Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Let the triangle be PQR.
PS QR, the straight line joining vertex P
to the line QR.
To prove : PQ > PT and PR > PT
In ΔPSQ,
PS2 + SQ2 = PQ2 ....(Pythagoras theroem)
PS2 = PQ2 - SQ2 ....(i)
In ΔPST,
PS2 + ST2 = PT2 ....(Pythagoras theroem)
PQ2 - SQ2 = PT2 - ST2 ....(ii)
PQ - (ST + TQ)2 = PT2 - ST2 ....[from (i) and (ii)]
PQ2 - (ST2 - 2ST x TQ + TQ2) = PT2 - ST2
PQ2 - (ST2 - 2ST x TQ - TQ2 = PT2 - ST
PQ2 - PT2 = TQ2 + 2ST x TQ
PQ2 - PT2 = TQ x (2ST + TQ)
As, TQ x (2ST + TQ) > 0 always.
PQ2 - PT2 > 0
PQ2 > PT2
PQ > PT
Also, PQ = PR
PR > PT.
APPEARS IN
संबंधित प्रश्न
Show that in a right angled triangle, the hypotenuse is the longest side.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see the given figure). Show that ∠A > ∠C and ∠B > ∠D.
In the given figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR >∠PSQ.
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 the following figure, ∠BAC = 60o and ∠ABC = 65o.
Prove that:
(i) CF > AF
(ii) DC > DF
In the following figure ; AC = CD; ∠BAD = 110o and ∠ACB = 74o.
Prove that: BC > CD.
In the following figure, write BC, AC, and CD in ascending order of their lengths.
In a triangle ABC, BC = AC and ∠ A = 35°. Which is the smallest side of the triangle?
ABCD is a trapezium. Prove that:
CD + DA + AB + BC > 2AC.
In ΔPQR, PS ⊥ QR ; prove that: PQ > QS and PQ > PS
