Advertisements
Advertisements
प्रश्न
S is any point on side QR of a ∆PQR. Show that: PQ + QR + RP > 2PS.
Advertisements
उत्तर
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
संबंधित प्रश्न
The angles of a triangle are (x − 40)°, (x − 20)° and `(1/2x-10)^@.` find the value of x
Is the following statement true and false :
An exterior angle of a triangle is less than either of its interior opposite angles.
Define a triangle.
In ΔRST (See figure), what is the value of x?
Find the value of the angle in the given figure:
Find x, if the angles of a triangle is:
x°, 2x°, 2x°
Classify the following triangle according to angle:
Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than `2/3` of a right angle.
In the following figure, points lying in the interior of the triangle PQR are ______, that in the exterior are ______ and that on the triangle itself are ______.
How many sides does a triangle have?
