A square unit is the surface enclosed by a square whose side is 1 unit (Fig. 9-1).
Fig. 9-1
The area of a closed plane figure, such as a polygon, is the number of square units contained in its surface. Since a rectangle 5 units long and 4 units wide can be divided into 20 unit squares, its area is 20 square units (Fig. 9-2).
Fig. 9-2
The area of a rectangle equals the product of the length of its base and the length of its altitude (Fig. 9-3). Thus if b = 8 in and h = 3 in, then A = 24 in2.
Fig. 9-3
The area of a square equals the square of the length of a side (Fig. 9-4). Thus if s = 6, then A = s2 = 36.
Fig. 9-4
It follows that the area of a square also equals one-half the square of the length of a diagonal. Since A = s2 and s = d/ 
, A = 
d2.
Note that we sometimes use the letter A for both a vertex of a figure and its area. You should have no trouble determining which is meant.
The reader should feel free to use a calculator for the work in this chapter.
9.1 Area of a rectangle
(a) Find the area of a rectangle if the base has length 15 and the perimeter is 50.
(b) Find the area of a rectangle if the altitude has length 10 and the diagonal has length 26.
(c) Find the lengths of the base and altitude of a rectangle if its area is 70 and its perimeter is 34.
Solutions
See Fig. 9-5.
Fig. 9-5
(a) Here p = 50 and b = 15. Since p = 2b + 2h, we have 50 = 2(15) + 2h so h = 10. Hence, A = bh = 15(10) = 150.
(b) Here d = 26 and h = 10. In right ΔACD, d2 = b2 + h2, so 262 = b2 + 102 or b = 24. Hence, A = bh = 24(10) = 240.
(c) Here A = 70 and p = 34. Since p = 2b + 2h, we have 34 = 2(b + h) or h = 17–b. Since A = bh, we have 70 = b(17–b), so b2–17b + 70 = 0 and b = 7 or 10. Then since h = 17–b, we obtain h = 10 or 7.
Ans. 10 and 7, or 7 and 10.
9.2 Area of a Square
(a) Find the area of a square whose perimeter is 30.
(b) Find the area of a square if the radius of the circumscribed circle is 10.
(c) Find the side and the perimeter of a square whose area is 20.
(d) Find the number of square inches in a square foot.
Solutions
(a) Since p = 4s = 30 in Fig. 9-6(a), s = 7. Then A = s2 = (7)2 = 56
.
Fig. 9-6
(b) Since r = 10 in Fig. 9-6(b), d = 2r = 20. Then A = d2 = (20)2 = 200.
(c) In Fig. 9-6(a), A = s2 = 20; hence, s = 2 
. Then perimeter = 4s = 8 .
(d) A = s2. Since 1ft = 12 in, 1 ft2 = 1ft × 1ft = 12 in × 12 in = 144 in2.
The area of a parallelogram equals the product of the length of a side and the length of the altitude to that side. (A proof of this theorem is given in Chapter 16.) Thus in 
ABCD (Fig. 9-7), if b = 10 and h = 2.7, then A = 10(2.7) = 27.
Fig. 9-7
9.3 Area of a parallelogram
(a) Find the area of a parallelogram if the area is represented by x2–4, the length of a side by x + 4, and the length of the altitude to that side by x–3.
(b) In a parallelogram, find the length of the altitude if the area is 54 and the ratio of the altitude to the base is 2:3.
Solutions
See Fig. 9-7.
(a) A = x2–4, b = x + 4, h = x–3. Since A = bh, x2–4 = (x + 4)(x–3) or x2–4 = x2 + x–12 and x = 8. Hence, A = x2–4 = 64–4 = 60.
(b) Let h = 2x, b = 3x. Then A = bh or 54 = (3x)(2x) = 6x2, so 9 = x2 and x = 3. Hence, h = 2x = 2(3) = 6.
The area of a triangle equals one-half the product of the length of a side and the length of the altitude to that side. (A proof of this theorem is given in Chapter 16.)
Fig. 9-8
9.4 Area of a triangle
Find the area of the triangle in Fig. 9-9.
Fig. 9-9
Solution
Here, b = 15 and h = 4. Thus, A = bh = (15)(4) = 30.
