Question

The area of a triangular block is 64 square inches. If the base of the triangle is twice the height, how long are the base and height of the triangle?

Answer 1

X=long of the base
y=long of ht height.

Area (triangle)=(b*h)/2

we suggest this system equations:
xy/2=64
x=2y
we solve this system by substitution method:
(2y)y/2=64
y²=64
y=⁺₋√64
Then, we have two solutions:
y₁=-8; invalid solution
y₂=8;    ⇒x=2y=2*8=16:

solution: 
base=16 in
height=8 in

Answer 2

Answer:

Area of the triangle(A) is given by:

$A = \frac{1}{2}bh$             ....[1]

where,

b is the base and h is the height of the triangle.

As per the statement:

The area of a triangular block is 64 square inches.

⇒$A= 64 in^2$

It is given that:  If the base of the triangle is twice the height

⇒$b = 2h$          .....[2]

Substitute these in [1] we have;

$64 = \frac{1}{2} \cdot (2h) \cdot h$

⇒$64 = h^2$

or

$h^2 = 64$

⇒$h = \sqrt{64} = 8$ inches.

Substitute h =8 in [2] we have;

$b = 2(8) = 16$ in.

therefore, the base and height of the triangle are:  16 inches and 8 inches.