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Geometry and Measurement - The Pythagorean Theorem and its Converse

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Pythagorean Theorem states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This is expressed as a2+b2=c2a^2 + b^2 = c^2.

A right-angled triangle showing sides a, b and hypotenuse c.
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The Converse of the Pythagorean Theorem is used to determine if a triangle is right-angled. If the side lengths satisfy a2+b2=c2a^2 + b^2 = c^2, then the angle opposite the longest side is exactly 90∘90^\circ.

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Pythagorean triples are sets of three positive integers (a,b,c)(a, b, c) that perfectly satisfy the theorem, such as (3,4,5)(3, 4, 5), (5,12,13)(5, 12, 13), and (8,15,17)(8, 15, 17).

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The distance between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) on a coordinate plane is an application of the theorem, where the distance dd is the hypotenuse of a right triangle with legs ∣x2−x1∣|x_2 - x_1| and ∣y2−y1∣|y_2 - y_1|.

📐Formulae

a2+b2=c2a^2 + b^2 = c^2

c=a2+b2c = \sqrt{a^2 + b^2}

a=c2−b2a = \sqrt{c^2 - b^2}

d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

💡Examples

Problem 1:

A ladder is leaning against a wall. The base of the ladder is 55 meters away from the wall, and the ladder reaches a height of 1212 meters up the wall. How long is the ladder?

Solution:

  1. Identify the given sides: The base distance (a=5a = 5 m) and the height (b=12b = 12 m) are the legs of a right triangle.
  2. We need to find the length of the ladder, which is the hypotenuse (cc).
  3. Apply the formula: c2=a2+b2c^2 = a^2 + b^2
  4. Substitute the values: c2=52+122c^2 = 5^2 + 12^2
  5. Calculate: c2=25+144=169c^2 = 25 + 144 = 169
  6. Solve for cc: c=169=13c = \sqrt{169} = 13 meters.

Explanation:

Since the wall and the ground form a 90∘90^{\circ} angle, we treat the ladder as the hypotenuse. We square both known sides, sum them, and take the square root to find the total length.

Problem 2:

Determine if a triangle with side lengths 77 cm, 2424 cm, and 2626 cm is a right-angled triangle.

Solution:

  1. Identify the longest side: c=26c = 26. The other sides are a=7a = 7 and b=24b = 24.
  2. Calculate the square of the longest side: c2=262=676c^2 = 26^2 = 676.
  3. Calculate the sum of the squares of the shorter sides: a2+b2=72+242=49+576=625a^2 + b^2 = 7^2 + 24^2 = 49 + 576 = 625.
  4. Compare the results: 625≠676625 \neq 676.
  5. Conclusion: Since a2+b2≠c2a^2 + b^2 \neq c^2, the triangle is not a right-angled triangle.

Explanation:

Using the Converse of the Pythagorean Theorem, we check if the relationship holds true. Because the sum of the squares of the legs does not equal the square of the longest side, the triangle does not contain a right angle.

Problem 3:

Calculate the length of the diagonal of a rectangle that has a width of 88 cm and a length of 1515 cm.

A rectangle with a diagonal line forming two right triangles.

Solution:

d2=82+152d^2 = 8^2 + 15^2 d2=64+225d^2 = 64 + 225 d2=289d^2 = 289 d=289d = \sqrt{289} d=17 cmd = 17\text{ cm}

Explanation:

A rectangle's diagonal splits it into two right-angled triangles. We use the width and length as the two legs (aa and bb) to find the hypotenuse (dd).

Problem 4:

An isosceles triangle has two equal sides of length 1010 cm and a base of 1212 cm. Find the perpendicular height (hh) of the triangle.

An isosceles triangle with a perpendicular height line drawn to the base.

Solution:

The height bisects the base into two segments of 6 cm6\text{ cm}. h2+62=102h^2 + 6^2 = 10^2 h2+36=100h^2 + 36 = 100 h2=100−36h^2 = 100 - 36 h2=64h^2 = 64 h=8 cmh = 8\text{ cm}

Explanation:

In an isosceles triangle, the altitude (height) to the base creates two congruent right-angled triangles. The base of each right triangle is half the total base (12/2=612/2 = 6). We then solve for the missing leg.