Question

The area of a triangle is `2x^2-11x+15`. If the triangle's base is `2x-5`, what is its height?

Answer 1

`"height" = color(red)(2)(x - 3)`

Explanation:

Recall;

Area of triangle `= 1/2 "base" xx "height"`

Area of triangle `= 2x^2 - 11x + 15`

base `=2x - 5`

Plugging in the given;

`2x^2 - 11x + 15 = 1/2 (2x - 5) xx "height"`

`2x^2 - 11x + 15 = (2x - 5)/2 xx "height"`

`(2x^2 - 11x + 15)/1 = (2x - 5)/2 xx "height"`

`2(2x^2 - 11x + 15) = (2x - 5) xx "height"`

`4x^2 - 22x + 30 = (2x - 5) xx "height"`

`(4x^2 - 22x + 30)/(2x - 5) = "height"`

Resolving the quadratic equation;

`(4x^2 - 22x + 30)`

Simplifying;

`(4x^2)/2 - (22x)/2 + 30/2`

`2x^2 - 11x + 15`

Using Factorization Method..

`6 and 5` are factors..

`2x^2 - 6x - 5x + 15`

Grouping;

`(2x^2 - 6x) (- 5x + 15)`

`2x(x - 3) - 5(x - 3)`

`(x - 3) (2x - 5)`

Therefore;

`(4x^2 - 22x + 30)/(2x - 5) = "height"`

`color(white)(xxxxx)darr`

`(color(red)(2)(x - 3) (2x - 5))/(2x - 5) = "height"`

`(color(red)(2)(x - 3) cancel(2x - 5))/cancel(2x - 5) = "height"`

`color(red)(2)(x - 3) = "height"`

Answer 2

`color(maroon)("Height of the triangle " h = 2 (x - 3)`

Explanation:

`"Area of triangle = (1/2) * (base * height)"`

`A_t = (1/2) b h`

`"Given : " A_t = 2x^2 - 11x + 15, b = (2x - 5)`

`h = ((2 * A_t) / b)`

`h = (2 * (2x^2 - 11x + 15)) / (2x - 5)`

`h = (2 * cancel(2x - 5) * (x - 3)) / cancel(2x - 5)`

`h = 2 * (x - 3)`