Solving Problems Involving Radicals
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This document provides examples and solutions for problems involving radicals. It begins by explaining how radicals are used in the Pythagorean theorem and distance formula. Several worked examples are then shown that apply these concepts to find lengths, areas, velocities, and unknown numbers. The examples demonstrate how to set up and solve equations using radicals to model real-world scenarios like finding the distance between objects or the speed of a falling object.
Solving Problems Involving Radicals
- 1. MATHEMATICS 9 Quarter 2: Module 9 Week 9 PROBLEMS INVOLVING RADICALS
- 2. Problems Involving Radicals Some of the important applications of radicals are problems involving the Pythagorean Theorem and the distance formula. Radicals are also often used in science.
- 3. Ancient Greek mathematicians studied a number of properties of right triangles. One result states that in a right triangle, the sum of the squares of the legs is equal to the square of the hypotenuse. This property is called the Pythagorean Theorem, in honor of Pythagoras (circa 590 B.C.). In symbols, the theorem says that (hypotenuse)2 = (leg)2 + (leg)2. From this equation, the hypotenuse c is equal to 𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆 = (𝒍𝒆𝒈)𝟐+(𝒍𝒆𝒈)𝟐 leg leg hypotenuse
- 4. STEPS IN SOLVING WORD PROBLEMS 1. Read and analyze the problem carefully. 2. Illustrate the problem. 3. Develop the plan to solve it. 4. Solve the problem 5. Looking back.
- 6. Example 1: Find the length of the hypotenuse of a right triangle whose legs are 5 cm and 12 cm. Solutions: The hypotenuse of a right triangle is the side opposite the right angle. the other sides are the legs. Let c = hypotenuse 𝑐 = (5) 2+ 12 2 𝑐 = 25 + 144 𝑐 = 169 𝒄 = 𝟏𝟑 5 cm 12 cm c Therefore, the length of the hypotenuse of the triangle is 13 cm.
- 7. Example 2: A small plane and a helicopter left an airport at the same time. After several minutes, the plane is 7 miles due north of the airport while the helicopter is 5 miles due east of the airport. How far apart are they at that time? Solutions: Let x = distance between the plane and helicopter 𝑥 = (7) 2+ 5 2 𝑥 = 49 + 25 𝑥 = 74 𝒙 ≈ 𝟖. 𝟔 7 mi 5 mi x Plane (North) Helicopter (East) Airport Therefore, the distance between the plane and helicopter that time is approximately 8.6 miles.
- 8. Example 3: Find the diagonal of a square whose area is 40 square meters. Solutions: The area of a square with side s is s2. Moreover, the diagonal d forms a right triangle with two sides each of whose length is s. 𝑑 = 𝑠 2 + 𝑠2 𝑑 = 2𝑠2 𝑑 = 2 40 𝑑 = 80 𝒙 = 4 𝟓 d s s Therefore, the length of the diagonal of the square is 4 𝟓 m.
- 9. Example 4: Find the area of a triangle with sides 8 cm, 9 cm, and 11 cm, respectively. Solutions: The area of a triangle can be found using the formula 𝑨 = 𝒔(𝒔 − 𝒂)(𝒔 − 𝒃)(𝒔 − 𝒄) where A is the area of a triangle, a, b, and c are the lengths of the sides, and 𝒔 = 𝟏 𝟐 𝒂 + 𝒃 + 𝒄 . Since, a = 8, b = 9, and c = 11 𝑠 = 1 2 (8 + 9 + 11) 𝑠 = 1 2 28 𝒔 = 𝟏𝟒 Heron’s Formula 𝑨 = 𝒔(𝒔 − 𝒂)(𝒔 − 𝒃)(𝒔 − 𝒄)
- 10. Example 4: Find the area of a triangle with sides 8 cm, 9 cm, and 11 cm, respectively. Solutions: And the area is, 𝐴 = 𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐) 𝐴 = 14(14 − 8)(14 − 9)(14 − 11) 𝐴 = 14(6)(5)(3) 𝐴 = 1260 𝐴 = (36)(35) 𝑨 = 𝟔 𝟑𝟓 Therefore, the area of the triangle is 𝟔 𝟑𝟓.
- 11. Example 5: Galileo Galilei performed many experiments on freely falling bodies at the Leaning Tower of Pisa. His experiments led to the formulation of relation between the velocity of a freely falling body and the distance travelled by the body. This relationship is expressed by the formula 𝒗 = 𝟐𝒈𝒉 where v is the velocity of the object in ft/s, h is the distance and g is the acceleration due to the gravity whose approximate value on Earth is 32 ft/s2. Find the velocity of a stone after it has fallen 15 ft.
- 12. Continuation: Example 5: Find the velocity of a stone after it has fallen 15 ft. Solutions: The distance (h) is 15 ft and the acceleration due to gravity is (g) is 32 ft/s2, hence, the velocity (v) is: 𝑣 = 2𝑔ℎ 𝑣 = 2 32 ൗ 𝑓𝑡 𝑠2 (15𝑓𝑡) 𝑣 = 960 ൗ 𝑓𝑡2 𝑠2 ≈ 𝟑𝟎. 𝟗𝟖 Τ 𝒇𝒕 𝒔 Therefore, the velocity of the stone is approximately 𝟑𝟎. 𝟗𝟖 Τ 𝒇𝒕 𝒔.
- 13. Example 6: The square root of the sum of a number and 5 is 3. Find the number. Solutions: Let x be the number. Checking: x = 4 Equation: 𝑥 + 5 = 3 𝑥 + 5 = 3 Solve for x. 4 + 5 = 3 𝑥 + 5 = 3 9 = 3 𝑥 + 5 2 = 3 2 3 = 3 TRUE 𝑥 + 5 = 9 𝑥 = 9 − 5 𝒙 = 𝟒 Therefore, the number is 4.
- 14. Example 7: The square root of the difference between a number and 5, plus the square root of the number is equal to 5. Find the number. Solutions: Let x be the number. Checking: x = 9 Equation: 𝑥 − 5 + 𝑥 = 5 𝑥 − 5 + 𝑥 = 5 Solve for x. 9 − 5 + 9 = 5 𝑥 − 5 = 5 − 𝑥 4 + 3 = 5 𝑥 − 5 2 = 5 − 𝑥 2 2+3 = 5 𝑥 − 5 = 25 − 10 𝑥 + 𝑥 5 = 5 TRUE 10 𝑥 = 30 𝑥 = 3 𝒙 = 𝟗 Therefore, the number is 9.