ABC is a triangle right angled at B. Let M and N be two points on AB such that AM=MN=NB. Let P and Q be two points on AC such that PM is parallel to QN and QN is parallel to CB. If BC=12~\text{cm}, then what is (PM+QN) equal to?

Explanation

Since PM \parallel QN \parallel BC, by similar triangles, \frac{PM}{BC} = \frac{AM}{AB} and \frac{QN}{BC} = \frac{AN}{AB}. Given AM=MN=NB, we have AM = \frac{1}{3}AB and AN = \frac{2}{3}AB. Thus, PM = \frac{1}{3} \times 12 = 4~\text{cm} and QN = \frac{2}{3} \times 12 = 8~\text{cm}. (PM + QN) = 4 + 8 = 12~\text{cm}.

Related questions on Geometry

• In a triangle ABC, \angle A = 30^\circ, AB = 7 cm and AC = 12 cm. What is the area of the triangle ABC?
• ABC is a triangle right angled at B. D is a point on AC such that BD is perpendicular to AC. If AB = p and BC = \sqrt{3}p, then what is BD...
• The difference between an interior angle and an exterior angle of a regular polygon is 120°. What is the number of sides of the polygon?
• An angle q is exactly one-fourth of its complementary angle. What is the value of angle q?
• The sides of a triangle are 11 cm, 60 cm and 61 cm. What is the area of the triangle formed by joining the mid-points of the sides of the tr...