, so that the area of AFGC exceed the area of ADEB by 27?​

Answer 1

Answer and Step-by-step explanation: Area of a right triangle, (as any other triangle), is calculated as:
`A1=((base)(height))/(2)`

Area of a rectangle is calculated as:
`A2=(side)(side)`

Area of a right trapezoid is:
`A3=((a+b)h)/(2)`, where:

a is short base

b is long base

h is height

1) Expressing areas in terms of x:

Area of triangle S1:

`S1=((2x-3)(4x-6))/(2)`

`S1=4x^(2)-12x+9`

Area of rectangle S2:

`S2 = (4x-6)(3x-2)`

`S2=12x^(2)-26x+12`

Area of trapezoid S3:

`S3=((2x+3+4x+1)(2x-3))/(2)`

`S3=((6x+4)(2x-3))/(2)`

`S3=6x^(2)-5x-6`

2) a)
`S=4x^(2)-12x+9+12x^(2)-26x+12-(6x^(2)-5x-6)`

`S=4x^(2)-12x+9+12x^(2)-26x+12-6x^(2)+5x+6`

`S=10x^(2)-33x+37`

Which is the same as S = (2x-3)(5x-9)

b) For the areas to be the same:

`((3x-2+3x-2+2x-3)(4x-6))/(2)=((6x+4)(2x-3))/(2)`

`((8x-7)(4x-6))/(2)=((6x+4)(2x-3))/(2)`

`32x^(2)-48x-28x+42=12x^(2)+8x-18x-12`

`20x^(2)-66x+54=0`

Using Bhaskara to solve the second degree equation:

`\frac{66+\sqrt{(-66)^(2)-(4.20.54)} }{2(20)}`

`x_(1)=(66+6)/(40)` = 1.8

`x_(2)=(66-6)/(40)` = 1.5

For the areas of AFGC and ADEB to be equal, x has to be 1.5 or 1.8.

c) Expand a polynomial (or equation) is to multiply all the terms, remiving the parenthesis. Reduce a polynomial (or equation) is to combine terms alike,e.g.:

`S=(2x-3)(5x-9)`

`S=10x^(2)-18x-15x+27` (expand)

`S=10x^(2)-33x+27` (reduce)

d) For area of AFCG to be bigger than area of ADEB by 27:

`32x^(2)-48x-28x+42=12x^(2)+8x-18x-12+27`

`32x^(2)-48x-28x+42=12x^(2)+8x-18x+15`

`20x^(2)-66x+27=0`

Solving:

`\frac{66+\sqrt{(-66)^(2)-(4.20.27)} }{2(20)}`

`(66+46.86)/(40)`

`x_(1)=(66+46.86)/(40)=` 2.82

`x_(2)=(66-46.86)/(40)` = 0.48

According to the enunciation, x cannot be less than 1.5, then, the value of x so that area AFGC exceeds the area ADEB by 27 is 2.82