; (4) as a decimal, 0.75; and (5) as a percent, 75%.Ratios are used to compare quantities by division. The ratio of two quantities is the first divided by the second. A ratio is an abstract number, that is, a number without a unit of measure. Thus, the ratio of 10 ft to 5 ft is 10 ft ÷ 5 ft, which equals 2.
A ratio can be expressed in the following ways: (1) using a colon, as in 3:4; (2) using “to” as in 3 to 4; (3) as a common fraction, as in ; (4) as a decimal, 0.75; and (5) as a percent, 75%.
The quantities involved in a ratio must have the same unit. A ratio should be simplified by reducing to lowest terms and eliminating fractions. Thus to find the ratio of 1 ft to 4 in, we first change the foot to 12 in, and then take the ratio of 12 in to 4 in; the result is a ratio of 3 to 1, or 3. Also, the ratio of 2
: would be restated as 5:1 or 5.
The ratio of three or more quantities may be expressed as a continued ratio. Thus, the ratio of $2 to $3 to $5 is the continued ratio 2:3:5. This enlarged ratio is a combination of three separate ratios; these are 2:3, 3:5, and 2:5.
Throughout this chapter, readers should use a calculator whenever they choose.
7.1 Ratio of two quantities with the same unit
Express each of the following ratios in lowest terms: (a) 15° to 3°; (b) $1.25 to $5; (c) 2 years to 2 years.
Solutions
(a) 
(b) 
(c) 
7.2 Ratio of two quantities with different units
Express each of the following ratios in lowest terms: (a) 2 years to 3 months; (b) 80 cents to $3.20.
Solutions
(a) 2 years to 3 months = 24 months to 3 months 
(b) 80 cents to $3.20 = 80 cents to 320 cents 
7.3 Continued ratio of three quantities
Express each of the following ratios in lowest terms: (a) 1 gal to 2 qt to 2 pt; (b) 1 ton to 1lb to 8 oz.
Solutions
(a) 1 gal to 2 qt to 2 pt = 4 qt to 2 qt to 1 qt = 4:2:1
(b) 1 ton to 1 lb to 8 oz = 2000 lb to 1 lb to lb = 2000:1: = 4000:2:1
7.4 Numerical and algebraic ratios
Express each of the following ratios in lowest terms: (a) 50 to 60; (b) 6.3 to 0.9; (c) 12 to 
; (d) 2x to 5x; (e) 5s2 to s3; (f) x to 5x to 7x.
Solutions
(a) 
(b) 
(c) 
(d) 
(e) 
(f) x:5x:7x=1:5:7
7.5 Using ratios in angle problems
If two angles are in the ratio of 3:2, find the angles if (a) they are adjacent and form an angle measuring 40°; (b) they are acute angles of a right triangle; (c) they are two angles of a triangle whose third angle measures 70°.
Solutions
Let the measures of the angles be 3x and 2x. Then:
(a) 3x + 2x = 40, so that 5x = 40 or x = 8; hence, the angles measure 24° and 16°.
(b) 3x + 2x = 90, so 5x = 90 or x = 18; hence, the angles measure 54° and 36°.
(c) 3x + 2x + 70 = 180, so 5x = 110 or x = 22; hence, the angles measure 66° and 44°.
7.6 Three angles having a fixed ratio
Three angles are in the ratio of 4:3:2. Find the angles if (a) the first and the third are supplementary; (b) the angles are the three angles of a triangle.
Solutions
Let the measures of the angles be 4x, 3x, and 2x. Then:
(a) 4x + 2x = 180, so that 6x = 180 for x = 30; hence, the angles measure 120°, 90°, and 60°.
(b) 4x + 3x + 2x = 180, so 9x = 180 or x = 20; hence, the angles measure 80°, 60°, and 40°.
A proportion is an equality of two ratios. Thus, 2:5 = 4:10 (or 
) is a proportion.
The fourth term of a proportion is the fourth proportional to the other three taken in order. Thus in 2:3 = 4: x, x is the fourth proportional to 2, 3, and 4.
The means of a proportion are its middle terms, that is, its second and third terms. The extremes of a proportion are its outside terms, that is, its first and fourth terms. Thus in a:b = c:d, the means are b and c, and the extremes are a and d.
If the two means of a proportion are the same, either mean is the mean proportional between the first and fourth terms. Thus in 9:3 = 3:1, 3 is the mean proportional between 9 and 1.
PRINCIPLE 1: In any proportion, the product of the means equals the product of the extremes.
Thus if a:b = c:d, then ad = bc.
PRINCIPLE 2: If the product of two numbers equals the product of two other numbers, either pair may be made the means of a proportion and the other pair may be made the extremes.
Thus if 3x = 5y, then x:y = 5:3 or y:x = 3:5 or 3:y = 5:x or 5:x = 3:y.
PRINCIPLE 3: (Inversion method) A proportion may be changed into an equivalent proportion by inverting each ratio.
Thus if 
, then 
PRINCIPLE 4: (Alternation method) A proportion may be changed into an equivalent proportion by interchanging the means or by interchanging the extremes.
Thus if 
, then 
PRINCIPLE 5: (Addition method) A proportion may be changed into an equivalent proportion by adding terms in each ratio to obtain new first and third terms.
Thus if 
, then 
. If 
, then 
.
PRINCIPLE 6: (Subtraction method) A proportion may be changed into an equivalent proportion by subtracting terms in each ratio to obtain new first and third terms.
Thus if 
, then 
. If 
, then 
.
PRINCIPLE 7: If any three terms of one proportion equal the corresponding three terms of another proportion, the remaining terms are equal.
Thus if 
and 
, then y = 4.
PRINCIPLE 8: In a series of equal ratios, the sum of any of the numerators is to the sum of the corresponding denominators as any numerator is to its denominator.
Thus if 
, then 
If 
, then 
= 
or 
7.7 Finding unknowns in proportions
Solve the following proportions for x:
(a) x:4 = 6:8
(b) 3:x = x:27
(c) x:5 = 2x:(x + 3)
(d) 
(e) 
(f) 
(a) Since 4(6) = 8x, 8x = 24 or x = 3.
