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CAT Pythagoras Theorem - Practice Questions & MCQ

Edited By admin | Updated on Oct 04, 2023 04:20 PM | #CAT

Quick Facts

3 Questions around this concept.

Solve by difficulty

A vertical stick 3.6 m long casts a shadow 45 cm long on the ground. At the same time, what is the length of the shadow of a pole 6 m high?

50cm

60cm

75cm

None of the above

View Solution

The height of an equilateral triangle having each side 6 cm, is

$3\sqrt{2}$

$2\sqrt{3}$

$2\sqrt{2}$

$3\sqrt{3}$

View Solution

Concepts Covered - 1

Pythagoras Theorem

This theorem was earlier given by an ancient Indian mathematician Baudhayan (about 800 B.C.E.) in the following form :

The diagonal of a rectangle produces by itself the same area as produced by its both sides (i.e., length and breadth). For this reason, this theorem is sometimes also referred to as the Baudhayan Theorem.

Theorem : In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Given: $\triangle A B C \text { in which } \angle A B C=90^{\circ}$

To Prove: $A C^{2}=A B^{2}+B C^{2}$

Construction: $\text{Draw }B D \perp A C.$

Proof: In $\triangle A D B$ and $\triangle A B C$ , we have

$\angle A=\angle A \qquad(\text { common })\\$

$\left.\angle A D B=\angle A B C \qquad \text { [each equal to } 90^{\circ}\right]\\$

$\therefore \quad \triangle A D B \sim \triangle A B C \qquad\text { [by AA-similarity] } \\$

$\Rightarrow \quad \frac{A D}{A B}=\frac{A B}{A C} \\$

$\Rightarrow \quad A D \times A C=A B^{2}\qquad\qquad\ldots(1)$

In $\triangle B D C$ and $\triangle ABC$ , we have

$\angle C=\angle C \qquad(\text { common })\\$

$\left.\angle BDC=\angle A B C \qquad \text { [each equal to } 90^{\circ}\right]\\$

$\therefore \quad \triangle BDC \sim \triangle A B C \qquad\text { [by AA-similarity] } \\$

$\Rightarrow \quad \frac{D C}{B C}=\frac{B C}{A C} \\$

$\Rightarrow \quad D C \times A C=B C^{2}\qquad\qquad\ldots(2)$

${\color{Blue} \text{From (i) and (ii), we get}}$

${\color{Blue} A D \times A C+D C \times A C=A B^{2}+B C^{2} }$

${\color{Blue} \\\Rightarrow \quad(A D+D C) \times A C=A B^{2}+B C^{2} }$

${\color{Blue} \Rightarrow \quad A C \times A C=A B^{2}+B C^{2} \Rightarrow A C^{2}=A B^{2}+B C^{2}}$

The converse of Pythagoras Theorem

In a triangle, if the square of one side is equal to the sum of the squares of the other two sides then the angle opposite to the first side is a right angle.

Given: $\triangle A B C \text { in which } A C^{2}=A B^{2}+B C^{2}$

To Prove: $\angle B=90^{\circ}$

Construction: $\text{Draw a } \triangle D E F\text{ such that } D E=A B, E F=B C\text{ and }\angle E=90^{\circ}$

Proof: $\text{In } \triangle D E F,\text{ we have } \angle E=90^{\circ}$

So, by Pythagoras’ theorem, we have

$D F^{2}=D E^{2}+E F^{2}$

$\\ \Rightarrow \quad D F^{2}=A B^{2}+B C^{2}\qquad\ldots(1) \quad[\because D E=A B \text { and } E F=B C]$

$\text{But, } A C^{2}=A B^{2}+B C^{2} \qquad\ldots(2)$

$\text{From (1) and (2), we get } {\color{Blue} A C^{2}=D F^{2} \Rightarrow A C=D F}$

$\text{Now, in } \triangle A B C\text{ and } \triangle D E F,\text{ we have}$

$A B=D E, B C=E F \text { and } A C=D F . \\$

$\therefore A B C \cong \triangle D E F \\$

$\text { Hence, } \angle B=\angle E=90^{\circ}$

Distinguish between right angle, acute angle and obtuse angle triangle (when sides of the triangle are given)

The Pythagorean Theorem for right triangles states a relationship between the three sides. Let a and b be the lengths of the two sides, and let c be the length of the hypotenuse. Then,

$\mathrm{a^2+b^2=c^2}$

However, if a triangle is not a right triangle, then this equation is not true. There are certain inequalities that will hold for acute and obtuse triangles that are very similar to the Pythagorean Theorem.

Pythagorean Theorem for Acute Triangles

$\mathrm{a^2+b^2>c^2}$

Pythagorean Theorem for Obtuse Triangles

$\mathrm{a^2+b^2<c^2}$

Note that, there is no definitive hypotenuse for a triangle that is acute or obtuse. Therefore we can assign the side lengths of these triangles to the three variable slots in the formula (a, b, and c) in any configuration we wish.