<span><span><span>Let the original number be "x".
(x-10)/4 = 2
x-10 = 8
x = 18 (original number of pieces of candy.)</span></span></span>
Answer: "The base of the ∆ is 8 units."
"The height of the ∆ is 4 units."
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Explanation:
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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 " ;
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So; Taking the formula for the Area, "A" , of a triangle:
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⇒ A = (1/2) × Base × Height ;
Let us plug in our given and known/calculated values into the formula:
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⇒ 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 ;
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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 " .
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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 " .
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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 .
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So, 1) when: "a + 8 = 0 ", what does "a" equal?
Add " 8 " to both sides:
⇒ a + 8 − 8 = 0 − 8 ;
to get: "a = - 8 " .
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So, 2) when: "a - 4 = 0 ", what does "a" equal?
Add " 4 " to both sides:
⇒ a - 4 + 4 = 0 + 4 ;
to get: "a = 4 " .
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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 " .}.
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" The base of the ∆ is 8 units. "
" The height of the ∆ is 4 units. "
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