Question

PLEAZZZZZZZZZZ HELP ASAP

One base of the trapizoid is three times the lengh of the second base. The height of the trapizoid is 2 in. smaller than the second base. If the area of the trapizoid is 30in^2, find the lengths of the bases and the height of the trapizoid .



A student is painting an accent wall in his room where the length of the wall is 3ft^2 more than its width. The wall in his room where the length is 3ft more than its width . The wall has an area of 130ft^2. What are the length and the width, in feet.

Find two consecutive even integers whose product is 80. (there are two pairs, and only an algebraic solution will be excepted)

Answer 1

QUESTION 1

GIVEN:


`long \: base = 3 * (short \: base)`


`height = (short \: base) - 2`


`area = 30 {in}^(2)`

IMPORTANT EQUATIONS


`trapezoid \: area = (sum \: of \: bases)/(2) * height`

SOLVE:


`30 = (short + long)/(2) (short - 2)`

`30 = ((short + 3short))/(2) * (short - 2)`


`30 = (4short)/(2) (short - 2)`


`30 = 2{(short)}^(2) - 4(short)`


`{short}^(2) - 2(short) - 15 = 0`

factor:


`(short - 5)(short + 3) = 0`


`short = 3 \: and \: - 5`
Negative five is not a reasonable answer, so we focus on the positive three and say that is the length of the short base.

so, the answers:


`short \: base = 3 \: in`


`long \: base = 3 * 3 = 9 \: in`


`height = 3 - 2 = 1 \: in`

QUESTION 2

GIVEN
convert 3ft to inches.


`l = 32 + w`


`area = lw = 130 {in}^(2)`
or


`w = (130)/(l)`

substitute


`l = 32 + (130)/(l)`

solve for l:


`{l}^(2) - 32 l- 130 = 0`
QUESTION 3


`x(x + 2) = 80`

solve for X


`{x}^(2) + 2x - 80 = 0`

`(x - 10)(x + 8) = 0`


`x = 10 \: and \: - 8`

take the positive value since it's the only one that makes sense in this context.

so the two values are:


`10 \: and \: 8`