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vekshin1
4 years ago
6

Triangles △ABC and △DFG are similar. The lengths of the two corresponding sides are 1.4m , and 56 cm. What is the ratio of the p

erimeters of these triangles ?
Mathematics
2 answers:
Marina CMI [18]4 years ago
6 0

Answer:

5:2

Step-by-step explanation:

We have been given that triangles △ABC and △DFG are similar. The lengths of the two corresponding sides are 1.4 m and 56 cm.

Since both triangles are similar, therefore all corresponding sides will have same proportion.

Let us find proportion of corresponding sides of both triangles.

1 meter = 100 centimeter

1.4 meter = 1.4* 100 centimeters = 140 centimeters.

\frac{\text{Side of triangle ABC}}{\text{Side of triangle DFG}}=\frac{140}{56}

\frac{\text{Side of triangle ABC}}{\text{Side of triangle DFG}}=\frac{5}{2}

The ratio of sides of △ABC to sides of△DFG is 5:2.

Since perimeter of a triangle is sum of lengths of three sides of the triangle and all sides of both triangle have the ratio 5:2, therefore, their perimeters will be in same ratio, that is 5:2.

Feliz [49]4 years ago
3 0
<h3>Answer:</h3>

5/2

<h3>Step-by-step explanation:</h3>

The ratio of perimeters is the same as the ratio of corresponding sides:

... (140 cm)/(56 cm) = 5/2

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\begin{gathered} cos^2\theta=sin^2\theta-1 \\ Note\text{ that} \\ sin^2\theta+cos^2\theta=1 \\ cos^2\theta=1-sin^2\theta \\ Therefore \\ cos^2\theta sin^2\theta-1,NOT\text{ }IDENTITIES \end{gathered}

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\begin{gathered} sin\theta=\frac{1}{csc\theta} \\ Note\text{ that} \\ csc\theta=\frac{1}{sin\theta} \\ sin\theta\times csc\theta=1 \\ sin\theta=\frac{1}{csc\theta} \\ Therefore \\ sin\theta=\frac{1}{csc\theta},is\text{ an identities} \end{gathered}

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\begin{gathered} sec\theta=\frac{1}{cot\theta} \\ note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ tan\theta cot\theta=1 \\ tan\theta=\frac{1}{cot\theta} \\ Therefore, \\ sec\theta\ne\frac{1}{cot\theta},NOT\text{ IDENTITY} \end{gathered}

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