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Paha777 [63]
1 year ago
14

Challenge Yossi used to have a square garage with 246 ft of floor space. recently built an addition to it. The garage is still a

​ square, but now it has​ 50% more floor space. What was the length of one side of the garage​ originally? What is the length of one side of the garage​ now? What was the percent increase in the length of one​ side?
Mathematics
1 answer:
irina [24]1 year ago
4 0

Considering the perimeter of a square, it is found that:

  • The length of one side of the garage​ originally was of 61.5 ft.
  • The length of one side of the garage​ now is of 92.25 ft.
  • The percent increase in the length of one​ side was of 50%.

<h3>What is the perimeter of a square?</h3>

The perimeter of a square of side length s is given by four times the length, that is:

P = 4s.

Before the change, the perimeter was of 246 ft, hence:

4s = 246

s = 246/4

s = 61.5.

The length of one side of the garage​ originally was of 61.5 ft.

After that, the perimeter increased by 50%, hence:

P = 246 x 1.5 = 369.

4s = 369

s = 369/4

s = 92.25.

The length of one side of the garage​ now is of 92.25 ft.

The percent increase is the increase divided by the initial value, hence:

(92.25 - 61.5)/61.5 = 50%.

The percent increase in the length of one​ side was of 50%.

More can be learned about the perimeter of a square at brainly.com/question/10489198
#SPJ1

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Lubov Fominskaja [6]

Answer:

See Below.

Step-by-step explanation:

We are given the isosceles triangle ΔABC. By the definition of isosceles triangles, this means that ∠ABC = ∠ACB.

Segments BO and CO bisects ∠ABC and ∠ACB.

And we want to prove that ΔBOC is an isosceles triangle.

Since BO and CO are the angle bisectors of ∠ABC and ∠ACB, respectively, it means that ∠ABO = ∠CBO and ∠ACO = ∠BCO.

And since ∠ABC = ∠ACB, this implies that:

∠ABO = ∠CBO =∠ACO = ∠BCO.

This is shown in the figure as each angle having only one tick mark, meaning that they are congruent.

So, we know that:

\angle ABC=\angle ACB

∠ABC is the sum of the angles ∠ABO and ∠CBO. Likewise, ∠ACB is the sum of the angles ∠ACO and ∠BCO. Hence:

\angle ABO+\angle CBO =\angle ACO+\angle BCO

Since ∠ABO =∠ACO, by substitution:

\angle ABO+\angle CBO =\angle ABO+\angle BCO

Subtracting ∠ABO from both sides produces:

\angle CBO=\angle BCO

So, we've proven that the two angles are congruent, thereby proving that ΔBOC is indeed an isosceles triangle.

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