CBSE Class 7 Mathematics Geometric Twins Worksheet Set 01

Long Answer Questions

Question 1. A school has to put up a second notice board carrying exactly the same symbol as the first one. The symbol is too large to be traced.
(i) What does it mean to say that the two symbols are congruent?
(ii) Describe the tracing-paper test for congruence, and explain why a figure may be rotated or flipped during this test.
(iii) Write the measurements that are enough to redraw each of these exactly: a circle, a rectangle, and a symbol made of two straight arms meeting at a point. Give a reason in each case.
Answer:
(i) The two symbols are congruent when they are exact copies, with the same shape and the same size, so that one can be placed on the other and fits exactly.
(ii) Trace the first symbol on tracing paper and place the tracing on the second symbol, moving it until it covers it. If it fits exactly, the symbols are congruent; if some part does not cover, they are not.
The tracing may be rotated or flipped, because turning or flipping changes only the position of the figure, not its shape or size.
(iii) Circle: the radius (or diameter) alone, since the radius decides the size of a circle completely.
Rectangle: the length and the breadth, since all four angles are already right angles.
Two-arm symbol: the lengths of both arms and the angle between them. Lengths alone allow the arms to open by different amounts; the angle fixes the opening, and so the shape and size.
In simple words: Congruent means exact copies; tracing (turning allowed) tests it; a circle needs its radius, a rectangle two sides, a two-arm symbol two arms and the angle.

Teacher's Note:
1. Each kind of figure needs its own set of measurements.
2. Fix only what can still change.
3. Turning or flipping never changes size.

Question 2. Two triangles can be shown to be congruent without comparing all six parts.
(i) Write the full form of each of the five conditions that guarantee the congruence of two triangles.
(ii) For SSS and for SAS, give one pair of measurement sets of your own that satisfies the condition.
(iii) Name one arrangement of three parts that does not guarantee congruence and state why it fails.
Answer:
(i) SSS: Side Side Side; SAS: Side Angle Side; ASA: Angle Side Angle; AAS: Angle Angle Side; RHS: Right angle Hypotenuse Side.
(ii) SSS: △ ABC with sides 4 cm, 6 cm, 7 cm and △ DEF with sides 4 cm, 6 cm, 7 cm. All three sides match, so the triangles are congruent.
SAS: △ ABC with AB = 5 cm, AC = 8 cm, ∠A = 60°, and △ DEF with DE = 5 cm, DF = 8 cm, ∠D = 60°. Two sides and the angle between them match, so the triangles are congruent.
(iii) SSA (two sides and an angle not between them). While constructing, the arc of the second side can cut the ray at two points, giving two triangles with the same three measurements but different shapes. So SSA does not guarantee congruence.
In simple words: Five conditions (SSS, SAS, ASA, AAS, RHS) work; SSA fails because it can give two triangles.

Teacher's Note:
1. Any correct examples are acceptable in (ii).
2. AAA also fails, because size is not fixed.

Question 3. In △ ABC, AB = AC. The altitude AD is drawn from A to BC.
(i) List the three pairs of equal parts of △ ADB and △ ADC, with a reason for each.
(ii) Name the congruence condition satisfied and write the congruence statement.
(iii) Hence state the property of an isosceles triangle that follows, and use it to find ∠B and ∠C when ∠A = 80°.
Answer:
(i) AB = AC (given)
∠ADB = ∠ADC = 90° (AD is an altitude to BC)
AD = AD (common side)
(ii) Each triangle has a right angle at D, the hypotenuses AB and AC are equal, and one more side AD is common. So the RHS condition holds and
△ ADB ≅ △ ADC
(iii) Corresponding angles of congruent triangles are equal, so ∠B = ∠C. This is the property: angles opposite to equal sides of a triangle are equal.
If ∠A = 80°, then ∠B + ∠C = 180° − 80° = 100°. As ∠B = ∠C, ∠B = ∠C = 50°.
In simple words: RHS makes the two halves congruent, so the base angles are equal; with ∠A = 80° each is 50°.

