Week 8 · Part 2 — Parallelogram & Trapezoid Basics0%
STEP 1 OF 24 · Lesson Opening
Today: Parallelogram & Trapezoid — the Basics
Two of the four core quadrilaterals. Almost every AIMO Q1–Q4 area chase decomposes into a parallelogram, a trapezoid, or a triangle — so the reflexes you build today pay back every year.
📌 What you will learn today
Topic
Parallelogram opposite-side / opposite-angle properties, area $= bh$, trapezoid area $= \tfrac{1}{2}(a+b)h$, midsegment $= \tfrac{1}{2}(a+b)$, and "diagonals bisect each other" inside a parallelogram.
Plus dozens of warm-up Q1 / Q2 area-of-quadrilateral questions across the past 25 years.
AoPS Reference
Introduction to Geometry by Richard Rusczyk, Chapter 7 (Quadrilaterals).
Why this matters
Every irregular shape on AIMO is eventually cut into rectangles, parallelograms and trapezoids. Master the area formulas plus the diagonal-bisection property today and you have a clean attack on AIMO Q1 (rhombus area), AIMO Q4 (square minus four corner triangles), and the foundation for Part 3 (cyclic quadrilaterals) next week.
Time required
About 70–90 minutes for the full lesson, plus practice papers.
Step 24: Bridge forward to Part 3 (Quadrilaterals II — cyclic & tangential).
Pedagogy: Pass 1 only — we deliberately defer angle bisectors, rhombus depth proofs, and the diagonal-product area rule (those land in Quadrilaterals II). Today is about getting the five core formulas burned into reflex.
STEP 2 OF 24 · Phase 0 · Prerequisite
Prerequisite ① — Parallelogram, rectangle, rhombus, square
Four names, one family tree. Get the hierarchy clear and half the lesson is already done.
The definitions, in one breath
Parallelogram: a quadrilateral whose both pairs of opposite sides are parallel. (Equivalent: both pairs of opposite sides are equal.)
Rectangle: a parallelogram with all four angles right angles.
Rhombus: a parallelogram with all four sides equal.
Square: both at once — rectangle and rhombus.
All four are parallelograms. Rectangle adds "right angles", rhombus adds "equal sides", square adds both.
🔑 The hierarchy. Anything true of a parallelogram is automatically true of a rectangle, rhombus and square. So when you "drop down" from the general formula Area = bh, every special case inherits it for free.
The five parallelogram facts you must own
Opposite sides equal: $AB = CD$ and $BC = DA$.
Opposite angles equal: $\angle A = \angle C$ and $\angle B = \angle D$.
Adjacent angles supplementary: $\angle A + \angle B = 180^\circ$ (and so on around).
Diagonals bisect each other: if diagonals $AC$ and $BD$ meet at $M$, then $AM = MC$ and $BM = MD$.
Each diagonal splits it into two congruent triangles.
In parallelogram $ABCD$, $\angle A = 70^\circ$. Find $\angle B$.
In parallelogram $ABCD$, $AB = 8$. What is $CD$?
STEP 3 OF 24 · Phase 0 · Prerequisite
Prerequisite ② — Trapezoid & the diagonal idea
A trapezoid (Australian: trapezium) is even simpler — only one pair of parallel sides is required.
Trapezoid definition
A quadrilateral with exactly one pair of parallel sides. The two parallel sides are called the bases, the other two are the legs.
Isosceles trapezoid: the two legs are equal.
Right trapezoid: one leg is perpendicular to the bases.
⚠ Convention warning. Some textbooks define a trapezoid as "at least one pair of parallel sides" — making every parallelogram also a trapezoid. AIMO follows the stricter "exactly one pair" definition, but treat the inclusive definition as harmless: any parallelogram formula is also a valid trapezoid formula with $a = b$.
Bases $a$ and $b$ are parallel. The height $h$ is measured perpendicular between them — never along a leg.
What is a "diagonal"?
A diagonal of a quadrilateral is any line segment joining two non-adjacent vertices. Every quadrilateral has exactly two diagonals.
In a parallelogram $ABCD$ the diagonals are $AC$ and $BD$.
They always bisect each other (this is one of the five parallelogram facts above).
In a rectangle, they are also equal in length.
In a rhombus, they are perpendicular.
In a square, they are equal and perpendicular.