9.5 Formulas for the area of an equilateral triangle
Derive the formula for the area of an equilateral triangle (a) whose side has length s; (b) whose altitude has length h.
See Fig. 9-10.
Fig. 9-10
(a) Here A = bh, where b = s and h2 = s2– or h = s 
.
Then A = bh = S(S) = s2S2 .
(b) Here A = bh, where b = s and h = S or 
Then 
9.6 Area of an equilateral triangle
In Fig. 9-11, find the area of (a) an equilateral triangle whose perimeter is 24; (b) a rhombus in which the shorter diagonal has length 12 and an angle measures 60°; (c) a regular hexagon with a side of length 6.
Fig. 9-11
Solutions
(a) Since p = 3s = 24, s = 8. Then 
(b) Since m∠A = 60°, ΔADB is equilateral and s = d = 12. The area of the rhombus is twice the area of ΔABD. Hence 
(c) A side s of the inscribed hexagon subtends a central angle of measure 
(360°) = 60°. Then, since OA = OB = radius R of the circumscribed circle, m∠OAB = m∠OBA = 60°. Thus ΔAOB is equilateral.
Area of hexagon = 6(area of ΔAOB) = 
The area of a trapezoid equals one-half the product of the length of its altitude and the sum of the lengths of its bases. (A proof of this theorem is given in Chapter 16.) Thus if h = 20, b = 27, and b′ = 23 in Fig. 9-12, then A = (20)(27 + 23) = 500.
Fig. 9-12
The area of a trapezoid equals the product of the lengths of its altitude and median. Since A = h(b + b′) and m = (b + b′), A = hm.
9.7 Area of a trapezoid
(a) Find the area of a trapezoid if the bases have lengths 7.3 and 2.7, and the altitude has length 3.8.
(b) Find the area of an isosceles trapezoid if the bases have lengths 22 and 10, and the legs have length 10.
(c) Find the bases of an isosceles trapezoid if the area is 52, the altitude has length 4, and each leg has length 8.
Solutions
See Fig. 9-13.
Fig. 9-13
(a) Here b= 7.3, b′ = 2.7, h = 3.8. Then A = h (b + b′) = (3.8 + 2.7) = 19.
(b) Here b = 22, b′ = 10, AB = 10. Also EF = b′ = 10 and AE = (22-10) = 6.
In ΔBEA, h2 = 102–62 = 64 so h = 8. Then A = h(b + b′) = (8)(22 + 10) = 128.
(c) AE = 
= = 4. Also FD = AE = 4, and b′ = b–(AE + FD) = b–8. Then 
or 
, from which 26 = 2b–8 or b = 17. Then b′= b–8 = 17 - 8 = 9.
The area of a rhombus equals one-half the product of the lengths of its diagonals.
Since each diagonal is the perpendicular bisector of the other, the area of triangle I in Fig. 9-14 is 
. Thus the rhombus, which consists of four triangles congruent to ΔI, has an area of 
.
Fig. 9-14
9.8 Area of a rhombus
(a) Find the area of a rhombus if one diagonal has length 30 and a side has length 17.
(b) Find the length of a diagonal of a rhombus if the other diagonal has length 8 and the area of the rhombus is 52.
Solutions
See Fig. 9-15.
Fig. 9-15
(a) In right 
Then d = 8 and d = 16. Now 
(b) We have d′ = 8 and A = 52. Then dd′ or 52 = (d)(8) and d = 13.
Figure 9-16 shows what we mean when we say that two polygons are of equal area, or are similar, or are congruent.
Fig. 9-16
PRINCIPLE 1: Parallelograms have equal areas if they have congruent bases and congruent altitudes.
Thus, the two parallelograms shown in Fig. 9-17 are equal.
Fig. 9-17
PRINCIPLE 2: Triangles have equal areas if they have congruent bases and congruent altitudes.
Thus in Fig. 9-18, the area of ΔCAB equals the area of ΔCAD.
Fig. 9-18
PRINCIPLE 3: A median divides a triangle into two triangles with equal areas.
Thus in Fig. 9-19, where 
is a median, the area of ΔAMB equals the area of ΔBMC since they have congruent bases (
≅ 
) and common altitude 
.
