Advertisements
Advertisements
प्रश्न
In the given figure, T is a point on the side PR of an equilateral triangle PQR. Show that RT < QT
Advertisements
उत्तर
In ΔPQR,
PQ = QR = PR
⇒ ∠P = Q = ∠ = 60°
In ΔTQR,
∠TQR < 60°
∴ ∠TQR < ∠R
∴ RT < QT.
APPEARS IN
संबंधित प्रश्न
In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
In the following figure, write BC, AC, and CD in ascending order of their lengths.
Arrange the sides of the following triangles in an ascending order:
ΔABC, ∠A = 45°, ∠B = 65°.
Arrange the sides of the following triangles in an ascending order:
ΔDEF, ∠D = 38°, ∠E = 58°.
Name the smallest angle in each of these triangles:
In ΔXYZ, XY = 6.2cm, XY = 6.8cm and YZ = 5cm
Prove that the perimeter of a triangle is greater than the sum of its three medians.
In ABC, P, Q and R are points on AB, BC and AC respectively. Prove that AB + BC + AC > PQ + QR + PR.
In ΔABC, BC produced to D, such that, AC = CD; ∠BAD = 125° and ∠ACD = 105°. Show that BC > CD.
In ΔABC, AE is the bisector of ∠BAC. D is a point on AC such that AB = AD. Prove that BE = DE and ∠ABD > ∠C.
