PQR is an equilateral triangle. MN is drawn parallel to QR such that M – McqOptions

MCQOPTIONS

1.	PQR is an equilateral triangle. MN is drawn parallel to QR such that M is on PQ and N is on PR. If PN = 6 cm, then the length of MN is:
A.	cm
B.	cm
C.	2 cm
D.	.5 cm
Answer» C. 2 cm

Discussion

No Comment Found

Related MCQs

In a triangle ABC, ∠A = 90° and D is mid-point of AC. The value of BC2 - BD2 is equal to
Let ΔABC be a triangle whose area is 10√3 units with side lengths |AB| = 8 units and |AC| = 5 units. Find possible values of the angle A.
In ΔABC, AB = 5 cm BC = 6 cm, and CA = 7 cm a transversal is drawn to cut the sides AB at F, BC produced at D and CA at E so that AF = 2 cm, AE = 4 cm applying Menelaus theorem the length of BD is:
If in a ΔABC, a = 8, b = 15 and c = 17, then the value of cos A will be
Consider the following statements in respect of the points (p, p - 3), (q + 3, q) and (6, 3):1. The points lie on a straight line.2. The points always lie in the first quadrant only for any value of p and q.Which of the above statements is/are correct?
If perpendicular of a right angled triangle is 8 cm and its area is 20 cm2, the length of base is?
In a triangle ABC, AB = AC and the perimeter of ΔABC is 8(2 + √2) cm. If the length of BC is √2 times the length of AB, then find the area of ΔABC.
In a triangle ΔABC, if AB = 5, BC = 3 and CA = 4, then the value of \(\sin \frac A 2 + \tan \frac A 2\) is:
In Δ ABC, the coordinates of B are (0, 0), AB = 2, ∠ABC = π/3 and the middle point of BC has the coordinates (2, 0). The centroid of triangle is:
In a lake, the tip of a bud of lotus is seen 10 cm above the surface of water. Forced by the wind, it gradually moved, and just submerged at a distance of 30 cm. The depth of water at the root of the lotus plant will be