Fig. 9-19
PRINCIPLE 4: Triangles are equal in area if they have a common base and their vertices lie on a line parallel to the base.
Thus in Fig. 9-20, the area of ΔABC is equal to the area of ΔADC.
Fig. 9-20
9.9 Proving an equal-areas problem
9.10 Proving an equal-areas problem stated in words
Prove that if M is the midpoint of diagonal 
in quadrilateral ABCD, and and 
are drawn, then the area of quadrilateral ABMD equals the area of quadrilateral CBMD.
Solution
PROOF:
The areas of similar polygons are to each other as the squares of any two corresponding segments.
Thus if ΔABC ~ 
A′B′C′ and the area of ΔABC is 25 times the area of ΔA′B′C′, then the ratio of the lengths of any two corresponding sides, medians, altitudes, radii of inscribed or circumscribed circles, and such is 5:1.
9.11 Ratio of areas and segments of similar triangles
Find the ratio of the areas of two similar triangles (a) if the ratio of the lengths of two corresponding sides is 3:5; (b) if their perimeters are 12 and 7. Find the ratio of the lengths of a pair of (c) corresponding sides if the ratio of the areas is 4:9; (d) corresponding medians if the areas are 250 and 10.
(a) 
(b) 
(c) 
(d) 
9.12 Proportions derived from similar polygons
(a) The areas of two similar polygons are 80 and 5. If a side of the smaller polygon has length 2, find the length of the corresponding side of the larger polygon.
(b) The corresponding diagonals of two similar polygons have lengths 4 and 5. If the area of the larger polygon is 75, find the area of the smaller polygon.
Solutions
(a) 
so 
Then 
and s = 8
(b) 
so 
Then 
9.1. Find the area of a rectangle
(9.1)
(a) If the base has length 11 in and the altitude has length 9 in
(b) If the base has length 2 ft and the altitude has length 1 ft 6 in
(c) If the base has length 25 and the perimeter is 90
(d) If the base has length 15 and the diagonal has length 17
(e) If the diagonal has length 12 and the angle between the diagonal and the base measures 60°
(f) If the diagonal has length 20 and the angle between the diagonal and the base measures 30°
(g) If the diagonal has length 25 and the lengths of the sides are in the ratio of 3:4
(h) If the perimeter is 50 and the lengths of the sides are in the ratio of 2:3
9.2. Find the area of a rectangle inscribed in a circle
(9.1)
(a) If the radius of the circle is 5 and the base has length 6
(b) If the radius of the circle is 15 and the altitude has length 24
(c) If the radius and the altitude both have length 5
(d) If the diameter has length 26 and the base and altitude are in the ratio of 5:12
9.3. Find the base and altitude of a rectangle
(9.1)
(a) If its area is 28 and the base has a length of 3 more than the altitude
(b) If its area is 72 and the base is twice the altitude
(c) If its area is 54 and the ratio of the base to the altitude is 3:2
(d) If its area is 12 and the perimeter is 16
(e) If its area is 70 and the base and altitude are represented by 2x and x + 2
(f) If its area is 160 and the base and altitude are represented by 3x–4 and x
9.4. Find the area of (a) a square yard in square inches; (b) a square meter in square decimeters (1 m = 10 dm).
(9.2)
9.5. Find the area of a square if (a) a side has length 15; (b) a side has length 3; (c) a side has length 1.8; (d) a side has length 8a; (e) the perimeter is 44; (f) the perimeter is 10; (g) the perimeter is 12b; (h) the diagonal has length 8; (i) the diagonal has length 9; (j) the diagonal has length 8 .
(9.2)
9.6. Find the area of a square if (a) the radius of the circumscribed circle is 8; (b) the diameter of the circumscribed circle is 12; (c) the diameter of the circumscribed circle is 10; (d) the radius of the inscribed circle is 3; (e) the diameter of the inscribed circle is 20.
(9.2)
9.7. If a floor is 20 m long and 80 m wide, how many tiles are needed to cover it if (a) each tile is 1 m2; (b) each tile is a square 2 m on a side; (c) each tile is a square 4 m on a side.
(9.2)
9.8. If the area of a square is 81, find the length of (a) its side; (b) its perimeter; (c) its diagonal; (d) the radius of the inscribed circle; (e) the radius of the circumscribed circle.
