10 Square Roots and Pythagoras’ Theorem 10.1 Square Roots and Surds 10.2 Pythagoras’ Theorem and Its Proofs 10.3 Applications of Pythagoras’ Theorem 10.4 Converse of Pythagoras’ Theorem and Its Applications 10.5 Rational and Irrational Numbers 10.6 Properties and Operations of Surds

10.1 Square Roots and Surds A. Square Roots

10.1 Square Roots and Surds A. Square Roots

Example 1T Solution: 10 Square Roots and Pythagoras’ Theorem Find the square roots of each of the following numbers. (a) 64 (b) 0.81 (c) Solution: (a) ∵ 8  8  64 and (–8)  (–8)  64 ∴ The square roots of 64 are 8. (b) ∵ 0.9  0.9  0.81 and (–0.9)  (–0.9)  0.81 ∴ The square roots of 0.81 are 0.9. (c) ∵ and ∴ The square roots of are .

Example 2T Solution: 10 Square Roots and Pythagoras’ Theorem Find the value of . Solution:

10.1 Square Roots and Surds A. Square Roots Calculators can only give positive square roots.

10.1 Square Roots and Surds B. Surds

10.1 Square Roots and Surds B. Surds 4( 22) and 9( 32) are two consecutive square numbers.

Example 3T Solution: 10 Square Roots and Pythagoras’ Theorem Without using a calculator, between which two consecutive integers does each of the following surds lie? (a) (b) Solution: (a) ∵ 64  72  81 (b) ∵ 169  183  196 ∴ ∴ ∴ lies between 8 and 9. ∴ lies between 13 and 14.

Example 4T Solution: 10 Square Roots and Pythagoras’ Theorem Arrange 15, , 14, and in ascending order. Solution: ∵ 187  196  205  225  260 ∴ ∴ , 14, , 15, are in ascending order.

10.2 Pythagoras’ Theorem and Its Proofs A. Pythagoras’ Theorem

Example 5T Solution: 10 Square Roots and Pythagoras’ Theorem The figure shows a right-angled triangle PQR with Q = 90, PQ = 12 and QR = 9. Find the value of q. Solution: In PQR, PQ2 + QR2  PR2 (Pyth. theorem) 122 + 92  q2 q2 = 144 + 81 = 225 ∴

Example 6T Solution: 10 Square Roots and Pythagoras’ Theorem In XYZ, X = 90, XY = 2 and YZ = 3. Find the value of f. (Give the answer in surd form.) Solution: In XYZ, XY2 + XZ2  YZ2 (Pyth. theorem) 22 + f 2  32 4 + f 2  92 f 2  5 ∴

10.2 Pythagoras’ Theorem and Its Proofs A. Pythagoras’ Theorem

Example 7T Solution: 10 Square Roots and Pythagoras’ Theorem In the figure, ABCD is a right-angled trapezium. Find AD. Solution: Construct a line CE such that DA // CE. ADCE is a rectangle, i.e., AD  CE and AE  DC  10 cm ∴ BE  (16 – 10) cm  6 cm Since BCE is a right-angled triangle, BE2 + CE2  BC2 (Pyth. theorem) ∴ ∴ AD  CE

Example 8T Solution: 10 Square Roots and Pythagoras’ Theorem In the figure, XYZ is a right-angled triangle with Y = 90. WZ = 4YW. WZ = 8 cm and XZ = 11 cm. (a) Find XY. (b) Find perimeter of XWZ. (Give the answer correct to 3 significant figures.) Solution: (a) 4YW  WZ In XYZ, 4YW  8 cm XY2 + YZ2  XZ2 (Pyth. theorem) YW  2 cm ∴ ∴ YZ  (2 + 8) cm  10 cm

Example 8T Solution: 10 Square Roots and Pythagoras’ Theorem In the figure, XYZ is a right-angled triangle with Y = 90. WZ = 4YW. WZ = 8 cm and XZ = 11 cm. (a) Find XY. (b) Find perimeter of XWZ. (Give the answer correct to 3 significant figures.) Solution: (b) In WXY, WX2 = WY2 + XY2 (Pyth. theorem) ∴ Perimeter of XWZ  (5 + 8 + 11) cm