Teacher's Note:
1. The altitude splits the isosceles triangle into two congruent halves.
2. Equal base angles share 180° − ∠A.

Question 4. It is given that △ MAT ≅ △ COW.
(i) Write the corresponding vertices, the corresponding sides and the corresponding angles.
(ii) Explain why it would be wrong to write △ AMT ≅ △ COW for the same pair of triangles.
(iii) Write two other correct ways of expressing this congruence, and state the convention you used.
Answer:
(i) Vertices: M ↔ C, A ↔ O, T ↔ W.
Sides: MA and CO, AT and OW, MT and CW.
Angles: ∠M and ∠C, ∠A and ∠O, ∠T and ∠W.
(ii) △AMT ≅ △COW would mean A ↔ C and M ↔ O. But actually A ↔ O and M ↔ C. It would make AT match CW and MT match OW, and such sides need not be equal. So the statement is wrong.
(iii) △ AMT ≅ △ OCW and △ TAM ≅ △ WOC.
Convention: both names are rearranged in exactly the same way, so first letter matches first, second matches second and third matches third.
In simple words: Read pairs by position; any reordering must be done to both names together.

Teacher's Note:
1. Write the vertex pairs before anything else.
2. One reordered name alone breaks the matching.

Question 5. ABCD is a square and AC is a diagonal.
(i) Show that △ ABC ≅ △ ADC, naming the condition used.
(ii) Check whether △ ABC ≅ △ CDA is also a correct statement for the same pair of triangles, giving the matching of sides.
(iii) Explain how one triangle can be congruent to another in two different ways, and state what this tells you about the diagonal AC of the square.
Answer:
(i) In △ ABC and △ ADC: AB = AD (sides of a square), BC = DC (sides of a square), AC = AC (common diagonal).
All three pairs of sides are equal, so by the SSS condition, △ ABC ≅ △ ADC.
(ii) In △ABC ≅ △CDA: A ↔ C, B ↔ D, C ↔ A. Then AB matches CD (equal sides of the square), BC matches DA (equal), and CA matches AC (the same segment).
Every matched pair is equal, so this statement is also correct.
(iii) Both work because △ABC is itself isosceles (AB = BC). It can be fitted on △ADC directly or after being flipped.
From △ABC ≅ △ADC, ∠BAC = ∠DAC and ∠BCA = ∠DCA. So the diagonal AC divides the right angles at A and C into two equal parts of 45° each.
In simple words: SSS gives the congruence in two ways because the triangle is isosceles; AC halves the corner angles into 45° each.

Teacher's Note:
1. A square has all four sides equal.
2. An isosceles triangle can be matched to its copy in two ways.

Question 6. In the figure, AB is parallel to CD and AB = CD. The segments AC and BD cross at O.
(i) Write two pairs of equal angles in △ AOB and △ COD, with reasons.
(ii) Name the congruence condition satisfied and express the congruence.
(iii) Write all the conclusions that follow about the point O, and justify them.
Answer:
(i) AB ∥ CD. AC is a transversal, so ∠OAB = ∠OCD (alternate angles).
BD is also a transversal, so ∠OBA = ∠ODC (alternate angles).
(ii) In △AOB and △COD: ∠OAB = ∠OCD, AB = CD (given), ∠OBA = ∠ODC.
The equal side AB lies between the two equal angles, so the ASA condition holds and △ AOB ≅ △ COD.
(iii) Corresponding sides are equal: OA = OC and OB = OD.
O lies on AC and on BD, so O is the midpoint of AC and also the midpoint of BD; the two segments cut each other into equal halves at O.
In simple words: Alternate angles and AB = CD give ASA, so O cuts both AC and BD in half.

Teacher's Note:
1. Parallel lines give equal alternate angles.
2. Equal halves on a segment mean a midpoint.

Question 7. Rabia draws a triangle with BC = 6 cm, ∠B = 50° and ∠C = 40°. Meera draws a triangle with angles 50°, 40° and 90° but takes no side length.
(i) Explain why every student who follows Rabia’s three measurements gets the same triangle.
(ii) Explain, with an example of two different triangles, why Meera’s three measurements do not fix the triangle.
(iii) State clearly which of the two sets of measurements is a condition for congruence, and what the other set guarantees instead.
Answer:
(i) Draw BC = 6 cm, a ray from B at 50° and a ray from C at 40°. The two rays meet at exactly one point, the vertex A. There is no choice anywhere, so everyone gets the same triangle.
(ii) One student may start with a base of 4 cm and another with a base of 8 cm; both can complete triangles with angles 50°, 40° and 90°. The second is a bigger copy of the first: same shape, different size. So they are not congruent.
(iii) Rabia’s set is the ASA condition, a condition for congruence. Meera’s set (three angles, AAA) is not; it only guarantees that the triangles have the same shape.
In simple words: A side with the angles at its ends fixes the triangle; three angles fix only its shape.

Teacher's Note:
1. Check that 50° + 40° + 90° = 180°.
2. AAA gives the same shape, not the same size.

Question 8. In △ PQR and △ XYZ, it is given that ∠P = ∠X = 50°, ∠R = ∠Z = 60° and PQ = XY = 6 cm.
(i) Find ∠Q and ∠Y, showing the working.
(ii) Show that the two triangles are congruent, naming the condition you finally use, and express the congruence.
(iii) Name the condition that the given data itself represents, and write the remaining pairs of equal parts of the two triangles.
Answer:
(i) In △PQR, ∠Q = 180° − 50° − 60° = 70°. In △XYZ, ∠Y = 180° − 50° − 60° = 70°. So ∠Q = ∠Y.
(ii) ∠P = ∠X = 50°, PQ = XY = 6 cm, ∠Q = ∠Y = 70°.
The side PQ lies between the two equal angles, so by the ASA condition, △ PQR ≅ △ XYZ.
(iii) In the given data, PQ is not between ∠P and ∠R, so the data is the AAS condition, which also guarantees congruence (as shown above using the angle sum).
Remaining equal parts: QR = YZ, PR = XZ and ∠Q = ∠Y.
In simple words: The third angles are 70° each, so AAS becomes ASA and the triangles are congruent.

Teacher's Note:
1. AAS works because two angles fix the third.
2. Corresponding parts give the remaining equalities.

Question 9. In △ ABC, AB = AC. The point D lies on BC and AD divides ∠BAC into two equal parts.
(i) Show that △ ABD ≅ △ ACD, naming the condition used.
(ii) Using the congruence, prove that D is the midpoint of BC and that AD is perpendicular to BC.
(iii) If ∠BAC = 50°, find ∠B, ∠C and ∠BAD.
Answer:
(i) In △ABD and △ACD: AB = AC (given), ∠BAD = ∠CAD (AD halves ∠BAC), AD = AD (common).
The equal angles lie between the equal sides, so by the SAS condition, △ ABD ≅ △ ACD.
(ii) Corresponding parts are equal. BD = CD, and D lies on BC, so D is the midpoint of BC.
Also ∠ADB = ∠ADC. As B, D, C lie on a straight line, these two angles add up to 180°, so each is 180° ÷ 2 = 90°. Hence AD ⊥ BC.
(iii) AB = AC, so ∠B = ∠C. ∠B + ∠C = 180° − 50° = 130°, so ∠B = ∠C = 65°.
AD halves ∠BAC, so ∠BAD = 50° ÷ 2 = 25°. (Check in △ABD: 65° + 25° + 90° = 180°.)
In simple words: SAS makes the two halves congruent, so D is the midpoint, AD is perpendicular, and the angles are 65°, 65° and 25°.

Teacher's Note:
1. Two equal angles on a straight line are 90° each.
2. Always check with the angle sum.

Question 10. Each set below gives measurements taken from △ PQR and △ XYZ. In each case decide whether the triangles must be congruent, name the condition, and express the congruence when it holds.
(i) PQ = XY, QR = YZ, RP = ZX
(ii) PQ = XY, ∠P = ∠X, PR = XZ
(iii) PQ = XY, ∠Q = ∠Y, PR = XZ
(iv) ∠P = ∠X, ∠Q = ∠Y, ∠R = ∠Z
(v) ∠Q = ∠Y = 90°, PR = XZ, QR = YZ
Answer:
(i) All three sides match: SSS. Congruent: △ PQR ≅ △ XYZ.
(ii) ∠P lies between PQ and PR, and ∠X between XY and XZ: SAS. Congruent: △ PQR ≅ △ XYZ.
(iii) The equal angle is at Q, which is not between PQ and PR: this is SSA. Two different triangles can be drawn, so congruence is not guaranteed.
(iv) Only angles match (AAA). Same shape but possibly different sizes, so not necessarily congruent.
(v) Right angles at Q and Y; PR and XZ are the hypotenuses; QR = YZ: RHS. Congruent: △ PQR ≅ △ XYZ.
In simple words: SSS, SAS and RHS give congruence; SSA and AAA do not.

Teacher's Note:
1. Locate the angle with respect to the given sides.
2. In (v) the hypotenuse is opposite the right angle.

Case Study Based Questions

Question 11. Two identical signboards for a park
A municipal office ordered two identical signboards for the two gates of a city park. Each board carries a symbol made of two straight arms meeting at a corner. The symbol on the first board is far too large to be traced on paper, so the painter decides to take measurements instead. He names the corner points A, B and C, and finds that the arm AB is 40 cm long and the arm BC is 70 cm long. Using only these two lengths he draws the symbol on the second board, but the manager says that the new symbol does not look like the first one at all. The painter then goes back to the first board and takes one more measurement.
(i) When will the two painted symbols be called congruent?
(ii) Why were the two arm lengths alone not enough to recreate the symbol?
(iii) Which third measurement should the painter take, and why does it fix the symbol?
(iv) The second board is finally fixed on the gate the other way round, so that the symbol appears flipped. Is it still congruent to the first symbol? Give a reason.
Answer:
(i) When they are exact copies: the same shape and the same size, so that one fits exactly over the other.
(ii) With AB = 40 cm and BC = 70 cm, the arms can be opened out by different amounts at B, so many different symbols are possible.
(iii) He should measure the angle ∠ABC between the arms. Two arm lengths with the angle between them fix the shape and size, so an exact copy can be drawn.
(iv) Yes, it is still congruent. Flipping changes only the way the figure faces, not its shape or size.
In simple words: The angle at B was missing; with it the symbol is fixed, and flipping does not spoil congruence.

Teacher's Note:
1. Two lengths plus the angle between them fix a two-arm figure.
2. Flipped copies are still congruent.

Question 12. Cut-out shapes in the craft period
In the craft period a teacher gives every child a sheet of paper and asks them to cut out shapes. Asha folds her sheet once and cuts a shape through both layers, so that two cut-outs come out together. Bhanu cuts two rectangles separately, one measuring 8 cm by 5 cm and the other 8 cm by 6 cm. Chandu traces one of Asha’s cut-outs on tracing paper, turns the tracing paper over and places it on the other cut-out; the two fit exactly. The teacher then asks the class which pairs of cut-outs on the table are congruent, and how they can be sure.
(i) What are Asha’s two cut-outs said to be, and why?
(ii) Chandu had to turn the tracing paper over. Does this spoil the test of congruence? Give a reason.
(iii) Are Bhanu’s two rectangles congruent? Give a reason.
(iv) Write the measurements a child must take to cut a rectangle congruent to a given rectangle.
Answer:
(i) They are congruent: they were cut together through both layers, so they have exactly the same shape and size.
(ii) No. A figure may be flipped before superimposing; turning the tracing over does not change its shape or size, so the test is still valid.
(iii) No. Both have a side of 8 cm, but the other sides are 5 cm and 6 cm, so they are of different sizes and cannot fit exactly.
(iv) The length and the breadth. All angles are right angles already, so these two measurements fix the rectangle.
In simple words: Asha’s pieces are congruent, flipping is allowed, Bhanu’s rectangles differ in breadth, and a rectangle needs length and breadth.

Teacher's Note:
1. Cutting through a fold gives congruent pieces.
2. Both sides of rectangles must match.

Question 13. Matching triangular tiles
A flooring shop sells triangular tiles. A customer brings one broken tile from her old floor and asks for twenty more tiles exactly like it. The shopkeeper measures the three sides of the broken tile and finds them to be 15 cm, 20 cm and 25 cm. He does not measure a single angle. He then picks tiles from his stock whose three sides also measure 15 cm, 20 cm and 25 cm, and says that they will match the old floor perfectly. The customer asks how he can be so sure without checking the corners, and whether a tile with sides 15 cm, 20 cm and 26 cm would also do.
(i) Name the congruence condition the shopkeeper is using.
(ii) Why is it not necessary for him to measure the angles of the tiles?
(iii) The broken tile is △ABC with AB = 15 cm, BC = 20 cm, CA = 25 cm, and a new tile is △PQR with PQ = 15 cm, QR = 20 cm, RP = 25 cm. Express the congruence.
(iv) Will a tile with sides 15 cm, 20 cm and 26 cm match the old floor? Give a reason.
Answer:
(i) The SSS (Side Side Side) condition.
(ii) SSS guarantees congruence: when all three sides match, the triangles are exact copies, so the angles are automatically equal.
(iii) AB = PQ, BC = QR, CA = RP, so A ↔ P, B ↔ Q, C ↔ R and △ ABC ≅ △ PQR.
(iv) No. Its third side is 26 cm, not 25 cm, so SSS is not satisfied and the tile will not be an exact copy.
In simple words: Equal three sides (SSS) fix the tile; a 26 cm side spoils the match.

Teacher's Note:
1. Three sides fix a triangle, so angles need no checking.
2. Even one unequal side breaks congruence.

Question 14. A rangoli made of triangles
For a festival, students draw a large rangoli on the school floor. The whole design is made of triangles only. In the middle there are six triangles, each having all three sides of length 30 cm. Around them are eight more triangles, each having two sides of 25 cm and a third side of 40 cm. A visitor asks the students what the angles of these triangles are. The students have no protractor with them, but their teacher tells them that a lot can be said about the angles from the side lengths alone, using what they have learnt about triangles with equal sides.
(i) What is the special name given to each of the six middle triangles?
(ii) Find the measure of each angle of one middle triangle, with reason.
(iii) In one of the eight outer triangles, which two angles are equal? Give a reason.
(iv) In an outer triangle the angle between the two 25 cm sides measures 106°. Find the other two angles.
Answer:
(i) Each is an equilateral triangle, since all three sides are equal.
(ii) All sides are equal, so all angles opposite them are equal. They add up to 180°, so each angle = 180° ÷ 3 = 60°.
(iii) The two angles at the ends of the 40 cm side are equal, because they lie opposite the two equal 25 cm sides.
(iv) The other two angles add up to 180° − 106° = 74° and are equal, so each is 74° ÷ 2 = 37°.
In simple words: Middle triangles are equilateral (60° each); in the outer ones the base angles are equal, 37° each.

Teacher's Note:
1. Equal sides give equal opposite angles.
2. Check: 106° + 37° + 37° = 180°.

Question 15. Two trusses for a school roof
A carpenter is making two wooden trusses for the sloping roof of a village school, and each truss is a triangle. For the first truss ABC he cuts the two rafters AB and AC each 3 m long and joins them at the ridge A so that the angle at A is 40°. For the second truss PQR he again cuts PQ and PR each 3 m long and joins them at P with the same angle of 40°. He does not measure the bottom beams BC and QR at all, yet he is confident that the two trusses will be exactly alike and that the roof will sit level on them.
(i) Which congruence condition makes the carpenter so sure?
(ii) Express the congruence of the two trusses.
(iii) Find ∠B and ∠C of the first truss, with reason.
(iv) Explain why he need not measure BC and QR separately.
Answer:
(i) The SAS condition: two sides (3 m and 3 m) and the included angle (40°) are the same in both trusses.
(ii) △ ABC ≅ △ PQR.
(iii) AB = AC, so ∠B = ∠C. ∠B + ∠C = 180° − 40° = 140°, so ∠B = ∠C = 70°.
(iv) SAS already fixes the whole triangle. Once the two rafters and the angle between them are equal, the bottom beams must come out equal too, so measuring them is not needed.
In simple words: Same two rafters and the same angle between them (SAS) make the trusses identical.

Teacher's Note:
1. SAS fixes the third side automatically.
2. Base angles share 140° equally.

Teacher's Note:
1. RHS needs all three: right angle, hypotenuse and a side.
2. Two sides alone never fix a triangle.

Question 17. Triangular flags for a sports meet
A tailor is asked to stitch thirty triangular flags for a sports meet. The organiser sends him a drawing in which the three angles of the flag are marked 40°, 60° and 80°, but no side length is written anywhere. The tailor cuts the first batch of flags using these angles only. When the flags are hung on the rope, the teachers find that the flags are of many different sizes, although every flag has the same shape. The organiser then sends a second note which gives the length of one side of the flag as well, and after that all the flags come out alike.
(i) Why did the first batch of flags come out in different sizes?
(ii) Do three equal angles guarantee that two triangles are congruent? Give a reason.
(iii) The second note says that the side lying between the 40° and 60° angles is 24 cm. Name the condition that now fixes the flag.
(iv) Suppose the note had given the side opposite the 80° angle instead. Would the flags still be congruent? Name the condition.
Answer:
(i) Only the three angles were given. Angles fix the shape but not the size, so each flag could be cut with a different base length.
(ii) No. Three equal angles (AAA) is not a congruence condition; the triangles have the same shape but may differ in size.
(iii) The 24 cm side lies between the 40° and 60° angles, so it is the ASA condition, which fixes the flag completely.
(iv) Yes. The side opposite the 80° angle is the side between the 40° and 60° angles, the same side as in (iii). So this is again ASA (it can also be seen as AAS), and the flags would be congruent.
In simple words: Angles alone gave different sizes; one side with the angles fixes every flag.

Teacher's Note:
1. AAA fixes shape only.
2. Any one side together with the angles fixes the triangle.

Question 18. A bracket for a shop signboard
A shop signboard is held to the wall by a triangular iron bracket DEF. Checking it with his tape, the fitter finds that the two slanting bars DE and DF are of exactly the same length. He then puts his protractor at one corner only and reads ∠E = 55°. From this single reading he works out the other two angles of the bracket without measuring them, and tells the shopkeeper that he can order a second bracket of exactly the same size if he also notes the length of the two slanting bars and the angle between them.
(i) Which two angles of the bracket are equal in size? Name the property used.
(ii) Find ∠D.
(iii) A second bracket has DE and DF equal to those of the first bracket, and the angle between these two bars is also equal. Name the condition that shows the two brackets are congruent.
(iv) What should the reading at E have been if the bracket were to be an equilateral triangle?
Answer:
(i) ∠E and ∠F are equal. Property: angles opposite equal sides are equal; ∠E is opposite DF and ∠F is opposite DE, and DE = DF.
(ii) ∠F = ∠E = 55°, so ∠D = 180° − 55° − 55° = 70°.
(iii) Two equal bars and the equal angle between them: the SAS condition.
(iv) 60°, because each angle of an equilateral triangle is 60°.
In simple words: Equal bars give ∠E = ∠F = 55°, so ∠D = 70°; equal bars and angle D give SAS.