(9.2)
9.9. (a) Find the length of the side of a square whose area is 6.
(9.2)
(b) Find the perimeter of a square whose area is 169.
(c) Find the length of the diagonal of a square whose area is 50.
(d) Find the length of the diagonal of a square whose area is 25.
(e) Find the radius of the inscribed circle of a square whose area is 144.
(f) Find the radius of the circumscribed circle of a square whose area is 32.
9.10. Find the area of a parallelogram if the base and altitude have lengths, respectively, of (a) 3 ft and 5
ft; (b) 4 ft and 1 ft 6 in; (c) 20 and 3.5; (d) 1.8 m and 0.9 m.
(9.3)
9.11. Find the area of a parallelogram if the base and altitude have lengths, respectively, of (a) 3x and x; (b) x + 3 and x; (c) x–5 and x + 5; (d) 4x + 1 and 3x + 2.
(9.3)
9.12. Find the area of a parallelogram if
(a) The area is represented by x2, the base by x + 3, and the altitude by x–2
(b) The area is represented by x2–10, the base by x, and the altitude by x–2
(c) The area is represented by 2x2–34, the base by x + 3, and the altitude by x–3
9.13. In a parallelogram, find
(9.3)
(a) The base if the area is 40 and the altitude has length 15
(b) The length of the altitude if the area is 22 and the base has length 1.1
(c) The length of the base if the area is 27 and the base is three times the altitude
(d) The length of the altitude if the area is 21 and the base has length four more than the altitude
(e) The base if the area is 90 and the ratio of the base to the altitude is 5:2
(f) The length of the altitude to a side of length 20 if the altitude to a side of length 15 is 16
(g) The length of the base if the area is 48, the base is represented by x + 3, and the altitude is represented by x + 1
(h) The length of the base if the area is represented by x2 + 17, the base by 2x–3, and the altitude by x + 1
9.14. Find the area of a triangle if the lengths of the base and altitude are, respectively, (a) 6 in and 3 in; (b) 1 yd and 2 ft; (c) 8 and x–7; (d) 5x and 4x; (e) 4x and x + 9; (f) x + 4 and x–4; (g) 2x–6 and x + 3.
(9.4)
9.15. Find the area of
(9.4)
(a) A triangle if two sides have lengths 13 and 15 and the altitude to the third side has length 12
(b) A triangle whose sides have lengths 10, 10, and 16
(c) A triangle whose sides have lengths 5, 12, and 13
(d) An isosceles triangle whose base has length 30 and whose legs each have length 17
(e) An isosceles triangle whose base has length 20 and whose vertex angle measures 68°
(f) An isosceles triangle whose base has length 30 and whose base angle measures 62°
(g) A triangle inscribed in a circle of radius 4 if one side is a diameter and another side makes an angle measuring 30° with the diameter
(h) A triangle cut off by a line parallel to the base of a triangle if the base and altitude of the larger triangle have lengths 10 and 5, respectively, and the line parallel to the base is 6
9.16. Find the altitude of a triangle if
(9.4)
(a) Its base has length 10 and the triangle is equal in area to a parallelogram whose base and altitude have lengths 15 and 8.
(b) Its base has length 8 and the triangle is equal in area to a square whose diagonal has length 4.
(c) Its base has length 12 and the triangle is equal in area to another triangle whose sides have lengths 6, 8, and 10.
9.17. In a triangle, find the length of
(9.4)
(a) A side if the area is 40 and the altitude to that side has length 10
(b) An altitude if the area is 25 and the side to which the altitude is drawn has length 5
(c) A side if the area is 24 and the side has length 2 more than its altitude
(d) A side if the area is 108 and the side and its altitude are in the ratio 3:2
(e) The altitude to a side of length 20, if the sides of the triangle have lengths 12, 16, and 20
(f) The altitude to a side of length 12 if another side and its altitude have lengths 10 and 15
(g) A side represented by 4x if the altitude to that side is represented by x + 7 and the area is 60
(h) A side if the area is represented by x2–55, the side by 2x–2, and its altitude by x–5
9.18. Find the area of an equilateral triangle if (a) a side has length 10; (b) the perimeter is 36; (c) an altitude has length 6; (d) an altitude has length 5; (e) a side has length 2b; (f) the perimeter is 12x; (g) an altitude has length 3r.
