ICSE Class 7 > Class 7 Notes > Mathematics > Chapter Notes: Pythagoras Theorem
Chapter Notes: Pythagoras Theorem
Introduction
The Pythagoras Theorem is a fundamental principle in geometry that applies to right-angled triangles. It states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. Mathematically, for a right-angled triangle with sides a, b, and hypotenuse c, the theorem is expressed as:c2 = a2 + b2
This theorem is used to find the length of an unknown side in a right-angled triangle when the lengths of the other two sides are known. It also helps determine whether a triangle is right-angled by verifying if the side lengths satisfy the theorem.
Tip: Always identify the hypotenuse (the longest side opposite the right angle) before applying the Pythagoras Theorem.
Key Concepts
1. Understanding the Theorem
In a right-angled triangle, the hypotenuse is the side opposite the right angle, and the other two sides are called the legs. The theorem is applicable only when the triangle has a right angle (90°). It can be used to:- Calculate the length of the hypotenuse given the legs.
- Calculate the length of a leg given the hypotenuse and the other leg.
- Verify if a triangle is right-angled by checking if c2 = a2 + b2.
Note: The theorem assumes all side lengths are positive and typically measured in consistent units (e.g., cm, m).
2. Applications of Pythagoras Theorem
The theorem is widely used in real-world scenarios, such as calculating distances, determining heights, or solving problems involving right-angled triangles in construction, navigation, and physics.Warning: Ensure the triangle is right-angled before applying the theorem, as it does not hold for acute or obtuse triangles.
Examples with Solutions
Example 1: Finding the Hypotenuse
Q: In triangle ABC, right-angled at A, AB = 18 cm and AC = 24 cm. Find the length of BC.Sol: According to Pythagoras Theorem:
BC2 = AB2 + AC2
BC2 = 182 + 242
BC2 = 324 + 576 = 900
BC = √900 = 30 cm
Example 2: Finding a Leg
Q: In triangle XYZ, right-angled at Z, XY = 13 cm and XZ = 12 cm. Find the length of YZ.Sol: According to Pythagoras Theorem:
XY2 = XZ2 + YZ2
132 = 122 + YZ2
169 = 144 + YZ2
YZ2 = 169 - 144 = 25
YZ = √25 = 5 cm
Example 3: Finding a Leg with Decimals
Q: In triangle PQR, right-angled at R, PQ = 34 cm and QR = 33.6 cm. Find the length of PR.Sol: According to Pythagoras Theorem:
PQ2 = PR2 + QR2
342 = PR2 + 33.62
1156 = PR2 + 1128.96
PR2 = 1156 - 1128.96 = 27.04
PR = √27.04 = 5.2 cm
Example 4: Verifying a Right-Angled Triangle
Q: Determine if triangles with the following sides are right-angled:(i) 16 cm, 20 cm, 12 cm
Sol: Check if 202 = 162 + 122
202 = 400
162 + 122 = 256 + 144 = 400
Since 400 = 400, the triangle is right-angled.
(ii) 6 m, 9 m, 13 m
Sol: Check if 132 = 92 + 62:
132 = 169
92 + 62 = 81 + 36 = 117
Since 169 ≠ 117, the triangle is not right-angled.
Example 5: Finding the Hypotenuse in a Diagram
Q: In triangle ABC, angle BAC = 90°, AC = 400 m, and AB = 300 m. Find the length of BC.Sol: BC2 = AB2 + AC2
BC2 = 3002 + 4002
BC2 = 90000 + 160000 = 250000
BC = √250000 = 500 m
Example 6: Complex Triangle with Multiple Right Angles
Q: In a figure, angle ACP = angle BDP = 90°, AC = 12 m, BD = 9 m, PA = PB = 15 m. Find CP, PD, and CD.Sol: (i) In triangle ACP:
AP2 = AC2 + CP2
152 = 122 + CP2
225 = 144 + CP2
CP2 = 225 - 144 = 81
CP = √81 = 9 m
(ii) In triangle BPD:
PB2 = BD2 + PD2
152 = 92 + PD2
225 = 81 + PD2
PD2 = 225 - 81 = 144
PD = √144 = 12 m
(iii) CD = CP + PD = 9 + 12 = 21 m
Example 7: Verifying a Right-Angled Triangle
Q: Show that triangle ABC with AB = 9 cm, BC = 40 cm, and AC = 41 cm is right-angled.Sol: Check if AC2 = BC2 + AB2:
412 = 402 + 92
1681 = 1600 + 81 = 1681
Since 1681 = 1681, triangle ABC is right-angled.
