NCERT Solutions Class 9 Exercise 7.4 of chapter 7-Triangles

NCERT Solutions of Class 9 Exercise 7.4 of chapter 7-Triangles are created here for helping the students of class 9 in helping their preparations for CBSE board exams. All NCERT Solutions of Class 9 Exercise 7.4 of chapter 7-Triangles are solved by an expert of maths in such a way that every student can understand easily without the help of anybody.

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You can download the pdf of NCERT solutions of chapter 7

pdf of class 9 NCERT solutions of the chapter 7 -Triangle

Exercise 7.1- Triangles

Exercise 7.2- Triangle

Exercise 7.4 – Triangle

Q1.Show that in a right-angled triangle, the hypotenuse is the longest triangle.

Ans.

 

GIVEN: ΔABC is a right triangle,let ∠B =90°

TO PROVE: Hypotenuse is the largest side in a right triangle

PROOF: ∠B = 90°(given)

∠A + ∠B + ∠C  = 180° (angle sum property of the triangle)

∠A + 90° + ∠C  = 180°

∠A + ∠C  = 180°-90°= 90°

∠A and ∠C must be acute angle

∴∠A < 90° and ∠C < 90°

Therefore ∠B is the largest angle

So, ∠B >∠A  and ∠B>∠C

AC> BC and AC > AB (in a triangle side opposite to larger angle is longer)

Hence AC(hypotenuse)  is the largest side in a right triangle

Q2.In the given figure sides AB and AC of ΔABC are extended to points P and Q respectively .Also ∠PBC < ∠QCB. Show that AC > AB.

 

Ans.

GIVEN: In the fig. ∠PBC < ∠QCB

TO PROVE:AC > AB

PROOF: ∠PBC < ∠QCB(given)

∠ABC + ∠PBC = 180°(linear pair)

∠PBC = 180° – ∠ABC ….(i)

∠ACB + ∠QCB = 180°(linear pair)

∠QCB = 180° – ∠ACB ….(ii)

From equation (i) and equation (ii)

180° – ∠ABC < 180° – ∠ACB

∠ABC >∠ ACB

AC > AB (side opposite to larger angle is greater)

Q3.In the given figure , ∠B < ∠A, and ∠C < ∠D. Show that AD < BC.

 

Ans.

GIVEN: ∠B < ∠A, and ∠C < ∠D

TO PROVE:AD < BC

PROOF: ∠B < ∠A (given)

OA < OB….(i) (side opposite to larger angle is grater)

∠C < ∠D

OD < OC…(ii) (side opposite to larger angle is grater)

From equation (i) and equation (ii)

OA + OD < OB + OC

AD < BC (from figure), Hence proved

Q4.AB and CD are respectively smallest and the longest side of quadrilateral ABCD(see the given figure). Show that ∠A > ∠C and ∠B > ∠D.

 

Ans.

GIVEN: AB and CD are respectively smallest and the longest side of quadrilateral ABCD

COSTRUCTION: Joining A to C and B to D

TO PROVE: (i) ∠A > ∠C  (ii)  ∠B > ∠D.

(i) PROOF: In ΔABC

BC> AB (AB is the smallest side in ABCD)

∠BAC > ∠ACB ….(i) (side opposite to larger angle is grater)

In ΔADC

CD > AD (CD is the largest side in ABCD)

∠DAC > ∠ACD….(ii)

Adding equation (i) and (ii)

∠BAC + ∠DAC > ∠ACB + ∠ACD

∠A > ∠C, Hence proved

PROOF: (ii)  ∠B > ∠D

In ΔABD

AD> AB (AB is the smallest side in ABCD)

∠ABD > ∠ADB ….(i) (side opposite to larger angle is grater)

In ΔADC

CD > BC (CD is the largest side in ABCD)

∠DBC > ∠BDC….(ii)

Adding equation (i) and (ii)

∠ABD +∠DBC > ∠ADB + ∠BDC

∠B > ∠D, Hence proved

Q5.In the given figure , PR > PQ and PS bisects ∠QPR , prove that ∠PSR > PSQ.

 

Ans.

GIVEN:PR > PQ

PS bisects ∠QPR

∠QPS = ∠SPR

TO PROVE: ∠PSR >∠ PSQ

PROOF: PR > PQ (given)

∠Q > ∠R (side opposite to larger angle is grater)

In ΔPQS

∠Q + ∠QPS + ∠ PSQ = 180°(angle sum property of the triangle)

∠Q = 180° -(∠QPS + ∠ PSQ)…..(i)

In ΔPSR

∠R + ∠SPR + ∠ PSR = 180°(angle sum property of the triangle)

∠R = 180° -(∠SPR + ∠ PSR)…..(ii)

From (i) and (ii)

180° -(∠QPS + ∠ PSQ) > 180° -(∠SPR + ∠ PSR)

∠SPR + ∠ PSR >∠QPS + ∠ PSQ

Since PS bisects ∠P, ∠QPS = ∠SPR

∠SPR + ∠ PSR >∠SPR + ∠ PSQ

∠ PSR >∠ PSQ, Hence proved

Q6. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Ans. Let A is an external point out of the line and AC⊥ BC

Let BC is a line and AB is a line segment drawn from the point A on the line BC

In triangle ΔABC

∠B <∠C (since ∠B is acute angle and ∠C =90°)

AC < AB (side opposite to larger angle is grater)

Similarly all line segments drawn from A on the line BC are larger than AC

Therefore AC is the shortest side drawn from A on BC, Hence proved

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