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The ratio of the corresponding sides is the same for each pair.
Let the perimeter of the triangle MNO be x.Since the perimeters of similar polygons have the same ratio as any two corresponding sides, we have13/26 = 44/x (cross multiply)13x =1,144 (divide both sides by 13)x = 88Or since 13/26 = 1/2, the perimeter of the triangle MNO is twice the perimeter of the triangle HIJ, which is 88.
Their corresponding angles are equal, or the ratio of the lengths of their corresponding sides is the same.
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You need to find the perimeter of one by adding together the lengths of all its sides. The perimeter of the similar shape is the answer multiplied by the similarity ratio.
The perimeter of the larger polygon will have the same ratio to the perimeter of the smaller as the ratio of the corresponding sides. Therefore, the larger polygon will have a perimeter of 30(15/12) = 37.5, or 38 to the justified number of significant digits stated.
If two polygons are similar then the ratio of their perimeter is equal to the ratios of their corresponding sides lenghts?
There cannot be a similar polygon by itself. One polygon is similar to another if all of their corresponding angles are equal. This requires that the lengths of corresponding sides are in the same ratio: that is, if one polygon is a dilation of the other.
It is the same.
Theorem: If two similar triangles have a scalar factor a : b, then the ratio of their perimeters is a : bBy the theorem, the ratio of the perimeters of the similar triangles is 2 : 3.For rectangles, perimeter is 2*(L1 + W1). If the second rectangle's sides are scaled by a factor S, then its perimeter is 2*(S*L1 + S*W1) = S*2*(L1 + W1), or the perimeter of the first, multiplied by the same factor S.In general, if an N-sided polygon has sides {x1, x2, x3....,xN}, then its perimeter is x1 + x2 + x3 + ... + xN. If the second similar polygon (with each side (labeled y, with corresponding subscripts) scaled by S, so that y1 = S*x1, etc. The perimeter is y1 + y2 + ... + yN = S*x1 + S*x2 + ... + S*xN = S*(x1 + x2 + ... + xN ),which is the factor S, times the perimeter of the first polygon.
The perimeters of two similar polygons have the same ratio as the measure of any pair of corresponding sides. So the ratio of the measure of two corresponding sides of two similar kites with perimeter 21 and 28 respectively, is 21/28 equivalent to 3/4.
It is k times the perimeter of EFGH where k is the constant ratio of the sides of ABCD to the corresponding sides of EFGH.
n. in congruent polygons, the pairs of sides which can be superimposed on one another. In similar polygons, the ratio of the length of a side on the larger polygon to the length of its corresponding side on the smaller polygon is the same for all the sides.
It is k times the perimeter of abcd where k is the constant ratio of the sides of efgh to the corresponding sides of abcd.
It is k times the perimeter of eh where k is the constant ratio of the sides of abcd to the corresponding sides of efgh.
If two rectangles are similar, they have corresponding sides and corresponding angles. Corresponding sides must have the same ratio.
The ratio between corresponding sides or angles of similar triangles are equal