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kaheart [24]
4 years ago
10

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 h

eight of the triangle?
Mathematics
2 answers:
forsale [732]4 years ago
5 0
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
kkurt [141]4 years ago
4 0

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.

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