Mathematics, 09.07.2019 11:30 liubitsaghn2145

What theorem can be used to prove that the two triangles are congruent?

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To prove the congruence between two triangles there are several postulates, that can be applied depending on the congruent parts of the triangles.

The postulates are:

SAS (Side-Angle-Side): This postulate states that if two triangles have two corresponding sides congruent and one corresponding angle congruent, we can conclude that those triangles are completely congruent.ASA (Angle-Side-Angle): This postulate states that if two triangles have two corresponding angles congruent and one corresponding side congruent, then those triangles are congruent.SSS (Side-Side-Side): This postulate states that if two triangles have all three corresponding sides congruent, then triangles are congruent.AAS (Angle-Angle-Side): If two pairs of corresponding angles are congruent and one pair of corresponding sides is congruent, then both triangle are congruent each other.HL (Hypothenuse-Leg): this postulate is applied when we have right angles. It states that two right triangles are congruent if they have one pair of corresponding leg congruent, and the hypothenuses are also congruent. We could conclude that those right triangles are congruent.

From these postulate, we can demonstrate the congruence between two triangles. Then, from that congruence, we deduct the congruence between each corresponding parts of those triangles. For example, if we demonstrate the congruence by AAS, we deduct that the third angle is also congruent, and the other two sides are also congruent.

Answer from: Quest

10 * 563.67 = 5636.7

Answer from: Quest

$w\geq 14$

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

$2l+2w\geq 40$

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

$2\times 6+2w\geq 40$

now, we can solve for w

$12+2w\geq 40$

subtract both sides by 12

$12+2w-12\geq 40-12$

$2w\geq 28$

divide both sides by 2

and we get

$w\geq 14$

so,

the smallest possible measurement of the width is 14 feet

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