Shunya / Geometry & Constructions

Shunya · Maths lab

Geometry & constructions

Drag the corners of triangles and quadrilaterals and watch angles, sides, areas and centres update. Test the angle-sum, congruence, similarity and Pythagoras theorems on shapes you can bend, cross parallel lines with a transversal, and follow compass-and-ruler constructions one step at a time.

Lessons

Pick your class, follow the steps in the lab, and check what you see against the working under the drawing.

Class 9 — Lines and angles

Concept: when a transversal crosses two parallel lines, corresponding angles are equal, alternate interior angles are equal and co-interior angles add up to 180°. Vertically opposite angles are equal for any two crossing lines.

Open Lines & angles. With the tilt at 0° the two lines are parallel. Choose Corresponding: the highlighted pairs, such as ∠1 and ∠5, are equal.
Drag the dot on the transversal (or use the slider). Every pair changes, but the two in a pair always match.
Choose Alternate interior and Co-interior: equal pairs, then pairs that add to 180°.
Now tilt the lower line to 12°. The equal pairs stop being equal: the lines are no longer parallel. That is the converse: equal corresponding angles prove parallel lines.
Choose Vertical and Linear pairs and tilt again: these rules hold even when the lines are not parallel.

Takeaway: the angle rules for a transversal only hold when the lines are parallel, except vertically opposite angles and linear pairs, which always hold.

Class 9 — Triangles: angle sum and exterior angle

Concept: the angles of any triangle add up to $180^\circ$ . An exterior angle equals the sum of the two interior angles far from it, and the longest side faces the largest angle.

Open Triangle and drag A, B and C anywhere. Watch the Angle sum card: it stays at 180°.
Tick show the exterior angle at C and check that it equals A + B.
Press equilateral, isosceles, right 3-4-5 and obtuse to see each kind named.
Drag one corner to make the triangle very flat: the triangle inequality $a + b > c$ starts to fail as the corners line up.
Read the working: the sides come from the distance formula and the angles from the law of cosines.

Takeaway: whatever shape the triangle has, the angle sum is 180°, and it is the equal angles that reveal an isosceles or equilateral triangle.

Class 9 — Congruence of triangles

Concept: two triangles are congruent if one fits exactly on the other. You do not need all six parts: SSS, SAS, ASA, AAS and RHS are enough. SSA is not.

Open Congruence and pick SSS. Ticks on the sides show which parts are used to build the orange triangle PQR.
Drag the Slide the copy slider to 100%: PQR lands exactly on ABC, and the table says the other parts follow.
Try SAS, ASA and AAS: the angle must be between the two sides for SAS, and between the two angles for ASA.
Press right-angled and choose RHS: a hypotenuse and one leg fix the rest.
Press SSA has two answers and slide the copy: it does not land on ABC. Two different triangles fit the same two sides and angle, so SSA proves nothing.

Takeaway: corresponding parts of congruent triangles are equal (CPCT), and SSA is the trap to remember.

Class 9 — Quadrilaterals and the midpoint theorem

Concept: the angles of a quadrilateral add up to $360^\circ$ . Its name depends on which sides are equal or parallel. Joining the midpoints of the sides of any quadrilateral gives a parallelogram.

Open Quadrilaterals and press square, rectangle, rhombus and parallelogram. Read the diagonals card each time: equal, perpendicular, bisecting.
Press trapezium and kite, then arrowhead, whose one angle is more than 180°.
Drag one corner of the rhombus: the name changes as soon as the shape stops being one. The angle sum stays 360°.
Tick join the midpoints of the sides on the general quadrilateral. The green shape is always a parallelogram, with half the area.

Takeaway: a square is a rectangle and a rhombus at once, and every quadrilateral hides a parallelogram inside.

Class 10 — Similar triangles and the basic proportionality theorem

Concept: if a line DE is parallel to BC in triangle ABC, then $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ and triangle ADE is similar to ABC. The areas of similar triangles are in the ratio of the squares of their sides.

Open Similar triangles. Drag D along AB (or use the slider). The cards AD ÷ DB and AE ÷ EC always match.
Read Area ADE ÷ area ABC: it equals the ratio squared (for a ratio of 0.5 the area ratio is 0.25).
Untick keep DE parallel to BC and move E: the two ratios differ and the triangles are no longer similar.
Now set E so that AE ÷ EC equals AD ÷ DB: DE is parallel to BC again. That is the converse of Thales.

Takeaway: parallel lines make similar triangles, similar triangles make equal ratios, and areas go with the square of the ratio.

Class 10 — Pythagoras theorem and its converse

Concept: in a right triangle $a^2 + b^2 = c^2$ , so the square on the hypotenuse equals the two other squares added. The converse holds too: if the squares match, the angle is 90°.

