Egypt Preparatory 3 Mathematics, Week 7: The perimeters of similar polygons, and finding unknown sides and angles

Similarity of polygons

Week 7 · Similarity of polygons: conditions, similarity ratio, special shapes and perimeters · 3 lessons

The perimeters of similar polygons, and finding unknown sides and angles

Mathematics Prep 3 Term 1, Unit 3, Lesson 1

GoalUse the fact that the ratio of the perimeters equals k to find perimeters and sides, and use equal angles to find missing angles.

New words

perimeter · the total length around a polygonscale · the ratio of a drawing or model to the real objectenlargement · a similar figure bigger than the original (k < 1 if the original is first)sum of the angles · for an n-sided polygon, (n − 2) × 180°

Explanation

If ABCD ~ EFGH with ratio k, every side of the first is k times the matching side of the second. Adding all the sides, the perimeter of the first is k times the perimeter of the second: perimeter of ABCD ÷ perimeter of EFGH = k. So the ratio of the perimeters equals the similarity ratio. This lets us find a perimeter from one pair of sides, or find k from the two perimeters. To find an unknown side, write a proportion with the matching sides, for example AB ÷ EF = BC ÷ FG, and solve. To find an unknown angle use equal corresponding angles, or the angle sum of the polygon: (n − 2) × 180°. Always check that the first polygon is on top in every ratio.

Photo	Length	Width	Perimeter
original	20 cm	15 cm	70 cm
enlarged	32 cm	24 cm	112 cm
ratio	32 ÷ 20 = 8/5	24 ÷ 15 = 8/5	112 ÷ 70 = 8/5

Worked examples

A photo is 20 cm long and 15 cm wide. It is enlarged so that the long side is 32 cm. Find the new width and the two perimeters. The ratio is 32 ÷ 20 = 8/5. New width = 15 × 8/5 = 24 cm. Perimeters: 2 × (20 + 15) = 70 cm and 2 × (32 + 24) = 112 cm. Check: 112 ÷ 70 = 8/5 ✓.

ABCD ~ EFGH, AB = 12 cm, EF = 18 cm, and the perimeter of ABCD is 40 cm. Find the perimeter of EFGH. k = 12 ÷ 18 = 2/3. Perimeter of ABCD ÷ perimeter of EFGH = 2/3, so the perimeter of EFGH = 40 × 3/2 = 60 cm.

A pentagon has sides 3, 5, 6, 8 and 10 cm. A similar pentagon has perimeter 80 cm. Find its sides. The first perimeter is 3 + 5 + 6 + 8 + 10 = 32 cm. k = 32 ÷ 80 = 2/5, so each side of the big pentagon is 5/2 times the small one: 7.5, 12.5, 15, 20, 25 cm. Check: the sum is 80 ✓.

Class activity

Model pitch. A football pitch is 100 m long and 64 m wide. Draw a similar rectangle 25 cm long on paper. How wide must it be? What is the scale (cm of drawing to metres of pitch)?

Practice

ABCDE ~ FGHIJ, AB = 8 cm, FG = 12 cm and the perimeter of ABCDE is 35 cm. Find the perimeter of FGHIJ.

The pitch of 100 m by 64 m is drawn as a similar rectangle 25 cm long. Find the width.

Two similar polygons have perimeters 45 cm and 30 cm. Find k, and the side in the second polygon that matches a 12 cm side of the first.

ABCD ~ EFGH with ∠A = 75°, ∠B = 105°, ∠C = 100°. Find ∠H.

Two rhombuses have perimeters 24 cm and 40 cm. Find their sides and k.