Proportion · Grade 10 · 5 units
In Proportion (Grade 10 · 5 units) you practise 8 skills: Thales’ theorem · The converse of Thales’ theorem · Derived proportions · Dividing a segment in a ratio · Intersecting chords · Tangents and secants from a point · Proportion in a right triangle · Perimeters and areas of similar triangles. Thales’ theorem: parallel lines cutting the arms of an angle cut them in proportion. In a triangle with a segment parallel to one side, the ratio of the two parts of one arm equals the ratio of the matching parts of the other arm. The converse holds too: if the segments are in proportion, the lines are parallel. Other proportions follow from a proportion, for example by adding 1 to both sides.
When two chords of a circle cross, the product of the parts of one chord equals the product of the parts of the other. When a tangent and a secant are drawn to a circle from a point outside it, the tangent squared equals the whole secant times its outer part. For two secants from the same point the products are equal. In a right triangle the altitude to the hypotenuse cuts the hypotenuse into two projections. The altitude squared equals the product of the projections, and each leg squared equals the hypotenuse times that leg’s projection. In similar triangles any two corresponding lengths are in the ratio of similarity, and the areas are in the square of that ratio. A model proof of Thales’ theorem in a triangle, by areas: draw a segment parallel to the base, cutting the two other sides. Join each end of the segment to the corner at the far end of the base. The two triangles this makes stand on the segment, and the third corner of each is on the base, which is parallel to it, so they have the same height and the same area. Now compare each of them with the small triangle at the top, which shares a height with it: the ratio of the areas is the ratio of the parts of that side, the upper part over the lower part. Both ratios compare the small triangle with two triangles of equal area, so they are equal, and the parts of the two sides are in proportion.
Tip: Say the proportion in words before you put in numbers: the small part to the big part on one side, and in the same order on the other.
What you learn
Thales’ theorem
For example: In triangle ABC the segment DE is parallel to BC . In centimetres, AD = 1, DB = 5, AE = 2 . How long is EC ?
Answer and solution
The answer: 10 cm
By Thales’ theorem AD / DB = AE / EC , that is 1/5 = 2 / EC . So EC = 5 × 2 ÷ 1 = 10 .
The converse of Thales’ theorem
For example: In triangle ABC , D is on AB and E is on AC . In centimetres, AD = 1, DB = 5, AE = 2, EC = 11 . Is DE ∥ BC ?
- Yes
- No
Answer and solution
The answer: No
Compare AD / DB = 1/5 with AE / EC = 2/11 by cross multiplying: 1 × 11 = 11 against 5 × 2 = 10 . The products differ, so the lines are not parallel. No.
Derived proportions
For example: If a / b = 2/9 , what is (a + b) / b ?
Answer and solution
The answer: 11/9
Add 1: (a + b) / b = a / b + 1 = 2/9 + 9/9 = 11/9 .
Dividing a segment in a ratio
For example: A segment 14 cm long is divided into two parts in the ratio 2 : 5 . How long is the shorter part?
Answer and solution
The answer: 4 cm
There are 2 + 5 = 7 shares, each 14 ÷ 7 = 2 . The shorter part is 2 × 2 = 4 .
Intersecting chords
For example: The chords AB and CD cross at P . In centimetres, AP = 2, PB = 18, CP = 3 . How long is PD ?
Answer and solution
The answer: 12 cm
By the theorem 2 × 18 = 3 × PD , that is 36 = 3 × PD , so PD = 36 ÷ 3 = 12 .
Tangents and secants from a point
For example: From the point P there is a tangent PT and a secant meeting the circle at A and B . In centimetres, PA = 1, PB = 9 . How long is the tangent PT ?
Answer and solution
The answer: 3 cm
By the theorem PT² = 1 × 9 = 9 , so PT = √9 = 3 .
Proportion in a right triangle
For example: In the right triangle ABC the angle ∢C is right, and CD is the altitude to the hypotenuse AB . In centimetres, AD = 1, DB = 9 . How long is the altitude CD ?
Answer and solution
The answer: 3 cm
By the theorem CD² = AD × DB = 1 × 9 = 9 , so CD = √9 = 3 .
Perimeters and areas of similar triangles
For example: The triangles ABC and DEF are similar, and the side AB corresponds to DE . In centimetres, AB = 5, DE = 11 . What is the ratio of the perimeter of ABC to the perimeter of DEF ? Give a fraction in its lowest terms.
Answer and solution
The answer: 5/11
The perimeters are in the ratio of similarity: 5/11 = 5/11 .
Not on the site yet
Parts of this topic that are not built yet: The proof of Thales’ theorem and the proportion theorems.
The exercises are made by rules and the answers are worked out in code, so there are new exercises every time. The topic follows the Israeli Ministry of Education’s high school maths curriculum at 5 units (grades 10 to 12).