(b) Since x2 = 3(27), x2 = 81 or x = ±9.
(c) Since 5(2x) = x(x + 3), we have 10x = x2 + 3x. Then x2–7x = 0, so x = 0 or 7.
(d) Since 2x = 3(5), 2x = 15 or x = 7.
(e) Since 3(2x–3) = 5x, we have 6x–9 =5x, so x = 9.
(f) Since 4(7) = (x–2)(x + 2), we have 28 = x2–4. Then x2 =32, so x = ±4
.
7.8 Finding fourth proportionals to three given numbers
Find the fourth proportional to (a) 2, 4, 6; (b) 4, 2, 6; (c) , 3, 4; (d) b, d, c.
Solutions
(a) We have 2:4 = 6:x, so 2x = 24 or x = 12.
(b) We have 4: 2 = 6: x, so 4x = 12 or x = 3.
(c) We have :3 = 4:x, so x = 12 or x = 24.
(d) We have b:d = c:x, so bx = cd or x = cd/b.
7.9 Finding the mean proportional to two given numbers
Find the positive mean proportional x between (a) 5 and 20; (b) and 
.
Solutions
(a) We have 5:x = x: 20, so x2 = 100 or x = 10.
(b) We have 
:x = x:, so x2 = 
or x = 
.
7.10 Changing equal products into proportions
(a) Form a proportion whose fourth term is x and such that 2bx = 3s2.
(b) Find the ratio x to y if ay = bx.
Solutions
(a) 2b:3s = s:x or 2b: 3 = s2:x or 2b: s2 = 3: x (b) x: y = a:b
7.11 Changing proportions into new proportions
Use each of the following to form a new proportion whose first term is x:
(a) 
(b) 
(c) 
(d) 
Solutions
(a) By Principle 
(b) By Principle 
(c) By Principle 
(d) By Principle 
thus, 
7.12 Combining numerators and denominators of proportions
Use Principle 8 to find x in each of the following proportions:
(a) 
(b) 
(c) 
(a) Adding numerators and denominators yields 
or 
, So x = 8.
(b) Here we have 
, which gives 
, so x = 4.
(c) We use all three ratios to get 
or , so x = 6.
If two segments are divided proportionately, (1) the corresponding new segments are in proportion, and (2) the two original segments and either pair of corresponding new segments are in proportion.
Thus if 
and 
in Fig. 7-1 are divided proportionately by 
, we may write a proportion such as 
using the four segments; or we may write a proportion such as 
using the two original segments and two of their new segments.
Fig. 7-1
A proportion such as 
can be arranged in eight ways. To obtain the eight variations, we let each term of the proportion represent one of the new segments of Fig. 7-1. Two of the possible proportions are then obtained from each direction, as follows:
PRINCIPLE 1: If a line is parallel to one side of a triangle, then it divides the other two sides proportionately.
Thus in ΔABC of Fig. 7-2, if || 
, then 
Fig. 7-2
PRINCIPLE 2: If a line divides two sides of a triangle proportionately, it is parallel to the third side. (Principles 1 and 2 are converses.)
Thus in ΔABC (Fig. 7-2), if 
, then || .
PRINCIPLE 3: Three or more parallel lines divide any two transversals proportionately.
Thus if 
in Fig. 7-3, then .
Fig. 7-3
PRINCIPLE 4: A bisector of an angle of a triangle divides the opposite side into segments which are proportional to the adjacent sides.
Thus in ΔABC of Fig. 7-4, if 
bisects ∠C, then .
Fig. 7-4
7.13 Applying principle 1
Find x in each part of Fig. 7-5.
Fig. 7-5
Solutions
(a) || ; hence 
so that x = 24.
(b) We have EC = 7 and || ; hence, 
. Then 7x = 5x + 20 and x = 10.
7.14 Applying principle 3
Find x in each part of Fig. 7-6.
Fig. 7-6
(a) We have EC = 4 and ||
||; hence, 
and x = 6.
(b) 
; hence, 
from which 20x–20 = 14x + 7. Then 6x = 27 and x = 4.
7.15 Applying principle 4
Find x in each part of Fig. 7-7.
Fig. 7-7
Solutions
(a) 
bisects ∠B; hence, 
and x = 12.
(b) bisects ∠B; hence, 
Thus, 21x–7 = 20x and x = 7.
7.16 Proving a proportional-segments problem
PROOF:
Similar polygons are polygons whose corresponding angles are congruent and whose corresponding sides are in proportion. Similar polygons have the same shape although not necessarily the same size.
The symbol for “similar” is ~. The notation ΔABC ~ ΔA′B′C′ is read “triangle ABC is similar to triangle A-prime B-prime C-prime.” As in the case of congruent triangles, corresponding sides of similar triangles are opposite congruent angles. (Note that corresponding sides and angles are usually designated by the same letter and primes.)
In Fig. 7-8 ΔABC ~ ΔA′B′C because
m∠A = m∠A′ = 37° m∠B = m∠B′ = 53° m∠C = m∠C′ = 90°
and 
or 
Fig. 7-8
In Solved Problem 7.25, it is given that ABCD in a figure like Fig. 7-9 is a parallelogram, and we must prove that 
. To prove this proportion, it is necessary to find similar triangles whose sides are in the proportion. This can be done simply by selecting the triangle whose letters A, E, and F are in the numerators and the triangle whose letters B, C, and F are in the denominators. Hence, we would prove ΔAEF ~ ABCF.
Fig. 7-9
Suppose that the proportion to be proved is 
In such a case, interchanging the means leads to 
The needed triangles can then be selected based on the numerators and the denominators.
Suppose that the proportion to be proved is 
. Then our method of selecting triangles could not be used until the term AD were replaced by BC. This is possible, because 
and are opposite sides of the parallelogram ABCD and therefore are congruent.
PRINCIPLE 1: Corresponding angles of similar triangles are congruent (by the definition).
PRINCIPLE 2: Corresponding sides of similar triangles are in proportion (by the definition).
PRINCIPLE 3: Two triangles are similar if two angles of one triangle are congruent respectively to two angles of the other.