Open Pythagoras and press 3-4-5. The squares have areas 9, 16 and 25, and 9 + 16 = 25.
Drag A along the base or B up the side to change a leg. Look for whole-number triples such as 5-12-13.
Tick free triangle and drag a corner so the angle at C opens beyond 90°: the two small squares now add to less than the big one (obtuse). Close it below 90° and they add to more (acute).
Press 1-1-√2: a right triangle whose hypotenuse is not a whole number.

Takeaway: Pythagoras is a test for right angles: compare the sum of the two smaller squares with the largest.

Class 10 — Triangle centres (a step beyond the textbook)

Concept: a triangle has four famous centres: the centroid (medians), the circumcentre (perpendicular bisectors), the incentre (angle bisectors) and the orthocentre (altitudes). In each case three lines meet at one point.

Open Triangle centres. Tick the centroid: the three medians meet at G, which cuts each median 2 : 1.
Tick the circumcentre: O is the same distance from A, B and C, so a circle fits through all three.
Tick the incentre: I is the same distance from all three sides, so a circle fits inside.
Tick the orthocentre and the Euler line: O, G and H are always in line.
Press right and obtuse: watch where O and H go, and equilateral, where all four centres coincide.

Takeaway: the centroid and incentre are always inside the triangle; the circumcentre and orthocentre can leave it.

Class 9 — Geometrical constructions

Concept: a construction uses only a ruler (to draw straight lines) and a compass (to draw arcs of equal radius). Every step can be justified by a congruent-triangle or equal-distance argument.

Open Constructions and choose Perpendicular bisector. Press Next step to see each arc appear, or Play to watch it all.
Drag A or B: the whole construction redraws, and at the end the cards show MA − MB = 0 and a 90° angle.
Try Bisector of an angle (check the two halves are equal), An angle of 60° and Perpendicular from a point to a line.
Choose A triangle from its three sides and change the side lengths: when one side is longer than the other two added, no triangle exists.
Read Why it works under the last step of each construction.

Takeaway: constructions are proofs in disguise: equal compass widths give equal lengths, and equal lengths give congruent triangles.

Class 10 — Constructions: dividing a line and drawing tangents

Concept: a segment can be divided in the ratio m : n using parallel lines and the basic proportionality theorem, and the tangents from an outside point use the fact that the angle in a semicircle is 90°.

Open Constructions and choose Divide a segment in the ratio m : n. Set m = 2 and n = 3 and press Play.
At the end the cards read AP ÷ PB = 0.667 and m ÷ n = 0.667. Change m and n and repeat.
Choose Tangents from an outside point. A circle on the diameter OP meets the given circle at T₁ and T₂.
Check the cards: the angle OT₁P is 90° and the two tangents are equal in length. Drag P closer to the circle: when it is inside, no tangent can be drawn.

Takeaway: a tangent is perpendicular to the radius at the point of contact, and two tangents from one point are equal.

The maths behind it

What each mode covers

Lines & angles — a transversal across two lines that you can tilt parallel or not, with all eight angles labelled and the five angle-pair rules checked live. Class 9 lines and angles.

Triangle — drag the corners for sides, angles, area, the angle sum, the exterior angle, the triangle inequality and the kind of triangle. Class 9 triangles.

Triangle centres — the centroid, circumcentre, incentre and orthocentre with their lines and circles, the 2 : 1 ratio and the Euler line. A step beyond class 9–10.

Congruence — build a second triangle from only the parts SSS, SAS, ASA, AAS or RHS give, slide it onto the first, and see why SSA fails. Class 9 triangles.

Similar triangles — a line DE across the triangle: drag it parallel to BC to see the Thales ratios, the area ratio and the perimeter ratio. Class 10 triangles.

Pythagoras — squares on the three sides, whole-number triples, and the converse for acute and obtuse triangles. Class 10 triangles.

Quadrilaterals — drag four corners and the shape is named live (square, rhombus, kite, trapezium…), with diagonals, angle sum and the midpoint parallelogram. Class 9 quadrilaterals.

Constructions — seven compass-and-ruler constructions you can step through, replay and drag: perpendicular bisector, angle bisector, 60°, perpendicular from a point, a triangle from three sides, dividing a segment in a ratio, and tangents from a point. Class 9 and 10 constructions.

Part of Shunya. See the syllabus map for the chapters these modes map to. Circles and their theorems have their own lab, coming next. Also try the Trigonometry lab, which builds on right triangles.

