Mensuration (2D)

Rectangle and square

• Rectangle: $K = l \times b$, $P = 2(l+b)$, diagonal $d = \sqrt{l^2+b^2}$.
• Square: $K = a^2 = \frac{d^2}{2}$, $P = 4a$, $d = a\sqrt2$. The rectangle questions rarely hand you both $l$ and $b$; they hand two of $\{K, P, d\}$ and expect the pair. The fastest trick: for a rectangle, $d^2 = (l+b)^2 - 2K$, and $d^2 = (l-b)^2 + 2K$ — so from $P$ and $d$ you get $l+b$ and then $K = \left(\frac{l+b}{2}\right)^2 - \left(\frac{d}{2}\right)^2$. Memorised example: $P = 34$, $d = 13$ → $l+b = 17$ → sides 12 and 5 → $K = 60$.

Rhombus

Diagonals bisect at right angles: $K = \frac{1}{2}d_1 d_2,\qquad \text{side} = \sqrt{\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2},\qquad P = 4 \times \text{side}$ Half-diagonals are engineered as triplets: (10, 24) → side 13; (16, 12) → side 10. Diagonal-ratio questions: write $d_1 = 5k$, $d_2 = 12k$, put into $K = \frac12 d_1 d_2$, find $k$.

Parallelogram

$K = \text{base} \times \text{height}$ (height is perpendicular to that base). A second form used when two adjacent sides and the included angle $\theta$ are known: $K = ab\sin\theta$. The $30^\circ$/$150^\circ$ pair ($\sin = \frac12$) is the favourite: sides 10, 8 at $30^\circ$ → $K = 40$.

Trapezium

$K = \frac{1}{2}(a+b)\,h$ Height-from-area: $h = \frac{2K}{a+b}$. If the parallel sides differ by $d$: write them $a$ and $a+d$; then $K = (a + \frac{d}{2})h$ solves for $a$ directly.

How the questions chain these

Watch the compound shells: a rectangle with $P = 34$ and $d = 13$ needs BOTH the half-perimeter and the diagonal — the triplet (5, 12, 13) finishes it in one glance, while the algebra route $K = \left(\frac{l+b}{2}\right)^2 - \left(\frac{d}{2}\right)^2 = 8.5^2 - 6.5^2 = 60$ works when no triplet fits. For a rhombus the usual chain is three steps: diagonals → side (Pythagoras on the halves) → perimeter ($\times 4$), or area → missing diagonal → side. With a ratio of diagonals, never guess the diagonals themselves — introduce $k$, use the area equation, then finish. Trapezium reverse questions hide the height: from $K$, $a$, $b$ get $h = \frac{2K}{a+b}$, and if the slant sides matter, drop perpendiculars to build a right triangle with half the difference $\frac{b-a}{2}$.

Quick revision

• Square: $K = \frac{d^2}{2}$; rectangle: $d^2 = (l+b)^2 - 2K$.
• Rhombus: $K = \frac12 d_1d_2$; side from half-diagonal triplet.
• Parallelogram: $bh$ or $ab\sin\theta$ ($30°$ → half).
• Trapezium: $\frac12(a+b)h$; sides differing by $d$: $K = (a + \frac{d}{2})h$.