Thus in Fig. 7-10, if ∠A ≐ ∠A′ and ∠B ≐ ∠B′, thenΔAABC ~ ΔA′B′C′.
Fig. 7-10
PRINCIPLE 4: Two triangles are similar if an angle of one triangle is congruent to an angle of the other and the sides including these angles are in proportion.
Thus in Fig. 7-10, if ∠C ∝ ∠C′ and 
, then ΔABC ~ ΔA′B′C′.
PRINCIPLE 5: Two triangles are similar if their corresponding sides are in proportion.
Thus in Fig. 7-10, if 
, then ΔABC ~ A′B′C′.
PRINCIPLE 6: Two right triangles are similar if an acute angle of one is congruent to an acute angle of the other (corollary of Principle 3).
PRINCIPLE 7: A line parallel to a side of a triangle cuts off a triangle similar to the given triangle.
Thus in Fig. 7-11, if ||, then ΔADE ~ΔABC.
Fig. 7-11
PRINCIPLE 8: Triangles similar to the same triangle are similar to each other.
PRINCIPLE 9: The altitude to the hypotenuse of a right triangle divides it into two triangles which are similar to the given triangle and to each other.
Thus in Fig. 7-12, ΔABC ~ ΔACD ~ ΔCBD.
Fig. 7-12
PRINCIPLE 10: Triangles are similar if their sides are respectively parallel to each other.
Thus in Fig. 7-13, ΔABC ~ ΔA′B′C′.
Fig. 7-13
PRINCIPLE 11: Triangles are similar if their sides are respectively perpendicular to each other.
Thus in Fig. 7-14, ΔABC ~ ΔA′B′C′.
Fig. 7-14
7.17 Applying principle 2
In similar triangles ABC and A′B′C′ (Fig. 7-15), find x and y if ∠A ≐ ∠A′ and ∠B ≐ ∠B′.
Fig. 7-15
Solution
Since ∠A ≐ ∠A’ and ∠B ≐ ∠B′, x and y correspond to 32 and 26, respectively. Hence, 
, from which x = 24; also 
so y = 19.
7.18 Applying principle 3
In each part of Fig. 7-16, two pairs of congruent angles can be used to prove the indicated triangles similar. Determine the congruent angles and state the reason they are congruent.
Fig. 7-16
Solutions
(a) ∠CBD ≐ ∠BDA and ∠BCA ≐ ∠CAD, since alternate interior angles of parallel lines are congruent (||). Also, ∠BEC and ∠AED are congruent vertical angles.
(b) ∠A ≐ ∠C and ∠B ≐ ∠D, since angles inscribed in the same arc are congruent. Also, ∠AED and ∠CEB are congruent vertical angles.
(c) ∠ABC ≐ ∠AED, since each is a supplement of ∠DEC. ∠ACB ≐ ∠ADE, since each is a supplement of ∠BDE. Also, ∠A ≐ ∠A.
7.19 Applying principle 6
In each part of Fig. 7-17, determine the angles that can be used to prove the indicated triangles similar.
Fig. 7-17
(a) ∠ACB and ∠ADC are right angles. ∠A ≐ ∠A.
(b) ∠AEC and ∠BDC are right angles. ∠B ≐ ∠ACE, since angles in a triangle opposite congruent sides are congruent.
(c) ∠ACB is a right angle, since it is inscribed in a semicircle. Hence, ∠AED ≐ ∠ACB. ∠D ≐ ∠B, since angles inscribed in the same arc are congruent.
7.20 Applying principle 4
In each part of Fig. 7-18, determine the pair of congruent angles and the proportion needed to prove the indicated triangles similar.
Fig. 7-18
Solutions
(a) ∠AEB ≐ ∠DEC;
(b) ∠A ≐ ∠A;
(c) ∠BAC ≐ ∠ACD;
7.21 Applying principle 5
In each part of Fig. 7-19, determine the proportion needed to prove the indicated triangles similar.
Fig. 7-19
(a) 
(b) 
(c) 
7.22 Proportions obtained from similar triangles
Find x in each part of Fig. 7-20.
Fig. 7-20
Solutions
(a) Since ||, ∠A ≐ ∠B and ∠C ≐ ∠D; hence, ΔAEC ~ ΔBED. Then 
and x = 15.
(b) Since 
||
, 
by Principle 7. Hence, 
and x = 4.
7.23 Finding heights using ground shadows
A tree casts a 15-ft shadow at a time when a nearby upright pole 6 ft high casts a shadow of 2 ft. Find the height of the tree if both tree and pole make right angles with the ground.
Solution
At the same time in localities near each other, the rays of the sun strike the ground at equal angles; hence, ∠B ≐ ∠B′ in Fig. 7-21. Since the tree and the pole make right angles with the ground, ∠C ≐ ∠C. Hence, ΔABC ~ ΔA′B′C′, so 
and h = 45 ft.
Fig. 7-21
7.24 Proving a similar-triangle problem stated in words
Prove that two isosceles triangles are similar if a base angle of one is congruent to a base angle of the other.
Solution
PROOF:
7.25 Proving a proportion problem involving similar triangles
PROOF:
PRINCIPLE 1: Corresponding sides of similar triangles are in proportion.
PRINCIPLE 2: Corresponding segments of similar triangles are in proportion.
PRINCIPLE 3: Corresponding segments of similar polygons are in proportion.
When segments replaces sides, Principle 1 becomes the more general Principle 2. When polygons replaces triangles, Principle 2 becomes the even more general Principle 3.
By segments we mean straight or curved segments such as altitudes, medians, angle bisectors, radii of inscribed or circumscribed circles, and circumferences of inscribed or circumscribed circles.
The ratio of similitude of two similar polygons is the ratio of any pair of corresponding lines.
Corollaries of Principles 2 and 3, such as the following, can be devised for any combination of corresponding lines:
1. Corresponding altitudes of similar triangles have the same ratio as any two correponding medians.
Thus if ΔABC ~ ΔA′B′C′ in Fig. 7-22, then 
Fig. 7-22
2. Perimeters of similar polygons have the same ratio as any two corresponding sides.
Thus in Fig. 7-23, if quadrilateral I ~ quadrilateral I’, then 
Fig. 7-23
(a) In two similar triangles, corresponding sides are in the ratio 3:2. Find the ratio of corresponding medians [see Fig. 7-24(a)].