10.2 Pythagoras’ Theorem and Its Proofs B. Different Proofs of Pythagoras’ Theorem

10.2 Pythagoras’ Theorem and Its Proofs B. Different Proofs of Pythagoras’ Theorem

10.2 Pythagoras’ Theorem and Its Proofs B. Different Proofs of Pythagoras’ Theorem

10.2 Pythagoras’ Theorem and Its Proofs B. Different Proofs of Pythagoras’ Theorem

10.2 Pythagoras’ Theorem and Its Proofs B. Different Proofs of Pythagoras’ Theorem

10.3 Applications of Pythagoras’ Theorem

Example 9T Solution: 10 Square Roots and Pythagoras’ Theorem A ladder of length 3.2 m leans against a vertical wall. The distance of its top from the ground is the same as the distance of its foot from the wall. Find the distance of the foot of the ladder from the wall. (Give the answer correct to 1 decimal place.) Solution: Let x m be the distance of the foot of the ladder from the wall. x2 + x2  3.22 (Pyth. theorem)  2.3 (cor. to 1 d. p.)  The distance of the foot of the ladder from the wall is 2.3 m.

Example 10T Solution: 10 Square Roots and Pythagoras’ Theorem A wire is bent to form a trapezium with AD // BC, A = B = 90. If AB = 12 cm, AD = 15 cm and BC = 20 cm, find the length of the wire. Solution: As shown in the figure, construct a line DE such that DE  BC to form a right-angled triangle CDE. From the figure, ABED is a rectangle.  DE  AB  12 cm and CE  (20  15) cm  5 cm

Example 10T Solution: 10 Square Roots and Pythagoras’ Theorem A wire is bent to form a trapezium with AD // BC, A = B = 90. If AB = 12 cm, AD = 15 cm and BC = 20 cm, find the length of the wire. Solution: In CDE, (Pyth. theorem)  Length of the wire = (12 + 20 + 13 + 15) cm

Example 11T Solution: 10 Square Roots and Pythagoras’ Theorem Peter and Lily left school at 4:00 p.m. Peter walked due east at a speed of 2.4 m/s to reach the library at 4.15 p.m. Lily walked due north at a speed of 2.25 m/s to reach the bookstore at 4:12 p.m. (a) How far did each of them walk? (b) Find the distance between the library and the bookstore. Solution: (a) As shown in the figure, AB = (2.4  15  60) m  2160 m AC  (2.25  12  60) m  1620 m ∴ Peter walked 2160 m, Lily walked 1620 m.

Example 11T Solution: 10 Square Roots and Pythagoras’ Theorem Peter and Lily left school at 4:00 p.m. Peter walked due east at a speed of 2.4 m/s to reach the library at 4.15 p.m. Lily walked due north at a speed of 2.25 m/s to reach the bookstore at 4:12 p.m. (a) How far did each of them walk? (b) Find the distance between the library and the bookstore. Solution: (b) In ABC, BC2 = AC2 + AB2 (Pyth. theorem) ∴ ∴ The distance between the library and the bookstore is 2700 m.

Example 12T Solution: 10 Square Roots and Pythagoras’ Theorem The figure shows the layout of a piece of land on a rectangular coordinate plane. A(–3, 2), B(–3, –5), C(2, –3) and D(6, 2) are the vertices. Suppose 1 unit represents 1 m. Find the perimeter of the land correct to 3 significant figures. Solution: First draw a horizontal line through C and meet AB at E(–3, –3), then draw a vertical line through C and meet AD at F(2, 2). AB  [2 – (–5)] units = 7 units AD  [6 – (–3)] units = 7 units BE  [–3 – (–5)] units = 2 units CF  [2 – (–3)] units = 5 units CE  [2 – (–3)] units = 5 units DF  (6 – 2) units = 4 units

Example 12T Solution: 10 Square Roots and Pythagoras’ Theorem The figure shows the layout of a piece of land on a rectangular coordinate plane. A(–3, 2), B(–3, –5), C(2, –3) and D(6, 2) are the vertices. Suppose 1 unit represents 1 m. Find the perimeter of the land correct to 3 significant figures. Solution: In BCE, BC2 = BE2 + CE2 (Pyth. theorem) ∴ In CDF, CD2 = CF2 + DF2 (Pyth. theorem) ∴ ∴ Perimeter of the land