Example 8: Complex Figure with Right Angles
Q: In a figure, angle ACB = angle ACD = 90°, AB = 10 m, BC = 6 cm, AD = 17 cm. Find AC and CD.Sol: (i) In triangle ABC:
AB2 = AC2 + BC2
102 = AC2 + 62
100 = AC2 + 36
AC2 = 100 - 36 = 64
AC = √64 = 8 cm
(ii) In triangle ACD:
AD2 = AC2 + CD2
172 = 82 + CD2
289 = 64 + CD2
CD2 = 289 - 64 = 225
CD = √225 = 15 cm
Example 9: Symmetry in a Right-Angled Triangle
Q: In a figure, angle ADB = 90°, AC = AB = 26 cm, BD = DC, AD = 24 cm. Find the length of BC.Sol: In triangle ADC:
AC2 = AD2 + DC2
262 = 242 + DC2
676 = 576 + DC2
DC2 = 676 - 576 = 100
DC = √100 = 10 cm
Since BD = DC, BD = 10 cm.
BC = BD + DC = 10 + 10 = 20 cm
Example 10: Multiple Right Angles in a Figure
Q: In a figure, angle ACD = angle ABC = 90°, AD = 13 cm, BC = 12 cm, AB = 3 cm. Find the length of DC.Sol: (i) In triangle ABC:
AC2 = AB2 + BC2
AC2 = 32 + 122 = 9 + 144 = 153
AC = √153 cm
(ii) In triangle ACD:
AD2 = AC2 + DC2
132 = (√153)2 + DC2
169 = 153 + DC2
DC2 = 169 - 153 = 16
DC = √16 = 4 cm
Example 11: Real-World Application (Ladder Problem)
Q: A ladder, 6.5 m long, rests against a vertical wall. The foot of the ladder is 2.5 m from the wall. Find the height the ladder reaches.Sol: Let BC be the ladder (6.5 m), AB the distance from the wall (2.5 m), and AC the height.
BC2 = AB2 + AC2
6.52 = 2.52 + AC2
42.25 = 6.25 + AC2
AC2 = 42.25 - 6.25 = 36
AC = √36 = 6 m
The ladder reaches a height of 6 m.
Example 12: Real-World Application (Distance Problem)
Q: A boy goes 5 m due north and then 12 m due east. Find the distance between his initial and final positions.Sol: Let AC = 5 m (north), AB = 12 m (east), and BC be the distance.
BC2 = AC2 + AB2
BC2 = 52 + 122 = 25 + 144 = 169
BC = √169 = 13 m
The distance is 13 m.
Example 13: Finding a Diagonal
Q: In a figure, AB = 20 cm, AO = 10 cm, BC = OD = 24 cm. Find the length of AD.Sol: In triangle AOD:
AD2 = AO2 + OD2
AD2 = 102 + 242 = 100 + 576 = 676
AD = √676 = 26 cm
Advanced Applications
The Pythagoras Theorem extends beyond simple triangles. It can be applied to complex figures by breaking them into right-angled triangles, as seen in examples with multiple right angles or shared sides. It is also used in coordinate geometry to calculate distances between points and in 3D geometry for finding diagonals of rectangular prisms.Tip: When dealing with complex figures, draw the diagram and identify all right-angled triangles to apply the theorem systematically.
The document Chapter Notes: Pythagoras Theorem is a part of the Class 7 Course Mathematics Class 7 ICSE.
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FAQs on Chapter Notes: Pythagoras Theorem
| 1. What is the Pythagorean theorem and how is it used in right triangles? |  |
Ans. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. This is expressed as a² + b² = c², where c is the length of the hypotenuse and a and b are the lengths of the other two sides. It is used to determine the length of one side of a right triangle when the lengths of the other two sides are known.
| 2. Can you provide a practical example of how to apply the Pythagorean theorem? | |
Ans. Certainly! If you have a right triangle where one side measures 3 meters and the other side measures 4 meters, you can find the hypotenuse. Using the theorem: a² + b² = c², we substitute the values: 3² + 4² = c² → 9 + 16 = c² → 25 = c². Therefore, c = √25 = 5 meters. The hypotenuse is 5 meters long.
| 3. What are some real-life applications of the Pythagorean theorem? | |
Ans. The Pythagorean theorem has various real-life applications, including construction (to ensure structures are square), navigation (to determine the shortest path), and in various fields like engineering and computer graphics for calculating distances. For instance, if you want to find the shortest distance across a park, you can treat the park's paths as sides of a right triangle and use the theorem to calculate the direct distance.
| 4. How can the Pythagorean theorem be used in determining distances on a coordinate plane? | |
Ans. In a coordinate plane, if you have two points, say (x₁, y₁) and (x₂, y₂), you can use the Pythagorean theorem to find the distance between them. The difference in x-coordinates (Δx = x₂ - x₁) and the difference in y-coordinates (Δy = y₂ - y₁) form the two sides of a right triangle. The distance (d) is then calculated using the theorem: d = √(Δx² + Δy²).
| 5. Is the Pythagorean theorem applicable in non-Euclidean geometry, and if so, how? | |
Ans. In non-Euclidean geometry, such as spherical geometry, the Pythagorean theorem does not hold in its classical form. However, there are analogous concepts. For example, on the surface of a sphere, the relationship between distances is described by spherical trigonometry. While the basic principles of triangles remain, the calculations differ due to the curvature of space.
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