(b) The sides of a triangle are 4, 6, and 7 [Fig. 7-24(b)]. If the perimeter of a similar triangle is 51, find its longest side.
(c) In ΔABC of Fig. 7-24(c), BC=25 and the measure of the altitude to is 10. A line segment terminating in the sides of the triangle is parallel to and 3 units from A. Find its length.
Fig. 7-24
Solutions
(a) IfΔABC ~ ΔA′B′C′ and 
, then 
(b) The perimeter of ΔA′B′C′ is 4 + 6 + 7 = 17. Since ΔABC ~ ΔA′B′C′, 
and s = 21.
(c) Since ΔADE ~ ABC, 
and DE = 7.
7.27 Line ratios from similar polygons
Complete each of the following statements:
(a) If corresponding sides of two similar polygons are in the ratio of 4:3, then the ratio of their perimeters is 
.
(b) The perimeters of two similar quadrilaterals are 30 and 24. If a side of the smaller quadrilateral is 8, the corresponding side of the larger is .
(c) If each side of a pentagon is tripled and the angles remain the same, then each diagonal is .
(a) Since the polygons are similar, 
(b) Since the quadrilaterals are similar, 
. Then 
and s = 10.
(c) Tripled, since polygons are similar if their corresponding angles are congruent and their corresponding sides are in proportion.
In a problem, to prove that the product of the lengths of two segments equals the product of the lengths of another pair of segments, it is necessary to set up the proportion which will lead to the two equal products.
7.28 Proving an equal-products problem
Prove that if two secants intersect outside a circle, the product of the lengths of one of the secants and its external segment equals the product of the lengths of the other secant and its external segment.
Solution
PROOF
7.7 Segments Intersecting Inside and Outside a Circle
PRINCIPLE 1: If two chords intersect within a circle, the product of the lengths of the segments of one chord equals the product of the lengths of the segments of the other.
Thus in Fig. 7-25, AE × EB = CE × ED.
Fig. 7-25
PRINCIPLE 2: If a tangent and a secant intersect outside a circle, the tangent is the mean proportional between the secant and its external segment.
Thus in Fig. 7-26, if 
is a tangent, then 
Fig. 7-26
PRINCIPLE 3: If two secants intersect outside a circle, the product of the lengths of one of the secants and its external segment equals the product of the lengths of the other secant and its external segment.
Thus in Fig. 7-27, AB × AD = AC × AE.
Fig. 7-27
7.29 Applying principle 1
Find x in each part of Fig. 7-28, if chords and intersect in E.
Fig. 7-28
Solutions
(a) ED = 4. Then 16x = 4(12), so that 16x = 48 or x = 3.
(b) AE = EB = x. Then x2 = 8(2), so x2 = 16 and x = 4.
(c) CE = 3 and AE = EB = x. Then x2 = 27(3) or x2 = 81, and x = 9.
7.30 Applying principle 2
Find x in each part of Fig. 7-29 if tangent 
and intersect at A.
Fig. 7-29
(a) AB = 9 + 9 + 6 = 24. Then x2 = 24(6) or x2 = 144, and x = 12.
(b) AB = 2x + 5. Then 5(2x + 5) = 100 and x = 7.
(c) AB = x + 6. Then x(x + 6) = 16 or x2 + 6x–16 = 0. Factoring gives (x + 8)(x–2) = 0 and x = 2.
7.31 Applying principle 3
Find x in each part of Fig. 7-30 if secants and intersect in A.
Fig. 7-30
Solutions
(a) AC = 12. Then 8x = 12(3) and x = 4.
(b) AC = 2x + 2 and AB = 12. Then 2(2x + 2) = 12(5) and x = 14.
PRINCIPLE 1: The length of the altitude to the hypotenuse of a right triangle is the mean proportional between the lengths of the segments of the hypotenuse.
Thus in right ΔABC (Fig. 7-31), 
Fig. 7-31
PRINCIPLE 2: In a right triangle, the length of either leg is the mean proportional between the length of the hypotenuse and the length of the projection of that leg on the hypotenuse.
Thus in right ΔABC (Fig. 7-31), 
and 
A proof of this principle is given in Chapter 16.
7.32 Finding mean proportionals in a right triangle
In each triangle in Fig. 7-32, find x and y.
Fig. 7-32
(a) By Principle 
, or x2=27, and 
By Principle 
, so y2 = 36 and y = 6.
(b) By Principle 
and x = 16. By Principle 
, so y2 = 80 and 
In a right triangle, the square of the length of the hypotenuse equals the sum of the squares of the lengths of the legs.
Thus in Fig. 7-33, c2 = a2 + b2.
Fig. 7-33
A proof of the Pythagorean Theorem is given in Chapter 16.
7.9A Tests for Right, Acute, and Obtuse Triangles
If c2 = a2 + b2 applies to the three sides of a triangle, then the triangle is a right triangle; but if c2 ≠ a2 + b2, then the triangle is not a right triangle.
In ΔABC, if c2 < a2 + b2 where c is the longest side of the triangle, then the triangle is an acute triangle.
Thus in Fig. 7-34, 92 < 62 + 82 (that is, 81 < 100); hence, ΔABC is an acute triangle.
Fig. 7-34
In ΔABC, if c2 > a2 + b2 where c is the longest side of the triangle, then the triangle is an obtuse triangle.
Thus in Fig. 7-35, 112 > 62 + 82 (that is, 121 > 100); hence, ΔABC is an obtuse triangle.
Fig. 7-35
7.33 Finding the sides of a right triangle
In Fig. 7-36, (a) find the length of hypotenuse c if a = 12 and b = 9; (b) find a if b = 6 and c = 8; (c) find b if a = 4
and c = 8.
Fig. 7-36
Solutions
(a) c2 = a2 + b2 = 122 + 92 = 225 and c = 15.
(b) a2 = c2–b2 = 82–62 = 28 and a = 2
.
(c) b2 = c2–a2 = 82–(4)2 = 64–48 = 16 and b = 4.