Formulas used in this lab

Lines and angles Class 9

Linear pair: $\angle 1 + \angle 2 = 180^\circ$
Vertically opposite angles: $\angle 1 = \angle 3$
Corresponding angles (l ∥ m): $\angle 1 = \angle 5$
Alternate interior angles (l ∥ m): $\angle 3 = \angle 5$
Co-interior angles (l ∥ m): $\angle 3 + \angle 6 = 180^\circ$
Converse: $\text{equal corresponding angles} \Rightarrow l \parallel m$

Triangles Class 9

Angle sum: $\angle A + \angle B + \angle C = 180^\circ$
Exterior angle: $\angle ACD = \angle A + \angle B$
Distance between two points: $AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Angle from the sides (law of cosines): $\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}$
Area from the coordinates: $\tfrac12\,\bigl|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\bigr|$
Heron's formula: $\text{Area} = \sqrt{s(s - a)(s - b)(s - c)},\qquad s = \dfrac{a + b + c}{2}$
Triangle inequality: $a + b > c$
Larger angle, longer side: $\angle A > \angle B \iff a > b$

Congruence of triangles Class 9

SSS: $AB = PQ,\ BC = QR,\ CA = RP$
SAS: $AB = PQ,\ \angle A = \angle P,\ AC = PR$
ASA: $\angle A = \angle P,\ AB = PQ,\ \angle B = \angle Q$
AAS: $\angle A = \angle P,\ \angle B = \angle Q,\ BC = QR$
RHS: $\angle C = \angle R = 90^\circ,\ AB = PQ,\ BC = QR$
Not a rule: $\text{SSA can fit two different triangles}$
Corresponding parts (CPCT): $\triangle ABC \cong \triangle PQR \Rightarrow \text{all six matching parts are equal}$

Quadrilaterals Class 9

Angle sum of a quadrilateral: $\angle A + \angle B + \angle C + \angle D = 360^\circ$
Parallelogram: $AB \parallel DC,\ BC \parallel AD \ \Rightarrow\ AB = DC,\ BC = AD$
Diagonals of a parallelogram: $\text{they bisect each other}$
Rhombus (a parallelogram with all sides equal): $AC \perp BD$
Rectangle (a parallelogram with a right angle): $AC = BD$
Midpoint theorem: $DE \parallel BC,\quad DE = \tfrac12 BC$
Midpoint quadrilateral (always a parallelogram): $\text{area} = \tfrac12\,[ABCD]$

Similar triangles Class 10

Basic proportionality theorem: $DE \parallel BC \ \Rightarrow\ \dfrac{AD}{DB} = \dfrac{AE}{EC}$
Ratio of the sides: $\dfrac{AD}{AB} = \dfrac{AE}{AC} = \dfrac{DE}{BC} = k$
Perimeters: $\dfrac{\text{perimeter of } ADE}{\text{perimeter of } ABC} = k$
Areas: $\dfrac{\text{area of } ADE}{\text{area of } ABC} = k^2$
AA similarity: $\angle A = \angle P,\ \angle B = \angle Q \ \Rightarrow\ \triangle ABC \sim \triangle PQR$

Pythagoras theorem Class 10

Pythagoras: $a^2 + b^2 = c^2$
Converse: $a^2 + b^2 = c^2 \ \Rightarrow\ \angle C = 90^\circ$
Acute triangle: $a^2 + b^2 > c^2$
Obtuse triangle: $a^2 + b^2 < c^2$
Pythagorean triples: $3\text{-}4\text{-}5,\quad 5\text{-}12\text{-}13,\quad 8\text{-}15\text{-}17$

Triangle centres Class 10

Centroid: $G = \left(\dfrac{x_1 + x_2 + x_3}{3},\ \dfrac{y_1 + y_2 + y_3}{3}\right)$
Centroid divides a median: $AG : GD = 2 : 1$
Circumradius: $R = \dfrac{abc}{4\,\text{Area}}$
Inradius: $r = \dfrac{\text{Area}}{s}$
Incentre: $I = \dfrac{aA + bB + cC}{a + b + c}$
Orthocentre: $H = A + B + C - 2O$
Euler line: $O,\ G,\ H \text{ are collinear},\qquad OG : GH = 1 : 2$

Constructions and tangents Class 10

Perpendicular bisector: $PA = PB \ \text{for every point } P \text{ on it}$
Angle bisector (by SSS): $\triangle VDF \cong \triangle VEF \Rightarrow \angle DVF = \angle EVF$
Dividing a segment: $\dfrac{AP}{PB} = \dfrac{m}{n}$
Tangent and radius: $OT \perp PT$
Tangent length: $PT^2 = OP^2 - r^2$
Two tangents from a point: $PT_1 = PT_2$
Angle in a semicircle: $\angle OTP = 90^\circ \ \text{when } OP \text{ is a diameter}$