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jenyasd209 [6]
3 years ago
7

The area of two similar triangles are 50dm^2 and 32dm^2. The sum of their perimeters is 117dm. What us the perimeters of each of

these triangle?
Mathematics
1 answer:
Kay [80]3 years ago
5 0

Answer:

Perimeter of one triangle is 65 dm

Perimeter of other triangle is 52 dm

Step-by-step explanation:

Please remember the concept

If sides are in the ratio of a:b

Then the area in the ratio of a^{2} :b^{2}

It is given sum of their perimeter is 117.

Let the small triangle has perimeter as x.

So, perimeter of big triangle is 117-x.

So, we can set up equation as

\frac{50}{32} =\frac{x^{2} }{(117-x)^{2} }

Cross multiply

50(117-x)^2 =32x^2

Expand the left side

50(13689 -234x+x^{2} )=32x^{2}

Distribute the left side

684450-11700x+50x^{2}=32x^{2}

Subtract both sides 32x^{2} and rewrite it 18x^{2} -11700x+684450=0

Solve this quadratic for x.

Divide both sides of the equation by 18 to simplify.

x^{2}-650 x+38025=0

Now, if possible let's factor

Find two integers whose multiplication is 38025 but adds to -650.

-65 and -585 works.

So, we can rewrite it as

(x -65)(x-585) =0

Solve them using zero product property

x=65, x=585

So, x=65 works here.

So, perimeter of one triangle is 65 dm

Perimeter of other triangle is 117-65= 52 dm

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The answer is 8 quarters, 16 nickles, and 32 dimes

Step-by-step explanation:

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  3. 10d+40d+25d/4 = 600 (Put all of that over 4, not first 25d)
  4. 75d= 600 x 4
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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

first simplify all the ratios.

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