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CAT Inequalities in a Triangle - Practice Questions & MCQ

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

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  • 5 Questions around this concept.

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In a quadrilateral ABCD, (A B+B C+C D+D A) < 2(B D+A C) 

Concepts Covered - 1

Inequalities in a Triangle

So far, you have been mainly studying the equality of sides and angles of a triangle or triangles. Sometimes, we do come across unequal objects, we need to compare them.

For example,

Line-segment AB is greater in length as compared to line segment CD

\angle A is greater than \angle B

 

Theorem : If two sides of a triangle are unequal, prove that the angle opposite to the longer side is greater.

Given: \triangle A B C\text{ in which }A C>A B.

To Prove: \angle A B C>\angle B C A

Construction: Mark a point D on AC such that AD = AB. Join BD

Proof: We know that in a triangle, the angles opposite to equal sides are equal.

\\\text{So, in }\triangle A B D,\text{ we have} \\\\

A B=A D=\angle B D A=\angle A B D

Now, in \triangleBCD, side CD has been produced to A, forming exterior angle \angleBDA.

\therefore \quad \angle B D A>\angle B C D\quad \text { [exterior angle is greater than int. opp. angle }]

\Rightarrow \quad \angle B D A=\angle B C A[\because \angle B C D=\angle B C A] \\

\Rightarrow \quad \angle A B D>\angle B C A\qquad[\text {using }(i)] \\

\Rightarrow \quad \angle A B C>\angle A B D>\angle B C A \qquad[\because \angle A B C>\angle A B D] \\

\Rightarrow \quad \angle A B C>\angle B C A

\\ \text { Hence, } \angle A B C>\angle B C A

 

Theorem : In any triangle, prove that the side opposite to the greater angle is longer.

Given: \triangle A B C\text{ in which }ABC>ACB.

To Prove: A C>A B

Proof: We have the following possibilities only.

(i) A C=A B\qquad (ii) A C<A B\qquad (iii) A C>A B

Out of these possibilities, exactly one must be true.

CASE 1:

If possible, let AC = AB.

We know that the angles opposite to equal sides of a triangle are equal.

\therefore \quad A C=A B \Rightarrow \angle A B C=\angle A C B

This contradicts the given hypothesis that \angle A B C>\angle A C B.

\therefore \quad A C \neq A B

CASE 2:

If possible, let A C<A B. Then, A B>A C.

Since the angle opposite to the longer side is larger, so

A B>A C \Rightarrow \angle A C B>\angle A B C

This contradicts the given hypothesis that \angle A B C>\angle A C B.

\therefore \quad A C\text{ cannot be less than }A B

CASE 3:

Now, we are left with the only possibility that A C>A B, which must be true.

Hence, A C>A B

Theorem : Prove that, of all the line segments that can be drawn to a given line, from a point not lying on it, the perpendicular line segment is the shortest.

Given: A line AB and a point P outside it. P M \perp A B and N is a point, other than M, on AB.

To Prove: PN>PM

Proof: 

\text{In }\triangle P M N,\text{ we have }\angle M=90^{\circ}

But, in a right-angled triangle, each one of the angles other than the right angle is an acute angle.

\therefore \quad \angle P N M<\angle P M N

But, the side opposite to the smaller angle in a triangle is shorter.

\therefore P M<P N

Hence, the perpendicular from P to the given line is shortest of all line segments from P to AB.

Note: The distance between a line and a point, not on it, is the length of perpendicular from the point to the given line.

The distance between a line and a point lying on it, is zero.

 

Theorem : Prove that the sum of any two sides of a triangle is greater than the third side.

Given: \Delta ABC.

To Prove: (i) A B+A C>B C\qquad (ii) A B+B C>A C\qquad (iii) B C+A C>A B

Construction: Produce BA to D such that AD = AC. Join CD

Proof: (i)\text{ In }\triangle A C D,\text{ we have}

A D=A C \qquad\text{ [by construction]}

\Rightarrow \angle A C D=\angle A D C\qquad[\angle\text{ opposite to equal sides]}

\Rightarrow \angle B C D>\angle A D C \qquad[\because \angle B C D>\angle A C D]

\Rightarrow \angle B C D>\angle AC D \qquad[\because \angle A D C=\angle B D C]

\Rightarrow B D>B C \qquad\text{ [side opposite to larger angle is larger]}

\Rightarrow \quad B A+A D>B C

\Rightarrow B A+A C>B C \qquad[\because A D=A C]

\therefore\quad A B+A C>B C

\text{Similarly, } A B+B C>A C\text{ and }B C+A C>A B

Theorem : Prove that the difference between any two sides of a triangle is less than its third side.

Given: A line AB and a point P outside it. P M \perp A B and N is a point, other than M, on AB.

To Prove: (i) A C-A B<B C\qquad (ii) B C-A C<A B\qquad (iii) B C-A B<A C

Construction: Let A C>A B. Then, along AC, set off AD = AB. Join BD.

Proof: A B=A D \Rightarrow \angle 1=\angle 2

Side CD of \triangle B C D has been produced to A

\therefore \quad \angle 2>\angle 4 \qquad.. (ii) \qquad[\because \text{ext. angle} > \text{each int. opp. angle}].

\text{Side }A D\text{ of }\triangle A B D\text{ has been produced to C}

\\\therefore \quad \angle 3>\angle 1\qquad\ldots(iii)\qquad [\text{ ext. angle} >\text { each int. opp. angle }]

\\ \Rightarrow \quad \angle 3>\angle 2 \qquad\ldots(iv) \qquad[\text { using }(i)]

\\ \text{From (ii) and (iv), we get } \\\angle 3>\angle 4

\text{Now, }\angle 4<\angle 3

\Rightarrow \quad C D<B C \\

\Rightarrow \quad A C-A D<B C \\

\Rightarrow \quad A C-A B<B C \quad[\because A D=A B] \\

\text { Hence, } A C-A B<B C \\

\text { Similarly, } B C-A C<A B \text { and } B C-A B<A C

 

Theorem : Prove that the sum of any two sides of a triangle is greater than twice the median drawn to the third side.

Given: \triangle ABC in which AD is a median.

To Prove: A B+A C>2 A D.

Construction: Let AD to E such that AD = DE. Join EC.

Proof: \text{In } \triangle A D B\text{ and }\triangle E D C,\text{ we have}

A D=D E \qquad\text{ (by construction) }

\angle A D B=\angle E D C\qquad \text { (vert. opp. angles) }

\\ B D=C D \qquad[\because D\text { is the midpoint } ]

\therefore \quad \triangle A D B \cong \triangle E D C \quad\qquad\text{ (by SAS-criteria)}

\therefore \quad A B=E C\qquad\text{[ c.p.c.t.]}

We know that the sum of any two sides of a triangle is greater than the third side. So, in \triangle A C E, we have

E C+A C>A E \\

\therefore \quad A B+A C>2 A D \quad[\because E C=A B \text { and } A E=2 A D] \\

\text { Hence, } A B+A C>2 A D

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