10.4 Converse of Pythagoras’ Theorem and Its Applications

10.4 Converse of Pythagoras’ Theorem and Its Applications

Example 13T Solution: 10 Square Roots and Pythagoras’ Theorem In the figure, XZ = 25, YZ = 30 and XW = 20. W is the mid-point of YZ. (a) Prove that XWZ = 90. (b) Prove that XYZ is an isosceles triangle. Solution: (a) In XWZ, XW2 + WZ2  202 + 152 WZ  30  2 = 400 + 225 = 15 = 625 XZ2  252 = 625 ∵ XW2 + WZ2  XZ2 ∴ XWZ  90 (converse of Pyth. theorem)

Example 13T Solution: 10 Square Roots and Pythagoras’ Theorem In the figure, XZ = 25, YZ = 30 and XW = 20. W is the mid-point of YZ. (a) Prove that XWZ = 90. (b) Prove that XYZ is an isosceles triangle. Solution: (b) In XWY, WY  30  2 = 15 XY2 = XW2 + WY2 (Pyth. theorem) ∵ XWZ = 90 ∴ ∴ XWY  90 ∴ XYZ is an isosceles triangle.

Example 14T Solution: 10 Square Roots and Pythagoras’ Theorem The figure shows a triangular piece of paper. It is known that XW  YZ, XW = 12 cm, YW = 9 cm, WZ = 16 cm. (a) Find XY and XZ. (b) Prove that the paper is in the shape of a right-angled triangle. Solution: (a) In XYW, XY2 = XW2 + WY2 (Pyth. theorem) ∴ In XWZ, XZ2 = XW2 + WZ2 (Pyth. theorem) ∴

Example 14T Solution: 10 Square Roots and Pythagoras’ Theorem The figure shows a triangular piece of paper. It is known that XW  YZ, XW = 12 cm, YW = 9 cm, WZ = 16 cm. (a) Find XY and XZ. (b) Prove that the paper is in the shape of a right-angled triangle. Solution: (b) In XYZ, XY2 + XZ2  152 + 202 = 225 + 400 = 625 ∵ XY2 + XZ2  YZ2 ∴ YXZ  90 (converse of Pyth. theorem) ∴ The paper is in the shape of a right-angled triangle.

10.5 Rational and Irrational Numbers A. Rational Numbers

10.5 Rational and Irrational Numbers A. Rational Numbers

Example 15T Solution: 10 Square Roots and Pythagoras’ Theorem Express the following numbers in the form , where x, y are integers and y  0. (a) 7 (b) (c) –3.55 Solution: (a) (b) (c)

Example 16T Solution: 10 Square Roots and Pythagoras’ Theorem Express in the form , where x, y are integers and y  0. Solution: Let = 1.272727... .............. (1) then 100x = 127.2727.......................... (2) (2) – (1): 100x – x  127.2727... – 1.272727... 99x  126

10.5 Rational and Irrational Numbers B. Irrational Numbers Numbers with radical signs may not be surds. For example,  3, which is not a surd.

Example 17T Solution: 10 Square Roots and Pythagoras’ Theorem Determine whether the following numbers are rational or irrational numbers. Explain your answers. (a) (b) (c) Solution: (a) ∵ ∴ is a rational number. (b) ∵ is a surd. ∴ is an irrational number. (c) ∵ which is neither a terminating decimal nor a recurring decimal. ∴ is an irrational number.

10.5 Rational and Irrational Numbers C. Representing Irrational Numbers on the Number Line

Example 18T Solution: 10 Square Roots and Pythagoras’ Theorem Represent on the number line. Solution: Step 1: Consider 13 as the sum of two square numbers, i.e., 22 + 32 = 13. Step 2: Construct a right-angled triangle OAB with A  90, OA  3 units and AB  2 units. Step 3: Draw an arc with centre O and radius OB to meet the number line at C. We have C

Example 19T Solution: 10 Square Roots and Pythagoras’ Theorem Represent on the number line. Solution: Step 1: Consider 11 as the sum of square numbers, i.e., 32 + 12 + 12 = 11. Step 2: Construct a right-angled triangle with legs 3 units and 1 unit. Then mark on the number line. Step 3: Continue to construct a right-angled triangle with legs units and 1 units. Then mark on the number line.