A trapezoid has bases $a = 6$ and $b = 10$ and height $h = 4$. Compute $\tfrac{1}{2}(a+b)$.
In parallelogram $ABCD$ the diagonals meet at $M$ and $AC = 18$. Find $AM$.
STEP 4 OF 24 · Phase 1 · Visual Intuition
Picture ① — push a rectangle sideways and you get a parallelogram
The cleanest way to see why the parallelogram area is still $bh$.
Push the top side of a rectangle horizontally — base $b$ and perpendicular height $h$ are unchanged, so the area is unchanged. The slanted side is not the height.
🔑 The slice argument. Cut a thin triangle off the right end of the parallelogram, slide it to the left end, and you get back the original rectangle. Area is preserved by the slide, so parallelogram area $= bh$.
⚠ The biggest beginner mistake: using the slanted leg as the height. The height is always the perpendicular distance between the two parallel sides, never the slanted leg.
A parallelogram has base $b = 9$ and (perpendicular) height $h = 4$. Find the area.
STEP 5 OF 24 · Phase 1 · Visual Intuition
Picture ② — a trapezoid is a triangle minus a smaller triangle
Extending the legs of a trapezoid until they meet gives a big triangle. The area follows automatically.
Extend the two legs of the trapezoid until they meet at apex $T$. The trapezoid is the big triangle minus the small (similar) top triangle.
From the picture to the formula
The big triangle has base $a$ and some height $H$. The small top triangle is similar to the big one with ratio $\tfrac{b}{a}$, so its height is $H \cdot \tfrac{b}{a}$ and the trapezoid's height is
$h \;=\; H - H \cdot \dfrac{b}{a} \;=\; H \cdot \dfrac{a - b}{a}.$
The trapezoid area is therefore
$[\text{trap}] \;=\; \tfrac{1}{2} a H \;-\; \tfrac{1}{2} b \cdot H\tfrac{b}{a} \;=\; \tfrac{H}{2a}(a^2 - b^2) \;=\; \tfrac{H}{2a}(a+b)(a-b).$Substitute $H \cdot \tfrac{a-b}{a} = h \Rightarrow H \cdot \tfrac{1}{a} = \tfrac{h}{a-b}$, so$[\text{trap}] \;=\; \tfrac{1}{2}(a + b) \cdot h.$
🎯 Slick mnemonic. The trapezoid area is "average of the two bases × height". The midsegment $\tfrac{1}{2}(a+b)$ is exactly that average, which is why $[\text{trap}] = \text{midsegment} \times h$.
A trapezoid has bases 4 and 10 and height 6. Find its area.
STEP 6 OF 24 · Phase 1 · Visual Intuition
Picture ③ — a diagonal of a parallelogram makes 2 congruent triangles
This single fact is the engine behind almost every parallelogram proof.
Diagonal $AC$ splits parallelogram $ABCD$ into $\triangle ABC$ (teal) and $\triangle ACD$ (yellow). The two are congruent — same shape, same size.
Why are they congruent?
Compare $\triangle ABC$ and $\triangle CDA$:
Side $AC$ is shared.
$AB = CD$ (opposite sides of a parallelogram).
$BC = DA$ (opposite sides of a parallelogram).
By SSS the two triangles are congruent. (Or use SAS with the alternate-interior-angle pair $\angle BAC = \angle DCA$ from $AB \parallel CD$.) ∎
🔑 Two corollaries you will use all the time.
The diagonal cuts the parallelogram into two equal-area pieces. Each triangle has area $\tfrac{1}{2} bh$.
The two diagonals together cut the parallelogram into four triangles, all of equal area $\tfrac{1}{4} bh$ (because diagonals bisect each other → each pair shares a base and has equal heights).
A parallelogram has area $48$. Diagonal $AC$ is drawn. What is the area of $\triangle ABC$?
Both diagonals of a parallelogram of area $48$ are drawn. They cut it into 4 triangles, each of area?
STEP 7 OF 24 · Phase 1.5 · Formula Handbook
Formula handbook — pin this page in your head
Five entries. Memorise.
① Parallelogram opposite sides & angles
$AB = CD,\quad BC = DA,\quad \angle A = \angle C,\quad \angle B = \angle D$
And
$\angle A + \angle B = 180^\circ$ (adjacent angles supplementary)
② Parallelogram area ★ core ★
$[\text{parallelogram}] \;=\; b \cdot h$
Where
$b$ = any side chosen as the base; $h$ = the perpendicular distance to the opposite parallel side.