7.34 Ratios in a right triangle
In a right triangle, the hypotenuse has length 20 and the ratio of the two arms is 3:4. Find each arm.
Solution
Let the lengths of the two arms be denoted by 3x and 4x. Then 202 = (3x)2 + (4x)2.
Multiplying out, we get 400 = 9x2 + 16x2 or 400 = 25x2, and x = 4; hence, the arms have lengths 12 and 16.
7.35 Applying the pythagorean theorem to an isosceles triangle
Find the length of the altitude to the base of an isosceles triangle if the base is 8 and the equal sides are 12.
Solution
The altitude h of an isosceles triangle bisects the base (Fig. 7-37). Then 
and 
Fig. 7-37
7.36 Applying the pythagorean theorem to a rhombus
In a rhombus, find (a) the length of a side s if the diagonals are 30 and 40; (b) the length of a diagonal d if a side is 26 and the other diagonal is 20.
Solution
The diagonals of a rhombus are perpendicular bisectors of each other; hence, 
in Fig. 7-38.
Fig. 7-38.
(a) If d = 30 and d’ = 40, then s2 = 152 + 202 = 625 or s = 25.
(b) If s = 26 and d’ = 20, then 
. Thus, d = 24 or d = 48.
7.37 Applying the pythagorean theorem to a trapezoid
Find x in each part of Fig. 7-39 if ABCD is a trapezoid.
Fig. 7-39
Solutions
The dashed perpendiculars in the diagrams are additional segments needed only for the solutions. Note how rectangles are formed by these added segments.
(a) EF = BC = 12 and AE = (22–12) = 5. Then x2 = 132 -52 = 144 or x = 12.
(b) b2 = 252 -l2 = 5l6 or b = 24; also, BE = b = 24 and CE = 1l–l = 10. Then x2 = 242 + 102or x = 26.
7.38 Applying the pythagorean theorem to a circle
(a) Find the distance d from the center of a circle of radius 17 to a chord whose length is 30 [Fig. 7-40(a)].
(b) Find the length of a common external tangent to two externally tangent circles with radii 4 and 9 [Fig. 7-40(b)].
Fig. 7-40
Solutions
(a) BC = (30) = 15. Then d2 = 172–152 = 64 and d = 8.
(b) 
≐ 
, RS = 4, OQ = 13, and SQ = 9–4 = 5. Then in right ΔOSQ, (OS)2 = 132–52 = 144 so OS = 12; hence PR = 12.
A 30°-60°-90° triangle is one-half an equilateral triangle. Thus, in right ΔABC (Fig. 7-41), a = c. Consider that c = 2; then a = 1, and the Pythagorean Theorem gives
Fig. 7-41
b2 = c2–a2 = 22–12 = 3 or b =
The ratio of the sides is then 
Principles of the 30°-60°-90° Triangle
PRINCIPLE 1: The length of the leg opposite the 30° angle equals one-half the length of the hypotenuse.
In Fig. 7-41, 
PRINCIPLE 2: The length of the leg opposite the 60° angle equals one-half the length of the hypotenuse times the square root of 3.
In Fig. 7-41, 
PRINCIPLE 3: The length of the leg opposite the 60° angle equals the length of the leg opposite the 30° angle times the square root of 3.
In Fig. 7-41, b = a .
Equilateral-Triangle Principle
PRINCIPLE 4: The length of the altitude of an equilateral triangle equals one-half the length of a side times the square root of 3. (Principle 4 is a corollary of Principle 2.)
In Fig. 7-41, 
A 45°-45°-90° triangle is one-half a square. In right triangle ABC (Fig. 7-42), c2 = a2 + a2 or c = a 
Hence, the ratio of the sides is a:a:c = 1:1: .
Fig. 7-42
Principles of the 45°-45°-90° Triangle
PRINCIPLE 5: The length of a leg opposite a 45° angle equals one-half the length of the hypotenuse times the square root of 2.
In Fig. 7-42, 
PRINCIPLE 6: The length of the hypotenuse equals the length of a side times the square root of 2.
In Fig. 7-42, c = a .
PRINCIPLE 7: In a square, the length of a diagonal equals the length of a side times the square root of 2.
In Fig. 7-42, d = s
7.39 Applying principles 1 to 4
(a) If the length of the hypotenuse of a 30°-60°-90° triangle is 12, find the lengths of its legs [Fig. 7-43(a)].
Fig. 7-43
(b) Each leg of an isosceles trapezoid has length 18. If the base angles are 60° and the upper base is 10, find the lengths of the altitude and the lower base [Fig. 7-43(b)].
Solutions
(a) By Principle 1, a = (12) = 6. By Principle 2, 
(b) By Principle 2, 
By Principle 1, AE = FD = (18) = 9; hence, b = 9 + 10 + 9 =28.
7.40 Applying principles 5 and 6
(a) Find the length of the leg of an isosceles right triangle whose hypotenuse has length 28 [Fig. 7-44(a)].
(b) An isosceles trapezoid has base angles measuring 45°. If the upper base has length 12 and the altitude has length 3, find the lengths of the lower base and each leg [Fig. 7-44(b)].
Fig. 7-44
Solutions
(a) By Principle 5, 
(b) By Principle 6, a = 3 , AE = BE = 3 and EF = 12; hence, b = 3 + 12 + 3 = 18.
