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 ΔABC which is right-angled at

B, i.e., ∠B 900,

Then ∠A + ∠B + ∠C = 180°

(by angle sum property)

∠A + 90° + ∠C = 180°

∠A + ∠C = 90°

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

and ∠B>∠C

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

Hence, AC, the hypotenuse is the longest side.

Question 2.

Infiguresides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

Solution:

∠PBC < ∠QCB (given)

⇒ 180°- ∠PBC>180°- ∠QCB

⇒∠ABC > ∠ACB

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

Question 3.

In the figure,∠B<∠A and ∠C<∠D. Show that AD∠BC

Solution:

∠B <∠A (given)

∠A >∠B

OB > OA ……….(1)

(Side opposite to greater angle is longer)

Now ∠C <∠D (given)

∴ ∠D > ∠C

∴ OC > OD ……….(2)

(Side opposite to greater angle is longer)

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

OB + OC > OA + OD

⇒ BC >AD

⇒ AD > ∠C

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:

Construction: Join AC.

Proof: In ΔABC,

AB AB

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

In ΔACD

CD>AD ( CDisthelongest side of quadrilateral)

∴ ∠CMD > ∠ACD ……(2) (Angle

opposite to longer side is greater)

Adding eqn. (1) and (2)

∠BAC + ∠CAD> ∠BCA + ∠ACD

⇒ ∠A > ∠C

Similarly, join B to D and we can prove

∠B > ∠D

Question 5.

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

Given:

In figure PR>PQ

and PS bisects ∠QPR.

To Prove:

∠PSR > ∠PSQ

Proof In ΔPQR,

PR > PQ (given)

∠PQR > ∠PRQ

∠PQS > ∠PRS ……..(1)

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

Adding ∠QPS on both sides in eqn. (1)

∠PQS + ∠QPS > ∠PRS + ∠QPS ……..(3)

From eqn. (2) and (3)

⇒ ∠PQS + ∠QPS> ∠PRS + ∠RPS …….(4)

⇒ ∠PSQ = ∠PRS + ∠RPS ……….(5) (By exterior angle theorem)

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

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

∠PSR > ∠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 ⊥ l and R is any point on l other than Q.

To Prove: PQ < PR

Proof: in Δ PQR

∠Q = 90º

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

⇒ ∠P + 90° + ∠R = 180°

⇒ ∠P + ∠R = 90°

Hence ∠R is an acute angle.

∴ ∠Q > ∠R

(Side opposite to greater angle is longer)

∴ PR >PQ

⇒ PQ<PR

Hence perpendicular line segment is the shortest.

Hence the perpendicular line segment is the shortest.