10.5 Rational and Irrational Numbers D. First Crisis of Mathematics

10.6 Properties and Operations of Surds A. Properties of Surds

Example 20T Solution: 10 Square Roots and Pythagoras’ Theorem Express in its simplest form. Solution:

Example 21T Solution: 10 Square Roots and Pythagoras’ Theorem (a) Express 396 and 5324 as the products of their prime factors in index notation. (b) If Solution: (a) (b)

10.6 Properties and Operations of Surds A. Properties of Surds

Example 22T Solution: 10 Square Roots and Pythagoras’ Theorem Express in its simplest form. Solution:

10.6 Properties and Operations of Surds B. Addition and Subtraction of Surds

Example 23T Solution: 10 Square Roots and Pythagoras’ Theorem Simplify the following expressions. (a) (b) Solution: (a) (b)

Example 24T Solution: 10 Square Roots and Pythagoras’ Theorem Simplify the following expressions. (a) (b) Solution: (a) (b)

10.6 Properties and Operations of Surds C. Multiplication and Division of Surds

Example 25T Solution: 10 Square Roots and Pythagoras’ Theorem Simplify the following expressions. (a) (b) Solution: (b) (a)

10.6 Properties and Operations of Surds D. Rationalization of Denominators

10.6 Properties and Operations of Surds D. Rationalization of Denominators

Example 26T Solution: 10 Square Roots and Pythagoras’ Theorem Rationalize the denominators of the following expressions. (a) (b) Solution: (b) (a)

10.6 Properties and Operations of Surds E. Mixed Operations of Surds (a + b)2  a2 + 2ab + b2

Example 27T Solution: 10 Square Roots and Pythagoras’ Theorem Simplify the following expressions. (a) (b) Solution: (a) (b)

Example 28T Solution: 10 Square Roots and Pythagoras’ Theorem Simplify . Solution:

Follow-up 1 Solution: 10 Square Roots and Pythagoras’ Theorem Find the square roots of each of the following numbers. (a) 36 (b) 1.44 (c) Solution: (a) ∵ 6  6  36 and (–6)  (–6)  36 ∴ The square roots of 36 are 6. (b) ∵ 1.2  1.2  1.44 and (–1.2)  (–1.2)  1.44 ∴ The square roots of 1.44 are 1.2. (c) ∵ and ∴ The square roots of are .

Follow-up 2 Solution: 10 Square Roots and Pythagoras’ Theorem Find the value of . Solution:

Follow-up 3 Solution: 10 Square Roots and Pythagoras’ Theorem Without using a calculator, between which two consecutive integers does each of the following surds lie? (a) (b) Solution: (b) ∵ 144  155  169 (a) ∵ 25  31  36 ∴ ∴ ∴ lies between 5 and 6. ∴ lies between 12 and 13.

Follow-up 4 Solution: 10 Square Roots and Pythagoras’ Theorem Arrange 11, 12, and in descending order. Solution: ∵ 144  137  122  121 ∴ ∴ 12, , , 11 are in decending order.

Follow-up 5 Solution: 10 Square Roots and Pythagoras’ Theorem The figure shows a right-angled triangle ABC with A = 90, AB = 12 and AC = 16. Find the value of a. Solution: In ABC, AB2 + AC2  BC2 (Pyth. theorem) 122 + 162  a2 a2 = 144 + 256 = 400 ∴

Follow-up 6 Solution: 10 Square Roots and Pythagoras’ Theorem In PQR, R = 90, PQ = 29 and QR = 20. Find the value of d. Solution: In PQR, PR2 + QR2  PQ2 (Pyth. theorem) d2 + 202  292 d2 + 400  841 d2 = 441 ∴

Follow-up 7 Solution: 10 Square Roots and Pythagoras’ Theorem In the figure, PQRS is a right-angled trapezium. Find SR. Solution: Construct a line PT such that PT // QR. PQRT is a rectangle, i.e., PT  QR  24 cm and TR  PQ  8 cm Since PST is a right-angled triangle, ST2 + PT2  PS2 (Pyth. theorem) ∴ ∴ SR  ST + TR  (7 + 8) cm

