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Geometry - Pythagoras' Theorem and its applications

Grade 7ICSE

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

🔑Concepts

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Pythagoras' 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 (BaseBase and PerpendicularPerpendicular). This is expressed as c2=a2+b2c^2 = a^2 + b^2.

Right-angled triangle labeling sides a, b, and c.
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A Pythagorean Triplet consists of three positive integers aa, bb, and cc, such that a2+b2=c2a^2 + b^2 = c^2. Common examples include (3,4,5)(3, 4, 5), (5,12,13)(5, 12, 13), and (8,15,17)(8, 15, 17).

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The Converse of Pythagoras' Theorem: If the square of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle must be a right-angled triangle, and the angle opposite the longest side is 90∘90^\circ.

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Applications include finding the diagonal of a rectangle. If a rectangle has length ll and breadth bb, its diagonal dd forms the hypotenuse of a right triangle: d=l2+b2d = \sqrt{l^2 + b^2}.

📐Formulae

c2=a2+b2c^{2} = a^{2} + b^{2}

Hypotenuse2=Base2+Perpendicular2Hypotenuse^{2} = Base^{2} + Perpendicular^{2}

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

a=c2−b2a = \sqrt{c^{2} - b^{2}}

b=c2−a2b = \sqrt{c^{2} - a^{2}}

💡Examples

Problem 1:

A ladder is placed against a wall such that its foot is 55 m away from the wall. If the ladder reaches a window 1212 m high on the wall, find the length of the ladder.

Solution:

  1. Identify the given values: Base (aa) = 55 m, Perpendicular (bb) = 1212 m. We need to find the length of the ladder, which is the Hypotenuse (cc).
  2. Use the formula: c2=a2+b2c^{2} = a^{2} + b^{2}
  3. Substitute the values: c2=52+122c^{2} = 5^{2} + 12^{2}
  4. Calculate the squares: c2=25+144c^{2} = 25 + 144
  5. Add the results: c2=169c^{2} = 169
  6. Find the square root: c=169=13c = \sqrt{169} = 13 m.

Explanation:

The wall and ground form a right angle. The ladder forms the hypotenuse. By squaring the distances from the wall and the height of the window, we find the square of the ladder's length.

Problem 2:

The hypotenuse of a right-angled triangle is 1010 cm and one of its legs is 88 cm. Find the length of the third side.

Solution:

  1. Identify the given values: Hypotenuse (cc) = 1010 cm, Leg (aa) = 88 cm. We need to find the other Leg (bb).
  2. Use the modified formula: b2=c2−a2b^{2} = c^{2} - a^{2}
  3. Substitute the values: b2=102−82b^{2} = 10^{2} - 8^{2}
  4. Calculate the squares: b2=100−64b^{2} = 100 - 64
  5. Subtract the values: b2=36b^{2} = 36
  6. Find the square root: b=36=6b = \sqrt{36} = 6 cm.

Explanation:

When we know the hypotenuse and one side, we subtract the square of the known side from the square of the hypotenuse to solve for the missing side.

Problem 3:

A rectangle has a length of 1515 cm and a diagonal of 1717 cm. Calculate the breadth of the rectangle.

Rectangle with diagonal 17 cm and length 15 cm.

Solution:

Diagonal2=Length2+Breadth2Diagonal^2 = Length^2 + Breadth^2 172=152+b217^2 = 15^2 + b^2 289=225+b2289 = 225 + b^2 b2=289−225b^2 = 289 - 225 b2=64b^2 = 64 b=64=8b = \sqrt{64} = 8 Therefore, the breadth is 88 cm.

Explanation:

In a rectangle, the diagonal divides it into two right-angled triangles. We apply Pythagoras' Theorem where the diagonal is the hypotenuse (1717 cm) and the length is the base (1515 cm). Solving for the height (breadth) gives 88 cm.

Problem 4:

Two poles of heights 99 m and 1414 m stand vertically on a plane ground. If the distance between their feet is 1212 m, find the distance between their tops.

Two vertical poles connected to form a right triangle at the top.

Solution:

Let the height of the shorter pole be AB=9AB = 9 m and the taller pole be CD=14CD = 14 m. The distance between feet BD=12BD = 12 m. Draw AE⊥CDAE \perp CD. Then AE=BD=12AE = BD = 12 m. CE=CD−DE=CD−AB=14−9=5CE = CD - DE = CD - AB = 14 - 9 = 5 m. In right triangle AECAEC: AC2=AE2+CE2AC^2 = AE^2 + CE^2 AC2=122+52AC^2 = 12^2 + 5^2 AC2=144+25=169AC^2 = 144 + 25 = 169 AC=169=13AC = \sqrt{169} = 13 The distance between the tops is 1313 m.

Explanation:

By drawing a horizontal line from the top of the shorter pole to the taller pole, we create a right-angled triangle. The base of this triangle is the distance between the poles (1212 m) and the height is the difference in pole heights (55 m). The hypotenuse is the distance between the tops.