Embedded figures questions give you a main (question) figure and ask you to identify which of the answer options is hidden or concealed within it — or they ask you to count how many figures of a given type exist inside the main figure. The embedded shape is always present in the main figure, either as a standalone region or as a shape formed by combining existing lines.
Think of it like finding a constellation in a star map. The stars are all there, but you have to mentally trace the right lines connecting them. The question figure's lines double as boundaries for multiple shapes simultaneously — the same line can belong to two adjacent triangles, for instance. Your job is to see past the "full picture" and isolate just the shape you are looking for.
Here is the analogy that works best in practice: imagine a tiling pattern on a floor. A single large hexagonal tile can conceal six equilateral triangles, two overlapping rectangles, and a central point all at once — depending on which lines you focus on. Embedded figures work on the same principle. The main figure is not one shape; it is a container of many.
In SSC CGL, two types of embedded figure questions appear:
Type 1 — Identification. A main figure is given. Four answer options show smaller or differently oriented figures. You must find which one is hidden within the main figure without rotating or flipping the answer option.
Type 2 — Counting. A main figure with internal lines is given. You must count how many times a particular shape (usually a triangle or rectangle) appears in it — at all sizes.
Both types reward systematic method over guessing. The student who randomly scans the figure takes 60–90 seconds and still gets it wrong. The student who applies a rule-based approach takes 15–25 seconds and gets it right. The gap between these two students in a 60-question Reasoning paper is enormous.
The core insight: in embedded figures, every intersection point (vertex) is potential corner of a hidden shape. Every line segment is a potential side. Start from those anchors.
Any polygon with n sides, when its internal lines (diagonals, midpoint connectors, altitudes) are drawn, subdivides the interior into smaller regions. These smaller regions — and combinations of them — form additional polygons. The key geometric relationships you must know cold:
Triangle family:
Rectangle family:
Regular polygon family (the must-knows for SSC CGL):
This is the method that cuts solving time in half. Instead of tracing every possible triangle manually:
Step 1: Label every distinct region created by the internal lines. Call them R1, R2, R3...
Step 2: Identify the smallest (unit) triangles — the ones with no internal lines inside them.
Step 3: Count combinations:
Step 4: Add all counts.
For a triangle with one median: 2 regions. Triangles = 2 small + 1 large = 3 total. For a triangle with all 3 altitudes: 6 small regions. Total triangles = 7 (trust this count — verify it in practice).
Look — this catches most people. In standard SSC CGL embedded figures (Type 1), the embedded shape must appear in the same orientation as shown in the answer option — no rotation, no reflection. So if the answer option shows a right-pointing triangle and the main figure has a left-pointing triangle, that is wrong.
Practical approach for Type 1:
Rather than scanning the entire figure, anchor on the distinctive feature of the shape you are looking for.
This reduces Type 1 solving time from ~40 seconds to ~15 seconds once practiced.
These come up repeatedly and knowing them saves counting time:
| Main Figure | Embedded Figure | |---|---| | Equilateral triangle (midpoints joined) | Smaller equilateral triangle | | Regular pentagon (all diagonals) | Smaller regular pentagon | | Five-pointed star | Regular pentagon at center | | Regular octagon (alternate vertices connected) | Square | | Rectangle (one diagonal) | 2 right-angled triangles | | Arrow shape (right-pointing) | Triangle (the arrowhead) |
Whenever you see a five-pointed star (pentagram) as the question figure, the embedded figure at the center is always a regular pentagon. No need to count lines or trace vertices — this is a fixed geometric truth. The five inner intersection points of a pentagram are always the five vertices of a regular pentagon.
Standard method (manually tracing all intersections): ~35 seconds. This pattern recall: ~3 seconds.
Micro-example: Question figure is a five-pointed star. Options are triangle, hexagon, pentagon, square. Immediately mark pentagon without any visual tracing.
For any regular octagon with lines connecting every alternate vertex (skip-one-vertex pattern), the embedded figure is always a square. The 8 vertices of a regular octagon, when alternate ones are connected, produce a square because they are equidistant and at 90° intervals.
Standard method (measuring angles, checking distances): ~45 seconds. This pattern recall: ~4 seconds.
Micro-example: Options are triangle, heptagon, pentagon, square. Main figure is an octagon with alternate vertices connected. Mark square immediately.
When a triangle has its three midpoints connected (medial triangle construction), the total number of triangles in the resulting figure is always 5: 3 corner triangles + 1 central triangle + 1 original large triangle. Do not recount. Do not second-guess. It is always 5.
Standard method (manually tracing all triangles): ~50 seconds with risk of miscounting. This direct recall: ~3 seconds.
Works for any triangle — equilateral, isosceles, scalene — the count of 5 is invariant as long as only the three midpoints are connected.
When all three altitudes of a triangle are drawn (creating the orthocenter), the total number of triangles of all sizes embedded in the figure is 7. This is a fixed count regardless of whether the triangle is acute, right, or obtuse (for the standard exam format where all altitudes land inside the triangle).