Follow-up 8 Solution: 10 Square Roots and Pythagoras’ Theorem In the figure, PQR is a right-angled triangle with R = 90. QSR is a straight line. QS = 2SR. PR = 12 cm and PS = 13 cm. (a) Find SQ. (b) Find PQ. (Give the answer correct to 3 significant figures.) Solution: (a) In PSR, PR2 + SR2  PS2 (Pyth. theorem) ∴ SQ  2SR  2  5 cm

Follow-up 8 Solution: 10 Square Roots and Pythagoras’ Theorem In the figure, PQR is a right-angled triangle with R = 90. QSR is a straight line. QS = 2SR. PR = 12 cm and PS = 13 cm. (a) Find SQ. (b) Find PQ. (Give the answer correct to 3 significant figures.) Solution: (b) In PQR, QR  SQ + SR (Pyth. theorem)  (10 + 5) cm ∴  15 cm

Follow-up 9 Solution: 10 Square Roots and Pythagoras’ Theorem A ladder of length 2 m leans against a vertical wall. Its foot is 1.6 m away from the wall. Find the vertical distance of the top of the ladder from the ground Solution: Let x m be the vertical distance of the top of the ladder from the ground. x2 + 1.62  22 (Pyth. theorem)  The vertical distance of the top of the ladder from the ground is 1.2 m.

Follow-up 10 Solution: 10 Square Roots and Pythagoras’ Theorem There are two vertical lampposts standing on a horizontal ground. One post is 6 m high and the other is 14 m high. If the distance between the tops of the lampposts is 17 m, find the horizontal distance between the two lampposts. Solution: As shown in the figure, construct a line BC such that CB // ED to form a right-angled triangle ABC. From the figure, CE  BD  6 m and AC  (14  6) m  8 m

Follow-up 10 Solution: 10 Square Roots and Pythagoras’ Theorem There are two vertical lampposts standing on a horizontal ground. One post is 6 m high and the other is 14 m high. If the distance between the tops of the lampposts is 17 m, find the horizontal distance between the two lampposts. Solution: In ABC,   The horizontal distance between the two lampposts is 15 m.

Follow-up 11 Solution: 10 Square Roots and Pythagoras’ Theorem Ships A and B left the port P at the same time. Ship A sailed due west and ship B sailed due south. After 3 hours, they reached ports S and T respectively. Suppose PT = 72 km and ST = 90 km. (a) Find the speeds of ships A and B. Solution: (a) In PST, PS2 + PT2  ST2 (Pyth. theorem) ∴

Follow-up 11 Solution: 10 Square Roots and Pythagoras’ Theorem Ships A and B left the port P at the same time. Ship A sailed due west and ship B sailed due south. After 3 hours, they reached ports S and T respectively. Suppose PT = 72 km and ST = 90 km. (a) Find the speeds of ships A and B. (b) After receiving a rescue message from ship A at port S, a helicopter was sent tout from port P at a speed of 81 km/h. How long did it take to reach port S? Solution: (b)

Follow-up 12 Solution: 10 Square Roots and Pythagoras’ Theorem The figure shows the layout of a triangular path on a rectangular coordinate plane. P (–7, 13), Q(17, –5) and R(–7, –5) are the vertices. Suppose 1 unit represents 1 m. (a) Find the total length of the path. Solution: (a) PR  [13 – (–5)] units (Pyth. theorem) = 18 units PQ  units QR  [17 – (–7)] units  units = 24 units  30 units 

Follow-up 12 Solution: 10 Square Roots and Pythagoras’ Theorem The figure shows the layout of a triangular path on a rectangular coordinate plane. P (–7, 13), Q(17, –5) and R(–7, –5) are the vertices. Suppose 1 unit represents 1 m. (a) Find the total length of the path. (b) If the cost of constructing a path is $420/m, find the total cost of constructing the path. Solution: (b) Total cost of constructing the path  $(72  420)

Follow-up 13 Solution: 10 Square Roots and Pythagoras’ Theorem In the figure, PR = 16, QR = 17 and SQ = 15. S is the mid-point of PR. Prove that QS is the perpendicular bisector of PR. Solution: In SQR, SR  16  2 = 8 SQ2 + SR2  152 + 82 ∵ SQ2 + SR2  QR2 = 225 + 64 ∴ QSR  90 (converse of Pyth. theorem) = 289 ∴ QS is the perpendicular bisector of PR. QR2  172 = 289

Follow-up 14 Solution: 10 Square Roots and Pythagoras’ Theorem The figure shows a piece of farmland PQRS. A partition QS is used to separate the chickens and the rabbits. It is known that , QR = 12 m, RS = 5 m, SP = 11 m and R  90. (a) Find the length of the partition QS. (b) Prove that PQS is a right-angled triangle. Solution: (b) In PQS, (a) In SQR, (Pyth. theorem)  121 + 48 ∴  169 QS2 = 169 ∵ PS2 + PQ2 = QS2 ∴ SPQ  90 (converse of Pyth. theorem) ∴ PQS is a right-angled triangle.

Follow-up 14 Solution: 10 Square Roots and Pythagoras’ Theorem The figure shows a piece of farmland PQRS. A partition QS is used to separate the chickens and the rabbits. It is known that , QR = 12 m, RS = 5 m, SP = 11 m and R  90. (a) Find the length of the partition QS. (b) Prove that PQS is a right-angled triangle. (c) Find the area of the piece of farmland. (Give the answer correct to 1 decimal place.) Solution: (c) Area of the piece of farmland = Area of PQS + Area of QRS

Follow-up 15 Solution: 10 Square Roots and Pythagoras’ Theorem Express the following numbers in the form , where x, y are integers and y ≠ 0. (a) –3 (b) (c) 1.975 Solution: (a) (b) (c)

Follow-up 16 Solution: 10 Square Roots and Pythagoras’ Theorem Express in the form , where x, y are integers and y ≠ 0. Solution: Let = 0.454545... .............. (1) then 100x = 45.4545..... ..................... (2) (2) – (1): 100x – x  45.4545... – 0.454545... 99x  45

Follow-up 17 Solution: 10 Square Roots and Pythagoras’ Theorem Determine whether the following numbers are rational or irrational numbers. Explain your answers. (a) (b) (c) π + 1 Solution: (a) ∵ is a surd. ∴ is an irrational number. (b) ∵ ∴ is a rational number. (c) ∵  + 1  3.1415… + 1  4.1415… which is neither a terminating decimal nor a recurring decimal. ∴  + 1 is an irrational number.

Follow-up 18 Solution: 10 Square Roots and Pythagoras’ Theorem Represent on the number line. Solution: Step 1: Consider 58 as the sum of two square numbers, i.e., 32 + 72 = 58. Step 2: Construct a right-angled triangle OAB with A  90, OA  7 units and AB  3 units. Step 3: Draw an arc with centre O and radius OB to meet the number line at C. We have

Follow-up 19 Solution: 10 Square Roots and Pythagoras’ Theorem Represent on the number line. Solution: Step 1: Consider 30 as the sum of square numbers, i.e., 52 + 12 + 22 = 30. Step 2: Construct a right-angled triangle with legs 5 units and 1 unit. Then mark on the number line. Step 3: Continue to construct a right-angled triangle with legs units and 2 units. Then mark on the number line.

Follow-up 20 Solution: 10 Square Roots and Pythagoras’ Theorem Express in its simplest form. Solution:

Follow-up 21 Solution: 10 Square Roots and Pythagoras’ Theorem (a) Express 500 and 720 as the products of their prime factors in index notation. (b) If Solution: (a) (b)

Follow-up 22 Solution: 10 Square Roots and Pythagoras’ Theorem Express in its simplest form. Solution:

Follow-up 23 Solution: 10 Square Roots and Pythagoras’ Theorem Simplify the following expressions. (a) (b) Solution: (a) (b)

Follow-up 24 Solution: 10 Square Roots and Pythagoras’ Theorem Simplify the following expressions. (a) (b) Solution: (a) (b)

Follow-up 25 Solution: 10 Square Roots and Pythagoras’ Theorem Simplify the following expressions. (a) (b) Solution: (a) (b)

Follow-up 26 Solution: 10 Square Roots and Pythagoras’ Theorem Rationalize the denominators of the following expressions. (a) (b) Solution: (b) (a)

Follow-up 27 Solution: 10 Square Roots and Pythagoras’ Theorem Simplify the following expressions. (a) (b) Solution: (a) (b)

Follow-up 28 Solution: 10 Square Roots and Pythagoras’ Theorem Simplify . Solution: