What is the area of a triangle where the height is (5a + 2)(5a+2) and the base is (4a - 1)(4a1)?

1 Answer
Jul 28, 2017

See a solution process below:

Explanation:

The formula for the area of a triangle is:

A = (hb)/2A=hb2

Where:

AA is the area of the triangle

hh is the height of the triangle from the base

bb is the length of the base

Substituting the expressions from the problem gives:

A = ((5a + 2)(4a - 1))/2A=(5a+2)(4a1)2

To multiply the two terms in the numerator you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

A = ((color(red)(5a) + color(red)(2))(color(blue)(4a) - color(blue)(1)))/2A=(5a+2)(4a1)2 becomes:

A = ((color(red)(5a) xx color(blue)(4a)) - (color(red)(5a) xx color(blue)(1)) + (color(red)(2) xx color(blue)(4a)) - (color(red)(2) xx color(blue)(1)))/2A=(5a×4a)(5a×1)+(2×4a)(2×1)2

A = (20a^2 - 5a + 8a - 2)/2A=20a25a+8a22

We can now combine like terms:

A = (20a^2 + [-5 + 8]a - 2)/2A=20a2+[5+8]a22

A = (20a^2 + 3a - 2)/2A=20a2+3a22

Or

A = (20a^2)/2 + (3a)/2 - 2/2A=20a22+3a222

A = 10a^2 + 3/2a - 1A=10a2+32a1