Exercise 25 Page 334

Practice makes perfect

a In Stage 1, the midsegments of the sides of the triangle shown in Stage 0 have been added. By the Triangle Midsegment Theorem, we know that the side lengths of the shaded triangle are half the side lengths of the original triangle.

Recall that the perimeter of a triangle is the sum of its side lengths. Let's calculate the perimeter of the shaded triangle. P=8+8+8 ⇒ P=24 units

b In Stage 2, three triangles were added. The sides of these triangles are the midsegments of equilateral triangles whose side lengths are 8. Therefore, by the Triangle Midsegment Theorem, the side lengths of the new triangles are all 4.

Let's find the perimeter of each of the new triangles.

4+4+4=12 Since we have three new triangles, we will multiply the above by 3 to obtain the perimeter of the three new triangles altogether. 12* 3= 36 From Part A we know that the perimeter of the shaded triangle in Stage 1 is 24. To obtain the perimeter of the shaded area in Stage 2, we add 36 and 24. P= 36+ 24 ⇒ P=60 units

c In Stage 3, nine triangles have been added. The sides of these new triangles are the midsegments of equilateral triangles whose side lengths are 4. Therefore, by the Triangle Midsegment Theorem, the side lenghts of the new triangles are all 2. Let's calculate the perimeter of one of these triangles.

2+2+2=6

Since we have nine new triangles, we will multiply the above by 9 to obtain the perimeter of the nine new triangles altogether. 6* 9= 54 From Part B we know that the perimeter of the shaded area in Stage 2 is 60. To obtain the perimeter of the shaded area in Stage 3, we add 54 and 60. P= 54+ 60 ⇒ P=114 units

Subchapter links

Communicate Your Answer p.329

Monitoring Progress p.330-332

Exercise 25 Page 334

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