Chapter 11.2 — Exercise 11.1 — Area of Rectangle

A Median Splits a Triangle Into Two Equal Halves

In right triangle ABC, ∠ABC = 90°, AB = 12 cm, BC = 6.5 cm, and D is the midpoint of AC; find the area of △ADB. First, the area of the whole triangle: since ∠B = 90° exactly, AB and BC are themselves already the base and height, so area = ½ × 6.5 × 12 = 39 cm² in total. Since AD = DC exactly, BD is a median, and a median always splits any triangle into two triangles of exactly equal area — so area of △ADB = area of △BDC = ½ × 39 = 19.5 cm².

The reason a median splits area evenly, not just length, is worth stating explicitly: △ADB and △BDC share the identical height (the perpendicular distance from B to line AC), and their bases AD and DC are equal by definition of a median — equal base, equal height, so their areas, each computed as ½ × base × height, come out equal too, entirely independent of where D happens to sit relative to the triangle's other angles. This same fact — a median always bisects a triangle's area, not just its opposite side — reappears constantly across the next two exercises, usually as a single quick justifying step buried inside a longer chain of reasoning rather than as the main result being proved.

A Quadrilateral Built From Two Right Triangles

Quadrilateral PQRS has ∠QPS = ∠SQR = 90°, PQ = 12 cm, PS = 9 cm, QR = 8 cm, SR = 17 cm; find its area. The diagonal QS splits PQRS into △PQS and △QRS. Since ∠QPS = 90°, area of △PQS = ½ × PQ × PS = ½ × 12 × 9 = 54 cm² right away. QS itself isn't given directly, but △PQS is right-angled at P, so the Pythagorean theorem gives QS² = PQ² + PS² = 144 + 81 = 225, so QS = 15 cm precisely. Since ∠SQR = 90° too, area of △QRS = ½ × QR × QS = ½ × 8 × 15 = 60 cm² exactly. Total area of PQRS = 54 + 60 = 114 cm² in all.

This problem genuinely needed the Pythagorean theorem purely as a stepping stone along the way — QS itself wasn't ever asked for on its own, but without recovering it first, the second triangle's area formula would have had no usable height to multiply against its base. Splitting the quadrilateral along QS specifically, rather than along PR, is what makes both resulting triangles right-angled and therefore directly solvable; a different diagonal choice here wouldn't hand over a clean right angle in either piece. Checking which diagonal produces usable right angles before committing to a split is a habit worth building deliberately: PR, the other diagonal available in this same figure, would instead produce two triangles with no stated right angle in either one, leaving no clean way to compute either piece's height without quite a bit of extra information the problem never actually supplies.

A Trapezium Hiding a Rectangle Inside It

Trapezium ABCD contains rectangle ADCE, with EC = 8 cm and BE = 3 cm (the same 3 cm also giving AD's own extension); find the area of ABCD. Area of ABCD = area of rectangle ADCE + area of △BCE. Area of rectangle ADCE = AD × AE = 8 × 3 = 24 cm². Area of △BCE = ½ × EC × BE = ½ × 8 × 3 = 12 cm². Total area = 24 + 12 = 36 cm² for the whole trapezium.

The hint "ABCD has two parts" is really pointing at the same decomposition strategy used in every problem in this exercise: a shape without its own direct formula (a trapezium built with a rectangle set inside it, in this case) is measured by identifying a clean split into two shapes that do have direct formulas, then simply adding the two resulting areas together. Rectangle ADCE was chosen as the "hidden" piece here specifically because it shares an entire side with the trapezium's own parallel sides — a deliberate construction, not a coincidence, since the whole point of drawing E in the first place is to isolate exactly the part of the figure a rectangle formula can already handle cleanly, leaving only the small triangular corner, △BCE, needing its own separate formula.

Congruent Triangles Guarantee Equal Areas

Parallelogram ABCD has diagonals AC and BD meeting at O; prove that area of △AOD = area of △BOC. In △AOD and △COB: AO = CO and BO = DO (a parallelogram's own diagonals always bisect each other), and AD = BC (opposite sides of any parallelogram are always equal) — SSS congruence, so △AOD ≅ △COB directly. Since any two congruent figures always have exactly equal areas without exception, area of △AOD = area of △BOC follows immediately and directly, with no separate numeric area calculation needed anywhere in this whole proof at all.

This is a genuinely different technique from the first three problems in this exercise: instead of computing two numeric areas and checking they match, congruence is used to guarantee the match without computing either area explicitly. "Congruent figures have equal area" is really a one-way street worth remembering carefully — congruent triangles are always equal in area, but two triangles can easily share the same area without being congruent at all, a distinction the next two exercises lean on heavily. This problem's own figure makes the point especially clearly: △AOD and △BOC don't look like mirror images of each other at first glance, sitting on opposite sides of the parallelogram's center rather than side by side, yet the SSS congruence above confirms they really are an exact match once rotated into alignment — a genuinely useful reminder that congruence is about matching measurements, never about two shapes already looking identical from a casual glance at the page.

Four Splits, Four Results

From Splitting Figures to Comparing Them

Every problem here computed one figure's exact numeric area, whether directly or by splitting it into simpler pieces first. Exercise 11.2 shifts the question entirely — instead of finding a specific area, it proves that two different parallelograms, or a parallelogram and a triangle, share exactly the same area whenever they sit on the same base between the same parallels, without ever needing to compute either area as a number at all — a proof style that turns out to be far more powerful for genuinely general figures, where an actual numeric answer often isn't even available to begin with.