Geometry and Coordinate Geometry at CAT Depth

Geometry is usually four to six of the twenty-two QA questions, and it is the topic where candidates most often know the formula and still lose the mark. The reason is that CAT rarely asks you to apply a formula; it asks what a configuration forces to be true. A typical question gives a triangle with a line drawn across it and asks for a ratio — no lengths at all. That is a similarity question wearing a mensuration costume. This chapter covers the results that answer such questions: similarity and the ratio theorems, the circle theorems including the power of a point, the special centres of a triangle, and coordinate geometry, which lets you replace a construction you cannot see with an algebraic calculation you can. The area and volume formulas themselves are in the foundation chapter and are not repeated here.

Similarity — the Most Asked Idea in CAT Geometry

Two triangles are similar when their angles match; their sides are then in a fixed ratio k. Three consequences follow, and the second is the one CAT tests:

  1. Corresponding sides are in the ratio k.
  2. Corresponding areas are in the ratio k² — not k. This is where most marks are lost in the topic.
  3. For similar solids, volumes are in the ratio k³.
  • The basic proportionality theorem (Thales). A line drawn parallel to one side of a triangle cuts the other two sides in the same ratio. So if DE ∥ BC in triangle ABC, then AD/DB = AE/EC. Conversely, if those ratios are equal then the line is parallel — which is how "prove DE ∥ BC" questions become one division.
  • Worked example (the k² trap). In triangle ABC, DE ∥ BC with AD : DB = 2 : 3. What fraction of the triangle's area is the small triangle ADE? The ratio of sides is AD/AB = 2/5, so the ratio of areas is (2/5)² = 4/25. A candidate who answers 2/5 has used the side ratio for an area — the single most common geometry error at this level. The trapezium DBCE therefore holds the remaining 21/25.
  • The angle bisector theorem. The bisector of angle A meets BC at D with BD/DC = AB/AC. A question that gives two sides and asks for a division of the third is almost always this theorem.
  • Median facts worth knowing. A median divides a triangle into two triangles of equal area (equal bases, same height). The three medians meet at the centroid, which divides each in the ratio 2 : 1 from the vertex, and the six small triangles they create all have equal area.

Triangles — the Results CAT Reuses

Worth knowing by heart
ResultWhat it gives you
Area = r × s, where r is the inradius and s the semi-perimeterThe inradius of any triangle from its area — no construction needed
In a right triangle, r = (a + b − c)/2 with c the hypotenuseThe fastest inradius there is, for the most common triangle in the paper
Circumradius of a right triangle = half the hypotenuseThe right angle sits on the circle — a very frequent hidden fact
Triangle inequality: any side is less than the sum of the other twoCounting how many integer-sided triangles are possible
Pythagorean triples: 3-4-5, 5-12-13, 8-15-17, 7-24-25, 9-40-41Recognising the answer before computing it, and all their multiples
  • Worked example (inradius). A right triangle has legs 6 and 8, so its hypotenuse is 10 and its area is 24. Semi-perimeter s = (6 + 8 + 10)/2 = 12, so r = Area/s = 24/12 = 2. The right-triangle shortcut agrees: (6 + 8 − 10)/2 = 2 ✓
  • Worked example (counting triangles). How many triangles with integer sides have perimeter 12? List with a ≤ b ≤ c and c < a + b: (2,5,5), (3,4,5), (4,4,4). Also (1,5,6)? No — 1 + 5 = 6 is not greater than 6, so those three points are collinear, not a triangle. Answer: 3. The inequality does the whole job; the trap is including the degenerate case.

Circles — Four Theorems That Answer Most Questions

  • The angle at the centre is twice the angle at the circumference standing on the same arc. Its most used special case: an angle in a semicircle is a right angle. If a question mentions a diameter and a point on the circle, a right angle has been handed to you.
  • A tangent is perpendicular to the radius at the point of contact, and the two tangents drawn from an external point are equal in length. Together these turn almost every tangent question into a right triangle.
  • Power of a point. For two chords crossing inside a circle at P: PA × PB = PC × PD. For two secants from an external point: PA × PB = PC × PD as well. And for a tangent PT with a secant PAB: PT² = PA × PB. This one family answers most "find the length" circle questions.
  • Cyclic quadrilateral. Opposite angles add to 180°. Conversely, if opposite angles of a quadrilateral add to 180° then its four vertices lie on one circle — which is how a question hides a circle it never draws.
  • Worked example (power of a point). From an external point P a tangent of length 8 touches a circle, and a secant through P cuts it at A and B with PA = 4. Find PB. PT² = PA × PB gives 64 = 4 × PB, so PB = 16, and the chord AB is 16 − 4 = 12.

Coordinate Geometry — Replace the Construction With Algebra

The value of coordinate geometry in CAT is not the handful of questions asked directly about lines. It is that any plane geometry question mentioning a right angle or a symmetric figure can be placed on axes and turned into arithmetic. Put the right angle at the origin and the legs along the axes, and lengths become subtractions.

The formulas you need, and no more
ForFormula
Distance between two points√[(x₂ − x₁)² + (y₂ − y₁)²]
Midpoint((x₁ + x₂)/2, (y₁ + y₂)/2)
Point dividing a segment in m : n((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n))
Area of a triangle from vertices½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
Slope, and the parallel/perpendicular testsm = (y₂ − y₁)/(x₂ − x₁); parallel if m₁ = m₂, perpendicular if m₁m₂ = −1
Circle with centre (h, k) and radius r(x − h)² + (y − k)² = r²
  • Collinearity is a zero area. Three points lie on a line exactly when the area formula returns 0. For (1,2), (3,6) and (5,10): ½|1(6 − 10) + 3(10 − 2) + 5(2 − 6)| = ½|−4 + 24 − 20| = 0, so they are collinear. This is quicker and safer than comparing two slopes, which can involve a division by zero.
  • Worked example (placing the figure). A right triangle has legs 6 and 8 along the axes. Where is its circumcentre? Place the vertices at (0,0), (6,0) and (0,8). The circumcentre of a right triangle is the midpoint of the hypotenuse, so it is at ((6 + 0)/2, (0 + 8)/2) = (3, 4), and the circumradius is the distance from there to the origin, √(9 + 16) = 5 — exactly half the hypotenuse of 10 ✓
  • Lines and intercepts. The line x/a + y/b = 1 cuts the axes at (a, 0) and (0, b), so the triangle it makes with the axes has area ½|ab|. Questions asking for "the line through (2, 3) cutting off equal intercepts" are answered by setting a = b and substituting: 2/a + 3/a = 1, giving a = 5 and the line x + y = 5.
🎯 Why "put it on axes" is worth practising
In a 40-minute section you cannot afford to stare at a figure hoping to see the construction. Placing the figure on axes is a method that always starts — it converts a problem you might not see into a calculation you certainly can do. It is slower than the elegant synthetic solution and far faster than not finding one.

Key takeaways

  • In similar figures sides scale by k, areas by k² and volumes by k³. Using k where k² is needed is the most common geometry error at this level.
  • A line parallel to one side of a triangle divides the other two proportionally. Most "find the ratio" questions with no lengths given are this theorem.
  • Area = r × s gives the inradius of any triangle, and r = (a + b − c)/2 gives it instantly for a right triangle.
  • An angle in a semicircle is a right angle, and the circumradius of a right triangle is half its hypotenuse. A mentioned diameter is a gifted right angle.
  • The power of a point — PA × PB for any line through P, and PT² for a tangent — answers most circle length questions in one equation.
  • Three points are collinear exactly when the coordinate area formula gives zero — safer than comparing slopes, which can divide by zero.
  • When a figure has a right angle or an axis of symmetry, put it on coordinate axes. A method that always starts beats an insight that might not arrive.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. In triangle ABC, DE is parallel to BC with D on AB and E on AC, and AD : DB = 2 : 3. The ratio of the area of triangle ADE to the area of triangle ABC is:

    1. 2 : 5
    2. 4 : 25
    3. 4 : 9
    4. 2 : 3
    Show answer

    Answer: B — 4 : 25

    DE ∥ BC makes ADE similar to ABC. The ratio of corresponding sides is AD : AB = 2 : (2 + 3) = 2 : 5, so the ratio of areas is the square of that, 4 : 25. Option A is the side ratio used as an area ratio; option C squares 2 : 3 instead, using DB rather than AB in the denominator.