③ Trapezoid area ★ core ★
$[\text{trap}] \;=\; \dfrac{1}{2}(a + b) \cdot h$
Read
"average of the two parallel sides, times the perpendicular height between them."
④ Trapezoid midsegment
$\text{midsegment} \;=\; \dfrac{1}{2}(a + b)$
Equivalently
$[\text{trap}] = \text{midsegment} \times h$. Joins the midpoints of the two legs and is parallel to both bases.
⑤ Parallelogram diagonals bisect each other
If $AC \cap BD = M$, then $AM = MC$ and $BM = MD$
Bonus
A diagonal also splits the parallelogram into two congruent triangles, each of area $\tfrac{1}{2} bh$.
Variable decoder
b = base of parallelogram, OR one of the parallel sides of a trapezoid (when both, called $a$ and $b$)
h = perpendicular height between the two parallel sides
a, b (in trapezoid) = the lengths of the two parallel sides (bases)
[X] = area of figure $X$
M = the intersection point of the two diagonals
Quick numerical checks to know cold
Shape
Inputs
Area
Parallelogram
$b=10$, $h=6$
$60$
Parallelogram
$b=12$, $h=5$
$60$
Rectangle
$8 \times 9$
$72$
Trapezoid
$a=4, b=10, h=6$
$42$
Trapezoid
$a=8, b=14, h=6$
$66$
Rhombus (via diagonals)
$d_1=6, d_2=8$
$24$
⚠ The biggest mistake. Using the slanted leg of a parallelogram or trapezoid as the "height". The height is always the perpendicular distance between the two parallel sides — drop a foot from the top side to the bottom side.
STEP 8 OF 24 · Phase 2 · Derivation ①
Derivation ① — parallelogram angles from $\angle A = 70^\circ$
Use opposite-equal + adjacent-supplementary to fill in all four angles.
Parallelogram $ABCD$ with $\angle A = 70^\circ$. Find $\angle B$, $\angle C$, $\angle D$.
$\angle C = \angle A = 70^\circ$ (opposite angles equal)$\angle B = 180^\circ - \angle A = 110^\circ$ (adjacent angles supplementary)$\angle D = \angle B = 110^\circ$ (opposite angles equal)
Opposite angles agree (70°/70° and 110°/110°). Adjacent angles add to 180°.
🎯 Sanity check. Sum of all four angles: $70 + 110 + 70 + 110 = 360^\circ$. ✓ (Always true for any quadrilateral.)
Parallelogram $PQRS$ with $\angle P = 115^\circ$. Find $\angle Q$.
STEP 9 OF 24 · Phase 2 · Derivation ②
Derivation ② — concrete trapezoid: bases 8 and 14, height 6
Equal-side ticks on $AB$ and $CD$; equal alternate angles $\alpha$ at $A$ and $C$.
🎯 Two corollaries you keep using.
Each triangle has area $\tfrac{1}{2}[\text{parallelogram}]$.
Both diagonals together cut the parallelogram into 4 triangles of equal area $\tfrac{1}{4}[\text{parallelogram}]$ (because diagonals bisect each other → opposing triangles share base + height).
Parallelogram of area $80$. Both diagonals drawn. Area of one of the four triangles?
STEP 11 OF 24 · Phase 3 · Worked Example 1 ⭐
Worked Example 1 — straight parallelogram area
WE 1 · ⭐ · skill Q-A2
WE1. A parallelogram has base $12$ and (perpendicular) height $5$. Find its area.
Type your answer (single number):
💡 Hints — open as needed
Direct application of the parallelogram area formula. The height $5$ is already given as perpendicular — no slanted-leg confusion.
Multiply: $b \times h = 12 \times 5$.
Answer: 60
Solution
$[\text{parallelogram}] = b \cdot h = 12 \cdot 5 = 60$.
Reflection
Pure pattern recognition. Whenever a problem hands you "base" and "perpendicular height" of a parallelogram, the answer is one multiplication away. The trap appears only when the slanted leg is given instead — then you must drop a foot first to find the true height.
Tried first?
STEP 12 OF 24 · Phase 3 · Worked Example 2 ⭐⭐
Worked Example 2 — straight trapezoid area
WE 2 · ⭐⭐ · skill Q-A3
WE2. A trapezoid has parallel sides $6$ and $14$ and (perpendicular) height $5$. Find its area.
Type your answer (single number):
💡 Hints — open as needed
Three numbers ($a = 6$, $b = 14$, $h = 5$) plug straight into the formula $\tfrac{1}{2}(a+b)h$.
Average the two parallel sides → $\tfrac{1}{2}(6 + 14) = 10$. Multiply by the height $5$.
The midsegment view is the same calculation in disguise: midsegment $= \tfrac{1}{2}(6+14) = 10$, area $= 10 \cdot 5 = 50$. Whichever framing feels faster on the day — they always agree.
Tried first?
STEP 13 OF 24 · Phase 3 · Worked Example 3 ⭐⭐⭐
Worked Example 3 — isosceles trapezoid: drop perpendiculars first
WE 3 · ⭐⭐⭐ · skill Q-A3 + Pythagoras
WE3. An isosceles trapezoid has parallel sides $8$ and $16$ and equal slant legs of length $5$. Find its area.
Drop perpendiculars from the top corners. Each cuts off a right triangle with hypotenuse $5$ and base $\tfrac{16-8}{2} = 4$, so $h = \sqrt{25 - 16} = 3$.
Type your answer (single number):
💡 Hints — open as needed
The height $h$ is not given — only the slant leg ($5$) and the two parallel sides ($8$ and $16$). Drop perpendiculars from each end of the short base to the long base.
By symmetry each perpendicular foot is offset by $\tfrac{16 - 8}{2} = 4$ from the corner of the long base. Pythagoras on the corner right triangle: $h = \sqrt{5^2 - 4^2} = 3$. Then apply $\tfrac{1}{2}(a+b)h$.
Answer: 36
Solution
By symmetry the long base overhangs the short base by $\tfrac{16 - 8}{2} = 4$ on each side.Each corner right triangle: hypotenuse $5$ (slant leg), horizontal leg $4$, vertical leg $h$.Pythagoras: $h^2 = 5^2 - 4^2 = 25 - 16 = 9 \Rightarrow h = 3$.$[\text{trap}] = \tfrac{1}{2}(8 + 16) \cdot 3 = \tfrac{1}{2} \cdot 24 \cdot 3 = 36$.
Reflection
This is the standard "isosceles-trap unlock". Whenever you have an isosceles trapezoid with the slant leg given but not the height, drop the two perpendiculars to expose the $(\tfrac{a-b}{2}, h, \text{slant})$ Pythagoras triple. The corner right triangles are congruent by symmetry — that is the key.
Tried first?
STEP 14 OF 24 · Phase 3 · Worked Example 4 ⭐⭐⭐⭐
Worked Example 4 — reverse engineer the height from area + midsegment
WE 4 · ⭐⭐⭐⭐ · skills Q-A3 + Q-A4 (reversed)
WE4. A trapezoid has midsegment of length $11$ and area $88$. Find its (perpendicular) height.
Type your answer (single number):
💡 Hints — open as needed
You don't need $a$ and $b$ separately. The midsegment is $\tfrac{1}{2}(a+b)$, which is exactly the factor that multiplies $h$ in the area formula.
The midsegment is the right unit for trapezoid reasoning whenever $a$ and $b$ aren't both needed individually. Memorise: area = midsegment × height. It saves a line of algebra on every trapezoid problem.
Tried first?
STEP 15 OF 24 · Phase 3 · Worked Example 5 ⭐⭐⭐⭐⭐
Worked Example 5 — rhombus from its diagonals
WE 5 · ⭐⭐⭐⭐⭐ · skills Q-A5 + Pythagoras
WE5. A rhombus has diagonals of length $16$ and $24$. Find its area, and verify by also computing the side length using Pythagoras.
Diagonals are perpendicular and bisect each other at $M$. Each of the four small right triangles has legs $8$ and $12$.
Type the area (single number):
💡 Hints — open as needed
A rhombus is a parallelogram whose diagonals are perpendicular (and still bisect each other). The four small triangles are congruent right triangles with legs $\tfrac{16}{2} = 8$ and $\tfrac{24}{2} = 12$.
Two routes give the same area: (i) sum of four right triangles $4 \cdot \tfrac{1}{2}(8)(12) = 192$; (ii) any rhombus has area $\tfrac{1}{2} d_1 d_2$. Then verify the side: $s = \sqrt{8^2 + 12^2}$.