7.1. Express each of the following ratios in lowest terms:
(7.1)
(a) 20 cents to 5 cents
(b) 5 dimes to 15 dimes
(c) 30 lb to 25 lb
(d) 20° to 14°
(e) 27 min to 21 min
(f) 50% to 25%
(g) 15° to 75°
(h) 33° to 77°
(i) $2.20 to $3.30
(j) $.84 to $.96
(k) 
(l) 
(m) 
(n) 
(o) 
7.2. Express each of the following ratios in lowest terms:
(7.2)
(a) 1 year to 2 months
(b) 2 weeks to 5 days
(c) 3 days to 3 weeks
(d) 2 h to 20 min
(e) 2 yd to 2 ft
(f) 2 yd to 2 ft
(g) 1 ft to 9 in
(h) 2 g to 8 mg
(i) 100 lb to 1 ton
(j) $2 to 25 cents
(k) 2 quarters to 3 dimes
(l) 1 yd2 to 2 ft2
7.3. Express each of the following ratios in lowest terms:
(7.3)
(a) 20 cents to 30 cents to $1
(b) $3 to $1.50 to 25 cents
(c) 1 quarter to 1 dime to 1 nickel
(d) 1 day to 4 days to 1 week
(e) day to 9 h to 3 h
(f) 2 h to h to 15 min
(g) 1 ton to 200 lb to 40 lb
(h) 3 lb to 1 lb to 8 oz
(i) 1 gal to 1 qt to 1 pt
7.4. Express each of the following ratios in lowest terms:
(7.4)
(a) 60 to 70
(b) 84 to 7
(c) 65 to 15
(d) 125 to 500
(e) 630 to 105
(f) 1760 to 990
(g) 0.7 to 2.1
(h) 0.36 to 0.24
(i) 0.002 to 0.007
(j) 0.055 to 0.005
(k) 6.4 to 8
(l) 144 to 2.4
(m) 7 to 22
(n) 1 to 10
(o) 
(p) 
7.5. Express each of the following ratios in lowest terms:
(7.4)
(a) x to 3x
(b) 15c to 5
(c) 11d to 22
(d) 2 πr to πD
(e) πab to πa2
(f) 4S to S2
(g) S3 to 6S2
(h) 9r2 to 6rt
(i) x to 4x to 10x
(j) 15y to 10y to 5y
(k) x3 to x2 to x
(l) 12w to 10w to 8w to 2w
7.6. Use x as the common factor to represent the following numbers and their sum:
(7.4)
(a) Two numbers whose ratio is 5:4
(b) Two numbers whose ratio is 9:1
(c) Three numbers whose ratio is 2:5:11
(d) Five numbers whose ratio is 1:2:2:3:7
7.7. If two angles in the ratio of 5:4 are represented by 5x and 4x, express each of the following statements as an equation; then find x and the angles:
(7.5)
(a) The angles are adjacent and together form an angle measuring 45°.
(b) The angles are complementary.
(c) The angles are supplementary.
(d) The angles are two angles of a triangle whose third angle is their difference.
7.8. If three angles in the ratio of 7:6:5 are represented by 7x, 6x, and 5x, express each of the following statements as an equation; then find x and the angles:
(7.6)
(a) The first and second are adjacent and together form an angle measuring 91°.
(b) The first and third are supplementary.
(c) The first and one-half the second are complementary.
(d) The angles are the three angles of a triangle.
7.9. Solve the following proportions for x:
(7.7)
(a) x:6 = 8:3
(b) 5:4 = 20:x
(c) 9:x = x:4
(d) x:2 = 10:x
(e)(x + 4): 3 = 3: (x–4)
(f) (2x + 8):(x + 2) = (2x + 5): (x + 1)
(g) a:b = c: x
(h) x:2y = 18y:x
7.10. Solve the following proportions for x:
(7.7)
(a) 
(b) 
(c) 
(d) 
(e) 
(f) 
(g) 
(h) 
7.11. Find the fourth proportional for each of the following sets of numbers:
(7.8)
(a) 1, 3, 5
(b) 8, 6, 4
(c) 2, 3, 4
(d) 3, 4, 2
(e) 3, 2, 5
(f) , 2, 5
(g) 2, 8, 8
(h) b, 2a, 3b
7.12. Find the positive mean proportional between each of the following pairs of numbers:
(7.9)
(a) 4 and 9
(b) 12 and 3
(c) and 27
(d) 2b and 8b
(e) 2 and 5
(f) 3 and 9
(g) p and q
(h) a2 and b
7.13. From each of these equations, form a proportion whose fourth term is x:
(7.10)
(a) cx = bd;
(b) pq = ax;
(c) hx = a2;
(d) 3x = 7;
(e) x = ab/c.
7.14. In each of the following equations, find the ratio of x to y:
(7.10)
(a) 2x = y;
(b) 3y = 4x;
(c) x =1 y;
(d) ax = hy;
(e) x = by.
7.15. Which of the following is not a proportion?
(7.11)
(a) 
(b) 
(c) 
(d) 
(e) 
when x = 6.
7.16. From each of the following, form a new proportion whose first term is x. Then find x.
(7.11)
(a) 
(b) 
(c) 
(d) 
(e) 
7.17. Find x in each of these pairs of proportions:
(7.11)
(a) a:b = c: x and a: b = c: d
(b) 5:7 = x:42 and 5:7 = 35:42
(c) 2:3x = 4:5y and 2:15 = 4:5y
(d) 7:5x–2 = 14:3y and 7:18 = 14: 3y
7.18. Find x in each of the following proportions:
(7.12)
(a) 
(b) 
(c) 
7.19. Find x in each part of Fig. 7-45.
(7.13)
Fig. 7-45
7.20. In which parts of Fig. 7-46 is a line parallel to one side of the triangle?
(7.13)
Fig. 7-46
7.21. Find x in each part of Fig. 7-47.
(7.14)
Fig. 7-47
7.22. Find x in each part of Fig. 7-48.
(7.15)
Fig. 7-48
7.23. Prove that three or more parallel lines divide any two transversals proportionately.
(7.16)
7.24. In similar triangles ABC and A′B′C of Fig. 7-49, ∠B and ∠B′ are corresponding angles. Find m∠B if (a) m∠A′ = 120° and m∠C′= 25°; (b) m∠A′ + m∠C = 127°.
(7.17)
Fig. 7-49
7.25. In similar triangles ABC and A′B′C′ of Fig. 7-50, ∠A ≐ ∠A′ and ∠B ≐ ∠B′. (a) Find a if c = 24; (b) find b if a = 20; (c) find c if b = 63.
(7.17)
Fig. 7-50
7.26. In each part of Fig. 7-51, show that the indicated triangles are similar.
(7.18)
Fig. 7-51
7.27. In each part of Fig. 7-52, two pairs of congruent angles can be used to prove the indicated triangles similar. Find the congruent angles.