  2. A right triangle has legs of 6 cm and 8 cm. The radius of its inscribed circle is:

    1. 2 cm
    2. 2.4 cm
    3. 3 cm
    4. 5 cm
    Show answer

    Answer: A — 2 cm

    The hypotenuse is 10 (a 3-4-5 triple doubled), the area is ½ × 6 × 8 = 24 and the semi-perimeter is 12, so r = Area/s = 24/12 = 2. The right-triangle shortcut confirms it: r = (6 + 8 − 10)/2 = 2. Option D, 5, is the circumradius (half the hypotenuse), which is a deliberate distractor.

  3. How many triangles with integer sides have a perimeter of 12 units?

    1. 2
    2. 3
    3. 4
    4. 5
    Show answer

    Answer: B — 3

    List unordered triples adding to 12 that satisfy the triangle inequality: (2,5,5), (3,4,5) and (4,4,4). The candidates that fail are (1,5,6), where 1 + 5 = 6 is not greater than 6, and (2,4,6) and (1,4,7) and (3,3,6) for the same reason — equality means the three points are collinear, giving no triangle. So the answer is 3, and counting the degenerate cases is what produces the larger options.

  4. From an external point P, a tangent of length 8 cm touches a circle. A secant from P cuts the circle at A and B, with PA = 4 cm. The length of the chord AB is:

    1. 12 cm
    2. 16 cm
    3. 20 cm
    4. 8 cm
    Show answer

    Answer: A — 12 cm

    The power of the point gives PT² = PA × PB, so 64 = 4 × PB and PB = 16. But PB is measured from P and includes PA, so the chord itself is AB = PB − PA = 16 − 4 = 12. Option B is PB, which is what the question does not ask — a distinction worth reading twice for in the exam.

  5. The area of the triangle whose vertices are (1, 1), (5, 1) and (5, 4) is:

    1. 6 square units
    2. 12 square units
    3. 7.5 square units
    4. 10 square units
    Show answer

    Answer: A — 6 square units

    Two sides are axis-parallel: the base from (1,1) to (5,1) is 4 and the height from (5,1) to (5,4) is 3, so the area is ½ × 4 × 3 = 6. The formula agrees: ½|1(1 − 4) + 5(4 − 1) + 5(1 − 1)| = ½|−3 + 15 + 0| = 6. Option B is the product of base and height without halving.

  6. The points (1, 2), (3, 6) and (5, k) are collinear. The value of k is:

    1. 8
    2. 10
    3. 12
    4. 9
    Show answer

    Answer: B — 10

    The slope from (1,2) to (3,6) is (6 − 2)/(3 − 1) = 2. For collinearity the slope from (3,6) to (5,k) must also be 2: (k − 6)/(5 − 3) = 2, so k − 6 = 4 and k = 10. Check with the area formula: ½|1(6 − 10) + 3(10 − 2) + 5(2 − 6)| = ½|−4 + 24 − 20| = 0 ✓

  7. A line passes through (2, 3) and cuts equal intercepts on the two axes. Its equation is:

    1. x + y = 5
    2. x − y = −1
    3. 2x + 3y = 13
    4. 3x + 2y = 12
    Show answer

    Answer: A — x + y = 5

    Equal intercepts means the line is x/a + y/a = 1, that is x + y = a. Substituting the point: 2 + 3 = 5, so the line is x + y = 5. Check the intercepts: (5, 0) and (0, 5), equal ✓. Option B is a line through the point with slope 1, which cuts intercepts of −1 and 1 — equal in size but not the same intercept.

  8. Two similar cylinders have heights in the ratio 2 : 3. The ratio of their volumes is:

    1. 2 : 3
    2. 4 : 9
    3. 8 : 27
    4. 6 : 9
    Show answer

    Answer: C — 8 : 27

    For similar solids every linear dimension scales by the same factor, so the radii are also in 2 : 3 and the volumes scale by the cube: 2³ : 3³ = 8 : 27. Option B, 4 : 9, is the ratio of their surface areas — correct for areas, wrong for volumes. Note the word "similar": two cylinders of the same radius and different heights have volumes in 2 : 3, so the word is doing real work in this question.