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2 years ago

Indicate the method you would use to prove the two triangles congruent. If no method applies, enter “none”.

Mathematics

1 answer:

mr Goodwill [35]2 years ago

8 0

We have to determine the congruence criteria which would prove that the given triangles are congruent.

Consider the triangles RSV and VTU

VR = VU (Given)

$\angle RSV = \angle VTU$ (Given)

$\angle SVR = \angle TVU$ (Given)

Hence, the triangles RSV and VTU are congruent by AAS congruence criteria.

As, if two angles and a non-included side of one triangle are congruent to the corresponding parts of another triangle, then the triangles are congruent by AAS congruence criteria.

So, option 4 is the correct answer.

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What are two numbers that multiply for 54 and add to give you -3

cupoosta [38]

Answer:

9 and -6 multiply to -54 and add to 3

Step-by-step explanation:

when listing the factors of 54, we can find a pair that add to 3

1, 2, 3, 6, 9, 18, 27 and 54

9 and -6 is the pair.

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2 years ago

A farmer has a field in the shape of a triangle. The farmer has asked the manufacturing class at your school to build a metal fe

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Answer:

Fencing required = 1586 m

Step-by-step explanation:

The given statements can be thought of a triangle $\triangle ABC$ as shown in the diagram attached.

A be the 1st vertex, B be the 2nd vertex and C be the 3rd vertex.

Distance between 1st and 2nd vertex, AB = 435 m

Distance between 2nd and 3rd vertex, AC = 656 m

$\angle A =49^\circ$

To find:

Fencing required for the triangular field.

Solution:

Here, we know two sides of a triangle and the angle between them.

To find the fencing or perimeter of the triangle, we need the third side.

Let us use Cosine Rule to find the third side.

Formula for cosine rule:

$cos A = \dfrac{b^{2}+c^{2}-a^{2}}{2bc}$

Where

a is the side opposite to $\angle A$

b is the side opposite to $\angle B$

c is the side opposite to $\angle C$

$\Rightarrow cos 49^\circ = \dfrac{656^{2}+435^{2}-BC^{2}}{2\times 435\times 656}\\\Rightarrow BC^2 = 430336+189225-2 (435)(656)cos49^\circ\\\Rightarrow BC^2 = 430336+189225-570720\times cos49^\circ\\\Rightarrow BC^2 =619561-570720\times cos49^\circ\\\Rightarrow BC \approx 495\ m$

Perimeter of the triangle = Sum of three sides = AB + BC + AC

Perimeter of the triangle = 435 + 495 + 656 = 1586 m

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Answer:

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Step-by-step explanation:

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