(7.18)
Fig. 7-52
7.28. In each part of Fig. 7-53, determine the angles that can be used to prove the indicated triangles similar.
(7.19)
Fig. 7-53
7.29. In each part of Fig. 7-54, determine the pair of congruent angles and the proportion needed to prove the indicated triangles similar.
(7.20)
Fig. 7-54
7.30. In each part of Fig. 7-55, state the proportion needed to prove the indicated triangles similar.
(7.21)
Fig. 7-55
7.31. In each part of Fig. 7-56, prove the indicated proportion.
Fig. 7-56
7.32. In ΔABC (Fig. 7-57), ||.
(7.22)
Fig. 7-57
(a) Let a = 4, AB = 8, p = 10. Find q.
(b) Let c = 5, AC = 15, q = 24. Find p.
(c) Let a = 7, p = 11, q = 22. Find b.
(d) Let b = 9, p = 20, q = 35. Find a.
(e) Let a = 10, p = 24, q = 84. Find AB.
(f) Let c = 3,p = 4, q = 7. Find d.
7.33. Find x in each part of Fig. 7-58.
(7.22)
Fig. 7-58
7.34. A 7-ft upright pole near a vertical tree casts a 6-ft shadow. At that same time, find (a) the height of the tree if its shadow is 36 ft long; (b) the length of the shadow of the tree if its height is 77 ft.
(7.23)
7.35. Prove each of the following:
(7.23)
(a) In ΔABC, if and 
are altitudes, then AD: CE = AB: BC.
(b) In circle O, diameter and tangent are sides of ΔABC. If intersects the circle in D, then AD:AB = AB: AC.
(c) The diagonals of a trapezoid divide each other into proportional segments.
(d) In right ΔABC, is the altitude to the hypotenuse , then AC:CD = AB:BC.
7.36. Prove each of the following:
(7.24)
(a) A line parallel to one side of a triangle cuts off a triangle similar to the given triangle.
(b) Isosceles right triangles are similar to each other.
(c) Equilateral triangles are similar to each other.
(d) The bases of a trapezoid form similar triangles with the segments of the diagonals.
7.37. Complete each of the following statements:
(7.26)
(a) In similar triangles, if corresponding sides are in the ratio 8:5, then corresponding altitudes are in the ratio .
(b) In similar triangles, if corresponding angle bisectors are in the ratio 3:5, then their perimeters are in the ratio .
(c) If the sides of a triangle are halved, then the perimeter is , the angle bisectors are , the medians are , and the radii of the circumscribed circle are
7.38. (a) Corresponding sides of two similar triangles have lengths 18 and 12. If an altitude of the smaller has length 10, find the length of the corresponding altitude of the larger.
(7.26)
(b) Corresponding medians of two similar triangles have lengths 25 and 15. Find the perimeter of the larger if the perimeter of the smaller is 36.
(c) The sides of a triangle have lengths 5, 7, and 8. If the perimeter of a similar triangle is 100, find its sides.
(d) The bases of a trapezoid have lengths 5 and 20, and the altitude has length 12. Find the length of the altitude of the triangle formed by the shorter base and the nonparallel sides extended to meet.
(e) The bases of a trapezoid have lengths 11 and 22. Its altitude has length 9. Find the distance from the point of intersection of the diagonals to each of the bases.
7.39. Complete each of the following statements:
(7.27)
(a) If corresponding sides of two similar polygons are in the ratio 3:7, then the ratio of their corresponding altitudes is .
(b) If the perimeters of two similar hexagons are in the ratio of 56 to 16, then the ratio of their corresponding diagonals is .
(c) If each side of an octagon is quadrupled and the angles remain the same, then its perimeter is .
(d) The base of a rectangle is twice that of a similar rectangle. If the radius of the circumscribed circle of the first rectangle is 14, then the radius of the circumscribed circle of the second is .
7.40. Prove each of the following:
(7.28)
(a) Corresponding angle bisectors of two triangles have the same ratio as a pair of corresponding sides.
(b) Corresponding medians of similar triangles have the same ratio as a pair of corresponding sides.
7.41. Provide the proofs requested in Fig. 7-59.
(7.28)
Fig. 7-59
7.42. Prove each of the following:
(7.28)
(a) If two chords intersect in a circle, the product of the lengths of the segments of one chord equals the product of the lengths of the segments of the other.
(b) In a right triangle, the product of the lengths of the hypotenuse and the altitude upon it equals the product of the lengths of the legs.
(c) If in inscribed ΔABC the bisector of ∠A intersects in D and the circle in E, then BD × AC = AD × EC.
(7.29)
(a) Let AE = 10, EB = 6, CE = 12. Find ED.
(b) Let AB = 15, EB = 8, ED = 4. Find CE.
(c) Let AE = 6, ED = 4, CD = 13. Find EB.
(d) Let ED = 5, EB = 2(AE), CD = 15. Find AE.
Fig. 7-60
In Fig. 7-61, diameter ⊥ chord
(e) Let OD = 10, OE = 8. Find AB.
(f) Let AB = 24, OE = 5. Find OD.
(g) Let OD = 25, EC = 18. Find AB.
(h) Let AB = 8, OD = 5. Find EC.
Fig. 7-61
7.44. A point is 12 in from the center of a circle whose radius is 15 in. Find the lengths of the longest and shortest chords that can be drawn through this point. (Hint: The longest chord is a diameter, and the shortest chord is perpendicular to this diameter.)
(7.29)
7.45. In Fig. 7-62, is a tangent
(7.29)
(a) Let AC = 16, AD = 4. Find AB.
(b) Let CD = 5, AD = 4. Find AB.
(c) Let AB = 6, AD = 3. Find AC.
(d) Let AC = 20, AB = 10. Find AD.
(e) Let AB = 12, AD = 9. Find CD.
Fig. 7-62
In Fig. 7-63, is a diameter; is a tangent
(f) Let AD = 6, OD = 9. Find AB.
(g) Let AD = 2, AB = 8. Find CD.
(h) Let AD = 5, AB = 10. Find OD.
(i) Let AB = 12, AC = 18. Find OD.
(j) Let OD = 5, AB = 12. Find AD.
Fig. 7-63
(7.31)
(a) Let AB = 14, AD = 4, AE = 7. Find AC.
(b) Let AC = 8, AE = 6, AD = 3. Find BD.
(c) Let BD = 5, AD = 7, AE = 4. Find AC.
(d) Let AD = DB, EC = 14, AE = 4. Find AD.
Fig. 7-64
In Fig. 7-65, is a diameter
(e) Let OC = 3, AE = 6, AD = 8. Find AB.
(f) Let BD = 7, AD = 5, AE = 2. Find OC.
(g) Let OC = 11, AB = 15, AD = 5. Find AE.
(h) Let OC = 5, AE = 6, BD = 4. Find AD.
Fig. 7-65
7.47. is the altitude to hypotenuse in Fig. 7-66.
(7.32)
(a) If p = 2 and q = 6, find a and h.
(b) If p = 4 and a = 6, find c and h.
(c) If p = 16 and h = 8, find q and b.
(d) If b = 12 and q = 6, find p and h.
Fig. 7-66
7.48. In a right triangle whose arms have lengths a and b, find the length of the hypotenuse c when
(7.33)
(a) a = 15, b = 20
(b) a = 15, b = 36
(c) a = 5, b = 4
(d) a = 5, b = 5
(e) a = 7, b = 7
7.49. In the right triangle in Fig. 7-67, find the length of each missing arm when
(7.33)
(a) a = 12, c = 20
(b) b = 6, c = 8
(c) b = 15, c = 17
(d) a = 2, c = 4
(e) a = 5, c = 10
(f) a = 
, c = 2
Fig. 7-67
7.50. Find the lengths of the arms of a right triangle whose hypotenuse has length c if these arms have a ratio of (a) 3:4 and c = 15; (b) 5:12 and c = 26; (c) 8:15 and c = 170; (d) 1:2 and c = 10.
(7.34)
7.51. In a rectangle, find the length of the diagonal if its sides have lengths (a) 9 and 40; (b) 5 and 10.
(7.33)
7.52. In a rectangle, find the length of one side if the diagonal has length 15 and the other side has length (a) 9; (b) 5; (c) 10.
(7.33)
7.53. Of triangles having sides with lengths as follows, which are right triangles?
(a) 33, 55, 44
(b) 120, 130, 50
(c) 4, 7, 8
(d) 25, 7, 24
(e) 5 in, 1 ft, 1 ft 1 in
(f) 1 yd, 1 yd 1 ft, 1 yd 2 ft
(g) 11 mi, 60 mi, 61 mi
(h) 5 cm, 5 cm, 7 cm
7.54. Is a triangle a right triangle if its sides have the ratio of (a) 3:4:5; (b) 2:3:4?
7.55. Find the length of the altitude of an isosceles triangle if each of its two congruent sides has length 10 and its base has length (a) 12; (b) 16; (c) 18; (d) 10.
(7.35)
7.56. In a rhombus, find the length of a side if the diagonals have lengths (a) 18 and 24; (b) 4 and 8; (c) 6 and 6 .
(7.36)
7.57. In a rhombus, find the length of a diagonal if a side and the other diagonal have lengths, respectively, (a) 10 and 12; (b) 17 and 16; (c) 4 and 4; (d) 10 and 10 .
(7.36)
7.58. In isosceles trapezoid ABCD in Fig. 7-68
(7.37)
(a) Find a if b = 32, B′ = 20, and h = 8.
(b) Find h if b = 24, B′ = 14, and a = 13.
(c) Find b if a = 15, B′ = 10, and h = 12.
(d) Find B′ if a = 6, b = 21, and h = 3.
Fig. 7-68
7.59. In a trapezoid ABCD in Fig. 7-69
(7.37)
(a) Find d if a = 11, b = 3, and c = 15.
(b) Find a if d = 20, b = 12, and c = 36.
(c) Find d if a = 5, p = 13, and c = 14.
(d) Find p if a = 20, c = 28, and d = 17.
Fig. 7-69
7.60. The radius of a circle is 15. Find (a) the distance from its center to a chord whose length is 18; (b) the length of a chord whose distance from its center is 9.
(7.38)
7.61. In a circle, a chord whose length is 16 is at a distance of 6 from the center. Find the length of a chord whose distance from the center is 8.
(7.38)
7.62. Two externally tangent circles have radii of 25 and 9. Find the length of a common external tangent.
(7.38)
7.63. In a 30°-60°-90° triangle, find the lengths of (a) the legs if the hypotenuse has length 20; (b) the other leg and hypotenuse if the leg opposite 30° has length 7; (c) the other leg and hypotenuse if the leg opposite 60° has length 5 .
(7.39)
7.64. In an equilateral triangle, find the length of the altitude if the side has length (a) 22; (b) 2a. Find the side if the altitude has length (c) 24 ; (d) 24.
(7.39)
7.65. In a rhombus which has an angle measuring 60°, find the lengths of (a) the diagonals if a side has length 25; (b) the side and larger diagonal if the smaller diagonal has length 35.
(7.39)
7.66. In an isosceles trapezoid which has base angles measuring 60°, find the lengths of (a) the lower base and altitude if the upper base has length 12 and the legs have length 16; (b) the upper base and altitude if the lower base has length 45 and the legs have length 28.
(7.39)
7.67. In an isosceles right triangle, find the length of each leg if the hypotenuse has length (a) 34; (b) 2a. Find the length of the hypotenuse if each leg has length (c) 34; (d) 15 .
(7.40)
7.68. In a square, find the length of (a) the side if the diagonal has length 40; (b) the diagonal if the side has length 40.
(7.40)
7.69. In an isosceles trapezoid which has base angles of measure 45°, find the lengths of (a) the lower base and each leg if the altitude has length 13 and the upper base has length 19; (b) the upper base and each leg if the altitude has length 27 and the lower base has length 65; (c) each leg and the lower base if the upper base has length 25 and the altitude has length 15.
(7.40)
7.70. A parallelogram has an angle measuring 45°. Find the distances between its pairs of opposite sides if its sides have lengths 10 and 12.
(7.40)