Answer: area $= 192$, side $= \sqrt{208} = 4\sqrt{13} \approx 14.42$
Solution — area
Diagonals of a rhombus are perpendicular and bisect each other, so the rhombus splits into 4 congruent right triangles with legs $8$ and $12$.$[\text{rhombus}] \;=\; 4 \cdot \tfrac{1}{2} \cdot 8 \cdot 12 \;=\; 4 \cdot 48 \;=\; 192.$Equivalently, $[\text{rhombus}] = \tfrac{1}{2} d_1 d_2 = \tfrac{1}{2}(16)(24) = 192$.
Verification — side length via Pythagoras
Each right triangle has hypotenuse equal to the rhombus side $s$.$s^2 = 8^2 + 12^2 = 64 + 144 = 208$$s = \sqrt{208} = 4\sqrt{13} \approx 14.42.$
Reflection
Three skills fused: (i) diagonals bisect each other (Q-A5), (ii) the special rhombus property "diagonals perpendicular", (iii) Pythagoras for the side. Memorise the multi-tool fact rhombus area = $\tfrac{1}{2} d_1 d_2$ — it shows up in AIMO 2013 Q1 next.
Tried first?
STEP 16 OF 24 · Phase 4 · Five Practice Problems
Five practice problems — Hint / Answer / Solution on demand
Try each cold first. Open the Hint only after 60 seconds of thought.
P1 · ⭐ · Q-A1
P1. In parallelogram $ABCD$, $\angle A = 55^\circ$. Find $\angle B$, $\angle C$, $\angle D$.
Opposite angles equal; adjacent angles add to $180^\circ$.
$\angle B = 125^\circ$, $\angle C = 55^\circ$, $\angle D = 125^\circ$
$\angle C = \angle A = 55^\circ$. $\angle B = 180^\circ - 55^\circ = 125^\circ$, then $\angle D = \angle B = 125^\circ$.
P2 · ⭐⭐ · Q-A2
P2. A parallelogram has area $84$ and base $14$. Find its (perpendicular) height.
Area $= bh \Rightarrow h = \tfrac{\text{area}}{b}$.
$h = 6$
$h = \dfrac{84}{14} = 6$.
P3 · ⭐⭐ · Q-A3
P3. A trapezoid has parallel sides $9$ and $15$ and height $4$. Find its area.
AIMO 2013 · Q1 — rhombus area from a diagonal-difference clue
Exam-format attempt. Try first, then open hints.
AIMO 2013 · Q1 · [2 marks]
Find the area in cm$^2$ of a rhombus whose side length is $29$ cm and whose diagonals differ in length by $2$ cm.
Diagonals $d_1, d_2$ differ by $2$, satisfy $(d_1/2)^2 + (d_2/2)^2 = 29^2$. Solve to get $d_1 = 42$, $d_2 = 40$.
Your answer (cm²):
💡 Need help? Open these in order, only as needed:
(1) Observe — what to notice (5-step model)
1. 关键词识别 (Keywords): "rhombus", "side $29$", "diagonals differ by $2$". 2. 已知量 (Known): side $s = 29$, $d_1 - d_2 = 2$. 3. 未知量 (Unknown): area of the rhombus. 4. 中间量 (Intermediate): the two diagonal lengths $d_1$ and $d_2$. 5. 隐藏约束 (Hidden): rhombus diagonals are perpendicular and bisect each other → at the centre we get a right triangle with legs $d_1/2$ and $d_2/2$ and hypotenuse $s$. So $(d_1/2)^2 + (d_2/2)^2 = s^2$, equivalently $d_1^2 + d_2^2 = 4 s^2$.
(2) Strategy — how to think about it
Set up two equations in $d_1, d_2$: the difference equation and the Pythagoras-on-the-half-diagonals equation. Solve as a 2-by-2 system. Then use the rhombus area formula $\tfrac{1}{2} d_1 d_2$.
Why this technique generalises: any rhombus problem with "side + one extra diagonal clue" reduces to the same Pythagoras-on-the-halves identity. Memorise $d_1^2 + d_2^2 = 4 s^2$.
(3) Step 1 — set up the system
Let $d_1$ be the longer diagonal, so $d_1 = d_2 + 2$. The Pythagoras identity at the centre gives
Combine the two basic rhombus facts: (i) diagonals are perpendicular and bisect each other, so the four small triangles at the centre are right triangles with legs $d_1/2$ and $d_2/2$ and hypotenuse equal to the rhombus side; (ii) area equals $\tfrac{1}{2} d_1 d_2$.
