Chapter 7 Triangles Ex 7.4 Solutions
Question - 1 : - 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.
However,AC is the hypotenuse of ΔABC. Therefore, hypotenuse is the longest side in aright-angled triangle.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.)
Question - 3 : - Inthe given figure, ∠B < ∠A and ∠C< ∠D. Show that AD < BC.
Answer - 3 : -
InΔAOB,
∠B < ∠A
⇒ 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
AD< BC
Question - 4 : - 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)]
Question - 6 : - 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.
Therefore,it can be observed that of all line segments drawn from a given point not onit, the perpendicular line segment is the shortest.