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
Is the following statement true and false :
All the angles of a triangle can be equal to 60°.
Is the following statement true and false :
A triangle can have two obtuse angles.
Fill in the blank to make the following statement true:
Sum of the angles of a triangle is ....
In a Δ ABC, the internal bisectors of ∠B and ∠C meet at P and the external bisectors of ∠B and ∠C meet at Q, Prove that ∠BPC + ∠BQC = 180°.
In ∆ABC, ∠A = ∠B = 62° ; find ∠C.
Classify the following triangle according to angle:
The length of the three segments is given for constructing a triangle. Say whether a triangle with these sides can be drawn. Give the reason for your answer.
9 cm, 6 cm, 16 cm
Can 30°, 60° and 90° be the angles of a triangle?
In the following figure, l || m and M is the mid-point of a line segment AB. Show that M is also the mid-point of any line segment CD, having its end points on l and m, respectively.
The sides of a triangle must always be connected so that:
