Advertisements
Advertisements
Question
In the given figure, T is a point on the side PR of an equilateral triangle PQR. Show that RT < QT
Advertisements
Solution
In ΔPQR,
PQ = QR = PR
⇒ ∠P = Q = ∠ = 60°
In ΔTQR,
∠TQR < 60°
∴ ∠TQR < ∠R
∴ RT < QT.
APPEARS IN
RELATED QUESTIONS
Show that in a right angled triangle, the hypotenuse is the longest side.
In the given figure sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC.
If two sides of a triangle are 8 cm and 13 cm, then the length of the third side is between a cm and b cm. Find the values of a and b such that a is less than b.
In a triangle ABC, BC = AC and ∠ A = 35°. Which is the smallest side of the triangle?
D is a point on the side of the BC of ΔABC. Prove that the perimeter of ΔABC is greater than twice of AD.
In ΔPQR, PR > PQ and T is a point on PR such that PT = PQ. Prove that QR > TR.
ABCD is a trapezium. Prove that:
CD + DA + AB > BC.
Prove that in an isosceles triangle any of its equal sides is greater than the straight line joining the vertex to any point on the base of the triangle.
ΔABC in a isosceles triangle with AB = AC. D is a point on BC produced. ED intersects AB at E and AC at F. Prove that AF > AE.