Solution
Let the diagonals be $d_1$ (longer) and $d_2$ with $d_1 = d_2 + 2$.Pythagoras at the centre: $(d_1/2)^2 + (d_2/2)^2 = 29^2$, i.e. $d_1^2 + d_2^2 = 4 \cdot 841 = 3364$.$(d_2 + 2)^2 + d_2^2 = 3364 \Rightarrow 2 d_2^2 + 4 d_2 - 3360 = 0 \Rightarrow d_2^2 + 2 d_2 - 1680 = 0.$$d_2 = \dfrac{-2 + \sqrt{4 + 6720}}{2} = \dfrac{-2 + 82}{2} = 40$, so $d_1 = 42$.$[\text{rhombus}] = \tfrac{1}{2} d_1 d_2 = \tfrac{1}{2} (42)(40) = 840$ cm$^2$.
$[\text{rhombus}] = 840$ cm²
Why this technique generalises: "side + one diagonal clue" $\to$ rhombus is fully determined. Always reach for $d_1^2 + d_2^2 = 4 s^2$ first; combined with whatever extra equation the problem gives, you get a 2-by-2 system that is almost always linear after one substitution.
💡 Connecting back: Skill Q-A5 (diagonals bisect each other) plus the rhombus-specific bonus (diagonals perpendicular) plus Pythagoras. The same set-up reappears whenever a contest gives "rhombus, side known, one extra equation about diagonals".
STEP 18 OF 24 · Phase 5 · AIMO Past Paper
AIMO 2020 · Q4 — square minus four corner triangles
AIMO 2020 · Q4 · [3 marks]
$ABCD$ is a square of side $10$ cm. Points $E$, $F$, $G$, $H$ lie on $AB$, $BC$, $CD$, $DA$ respectively with $EB = FC$ and $CG = DH$ and $CG - EB = 4$ cm. Find the area of quadrilateral $EFGH$.
Inner quadrilateral $EFGH$ (yellow) sits inside the square. Subtract the four corner triangles to get its area.
Your answer (cm²):
💡 Need help? Open these in order, only as needed:
(1) Observe — what to notice (5-step model)
1. 关键词识别 (Keywords): "square side $10$", four points "on AB / BC / CD / DA", two equal pairs $EB = FC$ and $CG = DH$, and a difference $CG - EB = 4$. 2. 已知量 (Known): square side $10$, four right triangles in the corners with two known leg-pair lengths. 3. 未知量 (Unknown): area of $EFGH$. 4. 中间量 (Intermediate): let $EB = FC = x$ and $CG = DH = x + 4$. Then write the four corner-triangle areas in terms of $x$. 5. 隐藏约束 (Hidden): the corner-triangle areas add up cleanly — the $x$ terms cancel out, leaving a constant. So the answer does not depend on the choice of $x$ as long as the problem is consistent.
(2) Strategy — how to think about it
$[\text{EFGH}] = [\text{square}] - 4 \cdot [\text{corner triangle}]$. Compute each corner triangle's area in terms of $x$, sum them, simplify. The $x$-dependence cancels.
Why this technique generalises: any "inscribed quadrilateral via cuts on each side" reduces to "big shape minus four corner triangles". This is the standard AIMO Q3–Q4 area-chase template.
(3) Step 1 — name the four corner triangles
Let $EB = FC = x$ and $CG = DH = x + 4$. Then $AE = 10 - x$, $BF = 10 - x$, $DG = 10 - (x+4) = 6 - x$, $AH = 10 - (x+4) = 6 - x$.
The four corner right triangles are:
— $\triangle AEH$ at corner $A$: legs $AE = 10 - x$ and $AH = 6 - x$.
— $\triangle BFE$ at corner $B$: legs $BF = 10 - x$ and $BE = x$.
— $\triangle CGF$ at corner $C$: legs $CG = x + 4$ and $CF = x$.
— $\triangle DHG$ at corner $D$: legs $DH = x + 4$ and $DG = 6 - x$.
(4) Step 2 — sum the four areas
Each corner triangle is right-angled at the corner, so area $= \tfrac{1}{2} \cdot \text{leg}_1 \cdot \text{leg}_2$.
