Chapter 7 Triangles Ex-7.1 |
Chapter 7 Triangles Ex-7.2 |
Chapter 7 Triangles Ex-7.3 |
Chapter 7 Triangles Ex-7.5 |

Showthat in a right angled triangle, the hypotenuse is the longest side.

**Answer
1** :

Letus consider a right-angled triangle ABC, right-angled at B.

InΔABC,

∠A + ∠B + ∠C= 180° (Angle sum property of a triangle)

∠A + 90º + ∠C = 180°

∠A + ∠C = 90°

Hence,the other two angles have to be acute (i.e., less than 90º).

∴ ∠B is the largest angle inΔABC.

⇒ ∠B > ∠Aand ∠B > ∠C

⇒ AC > BC and AC > AB

[Inany triangle, the side opposite to the larger (greater) angle is longer.]

Therefore,AC is the largest side in ΔABC.

Inthe given figure sides AB and AC of ΔABC are extended to points P and Qrespectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

**Answer
2** :

Inthe given figure,

∠ABC + ∠PBC = 180° (Linear pair)

⇒ ∠ABC = 180° − ∠PBC… (1)

Also,

∠ACB + ∠QCB = 180°

∠ACB = 180° − ∠QCB … (2)

As∠PBC < ∠QCB,

⇒ 180º − ∠PBC > 180º − ∠QCB

⇒ ∠ABC > ∠ACB[From equations (1) and (2)]

⇒ AC > AB (Side opposite to the largerangle is larger.)

Inthe given figure, ∠B < ∠A and ∠C< ∠D. Show that AD < BC.

**Answer
3** :

InΔAOB,

⇒ AO < BO (Side opposite to smaller angleis smaller) … (1)

InΔCOD,

∠C < ∠D

⇒ OD < OC (Side opposite to smaller angleis smaller) … (2)

Onadding equations (1) and (2), we obtain

AO+ OD < BO + OC

ABand CD are respectively the smallest and longest sides of a quadrilateral ABCD(see the given figure). Show that ∠A > ∠Cand ∠B > ∠D.

**Answer
4** :

Letus join AC.

InΔABC,

AB< BC (AB is the smallest side of quadrilateral ABCD)

∴ ∠2 < ∠1(Angle opposite to the smaller side is smaller) … (1)

InΔADC,

AD< CD (CD is the largest side of quadrilateral ABCD)

∴ ∠4 < ∠3(Angle opposite to the smaller side is smaller) … (2)

Onadding equations (1) and (2), we obtain

∠2 + ∠4 < ∠1+ ∠3

⇒ ∠C < ∠A

⇒ ∠A > ∠C

Letus join BD.

InΔABD,

AB< AD (AB is the smallest side of quadrilateral ABCD)

∴ ∠8 < ∠5(Angle opposite to the smaller side is smaller) … (3)

InΔBDC,

BC< CD (CD is the largest side of quadrilateral ABCD)

∴ ∠7 < ∠6(Angle opposite to the smaller side is smaller) … (4)

Onadding equations (3) and (4), we obtain

∠8 + ∠7 < ∠5+ ∠6

⇒ ∠D < ∠B

⇒ ∠B> ∠D

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

**Answer
5** :

AsPR > PQ,

∴ ∠PQR > ∠PRQ(Angle opposite to larger side is larger) … (1)

PSis the bisector of ∠QPR.

∴∠QPS = ∠RPS … (2)

∠PSR is the exterior angle of ΔPQS.

∴ ∠PSR = ∠PQR+ ∠QPS … (3)

∠PSQ is the exterior angle of ΔPRS.

∴ ∠PSQ = ∠PRQ+ ∠RPS … (4)

Addingequations (1) and (2), we obtain

∠PQR + ∠QPS > ∠PRQ+ ∠RPS

⇒ ∠PSR > ∠PSQ[Using the values of equations (3) and (4)]

Showthat of all line segments drawn from a given point not on it, the perpendicularline segment is the shortest.

**Answer
6** :

Let us take a line *l* andfrom point P (i.e., not on line *l*), draw two line segments PN andPM. Let PN be perpendicular to line *l* and PM is drawn at someother angle.

InΔPNM,

∠N = 90º

∠P + ∠N + ∠M= 180º (Angle sum property of a triangle)

∠P + ∠M = 90º

Clearly,∠M is an acute angle.

∴ ∠M < ∠N

⇒ PN < PM (Side opposite to the smallerangle is smaller)

Similarly, by drawing different linesegments from P to *l*, it can be proved that PN is smaller incomparison to them.

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