Advertisements
Advertisements
Question
S is any point on side QR of a ∆PQR. Show that: PQ + QR + RP > 2PS.
Advertisements
Solution
Given: In ∆PQR, S is any point on side QR.
To show: PQ + QR + RP > 2PS
Proof: In ∆PQS,
PQ + QS > PS ...(i) [Sum of two sides of a triangle is greater than the third side]
Similarly, in ∆PRS,
SR + RP > PS ...(ii) [Sum of two sides of a triangle is greater than the third side]
On adding equations (i) and (ii), we get
PQ + QS + SR + RP > 2PS
⇒ PQ + (QS + SR) + RP > 2PS
⇒ PQ + QR + RP > 2PS ...[∵ QR = QS + SR]
APPEARS IN
RELATED QUESTIONS
Can a triangle have All angles more than 60°? Justify your answer in case.
If the bisector of the exterior vertical angle of a triangle be parallel to the base. Show that the triangle is isosce
In the given figure, AB || DE. Find ∠ACD.
Is the following statement true and false :
All the angles of a triangle can be less than 60°
Fill in the blank to make the following statement true:
An exterior angle of a triangle is always ......... than either of the interior opposite angles.
If the side BC of ΔABC is produced on both sides, then write the difference between the sum of the exterior angles so formed and ∠A.
Calculate the unknown marked angles of the following figure :
Find the unknown marked angles in the given figure:
The length of the sides of the triangle is given. Say what types of triangles they are 3.7 cm, 3.4 cm, 4 cm.
Identify three triangles in the figure.
