The base of a right triangle is 13 less than 2 times the height. The area of the triangle is 12 square cm. What is the height of the triangle in cm? *

The height of the triangle is 8 cm

Step-by-step explanation:

First we need to write down the question in a mathematical format and then solve.

let b be the base of the triangle and h be the height of the triangle

"The base of a right triangle is 13 less than 2 times the height" can be written as

b = 2h - 13 -----------------------------------------------------(1)

"The area of the triangle is 12 square cm" can be written as;

area of the triangle = 12 cm²

but area of a triangle = ×b × h

This implies that

×b × h = 12 cm²

bh = 24 cm² --------------------------------------------------(2)

substitute equation (1) in equation (2)

(2h - 13)h = 24

2h² - 13h = 24

2h² - 13h - 24 = 0 ----------------------------------------------(3)

The above is now a quadratic equation, so we will solve by completing the square method.

Find two numbers such that its product will give -48 and its sum will give -13

The two numbers are -16 and 3

Replace -13h by  (-16h +  3h) in equation (3)

2h² - 13h - 24 = 0

2h² - 16h + 3h - 24 = 0

2h(h - 8) + 3(h-8) = 0

(h-8)(2h+3) = 0

either h-8 = 0

           h = 8

OR

2h + 3 = 0

2h = -3

h =-3/8

There is no such thing as negative length, therefore h= 8 is the answer

The height of the triangle is 8 cm

3. parallelogram

step-by-step explanation:

we turn each of our sides into direction vectors

ab=b-a=(3 - 7, 1 - 0) = (-4, 1)

bc=c-b=(1,6)

cd=d-c=(4,-1)

da=a-d=(-1,-6)

we see how the opposite sides have negated direction vectors; this is characteristic of a parallelogram.   we have congruent sides ab=cd and bc=da with unequal lengths because 1^2+6^2 is different from (-4)^2+1^2.   so it's not a square or a rhombus.  

we check the dot product of direction vectors of adjacent sides.   we have

ab dot bc = -4(1) + 1(6) = 2

that's not zero so the sides are not perpendicular.   we've ruled out rectangle.

the direction vectors indicate two pairs of congruent parallel sides, so we choose parallelogram over trapezoid.