Geometry & Mensuration — CAT Quantitative Ability

Geometry and Mensuration reward accurate visualisation more than any other QA topic — a correctly drawn, roughly-to-scale figure often reveals the answer before a single formula is applied, while an inaccurate sketch actively misleads. The sub-topics are triangles, circles, polygons, coordinate geometry and the mensuration of solids, and CAT typically mixes two of these within a single question — a circle inscribed in a triangle, or a solid built by combining a cylinder and a hemisphere.

1. Triangles

$\text{Pythagoras: }{a}^2+b^2=c^2(\text{right triangle, }c\text{ the hypotenuse})$

A triangle's area has three common forms, and choosing the right one for the given data is the actual skill:

$\text{Area}=\frac{1}{2}\times \text{base}\times \text{height}\text{Area}=\frac{1}{2}ab\sin C\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}(\text{Heron’s, }s=\frac{a+b+c}{2})$

The choice is dictated entirely by what the question actually gives: a base and a perpendicular height call for the first form; two sides and the included angle call for the second; and Heron's formula is the fallback the moment only the three side lengths are known and no height or angle is given directly — it needs no additional construction, at the cost of a somewhat heavier final computation.

A triangle's area also connects to its inradius $r$ and circumradius $R$ through two further identities worth knowing: $\text{Area}=r\times s$ (the inradius times the semi-perimeter) and $\text{Area}=\frac{abc}{4R}$. These convert a question that supplies an inscribed or circumscribed circle's radius directly into an area question, without needing to locate the incentre or circumcentre geometrically.

Similar triangles (matching angles, proportional sides) have area ratio equal to the square of their side ratio — not the side ratio itself. Two similar triangles with sides in ratio $2:3$ have areas in ratio $4:9$, a fact frequently tested by giving one triangle's area and a side ratio and asking for the other's area. Similarity is established by AA (two equal angles), SAS (one equal angle between proportional sides) or SSS (all three sides proportional).

Trap. A perimeter ratio is the same as the side ratio, but an area ratio is its square — mixing these up is the most common similar-triangles error under time pressure.

Two right-triangle ratios are worth memorising directly, since they recur constantly and re-deriving them wastes time: a $30°$-$60°$-$90°$ triangle has sides in ratio $1:\sqrt{3}:2$, and a $45°$-$45°$-$90°$ triangle has sides in ratio $1:1:\sqrt{2}$. Recognising either ratio in a figure immediately gives every side without applying Pythagoras from scratch.

The midpoint theorem states that the segment joining the midpoints of two sides of a triangle is parallel to the third side and exactly half its length — the special case, at ratio $1:1$, of the more general basic proportionality theorem (a line parallel to one side of a triangle divides the other two sides proportionally).

The angle bisector theorem is a third recurring fact: the internal bisector of an angle divides the opposite side in the ratio of the two adjacent sides.

2. Circles

$\text{Circumference}=2\pi r,\text{Area}=\pi r^2$

$\text{Arc length}=\frac{\theta }{360}\times 2\pi r,\text{Sector area}=\frac{\theta }{360}\times \pi r^2$

A perpendicular from the centre to a chord always bisects the chord — this single fact, combined with Pythagoras, solves nearly every "distance of a chord from the centre" question without needing any circle-specific formula beyond it.

An angle inscribed in a semicircle is always a right angle, which is why a triangle inscribed in a circle with one side as the diameter is automatically right-angled — a fact that turns some coordinate-geometry-looking questions into a one-line triangle question. In a cyclic quadrilateral (one whose vertices all lie on a circle), opposite angles sum to $180°$, which is the fastest route to an unknown angle whenever a quadrilateral is stated or shown to be cyclic.

A tangent to a circle is always perpendicular to the radius drawn to the point of contact — the single fact behind nearly every tangent-length question. A direct consequence: the two tangent segments drawn from any external point to a circle are equal in length, which is what makes "find the perimeter of a triangle circumscribing a circle" questions solvable by simple addition once the equal tangent pairs are marked.

For two chords intersecting inside a circle, or two secants meeting outside it, the power of a point gives a fixed product relationship ($PA\times PB=PC\times PD$ for two chords through the same interior point $P$) that solves for an unknown chord segment without any angle-chasing at all.

Two circles of radii $r_1$ and $r_2$ with centres a distance $d$ apart have two distinct kinds of common tangent, and confusing them is a standard error:

$\text{Direct common tangent}=\sqrt{d^2-(r_1-r_2{)}^2}\text{Transverse common tangent}=\sqrt{d^2-(r_1+r_2{)}^2}$

A direct tangent exists whenever the circles do not sit one entirely inside the other; a transverse tangent (which crosses between the two circles) exists only when the circles are far enough apart that $d>r_1+r_2$ — the two circles must not overlap at all.

Quadrilateral areas beyond the rectangle are worth a direct formula each, since deriving them from scratch every time is slower than recall: a trapezium with parallel sides $a,b$ and height $h$ has area $\frac{1}{2}(a+b)h$; a rhombus with diagonals $d_1,d_2$ has area $\frac{1}{2}{d}_1{d}_2$, since the diagonals of a rhombus are always perpendicular bisectors of each other.

A parallelogram with adjacent sides $a,b$ and included angle $\theta$ has area $ab\sin \theta$ — structurally identical to the triangle's $\frac{1}{2}ab\sin C$ formula, doubled, since a parallelogram is two congruent triangles joined along a diagonal.

3. Polygons

$\text{Sum of interior angles of an }n\text{-gon}=(n-2)\times 180°$

The sum of the exterior angles of any convex polygon is always $360°$, regardless of the number of sides — a fact many candidates forget exists as a separate, simpler rule alongside the interior-angle formula. For a regular polygon, each interior angle is $\frac{(n-2)\times 180°}{n}$ and each exterior angle is $\frac{360°}{n}$.

The area of a regular polygon with $n$ sides of length $a$ is $\frac{1}{4}n{a}^2\cot \left(\frac{180°}{n}\right)$, which reduces to the familiar $\frac{\sqrt{3}}{4}{a}^2$ for an equilateral triangle ($n=3$) and to $a^2$ for a square ($n=4$) once the cotangent values are substituted — a useful check that the general formula is being applied correctly rather than memorised blindly.

4. Coordinate geometry

$\text{Distance}=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}\text{Section formula (ratio }m:n\text{)}:\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$

The area of a triangle given its three vertices is computed directly from coordinates, without ever drawing the figure, using the shoelace (determinant) formula:

$\text{Area}=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$

This single formula replaces the slower route of finding a base length, a height via a perpendicular distance, and then applying $\frac{1}{2}\times \text{base}\times \text{height}$ — and it works even for an obtuse or awkwardly-oriented triangle where identifying a convenient base visually is difficult.

A line's slope, $m=\frac{y_2-y_1}{x_2-x_1}$, decides both its direction and its relationship to other lines: two lines are parallel exactly when their slopes are equal, and perpendicular exactly when the product of their slopes is $-1$. The perpendicular distance from a point $(x_0,y_0)$ to a line $ax+by+c=0$ is

$d=\frac{|a{x}_0+b{y}_0+c|}{\sqrt{a^2+b^2}}$

which is the coordinate-geometry route to a triangle's height once its base lies along a known line — an alternative to the shoelace formula that is faster when one side's equation is already given rather than just its endpoints.

A triangle's centroid — the intersection of its three medians — has coordinates that are simply the average of the three vertices, $\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$, with no construction needed at all. The centroid always divides each median in the ratio $2:1$ from the vertex, a fact occasionally tested directly via the section formula rather than through the averaging shortcut.

5. Mensuration of solids

A "combined solid" question — a cylinder capped with a hemisphere, or a cone carved out of a cylinder — is solved by adding or subtracting the individual volumes and surface areas, never by inventing a new formula for the composite shape. The one recurring subtlety is surface area: the flat circular face where two solids join is internal and must be excluded from the combined solid's outer surface area, even though each solid's own formula would normally include it.

The slant height of a cone is not the same as its height, and using $h$ where $l$ belongs (or vice versa) is the most common cone error — the curved surface area formula $\pi rl$ needs the slant height specifically, found via $l=\sqrt{r^2+h^2}$ when only the vertical height is given.

A frustum is what remains of a cone after its top is sliced off by a plane parallel to the base, leaving two circular faces of radii $R$ (bottom) and $r$ (top) and a slant height $l$:

$\text{Volume}=\frac{\pi h}{3}\left(R^2+Rr+r^2\right),\text{Curved surface area}=\pi (R+r)l,l=\sqrt{h^2+(R-r{)}^2}$

The frustum formula reduces correctly to the cone formula as a sanity check: setting $r=0$ turns the volume expression into $\frac{\pi h}{3}{R}^2$, exactly the ordinary cone volume — a quick way to confirm the formula is being remembered correctly rather than confused with the cylinder or the full cone.

Worked Examples

Example 1 (triangles, Heron's — easy). Find the area of a triangle with sides 13, 14 and 15.

$s=\frac{13+14+15}{2}=21$. Area $=\sqrt{21\times 8\times 7\times 6}=\sqrt{7056}=84$.

Example 2 (similar triangles — medium). Two similar triangles have areas 50 cm² and 72 cm². If the smaller triangle's shortest side is 10 cm, find the corresponding side of the larger triangle.

