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DedPeter [7]
3 years ago
10

Imagine you had 40 one-metre sections of fencing.

Mathematics
1 answer:
Darya [45]3 years ago
8 0

Answer: If I’m not mistaken, the largest area is 75 m2 and the smallest is 2 m2

Step-by-step explanation:

For both : since they are asking for the area, you should calculate the perimeter.

So for the largest area, since you know that you have 40 fences, and each are 1 meter long, and they’re also asking for the largest rectangular area, this means that only the opposite sides will have the same length, so in order to divide the rectangle with the 40 one-meter fences, the biggest sides have to be 15 meters long (so for both : 15x2=30 meters). Then you deduct from 40, 30 : 40-30 = 10 which then you can devide by 2 and find 5 + 5 and that means that the 2 smallest sides will be 5 meters long each. Then finally to calculate the perimeter you should do the biggest side multiplied by the smallest side : 15 x 5 and you find 75 m2 (squaremeters), if I’m not mistaken.

And for the smallest area you take the smallest possibilities knowing that only the opposite sides should have the same length and not all of them, so you do 1 x 2 and you find 2 m2.

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Answer:

Δ PQT ~ Δ QRS  .....{S-S-S test for similarity}...Proof is below.

Step-by-step explanation:

Given:

In Δ PQT

PQ = 30 ft

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In Δ QRS

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To Prove:

Δ PQT ~ Δ QRS

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First we consider  the ratio of the sides

\frac{PQ}{QR}=\frac{30}{15} = \frac{2}{1}            ..............( 1 )

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Now in Δ PQT  and Δ QRS we have

\frac{PQ}{QR}=\frac{QT}{RS} = \frac{TP}{SQ}

Which are corresponding sides of a similar triangle in proportion.

∴ Δ PQT ~ Δ QRS  .....{S-S-S test for similarity}...Proved

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<em>Hi there!</em>

<em>~</em>

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