Label the 6 small regions R1 through R6 if you need to verify: 6 unit triangles + combinations that validly form larger triangles bring the total to 7. The combinations check takes time — instead, memorize 7 and spend zero seconds in the exam hall.
Standard method (label-and-count): ~60 seconds. Direct recall: ~3 seconds.
Any arrow figure (the classic rightward-pointing pentagon-like shape used in SSC CGL) is geometrically a rectangle attached to a triangle at the right side. This means: a triangle is always embedded in an arrow figure, and a rectangle is always embedded in the body portion.
When options include triangle for an arrow figure: pick triangle immediately without tracing. When options include circle, hexagon, or octagon for an arrow figure: eliminate immediately — none of these can be formed from straight sides of a simple arrow.
Time saved vs. manual tracing: ~30 seconds per question. 3-step elimination replaces visual search entirely.
In the exam hall, apply this decision tree in under 20 seconds:
Is it a counting question (how many triangles/shapes)?
Is it an identification question (which shape is embedded)?
If a question takes more than 25 seconds: mark your best guess, flag it, and return if time permits.
Why this question: Tests the most fundamental embedded figure relationship — midpoint construction on an equilateral triangle. The answer follows directly from geometric construction, not visual search.
Solving path: Equilateral triangle + midpoints joined → recall the pattern: smaller equilateral triangle is formed inside. Do not attempt to visually embed a circle, pentagon, or hexagon — straight-sided figures cannot contain curved shapes, and pentagons/hexagons require sides or angles that the triangle's construction does not provide. Answer: (A).
Why this question: Tests the rectangle-diagonal relationship. Foundational for both Type 1 and Type 2 counting questions.
Solving path: Rectangle + one diagonal = two right-angled triangles. The diagonal becomes the hypotenuse of both. Pentagon, hexagon, and octagon all require more than 4 sides to form — impossible within a simple rectangle's boundary. Answer: (B).
Why this question: Classic shape-in-star question. Tests whether you know the pentagram's geometric property.
Solving path: Five-pointed star → recall the STAR→PENTAGON pattern. The five inner intersection points of the star's sides form the five vertices of a regular pentagon. Zero tracing needed. Answer: (C).
Why this question: Tests the alternate-vertex property of a regular octagon, a recurring SSC CGL pattern.
Solving path: Regular octagon has 8 vertices spaced at 45° intervals. Connecting every alternate vertex (4 vertices total) = 4 points equidistant and at 90° intervals = a square. Recall the OCTAGON ALTERNATE VERTICES = SQUARE pattern. Eliminate triangle (needs 3 vertices), heptagon (7-sided, impossible with 4 connection points), and pentagon (5-sided). Answer: (D).
Why this question: Tests altitude-based triangle counting — the hardest sub-type because the counting is non-intuitive without the fixed-count shortcut.
Solving path: All three altitudes drawn inside a triangle → recall ALTITUDE TRIANGLE COUNT = 7. The 6 small regions created by the three altitudes combine into additional larger triangles, and summing all valid sizes gives 7. Answer: (C).
Why this question: Tests midpoint-triangle counting, a direct application of the memorized count of 5.
Solving path: Midpoints of all three sides connected → recall MIDPOINT TRIANGLE COUNT = 5. The 4 smaller congruent triangles formed plus the original large triangle = 5 total. Do not stop at 4 (forgetting the large outer triangle is the classic mistake). Answer: (D).
Forgetting the large outer triangle when counting. In any triangle-within-triangle configuration, students count the inner triangles only. Always add the original large triangle to your count. The midpoint construction gives 4 inner triangles + 1 outer = 5, not 4.
Allowing rotation when it is not permitted. In Type 1 identification questions, the embedded shape must appear in the same orientation as the answer option. If you find the shape but it is rotated 90°, that option is wrong. Check orientation before confirming an answer.
Trying to embed curved shapes in straight-line figures. If the main figure is made entirely of straight lines (all polygons), a circle cannot be an embedded figure in the geometric sense these questions use. Eliminate all circular options instantly when the main figure is a polygon.
Double-counting overlapping regions. When counting triangles in figures with multiple internal lines, the same unit region can contribute to several different larger triangles. This is correct — count each valid triangle once regardless of how many times its sub-regions appear in other triangles. Students often over-correct and under-count to avoid "double counting."
Stopping at the most obvious embedded shape. The question may have multiple valid embedded shapes, but the answer option list will contain only one correct one. Do not pick the first embedded shape you find — verify it against the specific answer option given.
Ignoring composite shapes. An arrow is not just a pentagon — it contains both a rectangle and a triangle as embedded shapes. When the answer option says "triangle" for an arrow figure, students sometimes reject it because "the whole figure has 5 sides, not 3." The triangle is the arrowhead portion; it is embedded within the composite figure.