Advertisements
Advertisements
प्रश्न
Q is a point on the side SR of a ∆PSR such that PQ = PR. Prove that PS > PQ.
Advertisements
उत्तर
In triangle PSR, Q is a point on the side SR such that PQ = PR.
To proof that PS > PQ

Proof: In triangle PRQ,
PQ = PR ...[Given]
∠R = ∠PQR ...(i) [Angle opposite to equal sides are equal]
∠PQR > ∠S ...(ii) [Exterior angle of a triangle is greater than each of the opposite interior angle]
Now, from equation (i) and (ii), we get
∠R > ∠S
PS > PR ...[Side opposite to greater angle is longer]
PS > PQ ...[PQ = PR]
APPEARS IN
संबंधित प्रश्न
Is the following statement true and false :
An exterior angle of a triangle is less than either of its interior opposite angles.
In Δ ABC, if u∠B = 60°, ∠C = 80° and the bisectors of angles ∠ABC and ∠ACB meet at a point O, then find the measure of ∠BOC.
In a triangle, an exterior angle at a vertex is 95° and its one of the interior opposite angle is 55°, then the measure of the other interior angle is
In a triangle ABC, ∠A = 45° and ∠B = 75°, find ∠C.
In a triangle PQR, ∠P = 60° and ∠Q = ∠R, find ∠R.
The angle of a vertex of an isosceles triangle is 100°. Find its base angles.
Classify the following triangle according to angle:

The angles of the triangle are 3x – 40, x + 20 and 2x – 10 then the value of x is
Can we have two acute angles whose sum is a reflex angle? Why or why not?
The sides of a triangle must always be connected so that:
