NCERT Solutions for Class 7 Maths Chapter 9 Geometric Twins (PDF) – 2027

One way is to trace the outline of the symbol on tracing paper and copy it on to the new board. For a big signboard this is difficult, so a better way is to take measurements of the symbol and redraw it from them.

Yes. Name the corner points of the symbol A, B and C as the book does. The symbol is just two straight arms joined at B.

Take these three measurements:

• the length of the arm AB,
• the length of the arm BC,
• the angle ∠ABC between the two arms.

No. Two arm lengths are not enough.

Take AB = 4 cm and BC = 8 cm. Keeping both lengths fixed, the arms can still be opened out wide or closed up narrow. Every opening gives a different symbol.

Yes — measure the angle ∠ABC between the two arms.

Yes. Here is the construction.

1. Draw BC = 8 cm with a ruler.
2. Place the protractor at B and mark 80° from BC. Draw a ray BX along that mark.
3. From B, cut off BA = 4 cm on the ray BX.
4. The bent line A–B–C is the required symbol.

No. Equal arm lengths are not enough.

We have already seen several symbols with AB = 4 cm and BC = 8 cm that are not copies of each other. They differ in the angle at B.

They are not congruent.

Each figure is two arms joined at a point. Measure both arms and the angle in each figure:

Circle the two teardrops and the two leaf pairs. These two pairs appear congruent; the clouds and the star bursts do not.

(a) Circle — one measurement is enough: the radius (or the diameter, which is twice the radius).

(b) Rectangle — two measurements: the length and the breadth.

The figure is made of two straight strokes that meet at a point. Three arms come out of that meeting point, so measure:

• the three arm lengths from the meeting point, and
• the angles between the arms.

If all of these match in the same order, the figures are congruent (a figure may be turned or flipped first).

The two pairs given in the book: both pairs are congruent. In each pair the three arms and the angles between them measure the same; one figure is only the other shifted to a new place on the page.

They should measure the frame and then build the cutout from those measurements.

With a measuring tape they can take the three sidelengths of the triangular frame. Those three numbers can be carried back to the desk, and a triangle with exactly those sides can be drawn on cardboard with a ruler and a compass.

Yes, Meera is right. The three sidelengths are enough.

With the three sides fixed, the ruler-and-compass construction always gives the same triangle (at most flipped over), so the angles come out the same by themselves.

Yes, it is sufficient. Here is the construction, drawn small enough to fit on a page.

1. Draw AB = 6 cm with a ruler.
2. With A as centre and radius 4 cm, draw an arc.
3. With B as centre and radius 8 cm, draw a second arc cutting the first at E.
4. Join AE and BE. ∆ABE has sides 6 cm, 4 cm and 8 cm.

Yes, they do have the same shape and size. ∆ABE and ∆ABF are congruent, so Rabia’s worry does not arise.

Both triangles are built from the same three lengths:

All three methods give the same answer: ∆ABE ≅ ∆ABF.

• Tracing: trace ∆ABE, flip the tracing paper over about the line AB, and it lands exactly on ∆ABF.
• Cutout: cut both triangles out. Turn one over and place it on the other — the three sides match.
• Symmetry: the arcs below AB were drawn with the same centres and the same radii as the arcs above it. So the lower half is the mirror image of the upper half in the line AB.

Place A over X, B over Y and C over Z.

Since the triangles now fit exactly, their angles also match:

For these triangles, no — there is only one way.

The three sides of the triangle have three different lengths, so each side has exactly one partner of the same length in the other triangle. That fixes the matching completely.

The diagonal BD cuts the rectangle into ∆ABD and ∆CDB, and these two are congruent.

So ∆ABD and ∆CDB are congruent.

No, this matching does not work. Cut the two triangles out of the rectangle and try it — they will not settle on each other.

The correct matching is A ↔ C, B ↔ D, D ↔ B.

Every matched pair is a pair of equal sides, so this correspondence is correct.

The statement ∆HEN ≅ ∆BIG fixes three pairs of corresponding vertices:

Any way of writing the same three pairs is correct. The three pairs can be listed in 6 orders, so besides the given one there are 5 other correct ways:

Each of these may also be written the other way round, for example ∆BIG ≅ ∆HEN or ∆IGB ≅ ∆ENH.

Yes, they are congruent, by the SSS condition.

Match the equal sides:

Reading the matching down the columns gives R ↔ J, E ↔ A, D ↔ M. So

Join AC. It splits the figure into ∆ABC and ∆ADC, and these two are congruent.

Does AC divide the two angles equally? Yes.

Because the triangles are congruent, their corresponding angles are equal:

The two triangles in the figure are congruent — but ∆DFE ≅ ∆GED is not the correct way to write it.

First, the congruence itself. Look at ∆DFE and ∆DGE:

Now test the order given in the question. Writing ∆DFE ≅ ∆GED pairs the vertices like this:

None of these three pairs is known to be equal, so that correspondence is wrong.

No. The three angles fix the shape but not the size.

Two triangles with the same three angles look alike, but one may be a big version of the other. They can be placed one over the other only if they also happen to be the same size, and the angles do not tell us that.

No. Many different triangles have these three angles.

Draw a base of 4 cm and put 30° at one end and 70° at the other; the third angle comes out as 80°. Now do it again with a base of 8 cm. Both triangles have angles 30°, 70°, 80°, but the second is twice as big as the first.

Yes, they are congruent.

Here the equal angle lies between the two equal sides — ∠A is the angle between AB and AC, and ∠X is the angle between XY and XZ.

Yes — every triangle in the class will be congruent (some may be flipped over).

The construction leaves no free choice:

1. Draw AB = 6 cm.
2. At A, draw a ray making 30° with AB. There is only one such ray on a given side of AB.
3. On that ray mark AC = 5 cm. There is only one such point C.
4. Join BC — there is only one segment joining two fixed points.

Not necessarily. Two different triangles can be drawn from these measurements.

∠B is not between the equal sides AB and AC, so this is the SSA case. Construct (P stands for B, Q for A):

1. Draw the base PQ = 6 cm.
2. From P draw a line l making 30° with PQ.
3. From Q draw a long arc of radius 4 cm. It cuts the line l at two points, R and S.

Both ∆PQR and ∆PQS have PQ = 6 cm, a 4 cm side from Q and ∠P = 30°, but they are different — so the triangles need not be congruent.

A point where the arc meets the line l gives the third vertex. But the arc meets l at two points, R and S — so the figure gives two triangles, ∆PQR and ∆PQS.

Yes, they are congruent, and no other triangle is possible with these measurements.

The equal side lies between the two equal angles, so this is the ASA case.

1. Draw BC = 5 cm.
2. At B draw a ray making 50° with BC.
3. At C draw a ray making 30° with CB, on the same side.
4. The two rays meet at exactly one point — that is A.

AB = CD. The two lengths are equal.

Look at ∆AOB and ∆DOC:

The correspondence is A ↔ D, O ↔ O, B ↔ C. So AB and DC are corresponding sides, and

Yes — ∠AOB = ∠DOC, because they are vertically opposite angles formed where the segments AD and BC cross at O.

Yes, they are congruent, by the SAS condition.

In each triangle the marked angle lies between the two marked sides:

So B corresponds to Z, A corresponds to X and C corresponds to Y:

In the figure the segments DB and CA cross at O, making ∆ODC and ∆OBA.

Step 1 — the equal angles. DC and AB are parallel, so:

Step 2 — the congruence. Together with the given equal sides:

Step 3 — the other equal parts. From the congruence:

• OD = OB
• OC = OA
• ∠DOC = ∠BOA (also vertically opposite angles)

Yes, the triangles are congruent, and that is what makes the two angles equal.

The segment BC is shared by ∆ABC (above it) and ∆DBC (below it).

The correspondence is A ↔ D, B ↔ B, C ↔ C. Hence the third pair of angles must also match:

Step 1 — add the given angles at B and at C:

Step 2 — find the congruent triangles. Look at ∆ABC and ∆DCB, which share the side BC:

Step 3 — list the equal parts.

• AB = DC and AC = DB (corresponding sides)
• ∠BAC = ∠CDB (corresponding angles)
• ∠ABC = ∠DCB, and of course BC = CB

Yes, they are congruent.

The given side BC is not between the two given angles ∠A and ∠C, so ASA cannot be used straight away. But the third angle can be found first.

Now use ∠B, BC and ∠C, which do fit the ASA pattern:

Both are 70°.

So ∠B = ∠Y.

Yes. Finding the third angle turns the problem into an ordinary ASA problem.

Now the equal side BC lies between the two equal angles ∠B and ∠C, so ASA applies and

Yes, they are congruent, and no different triangle can be drawn with these measurements.

Follow the construction with a rough diagram in front of you:

1. Draw the base QR = 4 cm.
2. At Q draw a line l perpendicular to QR.
3. From R cut an arc of radius 5 cm on the line l. Call the crossing point P.
4. Join PR. ∆PQR is the required triangle.

Yes, the arc does meet the line below QR — but no, the new triangle is not different.

Call the lower crossing point P′. Then ∆P′QR also has ∠Q = 90°, QR = 4 cm and P′R = 5 cm, exactly the required measurements.

The lower triangle ∆P′QR is the mirror image of ∆PQR in the line QR, so it fits exactly on ∆PQR when flipped over.

So both possible triangles are congruent to each other, and therefore every triangle with these measurements is congruent to ∆PQR.

∠B = ∠C. The angles opposite the two equal sides are equal.

Drop the altitude from A to BC, meeting BC at D. Then compare ∆ADB and ∆ADC:

So the corresponding angles ∠B and ∠C are equal.

Yes. Call each of them x.

All three angles are equal.

Use the fact just discovered, twice over, in ∆ABC:

Each angle is 60°.

Construct an equilateral triangle and measure its angles.

1. Draw AB = 5 cm.
2. With A as centre and radius 5 cm, draw an arc above AB.
3. With B as centre and the same radius 5 cm, draw another arc cutting the first at C.
4. Join AC and BC.
5. Measure ∠A, ∠B and ∠C with a protractor.

Congruent triangles appear in every one of the pictures.

Read the letters in the same order on both sides of the ≅ sign.

Four of the five cases give congruent triangles.

The segments AD and BC cross at O. Look at ∆AOB and ∆DOC.

Corresponding angles of congruent triangles are equal, so

AD is a transversal cutting the two lines AB and DC. The angles ∠DAB and ∠ADC are alternate angles for that transversal, and they are equal. Hence

Part 1 — ∆ABC ≅ ∆ADC:

Part 2 — is ∆ABC ≅ ∆CDA as well? Yes.

So ∆ABC ≅ ∆CDA as well (SSS): the triangles are congruent in two ways.

More examples (two ways): any isosceles ∆PQR (PQ = PR) and its copy ∆XYZ (XY = XZ): ∆PQR ≅ ∆XYZ and ∆PQR ≅ ∆XZY.

Six ways: an equilateral ∆PQR and its copy ∆XYZ:

∠B = ∠C = 30°.

B and C lie on the circle and A is its centre, so AB and AC are both radii:

Now use the angle sum, writing x for each of the equal angles:

The figure is a rectangle ACDB cut into 16 triangles. Four points inside it are not named in the book, so let us call them P, Q, S and T, as marked here.

The marked equal sides. Single ‘|’: CR = RV = VD = CU = UA = US = UT = VQ. Triple ‘|||’: RQ = AT = KL = LB. Double ‘||’: PF = PB = FB.

Working, step by step (each step uses the angle sum of a triangle, angles on a line or around a point, equal angles opposite equal sides, or one congruence):

All the angles, triangle by triangle.

The grid is 5 × 5, and the green square in the middle is not part of the region.

So each region must have 4 squares. Six congruent L-shaped pieces of 4 squares fill the white region. One way is shown below — the squares with the same number form one piece.