The area of a triangle is 16 square units. the base of the triangle is four more than the height of the triangle. what is the base and height of the triangle?

"The base of the ∆ is 8 units."

"The height of the ∆ is 4 units."



Explanation:



We are asked to find: 1) the base, "b" , and 2) the height, "h", of the ∆ .

Let the height, "h", of the ∆ , equal: "a" units .

We want to find the value for "a" {the height.}.

So; According to the question ,

The base, "b", is 4 units more than: the height, "h", of the ∆ ;

So; the base, "b" = " (a + 4)" units .

If we solve for "a" , the "height",

we can solve for the the base, "b" , which is: "(a + 4)" .

Now; note the formula for the Area, "A", of a triangle:

⇒ Area of ∆ , "A" = ( 1/2 ) × Base × Height ;

that is:

⇒ A = (1/2) × Base × Height .

We are given the Area, "A", of the triangle:

⇒ which is: " 16 unit² " ;



So; Taking the formula for the Area, "A" , of a triangle:



⇒ A = (1/2) × Base × Height ;

Let us plug in our given and known/calculated values into the formula:



⇒ 16 units² = (1/2) × Base × Height ;

⇒ 16 unit² = ( 1/2 ) × ( a + 4 ) × ( a ) unit² ;

So; we multiply both side of the equation by: " 2 " ;

to cancel out the "fraction" on the "right-hand side" of the equation:

as follows:

⇒ 2 × ( 16 unit² ) = 2 × ( 1/2 ) × ( a + 4 ) × ( a ) unit² ;

We see right-hand side that: " 2 * (1/2) = 1 " .

⇒ 32 units² = 1 × (a + 4) × a units² ;

Note: On the right-hand side, we eliminate the " 1 " ;

⇒ since any value(s), multiplied by " 1 " ,

result in those exact same value(s).

⇒ 32 units² = (a + 4) × a units² ;

⇔ (a + 4) × a units² = 32 units² ;

⇔ a (a + 4) units² = 32 units² ;



To simplify the left-hand side of the equation,

use the distributive property of multiplication;

So we use: a(b + c) = ab + ac ;

⇒ So ; a(a + 4) = a*a + a*4 = a² + 4a.

⇒ We take our equation:

a (a + 4) units² = 32 units² ;

and replace the: " a (a + 4) " ;

with: " (a² + 4a) " ;

to get: " (a² + 4a) = 32 " .

Rewrite as:

" a² + 4a = 32 " ;

Now, subtract " 32 " from BOTH SIDES of the equation:

" a² + 4a − 32 = 32 − 32 " ;

TO GET:

" a² + 4a − 32 = 0 " .



Now, let's factor the left hand side of the equation; and find the "zeros", or solutions for "a" , when the left-hand side of the equation is equal to "0" ;

⇒ " a² + 4a − 32 ";

To factor:

What factors of "-32" add up to "4" ?

-8 , 4 ? -8 + 4 = ? 4 ? ; -8 + 4 = - 4 ; -4 ≠ 4. So, "no".

8, -4 ? 8 + (- 4) = ? 4 ? ; 8 − 4 = 4 ? Yes!

So, rewrite:

" a² + 4a − 32 = 0 " .



as: " (a + 8) (a − 4) = 0 " ,

Any value, multiplied by "0" , is equal to "0" .

There are two (2) values being multiplied, that are equal to "0".

So, the equation holds true when:

1) (a + 8) = 0 ; or:

2) (a − 4) = 0 .



So, 1) when: "a + 8 = 0 ", what does "a" equal?

Add " 8 " to both sides:

⇒ a + 8 − 8 = 0 − 8 ;

to get: "a = - 8 " .



So, 2) when: "a - 4 = 0 ", what does "a" equal?

Add " 4 " to both sides:

⇒ a - 4 + 4 = 0 + 4 ;

to get: "a = 4 " .



So, the equation holds true when: "a = 4" or "a = -8" .

Now, we know that "a" refers to the height. The "height" cannot be a "negative" number ⇒ So we use " a = 4 " . The height is: " 4 units."

The base is "4 units more than the height" ; so the base is:

⇒ { "4 units" + "4 units" = " 8 units " .}.



" The base of the ∆ is 8 units. "

" The height of the ∆ is 4 units. "