How many triangles are there in the following figure ?
[FIGURE: A triangle drawn with its apex at the top, one vertex at the bottom-left and one at the bottom-right. Two straight lines are drawn insid

Question

How many triangles are there in the following figure ?
[FIGURE: A triangle drawn with its apex at the top, one vertex at the bottom-left and one at the bottom-right. Two straight lines are drawn inside it: one from the top apex down to a point on the bottom side, and one from the bottom-left vertex up to a point on the right-hand side. The two lines cross at a point inside the triangle. No numbers, letters or other labels are printed anywhere in the figure.]

UnlockIAS · Accepted Answer

Correct — B, 8. Name the figure to count it. Call the apex A, the bottom-left vertex B and the bottom-right vertex C. One inner line runs from A to a point D on the base BC, the other from B to a point E on the right-hand side AC, and the two cross at an interior point F. Now count by LINES rather than by regions, because regions are where miscounts come from. Exactly five straight lines are drawn: AB, BC, CA, AD and BE. Any triangle in the picture must be bounded by three of them, and three lines chosen from five can be picked in 5 × 4 × 3 ÷ 6 = 10 ways. Two of those ten choices fail, and both fail for the same reason — the three lines pass through a single point instead of enclosing an area. AB, CA and AD all meet at A; AB, BC and BE all meet at B. Every one of the remaining eight triples does bound a genuine triangle, and they are: ABC, the whole figure; ABD and ADC, the two halves made by the cevian AD; ABE and BCE, the two halves made by the cevian BE; and the three small ones round the crossing point — ABF on the base line AB, BDF in the bottom-right corner and AEF in the upper-right corner. Eight in all. The strongest thing about this method is what it does not need: nowhere did the position of D or of E enter the count. Slide either foot along its side and the same five lines still meet in the same pattern, so the answer stays 8. A line drawn from a vertex to the opposite side is called a cevian, and the general rule worth carrying out of this question is that TWO CEVIANS FROM TWO DIFFERENT VERTICES ALWAYS PRODUCE 8 TRIANGLES — which is exactly why this question was answerable even though the booklet printed no labels inside the figure at all.

Why the other options are wrong — (a) The count you get from the raw combinatorics without removing the degenerate cases — three lines chosen from five give ten triples, and a candidate who stops there answers 10. The two triples that must be struck out are the ones concurrent at a vertex: AB with CA and AD at A, and AB with BC and BE at B. Three lines through one point enclose nothing.  (c) Comes from treating the four-sided region FDCE as if it could be split into triangles. It cannot: no line is drawn between C and F, or between D and E, so that region stays a quadrilateral in this figure. Adding either of those two segments would genuinely raise the count, which is why the option looks plausible to someone reasoning about areas rather than about drawn lines.  (d) The answer to a richer figure than the one printed — 12 is what a rectangle with both diagonals and two half-medians yields, and it is the published answer to the count-the-triangles question on the 70th CCE paper of December 2024. Here it usually arises from double-counting the small regions around the crossing point F, or from counting a region and its mirror image twice.

Concept: Counting figures inside a figure is a standard reasoning item, and it has a reliable method that does not depend on eyesight or patience. Every triangle in a line drawing is the intersection of exactly three straight lines, so the count of triangles is the count of ways to choose three of the drawn lines, minus the choices that fail. A choice fails if the three lines are concurrent — they all pass through one point — or if they are parallel, or if the point where two of them cross lies outside the drawn segments. Work through the drawn lines, take the combination, then subtract the failures. For simple cevian figures the failures are always the concurrences at the vertices, and they are easy to list. A related standard result is worth memorising alongside: if n cevians are drawn from ONE vertex of a triangle to the opposite side, the number of triangles is (n + 1)(n + 2) ÷ 2 — three for one cevian, six for two, ten for three.

State PCS 2025 Q10 — Science & Technology PYQ with Answer

State PCS 2025 Q10 — Science & Technology PYQ with Answer