Solve algebra problem-A trangle has its base and altitude equal Find the base if an increase of two feet in?

Well,

we have :

(1/2)a^2 = S

and

(1/2)(a+2)(a+3) = S + 8

so :

(1/2)(a+2)(a+3) = (1/2)a^2 + 8

(a+2)(a+3) = a^2 + 16

a^2 + 5a + 6 = a^2 + 16

5a = 10

finally :

the dimensions of the initial triangle where : base = height = a where :

a = 2

hope it' l help !!

Let b be the base and h be the height. Initially:

b = h => Area = (bh)/2 = (b^2)/2 or (h^2)/2

After:

(b + 2)(h + 3)/2 = (bh)/2 + 8

(3b + 2h + 6)/2 = 8

3b + 2h = 10

5b = 10 => b = 2 ft

let b = base

let h = altitude

area of original TRIANGLE: A = (1/2)*b*h

but, h = b

so, A = (b^2)/2

AFTER INCREASES...

base = (b + 2)

altitude = (h + 3) = (b + 3)

Area (of increased triangle) = (1/2)(b + 2)(b + 3) = (b^2 + 5b + 6)/2 = [(b^2)/2] + 8

solve:

clear the denominator...

b^2 + 5b + 6 = b^2 + 16

5b = 10

b = 2

so, h = 2

check

area of original triangle = 2 units^2

area of increased triangle = (1/2)(4)(5) = 10 units^2

cat

area = x^2/2

base --> x + 2

heigth ---> x + 3

new area --> (x + 2)(x + 3)/2 = x^2 / 2 + 8

multiply all terms by 2

x^2 + 3x + 2x + 6 = x^2 + 16

5x = 10

x = 2

smaller area 2x2/2 = 2

(2 + 3)(2 + 2)/2 = 10 = 2 + 8

checked

A1=bh/2

b=h

A1=b^2/2

A2=(b+2)(b+3)/2

A2=(b^2+3b+2b+6)/2

A2=(b^2+5b+6)/2

A2-A1=8

8=(b^2+5b+6-b^2)/2

5b+6=16

5b=10

b=2

check

A1=bh/2 when h=b

A1=b^2/2 if b=2

A1=2

So A2 should equal 10 because it should be A1+8 or 10

A2=(b+2)(b+3)/2

A2=(2+2)(2+3)/2

A2=4(5)/2=20/2

A2=10

correct :)