(9.6)
9.19. Find the area of a rhombus having an angle of 60° if (a) a side has length 2; (b) the shorter diagonal has length 7; (c) the longer diagonal has length 12; (d) the longer diagonal has length 6 .
(9.6)
9.20. Find the area of a regular hexagon if (a) a side is 4; (b) the radius of the circumscribed circle is 6; (c) the diameter of the circumscribed circle is 20.
(9.6)
9.21. Find the side of an equilateral triangle whose area equals
(9.6)
(a) The sum of the areas of two equilateral triangles whose sides have lengths 9 and 12
(b) The difference of the areas of two equilateral triangles whose sides have lengths 17 and 15
(c) The area of a trapezoid whose bases have lengths 6 and 2 and whose altitude has length 9
(d) Twice the area of a right triangle having a hypotenuse of length 5 and an acute angle of measure 30°
9.22. Find the area of trapezoid ABCD in Fig. 9-21, if:
(9.7)
Fig. 9-21
(a) b = 25, b′ = 15, and h = 7
(b) m = 10 and h = 6.9
(c) AB = 30, m∠A = 30°, b = 24, and b′ = 6
(d) AB = 12, m∠A = 45°, b = 13, and b′ = 7
(e) AB = 10, m∠A = 70°, and b + b′ = 20
9.23. Find the area of isosceles trapezoid ABCD in Fig. 9-22, if
(9.7)
Fig. 9-22
(a) b′ = 17, l = 10, and h = 6
(b) b = 22, b′ = 12, and l = 13
(c) b = 16, b′ = 10, and m∠A = 45°
(d) b = 20, l = 8, and m∠A = 60°
(e) b = 40, b′ = 20, and m∠A = 28°
9.24. (a) Find the length of the altitude of a trapezoid if the bases have lengths 13 and 7 and the area is 40.
(9.7)
(b) Find the length of the altitude of a trapezoid if the sum of the lengths of the bases is twice the length of the altitude and the area is 49.
(c) Find the sum of the lengths of the bases and the median of a trapezoid if the area is 63 and the altitude has length 7.
(d) Find the lengths of the bases of a trapezoid if the upper base has length 3 less than the lower base, the altitude has length 4, and the area is 30.
(e) Find the lengths of the bases of a trapezoid if the lower base has length twice that of the upper base, the altitude has length 6, and the area is 45.
9.25. In an isosceles trapezoid
(9.7)
(a) Find the lengths of the bases if each leg has length 5, the altitude has length 3, and the area is 39.
(b) Find the lengths of the bases if the altitude has length 5, each base angle measures 45°, and the area is 90.
(c) Find the lengths of the bases if the area is 42, the altitude has length 3, and each base angle measures 60°.
(d) Find the length of each leg if the bases have lengths 24 and 32 and the area is 84.
(e) Find the length of each leg if the area is 300, the median has length 25, and the lower base has length 30.
9.26. Find the area of a rhombus if
(9.8)
(a) The diagonals have lengths 8 and 9.
(b) The diagonals have lengths 11 and 7.
(c) The diagonals have lengths 4 and 6.
(d) The diagonals have lengths 3x and 8x.
(e) One diagonal has length 10 and a side has length 13.
(f) The perimeter is 40 and a diagonal has length 12.
(g) The side has length 6 and an angle measures 30°.
(h) The perimeter is 28 and an angle measures 45°.
(i) The perimeter is 32 and the length of the short diagonal equals a side in length.
(j) A side has length 14 and an angle measures 120°.
9.27. Find the area of a rhombus to the nearest integer if (a) the side has length 30 and an angle measures 55°; (b) the perimeter is 20 and an angle measures 33°; (c) the side has length 10 and an angle measures 130°.
(9.8)
9.28. In a rhombus, find the length of
(9.8)
(a) A diagonal if the other diagonal has length 7 and the area is 35
(9.8)
(b) The diagonals if their ratio is 4:3 and the area is 54
(c) The diagonals if the longer is twice the shorter and the area is 100
(d) The side if the area is 24 and one diagonal has length 6
(e) The side if the area is 6 and one diagonal has length 4 more than the other
9.29. A rhombus is equal to a trapezoid whose lower base has length 26 and whose other three sides have length 10. Find the length of the altitude of the rhombus if its perimeter is 36.
