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Semmy [17]
3 years ago
6

Which pair of triangles can be proved congruent by the SAS postulate?

Mathematics
1 answer:
Olenka [21]3 years ago
5 0
C. â–łADE and â–łEBA

Let's look at the available options and see what will fit SAS.

A. â–łABX and â–łEDX
* It's true that the above 2 triangles are congruent. But let's see if we can somehow make SAS fit. We know that AB and DE are congruent, but demonstrating that either angles ABX and EDX being congruent, or angles BAX and DEX being congruent is rather difficult with the information given. So let's hold off on this option and see if something easier to demonstrate occurs later.

B. â–łACD and â–łADE
* These 2 triangles are not congruent, so let's not even bother.

C. â–łADE and â–łEBA
* These 2 triangles are congruent and we already know that AB and DE are congruent. Also AE is congruent to EA, so let's look at the angles between the 2 pairs of congruent sides which would be DEA and BAE. Those two angles are also congruent since we know that the triangle ACE is an Isosceles triangle since sides CA and CE are congruent. So for triangles â–łADE and â–łEBA, we have AE self congruent to AE, Angles DAE and BEA congruent to each other, and finally, sides AB and DE congruent to each other. And that's exactly what we need to claim that triangles ADE and EBA to be congruent via the SAS postulate.
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Using the side of a building as one side and fencing for the other three sides, a rectangular garden will be constructed. Given
scZoUnD [109]

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The dimensions that would create the garden of maximum area are 30 feet and 15 feet

The maximum area is 450 feet²

Step-by-step explanation:

The garden is fencing for three sides only, That means the length of the fencing is equal to the sum of the length of the three sides

Assume that garden is y feet long and x feet wide and the fencing will cover on side of y feet and two sides of x feet

∵ The sum of the length of the 3 sides = y + x + x

∴ The sum of the length of the 3 sides = y + 2x

- The length of the fencing is equal to the sum of the sides

∵ The fencing is 60 feet

- Equate y + 2x by 60

∴ y + 2x = 60

- Find y in terms of x by subtracting 2x from both sides

∴ y = 60 - 2x

<em>To find the dimensions which make the maximum area, find the area of the garden, then substitute y by x, and differentiate it with respect to x, then equate the differentiation by 0 to find the value of x, and substitute this value in the equation of y to find y and in the equation of the area to find the maximum area</em>

∵ The formula of the area of a rectangle is A = l × w

∵ l = y and w = x

∴ A = x y

- Substitute the value of y above in A

∵ A = x(60 - 2x)

- Multiply bracket by x

∴ A = 60x - 2x²

Now differentiate x with respect to x

∵ A' = 60(1) - 2(2)x

∴ A' = 60 - 4x

- Equate A' by 0 to find x

∴ 0 = 60 - 4x

- Add 4x to both sides

∴ 4x = 60

- Divide both sides by 4

∴ x = 15

- Substitute the value of x in the equation of y to find it

∵ y = 60 - 2(15)

∴ y = 30

The dimensions that would create the garden of maximum area are 30 feet and 15 feet

To find the maximum area substitute x by 15 in the equation of the area

∵ A = 60(15) - 2(15)²

∴ A = 900 - 450

∴ A = 450

The maximum area is 450 feet²

5 0
3 years ago
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