$[\text{EFGH}] = [\text{square}] - \sum [\triangle] = 100 - 42 = 58$ cm$^2$.
Notice the $x$ has cancelled — the answer doesn't depend on the specific value of $EB$, only on the difference $CG - EB = 4$.
Strategy
"Square minus four corner triangles." Parametrise the corner-triangle leg lengths by $x = EB$, sum the four areas, watch the $x$-terms cancel, and subtract from $100$.
Solution
Let $EB = FC = x$ and $CG = DH = x + 4$. Then $AE = 10 - x$, $BF = 10 - x$, $DG = 6 - x$, $AH = 6 - x$.The four corner right triangles have areas (in halves of $\text{leg}_1 \cdot \text{leg}_2$):$\triangle AEH = \tfrac{1}{2}(10-x)(6-x)$$\triangle BFE = \tfrac{1}{2}(10-x)(x)$$\triangle CGF = \tfrac{1}{2}(x+4)(x)$$\triangle DHG = \tfrac{1}{2}(x+4)(6-x)$$2 \sum [\triangle] = (10-x)(6-x) + (10-x)x + (x+4)x + (x+4)(6-x).$Expand and simplify: $60 - 16x + x^2 + 10x - x^2 + x^2 + 4x + 24 + 2x - x^2 = 84.$$\sum [\triangle] = 42$.$[\text{EFGH}] = [\text{square}] - \sum [\triangle] = 100 - 42 = 58$ cm$^2$.
$[\text{EFGH}] = 58$ cm²
Why this technique generalises: any inscribed quadrilateral defined by cuts of given lengths on the four sides of a known polygon reduces to "big polygon area − four corner triangles". The corner triangles are always right-angled at the original polygon's corners (when the polygon is a rectangle/square), so each is a one-line area calculation. The $x$-cancellation here also tells you the answer would have been the same had $EB$ been any other value.
💡 Connecting back: Skill Q-A2 (rectangle/square area $= bh$) for the outer $100$, plus repeated $\tfrac{1}{2} \cdot \text{leg} \cdot \text{leg}$ triangle areas. The cancellation pattern is worth remembering: when "the answer doesn't depend on $x$", you have the right set-up.
STEP 19 OF 24 · Phase 5.5 · Synthesis
Synthesis — midsegment cuts a trapezoid into two trapezoids
Q-A4 (midsegment) + Q-A3 (trapezoid area) + height bisection. Three skills, one problem.
SYNTHESIS · ⭐⭐⭐⭐
Trapezoid $ABCD$ has $AB \parallel CD$ with $AB = 12$, $CD = 20$, and (perpendicular) height $8$. Let $M$ and $N$ be the midpoints of legs $AD$ and $BC$ respectively.
(a) Find the length of $MN$.
(b) Find the area of trapezoid $ABNM$ (top half).
(c) Find the area of trapezoid $MNCD$ (bottom half).
(a) MN =
(b) Area of $ABNM$ =
(c) Area of $MNCD$ =
💡 Need help?
Hint — Observe
$MN$ is the midsegment of the original trapezoid, so $MN = \tfrac{1}{2}(AB + CD) = \tfrac{1}{2}(12 + 20) = 16$. Each "half-trapezoid" has its own pair of parallel sides (one of which is $MN = 16$) and height $\tfrac{8}{2} = 4$ (because the midsegment sits halfway between the two original bases).
Hint — Strategy
Apply $\tfrac{1}{2}(\text{base}_1 + \text{base}_2) \cdot h$ separately to each half. For the top half $ABNM$: bases $12$ and $16$, height $4$. For the bottom half $MNCD$: bases $16$ and $20$, height $4$.
Sanity check. Total area $= 56 + 72 = 128$. Original trapezoid area $= \tfrac{1}{2}(12 + 20) \cdot 8 = 16 \cdot 8 = 128$. ✓
Why this technique generalises: the midsegment is exactly the average of the two bases, so it sits at the "average height" between them. That single fact lets you split any trapezoid into two trapezoids whose areas you can compute independently — useful whenever a problem asks for a slice or a ratio.
STEP 20 OF 24 · Atomic Skills + Micro-validation
Atomic skill matrix & quick checks
Five atomic skills, two checks each. Aim for 10/10 before moving on.
Rectangle: diagonals also equal in length.