Teacher's Note:
1. Find the equal angles from the equal sides.
2. Check: 55° + 55° + 70° = 180°.

Question 19. The top of an iron gate
A welder is making the top part of an iron gate. It is a triangle PQR in which the two slanting bars PQ and PR are each 1.4 m long and the angle between them at P is 40°. To make the frame strong he welds one more bar from P to the point M, the midpoint of the bottom bar QR. Some weeks later the owner orders a second gate of exactly the same design, and the welder again cuts the two slanting bars 1.4 m long and sets the angle between them at 40°.
(i) Find ∠Q and ∠R of the gate top, with reason.
(ii) Name the condition that shows △ PQM ≅ △ PRM and express the congruence.
(iii) Find ∠PMQ, giving the reason.
(iv) Is the top of the second gate congruent to the top of the first gate? Name the condition used.
Answer:
(i) PQ = PR, so ∠Q = ∠R. ∠Q + ∠R = 180° − 40° = 140°, so ∠Q = ∠R = 70°.
(ii) In △PQM and △PRM: PQ = PR (given), QM = RM (M is the midpoint of QR), PM = PM (common bar). By the SSS condition, △ PQM ≅ △ PRM.
(iii) From the congruence, ∠PMQ = ∠PMR. They lie on the straight bar QR, so together they make 180°. Hence ∠PMQ = 180° ÷ 2 = 90°. (Check in △PQM: 70° + 20° + 90° = 180°.)
(iv) Yes. Both tops have two sides of 1.4 m with the same included angle of 40°, so by the SAS condition they are congruent.
In simple words: Base angles are 70°, SSS makes the halves congruent so PM is perpendicular, and SAS makes the two gate tops congruent.

Teacher's Note:
1. A midpoint gives two equal halves.
2. Two equal angles on a straight line are 90° each.

Question 20. A check that let a faulty plate pass
A workshop cuts triangular steel plates for a machine. The checker at the gate is told to pass a plate if it has a long edge of 9 cm, another edge of 6 cm, and an angle of 35° at one end of the 9 cm edge, the 6 cm edge being the one opposite that angle. Two plates of clearly different shapes were brought to him on the same day, and both of them satisfied all three of his measurements. One of the two plates fitted the machine and the other did not, yet the checker had passed both. The supervisor says the fault is not in the checker but in the set of measurements he was given.
(i) Name the condition the checker was using, and say whether it guarantees congruence.
(ii) Using the idea of drawing an arc, explain how two plates of different shapes can satisfy the same three measurements.
(iii) Which single measurement, taken instead of the 35° angle, would make the check reliable? Name the condition that would then apply.
(iv) If the angle measured had been 90° instead of 35°, with the 9 cm edge opposite it, would the check have been reliable? Name the condition.
Answer:
(i) He was using SSA (two sides and an angle not between them). SSA does not guarantee congruence.
(ii) Draw the 9 cm edge, a ray making 35° at one end, and an arc of radius 6 cm from the other end. The arc cuts the ray at two different points, giving two triangles of different shapes with the same three measurements.
(iii) The length of the third edge; then all three sides are known and the SSS condition applies. (Measuring the angle between the 9 cm and 6 cm edges would also work, by SAS.)
(iv) Yes. There would be a right angle, a hypotenuse of 9 cm opposite it and one more side of 6 cm. This is the RHS condition, which guarantees congruence.
In simple words: SSA let two shapes pass; the third side (SSS) or a right angle with the hypotenuse (RHS) would make the check reliable.

Teacher's Note:
1. Two crossing points of the arc mean two possible plates.
2. RHS is the safe special case of SSA.