(i) Exactly 3 distinct cyclic edge orderings can be formed. (ii) If the 4 points form a convex hull: 1 is convex and 2 are self-intersecting. If 1 point is inside: 0 are convex, 3 are non-convex and 0 self-intersect. Class ...
Discussion Forum Latest Questions
The segment EF connecting midpoints equals (3 + 5)/2 = 4 cm and is parallel to the bases. Since E and F bisect the non-parallel sides, both resulting trapeziums share identical heights. Hence, their area ratio is (3 + 4)/(4 ...
In triangle OBP and triangle ODQ, OB = OD, ∠POB = ∠QOD and alternate interior angles ∠OBP = ∠ODQ. By ASA congruence, triangle OBP ≅ triangle ODQ, proving OP = OQ; this direct triangle congruence is simplest. Cbse Class 9 Maths ...
(i) Diagonal BD meets EF at M. By Theorem 7, M and F bisect BD and BC. Adding segment lengths gives EF = (AB + CD)/2. (ii) Both methods work; using parallel line uniqueness is simpler. Class 9 Ganita Manjari Part ...
Folding A onto B creates the perpendicular bisector crease across AB, marking its midpoint. Since this crease is parallel to side BC, by the Converse of the Midpoint Theorem it also bisects AC. Class 9 Ganita Manjari part 2 Chapter 12 ...