Gujarat Board GSEB Solutions Class 9 Maths Chapter 7 Triangles Ex 7.4 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 9 Maths Chapter 7 Triangles Ex 7.4

Question 1.

Show that in a right-angled triangle, the hypotenuse is the longest side.

Solution:

Given: Right Î”ABC which is right-angled at

B, i.e., âˆ B 900,

Then âˆ A + âˆ B + âˆ C = 180Â°

(by angle sum property)

âˆ A + 90Â° + âˆ C = 180Â°

âˆ A + âˆ C = 90Â°

âˆ´ AC > BC (Side opposite to greater angle is longer)

and âˆ B>âˆ C

âˆ´ AC > AB (Side opposite to greater angle is longer)

Hence, AC, the hypotenuse is the longest side.

Question 2.

Infiguresides AB and AC of Î”ABC are extended to points P and Q respectively. Also, âˆ PBC < âˆ QCB. Show that AC > AB.

Solution:

âˆ PBC < âˆ QCB (given)

â‡’ 180Â°- âˆ PBC>180Â°- âˆ QCB

â‡’âˆ ABC > âˆ ACB

âˆ´ AC > AB (Side opposite to greater angle is longer)

Question 3.

In the figure,âˆ B<âˆ A and âˆ C<âˆ D. Show that ADâˆ BC

Solution:

âˆ B <âˆ A (given)

âˆ A >âˆ B

OB > OA ……….(1)

(Side opposite to greater angle is longer)

Now âˆ C <âˆ D (given)

âˆ´ âˆ D > âˆ C

âˆ´ OC > OD ……….(2)

(Side opposite to greater angle is longer)

Adding eqn. (1) and (2), we get

OB + OC > OA + OD

â‡’ BC >AD

â‡’ AD > âˆ C

Question 4.

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

Solution:

Construction: Join AC.

Proof: In Î”ABC,

AB AB

âˆ BAC > âˆ BCA …….. (1) (Angle opposite to longer side is greater)

In Î”ACD

CD>AD ( CDisthelongest side of quadrilateral)

âˆ´ âˆ CMD > âˆ ACD ……(2) (Angle

opposite to longer side is greater)

Adding eqn. (1) and (2)

âˆ BAC + âˆ CAD> âˆ BCA + âˆ ACD

â‡’ âˆ A > âˆ C

Similarly, join B to D and we can prove

âˆ B > âˆ D

Question 5.

In figure, PR > PQ and PS bisects âˆ QPR. Prove that âˆ PSR> âˆ PSQ.

Given:

In figure PR>PQ

and PS bisects âˆ QPR.

To Prove:

âˆ PSR > âˆ PSQ

Proof In Î”PQR,

PR > PQ (given)

âˆ PQR > âˆ PRQ

âˆ PQS > âˆ PRS ……..(1)

âˆ QPS = âˆ RPS …….(2) (PS is the angle bisector of âˆ QPR)

Adding âˆ QPS on both sides in eqn. (1)

âˆ PQS + âˆ QPS > âˆ PRS + âˆ QPS ……..(3)

From eqn. (2) and (3)

â‡’ âˆ PQS + âˆ QPS> âˆ PRS + âˆ RPS …….(4)

â‡’ âˆ PSQ = âˆ PRS + âˆ RPS ……….(5) (By exterior angle theorem)

and âˆ PSR = âˆ QPS + âˆ PQS …….(6) (by exterior angle theorem)

From eqn. (4), (5) and (6), we have

âˆ PSR > âˆ PSQ

Question 6.

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

Solution:

Given: A line I and P is a point not lying online l and PQ âŠ¥ l and R is any point on l other than Q.

To Prove:Â PQ < PR

Proof: in Î” PQR

âˆ Q = 90Âº

âˆ P + âˆ Q + âˆ R = 180Â° m(by angle sum property)

â‡’ âˆ P + 90Â° + âˆ R = 180Â°

â‡’ âˆ P + âˆ R = 90Â°

Hence âˆ R is an acute angle.

âˆ´ âˆ Q > âˆ R

(Side opposite to greater angle is longer)

âˆ´ PR >PQ

â‡’ PQ<PR

Hence perpendicular line segment is the shortest.

Hence the perpendicular line segment is the shortest.