The base of a triangle is 6cm greater than twice the height. THe area is 28cm^. Find the height and width.?

Question

Can anyone show me how to figure this out getting the answer Height 4cm, base 14 cm? Using the equation 1/2(base)(height). These questions confuse me. Thank you

Answer 1

Given:
Area of triangle = 28 cm^2
Let h = height of the triangle
The base (b) is 6cm greater than twice the height, meaning
base (b) = 2h + 6 cm or 2h + 6 (to simplify)

Formula for area of triangle
A = 1/2 times b times h
A = 1/2 bh
Substitute all the values in the formula
A = 1/2 bh
28= 1/2 ( 2h + 6 ) (h)
To get rid of 1/2, multiply both sides by 2
(2)(28) = 2 (1/2) (2h + 6) (h)
56 = (2h + 6) h
Multiply right sides by distributive
56 = 2h^2 + 6h
Make the equation in the form ax^2 + bx + c = 0, by adding -56 to both sides
56 -56 = 2h^2 + 6h - 56
0 = 2h^2 + 6h - 56
2h^2 + 6h - 56 = 0
You can solve h by factoring or using the quadratic formula
By factoring or completing the square

2(h^2 + 3h - 28) = 0
2( h + 7) (h - 4) = 0
Divide both sides by 2 to get
(h + 7) (h - 4) = 0
h has two values
h + 7 = 0
h = -7
and h - 4 = 0
h = 4

If h = -7
A = 1/2 b h
28 = 1/2 ( 2 (-7) + 6) (-7)
28 = 1/2 (-14 + 6) (-7)
28 =1/2(-8)(-7)
28 = 1/2 (56)
28 = 28

If h = 4
A = 1/2 b h
28 = 1/2(2(4) + 6) (4)
28 = 1/2 (8 + 6) (4)
28 = 1/2 (14) (4)
28 = 1/2 (56)
28 = 28

Back to our problem
The height is 4 cm or - 7cm
The width is
2(4) + 6 = 8 + 6 = 14 cm or
2(-7) + 6 = -14 + 6 = -8 cm

The pair (b,h) = 14 cm and 4 cm or
- 8 cm and - 7 cm
The first triangle is found in the first quadrant, while the second in the third quadrant.

Answer 2

Now, let h denotes the height of the triangle and x denotes the base of the triangle.

Then, we will have x = 6 + 2h

So, area = 28
=> (1/2)(x)(h) = 28
=> (6+2h)*(h) = 56
=> 2h^2 + 6h = 56
=> h^2 + 3h = 28
=> h^2 + 3h - 28 = 0
=> (h+7) * (h-4) = 0
=> h = -7 or h = 4

Since the height cannot be negative, h = 4 cm

Hence, base = 2(4)+6 = 14 cm

:)

Answer 3

lets call the proper 'h' and the bottom 'b' formula for the realm of a triangle: section = base x proper -------------------- 2 Base is 6 + 2h proper is h h(6 + 2h) ------------- = 28cm^2 2 Multiply out the brackets 6h + 2h^2 -------------- = 28 2 3h + h^2 = 28 let's make this a quadratic equation h^2 + 3h - 28 = 0 (h + 7)(h - 4) h + 7 = 0 OR h - 4 = 0 h = -7 h = 4 proper should be 4 cm because it really is not any longer possible to have a adverse huge style because the proper. so that you ought to discover the bottom you do 4 circumstances 2 that's 8, then upload 6 that's 14cm. proper is 4cm Base is 14cm examine it in case you do not trust me... 4 circumstances 14 is fifty six fifty six divided by technique of two is 28!!!

Answer 4

Form equations:

Let x be base
Let h be height

x = 6 + 2h

Also we know that Area = 1/2 * x * h

28 = 1/2 * (6 + 2h) * h
56 = h(6+2h)
56 = 6h + 2h^2
28 = 3h + h^2
0 = h^2 + 3h - 28
Now we factorise:
0 = (h - 4)(h + 7)

Therefore h = 4 or h = -7

Height can only be positive so height is 4 cm

Substitute into previous equation:

x = 6 + 2h
x = 6 + 2(4)
x = 6 + 8
x = 14

So we have height is 4 and Base is 14

Answer 5

OK, b= 6cm + 2h , a = 28 cm

a = b x 1/2 h = 28 , (6 + 2h)x 1/2 h = 28

3h + h² =28 , h²+3h -28 =0 , (h+7)(h-4) = 0....h = 4

h= -7 is ignored since height can't be negative.

b = 6+ 8 = 14 cm

Good Luck.

Answer 6

b = 2h + 6
Area = (1/2).b.h
28 = (1/2).(2h + 6).h
56 = 2h² + 6h
h² + 3h - 28 = 0
(h + 7).(h - 4) = 0
h = 4 cm (taking +ve value for h)
b = 2 x 4 + 6 = 14 cm
Dimensions are :-
base = 14 cm
height = 4 cm

Answer 7

Area = b*h/2 = 28

2h + 6 = b

So (2h + 6)*h/2 = 28

2h^2 + 6h - 56 = 0

h^2 + 3h - 28 = 0
(h + 7)(h - 4) = 0

h = 4 (h cannot = -7)
b = 2*4+6 = 14