Area ratio $=\frac{50}{72}=\frac{25}{36}$, so the side ratio is $\sqrt{\frac{25}{36}}=\frac{5}{6}$. If the smaller side is 10 cm and corresponds to the "5" part of the ratio, the larger triangle's side is $10\times \frac{6}{5}=12$ cm.

Example 3 (circles, sector — easy). Find the arc length and area of a sector of radius 21 cm with a central angle of $60°$ (use $\pi =\frac{22}{7}$).

Arc length $=\frac{60}{360}\times 2\times \frac{22}{7}\times 21=\frac{1}{6}\times 2\times 22\times 3=22$ cm.

Sector area $=\frac{60}{360}\times \frac{22}{7}\times 21^2=\frac{1}{6}\times \frac{22}{7}\times 441=\frac{1}{6}\times 1386=231$ cm².

Example 4 (circles, cyclic quadrilateral — medium). In a cyclic quadrilateral $ABCD$, $\angle A=3x+5°$ and $\angle C=2x+15°$. Find $x$ and $\angle A$.

Opposite angles of a cyclic quadrilateral sum to $180°$: $(3x+5)+(2x+15)=180\Rightarrow 5x+20=180\Rightarrow x=32$. So $\angle A=3(32)+5=101°$.

Example 5 (coordinate geometry, shoelace — medium). Find the area of the triangle with vertices $(1,1)$, $(4,5)$ and $(6,2)$.

$\text{Area}=\frac{1}{2}\left|1(5-2)+4(2-1)+6(1-5)\right|=\frac{1}{2}\left|3+4-24\right|=\frac{1}{2}\times 17=8.5$

Example 6 (solids, cylinder — easy). Find the volume and total surface area of a cylinder with radius 7 cm and height 10 cm (use $\pi =\frac{22}{7}$).

Volume $=\frac{22}{7}\times 7^2\times 10=\frac{22}{7}\times 490=1540$ cm³.

Total surface area $=2\times \frac{22}{7}\times 7\times (7+10)=44\times 17=748$ cm².

Example 7 (solids, cone slant height — medium). A cone has base radius 3 cm and height 4 cm. Find its curved surface area.

Slant height $l=\sqrt{3^2+4^2}=\sqrt{25}=5$ cm. Curved surface area $=\pi rl=\pi \times 3\times 5=15\pi \approx 47.12$ cm².

Example 8 (solids, combined figures — hard). A cone and a hemisphere share the same base radius $r$. If the cone's volume equals the hemisphere's volume, find the ratio of the cone's height to its radius.

Cone volume $=\frac{1}{3}\pi r^{2}h$; hemisphere volume $=\frac{2}{3}\pi r^3$. Setting them equal:

$\frac{1}{3}\pi r^{2}h=\frac{2}{3}\pi r^3⟹h=2r⟹\frac{h}{r}=2$

The cone must be twice as tall as its own radius to match a hemisphere of the same radius — a result worth remembering directly, since it recurs whenever a question compares a cone and a hemisphere on equal footing.

Example 9 (solids, frustum — hard). A bucket in the shape of a frustum has bottom radius 6 cm, top radius 3 cm, and height 4 cm. Find its volume and curved surface area.

Slant height $l=\sqrt{4^2+(6-3{)}^2}=\sqrt{16+9}=5$ cm.

$\text{Volume}=\frac{\pi \times 4}{3}\left(6^2+6\times 3+3^2\right)=\frac{4\pi }{3}(36+18+9)=\frac{4\pi }{3}\times 63=84\pi {\text{cm}}^3$

$\text{Curved surface area}=\pi (6+3)\times 5=45\pi {\text{cm}}^2$

As a sanity check, setting the top radius to $0$ would reduce this to an ordinary cone of base radius 6 and the same slant height, confirming the frustum formula is being applied correctly.

Summary

Draw the figure to scale before reaching for a formula — geometry rewards visualisation, and CAT often combines two sub-topics (a circle inside a triangle, a solid built from two shapes) in one question.

A triangle's area has three forms — base-height, $\frac{1}{2}ab\sin C$, and Heron's — and picking the one that matches the given data is the real skill; similar triangles have an area ratio equal to the square of their side ratio.

A perpendicular from a circle's centre bisects any chord; an angle in a semicircle is always $90°$; opposite angles of a cyclic quadrilateral sum to $180°$.

The exterior angles of any convex polygon always sum to $360°$, a separate and simpler fact from the interior-angle-sum formula.

The shoelace formula computes a coordinate-triangle's area directly from its vertices, without ever needing to identify a base and height visually.

For solids, add or subtract volumes/surface areas of combined shapes directly, remembering to exclude any internal joining face from the combined surface area, and never confuse a cone's slant height with its vertical height.