Triangles-9th-Math: 9 Math

Solution of NCERT Exercise 7.4: 9th math

Question (1) Show that in a right angled triangle, the hypotenuse is the longest side.

Solution:

Let, ABC is a right angled triangle, in which

∠ A = 90^o

And BC is the hypotenuse.

To show, that hypotenuse BC is the longest side.

We know that,

In a triangle, sum of all the three angles is equal to 180^o.

∴ In triangle ABC

∠ A + ∠ B + ∠ C = 180^o

⇒ 90^o + ∠ B + ∠ C = 180^o

⇒ ∠ B + ∠ C = 180^o – 90^o

⇒ ∠ B + ∠ C = 90^o

Clearly, ∠ B < 90⁰

And, ∠ C < 90⁰

Thus, in triangle ABC, ∠ A = 90^o is the largest angle.

Now, we know that, In a triangle, side opposite to the larger angle is larger.

Since, in triangle ABC, ∠ A is the largest angle, thus side (BC) which opposite to ∠ A will be largest.

Since, side BC is the hypotenuse, thus in a right angled triangle the hypotenuse is the largest side. Proved

Question (2) In figure sides AB and AC of Δ ABC are extended to points P and Q respectively. Also, ∠ PBC < ∠ QCB. Show that AC > AB.

Solution:

Given, ∠ PBC < ∠ QCB

Thus, to prove, AC > AB

In the given figure,

Since, ∠ ABC and ∠ PBC are linear pair of angles.

Thus, ∠ ABC + ∠ PBC = 180^o

∴ ∠ ABC = 180^o – ∠ PBC - - - - - (i)

Similarly, ∠ ACB and ∠ QCB are linear pair of angles.

Thus, ∠ ACB + ∠ QCB = 180^o

∴ ∠ ACB = 180^o – ∠ QCB - - - - - (ii)

Now, since, according to question, ∠ PBC < ∠ QCB

Thus, 180^o – ∠ PBC > 180^o – ∠ QCB

Therefore, from equation (i) and equation (ii), we get

∠ ABC > ∠ ACB

Now, we know that, In a triangle, the side opposite to larger angle is larger.

Thus, side AC which is opposite to ∠ ABC is larger than the side AB which is opposite to ∠ ACB

That is, AC > AB Proved

Question (3) In figure, ∠ B < ∠ A and ∠ C < ∠ D. Show that AD < BC.

Solution:

Given, ∠ B < ∠ A

And, ∠ C < ∠ D

∴ prove that AD < BC

We know that, In a triangle, side opposite to larger angle is larger and side opposite to smaller angle is smaller.

In triangle ABO,

Since, ∠ B < ∠ A

Thus, side AO which is opposite to smaller ∠ B is smaller than the side OB which is opposite to larger ∠ A

∴ AO < OB - - - - - - (i)

And, in triangle OCD

Since, ∠ C < ∠ D

Thus, side OD which is opposite to smaller ∠ C is smaller than the side OC which is opposite to larger ∠ D

∴ OD < OC - - - - - - (ii)

Now, after adding equation (i) and (ii) we get

AO + OD < OB + OC - - - - - - (iii)

Now, from figure, since AO + OD = AD and OB + OC = BC

Thus, equation (iii) can be written as

AD < BD Proved.

Question (4) AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that ∠ A > ∠ C and ∠ B > ∠ D.

Solution:

Given, in quadrilateral ABCD

AB is the smallest side

and CD is the longest side

Therefore, to prove ∠ A > ∠ C

And ∠ B > ∠ D

Let, point A and C are joined.

Again point B and D are joined together.

We know that, In a triangle angle opposite to longer side is greater and angle opposite to the shorter side is smaller.

Now, in triangle ABC

Since AB is the smallest side in quadrilateral ABCD,

Thus, Side AB < BC

∴ ∠ ACB < ∠ BAC - - - - - - (i)

And, in triangle ADC

Since, CD is the longest side [according to question]

i.e. AD < CD

∴ ∠ ACD < ∠ CAD - - - - - - - (ii)

After adding equation (i) and (ii), we get

∠ ACB + ∠ ACD < ∠ BAC + ∠ CAD

⇒ ∠ BCD < ∠ BAD

⇒ ∠ BAD > ∠ BCD

R ∠ A > ∠ C - - - - - - (iii)

Now, in triangle ABD

Since, AB is the smallest side of quadrilateral

Therefore, AB < AD

∴ ∠ ADB < ∠ ABD - - - - - (iv)

Now, in triangle BDC

Since CD is the longes side of given quadrilateral

∴ BC < CD

∴ ∠ BDC < ∠ DBC - - - - - (v)

Now, by adding (iv) and (v), we get

∠ ADB + ∠ BDC < ∠ ABD + ∠ DBC

⇒ ∠ ADC < ∠ ABC

⇒ ∠ ABC > ∠ ADC

⇒ ∠ B > ∠ D - - - - - - (vi)

Therefore, from equation (iii) and (vi)

∠ A > ∠ C and ∠ B > ∠ D Proved

Question (5) In figure PR > PQ and PS bisects ∠ QPR. Prove that ∠ PSR > ∠ PSQ.

Solution:

Given, PR > PQ

And PS bisects ∠ QPR

∴ ∠ QPS = ∠ RPS - - - - - - (i)

Thus, to prove ∠ PSR > ∠ PSQ

Now, In triangle PQR,

As given, PR > PQ

And, we know that, In a triangle angle opposite to larger side is larger.

Therefore, ∠ PQR > ∠ PRQ - - - - - - - (ii)

Now, in triangle QSP

∠ PSR is the exterior angle.

And we know that, An exterior angle of a triangle is equal to the sum of the opposite interior angles.

∴ ∠ PSR = ∠ QPS + ∠ PQR - - - - - - (iii)

Similarly, in triangle PSR

PSQ is the exterior angle

∴ ∠ PSQ = ∠ PRQ + ∠ RPS - - - - - - (iv)

Now, after addition of equation (i) and (ii), we get

∠ QPS + ∠ PQR > ∠ RPS + ∠ PRQ

Now, after substituting values from equation (iii) and (iv), we get

∠ PSR > ∠ PSQ

And hence Proved

Question (6) Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Solution:

Let there is a line segment m

Let there is a point P which is not on the line m.

Now, let from point P a line PR `_|_` line m is drawn.

And an another line PQ, which is not perpendicular to line m is drawn.

Now, to prove that, PR which is perpendicular to line m is the shortest

Now, in Δ PQR

∠ R = 90^o

[∵ PR is perpendicular to QR]

From the theorem of angle sum property of a triangle we know that, sum of all the three interior angles of a triangle is equal to 180^o.

Thus, ∠ R + ∠ P + ∠ Q = 180^o

⇒ 90^o + ∠ P + ∠ Q = 180^o

⇒ ∠ P + ∠ Q = 180^o – 90^o

⇒ ∠ P + ∠ Q = 90^o

Thus, it is clear that, ∠ Q is an acute angle,

i.e. ∠ Q < ∠ R

Now, we know that, In a triangle side opposite to the smaller angle is shorter.

Thus, PR < PQ Proved

In similar way by drawing other lines from P, which are not perpendicular to the line m, it can be proved that, PR is the shortest.

Thus, it can be shown that all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

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