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Geometry - Pythagoras Theorem and its Converse

Grade 9ICSE

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

🔑Concepts

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The Pythagoras Theorem states that in a right-angled triangle, the square of the hypotenuse (hh) is equal to the sum of the squares of the other two sides, namely the perpendicular (pp) and the base (bb). The hypotenuse is always the side opposite the 90∘90^{\circ} angle and is the longest side.

Right-angled triangle labeled with Base, Perpendicular, and Hypotenuse.
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The Converse of Pythagoras Theorem states that 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 angle opposite the longest side is a right angle (90∘90^{\circ}). If c2=a2+b2c^2 = a^2 + b^2, then the triangle is right-angled.

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Pythagorean Triplets are sets of three positive integers (a,b,c)(a, b, c) that satisfy the rule a2+b2=c2a^2 + b^2 = c^2. Common examples include (3,4,5)(3, 4, 5), (5,12,13)(5, 12, 13), and (7,24,25)(7, 24, 25).

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The theorem is used to find the length of the diagonal of a rectangle or square. For a rectangle with length ll and width ww, the diagonal dd creates two congruent right-angled triangles.

Rectangle showing a diagonal splitting it into two right-angled triangles.

📐Formulae

Pythagoras Theorem: h2=p2+b2h^2 = p^2 + b^2

Length of Hypotenuse: h=p2+b2h = \sqrt{p^2 + b^2}

Length of Perpendicular: p=h2−b2p = \sqrt{h^2 - b^2}

Length of Base: b=h2−p2b = \sqrt{h^2 - p^2}

Diagonal of a Rectangle: d=l2+w2d = \sqrt{l^2 + w^2}

Diagonal of a Square: d=a2d = a\sqrt{2}

Condition for Pythagorean Triplet: c2=a2+b2c^2 = a^2 + b^2

💡Examples

Problem 1:

A ladder 1313 m long reaches a window 1212 m above the ground. Find the distance of the foot of the ladder from the wall.

Solution:

Let the length of the ladder be the hypotenuse h=13h = 13 m. Let the height of the window be the perpendicular p=12p = 12 m. We need to find the base bb. Using Pythagoras Theorem: b2=h2−p2b^2 = h^2 - p^2 b2=132−122b^2 = 13^2 - 12^2 b2=169−144b^2 = 169 - 144 b2=25b^2 = 25 b=25=5b = \sqrt{25} = 5 m.

Explanation:

In this real-world scenario, the wall, the ground, and the ladder form a right-angled triangle. The ladder represents the hypotenuse because it is leaning opposite the 90∘90^\circ angle formed by the wall and the ground.

Problem 2:

Determine whether a triangle with sides 88 cm, 1515 cm, and 1717 cm is a right-angled triangle.

Solution:

Let the sides be a=8a = 8, b=15b = 15, and the longest side c=17c = 17. Calculate the sum of squares of the smaller sides: a2+b2=82+152=64+225=289a^2 + b^2 = 8^2 + 15^2 = 64 + 225 = 289 Calculate the square of the longest side: c2=172=289c^2 = 17^2 = 289 Since a2+b2=c2a^2 + b^2 = c^2 (289=289289 = 289)...

Explanation:

According to the Converse of Pythagoras Theorem, if the square of the longest side equals the sum of the squares of the other two sides, the triangle is right-angled. Since the values satisfy the equation, this is a right-angled triangle.

Problem 3:

A man drives 66 km North and then 88 km East. Calculate the shortest distance from his starting point to his finishing point.

A path showing movement North then East, forming a right triangle with the displacement.

Solution:

Let AA be the starting point. The man moves 66 km North to BB, and then 88 km East to CC. In right △ABC\triangle ABC, by Pythagoras Theorem: AC2=AB2+BC2AC^2 = AB^2 + BC^2 AC2=62+82AC^2 = 6^2 + 8^2 AC2=36+64AC^2 = 36 + 64 AC2=100AC^2 = 100 AC=100=10AC = \sqrt{100} = 10 So, the shortest distance is 1010 km.

Explanation:

The North and East directions are perpendicular to each other, forming a right-angled triangle where the shortest distance is the hypotenuse.

Problem 4:

An isosceles triangle has equal sides of length 1010 cm each and a base of 1212 cm. Find the altitude (height) of the triangle drawn to the base.

Isosceles triangle with an altitude dividing the base into two equal parts.

Solution:

In an isosceles △ABC\triangle ABC with AB=AC=10AB = AC = 10 cm and BC=12BC = 12 cm, let ADAD be the altitude to the base BCBC. In an isosceles triangle, the altitude to the base bisects the base. Therefore, BD=DC=122=6BD = DC = \frac{12}{2} = 6 cm. In right △ABD\triangle ABD: AB2=AD2+BD2AB^2 = AD^2 + BD^2 102=AD2+6210^2 = AD^2 + 6^2 100=AD2+36100 = AD^2 + 36 AD2=100−36=64AD^2 = 100 - 36 = 64 AD=64=8AD = \sqrt{64} = 8 Thus, the altitude is 88 cm.

Explanation:

By dropping an altitude in an isosceles triangle, we create two congruent right-angled triangles. We then apply Pythagoras theorem to one of these triangles.