Rhombus: diagonals also perpendicular; area $= \tfrac{1}{2} d_1 d_2$; side relates by $d_1^2 + d_2^2 = 4 s^2$.
Square: both — diagonals equal and perpendicular.
Drop perpendiculars from the short-base corners. The horizontal leg of each corner right triangle is $\tfrac{a-b}{2}$.
⑧ "Inscribed quadrilateral via cuts" template
Inner area $=$ outer polygon area $-$ four corner-triangle areas. Often the parameter cancels and the answer depends only on the differences (AIMO 2020 Q4 pattern).
STEP 22 OF 24 · Self-assessment
⭐ Self-assessment
Rate your confidence on each atomic skill. Three stars = "I can teach this back in my own words."
① Q-A1 · I can use opposite-equal / adjacent-supplementary to find every angle of a parallelogram from one given.
② Q-A2 · I apply parallelogram area $= b \cdot h$ correctly and never mistake the slanted leg for the height.
③ Q-A3 · I apply trapezoid area $= \tfrac{1}{2}(a+b)h$ in both directions (forward and backward).
④ Q-A4 · I know the midsegment is the average of the bases, sits halfway in height, and equals area$/h$.
⑤ Q-A5 · I use "diagonals bisect each other" and the rhombus bonus "diagonals perpendicular" + $d_1^2 + d_2^2 = 4 s^2$.
⭐ 0 / 15 — click stars
STEP 23 OF 24 · Error Book
📒 Your error book — Week 8, Part 2
Every wrong submission today has been silently logged here. Re-do these two more times before the Sunday Mock.
Logged errors this session
No errors yet — submit some AIMO problems to populate this list.
Errors are persisted in sessionStorage.aimoErrors_W8_P2. Reset by closing the browser tab.
Habit to build: when an item appears here twice, write the problem on a physical card and put it on your desk. The third time you re-attempt it cold, you should solve it under 2 minutes.
STEP 24 OF 24 · Wrap-up
🎉 Part 2 complete — Bridge to Part 3
Part 2 of Week 8 complete. You now hold the five pillars of AIMO Quadrilaterals I:
Parallelogram area $= bh$ — base times perpendicular height
Trapezoid area $= \tfrac{1}{2}(a+b)h$ — average of bases × height
Trapezoid midsegment $= \tfrac{1}{2}(a+b)$ — area $=$ midsegment × $h$
Diagonals bisect each other — and rhombus diagonals are also perpendicular
📌 What's next: Part 3 — Quadrilaterals II (Cyclic & Tangential)
Part 3 introduces the cyclic quadrilateral (all four vertices on a circle) and the tangential quadrilateral (all four sides tangent to a circle). You'll learn:
Cyclic quadrilateral: opposite angles sum to $180^\circ$.
Ptolemy's theorem: $AC \cdot BD = AB \cdot CD + AD \cdot BC$.
Brahmagupta's area formula for cyclic quadrilaterals.
Tangential quadrilateral: opposite sides sum to the same total.
The parallelogram and trapezoid skills you mastered today are still the workhorses — cyclic / tangential quadrilaterals just add circle constraints on top.
Before Sunday Mock:
Re-attempt every item in your error book until you get it right under 2 minutes cold.
Re-skim the Cheat Sheet (Step 21). Memorise the five formulas verbatim.
Make sure all five atomic skills are 3-star on your self-assessment.
▶ Next: Part 3 — Quadrilaterals II (Cyclic & Tangential)
💡 Stuck? Open this for guiding questions
Ask yourself:
Is the figure a parallelogram? (yes → opposite sides & angles equal, diagonals bisect each other)
Is the figure a trapezoid? (yes → area $= \tfrac{1}{2}(a+b)h$)
Am I given the slanted leg and need the height? (drop a perpendicular, use Pythagoras)
Is it an isosceles trapezoid? (the two corner right triangles are congruent — overhang $= \tfrac{a-b}{2}$)
Is the figure a rhombus? (diagonals are perpendicular → area $= \tfrac{1}{2} d_1 d_2$, and $d_1^2 + d_2^2 = 4 s^2$)
Is the figure a square? (rectangle + rhombus combined — diagonals equal, perpendicular, length $s\sqrt{2}$)
Is the inner region defined by cuts on each side? (use "outer area − four corner triangles")
Do I know the midsegment but not the bases? (area $= $ midsegment × $h$, no need to find $a, b$ separately)