Class 7 Maths Part 2 Chapter 1: Geometric Twins Notes

What is this chapter about?

Geometric Twins is about congruence: when are two figures exact copies of each other, and which measurements are enough to make or check a copy? The chapter starts with a symbol on a signboard, moves on to triangles, finds five sets of measurements that fix a triangle completely, and shows two sets that do not. At the end it uses congruence to discover the angles of isosceles and equilateral triangles.

Congruent figures

A symbol on a signboard, made of two straight arms AB and BC, has to be copied onto another board. Tracing works for small figures, but for a big board it is easier to take measurements. The two arm lengths are not enough. With AB = 4 cm and BC = 8 cm, the arms can open to many different angles, and each angle gives a different shape. Fixing ∠ABC as well, say at 80°, leaves only one shape.

Figures with the same shape and size are congruent. A tracing of one fits exactly over the other, and you may turn the tracing or flip it over to make it fit. To copy a circle you need only its radius; to copy a rectangle, its length and breadth. So two circles with equal radii are congruent, and so are two rectangles with the same length and the same breadth.

Congruent triangles: the SSS condition

Meera and Rabia must make a cardboard copy of a triangular frame that is too big to trace. They measure its sides: 40 cm, 60 cm and 80 cm. Meera says the angles are not needed. To test this on paper with 4 cm, 6 cm and 8 cm, Rabia draws AB = 6 cm and circles of radius 4 cm around A and 8 cm around B. The circles cross at two points, E and F, so there are two triangles, ΔABE and ΔABF. They are mirror images of each other across AB, so they are congruent, and Meera was right.

This is the SSS (Side Side Side) condition: if two triangles have the same three side lengths, they are congruent. See the steps and the picture in congruent figures and triangles.

Writing a congruence

If ΔABC fits on ΔXYZ with A on X, B on Y and C on Z, these are the corresponding vertices. Then AB and XY, BC and YZ, and CA and ZX are corresponding sides, and ∠A = ∠X, ∠B = ∠Y and ∠C = ∠Z. We write ΔABC ≅ ΔXYZ, and the order of the letters shows which vertex matches which. ΔACB ≅ ΔXZY says the same thing, but ΔACB ≅ ΔXYZ is wrong, because it pairs C with Y.

For example, the diagonal BD cuts rectangle ABCD into ΔABD and ΔCDB. AB = CD and AD = CB, because they are opposite sides of the rectangle, and BD is common to both, so the triangles are congruent by SSS. AB has to sit on CD, so A matches C, B matches D and D matches B: ΔABD ≅ ΔCDB.

Five conditions for congruent triangles

• SSS: all three sides are equal.
• SAS: two sides and the included angle, the angle between them. If AB = XY = 6 cm, AC = XZ = 5 cm and ∠A = ∠X = 30°, everyone who constructs the triangle gets the same one, so ΔABC ≅ ΔXYZ.
• ASA: two angles and the included side. BC = YZ = 5 cm, ∠B = ∠Y = 50° and ∠C = ∠Z = 30° fix the triangle.
• AAS: two angles and a side that is not between them. Since the angles of a triangle add up to 180°, the third angles are equal too, and ASA applies. If ∠A = ∠X = 35°, ∠C = ∠Z = 75° and BC = YZ = 4 cm, then ∠B = ∠Y = 180° − 110° = 70°.
• RHS: in two right-angled triangles, the hypotenuse (the side opposite the right angle) and one more side. With BC = YZ = 4 cm, ∠B = ∠Y = 90° and AC = XZ = 5 cm, the construction gives one triangle above the base and its mirror image below it, and the two are congruent.

Measurements that are not enough

Three angles are not enough. Triangles with angles 30°, 70° and 80° have the same shape but can be any size, so two such triangles need not be congruent.

Two sides and a non-included angle (SSA) are not enough. Take PQ = 6 cm, ∠P = 30° and QR = 4 cm. Draw PQ and a line l from P at 30°. An arc of radius 4 cm from Q cuts l at two points, R and S, so ΔPQR and ΔPQS both fit the measurements but are clearly different. RHS is a special case of SSA that does work, because the angle there is a right angle.

Using congruence

Congruence proves that lengths or angles are equal without measuring them. If O is the midpoint of AD and of BC, then AO = OD and BO = OC, and ∠AOB = ∠DOC because they are vertically opposite angles. By SAS, ΔAOB ≅ ΔDOC, with A matching D, O matching O and B matching C. So AB and DC are corresponding sides, and AB = DC.

Isosceles and equilateral triangles

In ΔABC, let AB = AC and ∠A = 80°. Draw the altitude AD from A to BC. The triangles ADB and ADC have a right angle at D, equal hypotenuses AB and AC, and the common side AD, so ΔADB ≅ ΔADC by RHS. Hence ∠B = ∠C, and each is (180° − 80°) ÷ 2 = 50°. In every triangle, angles opposite equal sides are equal.

In an equilateral triangle all three sides are equal, so all three angles are equal. They add up to 180°, so each angle is 60°. The full reasoning, with more examples, is in angles of isosceles and equilateral triangles.

Congruent triangles around us

Congruent triangles appear in buildings and designs, from ancient times to today. Look for them in the glass pyramid at the Louvre Museum in Paris, the Great Pyramid of Giza in Egypt, dome frames, rangoli patterns and the steel frame of Rabindra Setu (Howrah Bridge) in Kolkata. Answers to every question of the chapter are in the NCERT solutions for Part 2 Chapter 1.