(9.8)
9.30. Provide the proofs requested in Fig. 9-23.
(9.9)
Fig. 9-23
9.31. Provide the proofs requested in Fig. 9-24.
(9.9)
Fig. 9-24
9.32. Prove each of the following:
(9.10)
(a) A median divides a triangle into two triangles having equal areas.
(b) Triangles are equal in area if they have a common base and their vertices lie in a line parallel to the base.
(c) In a triangle, if lines are drawn from a vertex to the trisection points of the opposite sides, the area of the triangle is trisected.
(d) In trapezoid ABCD, base 
is twice base 
. If M is the midpoint of , then ABCM and BCDM are parallelograms which are equal in area.
9.33. (a) In ΔABC, E is a point on , the median to . Prove that area(ΔBEA) = area(ΔBEC).
(b) In ΔABC, Q is a point on , M is the midpoint of 
, and P is the midpoint of . Prove that area(ΔBQM) + area(ΔPQC) = area(quadrilateral ΔPQM).
(c) In quadrilateral ABCD, diagonal bisects diagonal . Prove that area(ΔABC) = area (ΔACD).
(d) Prove that the diagonals of a parallelogram divide the parallelogram into four triangles which are equal in area. (9.10)
(9.10)
9.34. Find the ratio of the areas of two similar triangles if the ratio of two corresponding sides is (a) 1:7; (b) 7:2; (c) 1: ; (d) a:5a; (e) 9: x; (f) 3: 
(g) s:s .
(9.11)
9.35. Find the ratio of the areas of two similar triangles
(9.11)
(a) If the ratio of the lengths of two corresponding medians is 7:10
(b) If the length of an altitude of the first is two-thirds of a corresponding altitude of the second
(c) If two corresponding angle bisectors have lengths 10 and 12
(d) If the length of each side of the first is one-third the length of each corresponding side of the second
(e) If the radii of their circumscribed circles are 7 and 5
(f) If their perimeters are 30 and 30
9.36. Find the ratio of any two corresponding sides of two similar triangles if the ratio of their areas is (a) 100:1; (b) 1:49; (c) 400:81; (d) 25:121; (e) 4:y2; (f) 9x2:1; (g) 3:4; (h) 1:2; (i) x2:5; (j) x:16. (9.11)
(9.11)
9.37. In two similar triangles, find the ratio of the lengths of
(a) Corresponding sides if the areas are 72 and 50
(b) Corresponding medians if the ratio of the areas is 9:49
(c) Corresponding altitudes if the areas are 18 and 6
(d) The perimeters if the areas are 50 and 40
(e) Radii of the inscribed circles if the ratio of the areas is 1:3
9.38. The areas of two similar triangles are in the ratio of 25:16. Find (9.11)
(a) The length of a side of the larger if the corresponding side of the smaller has length 80
(b) The length of a median of the larger if the corresponding median of the smaller has length 10
(c) The length of an angle bisector of the smaller if the corresponding angle bisector of the larger has length 15
(d) The perimeter of the smaller if the perimeter of the larger is 125
(e) The circumference of the inscribed circle of the larger if the circumference of the inscribed circle of the smaller is 84
(f) The diameter of the circumscribed circle of the smaller if the diameter of the circumscribed circle of the larger is 22.5
(g) The length of an altitude of the larger if the corresponding altitude of the smaller has length 16
9.39. (a) The areas of two similar triangles are 36 and 25. If a median of the smaller triangle has length 10, find the length of the corresponding median of the larger.
(9.12)
(b) Corresponding altitudes of two similar triangles have lengths 3 and 4. If the area of the larger triangle is 112, find the area of the smaller.
(c) Two similar polygons have perimeters of 32 and 24. If the area of the smaller is 27, find the area of the larger.
(d) The areas of two similar pentagons are 88 and 22. If a diagonal of the larger has length 5, find the length of the corresponding diagonal of the smaller.
(e) In two similar polygons, the ratio of the lengths of two corresponding sides is :1. If the area of the smaller is 15, find the area of the larger.