Chapter 1 exercise answers: Geometric Twins

Open NCERT p. 3Checked by computer

1. Each figure is two arms meeting at a corner, like the signboard symbol. Such a figure is fixed by three measurements: the two arm lengths and the angle between the arms.
2. Measure (or trace) the arms: the short arms are the same length in both figures, and the long arms are the same length too.
3. Now measure the angle between the arms: it is about 90° in the first figure and about 71° in the second, so the second figure is more closed.
4. Equal arm lengths are not enough: the chapter showed several different symbols with the same arm lengths. Because the angles are different, one figure cannot be laid exactly over the other, even after turning or flipping it.

AnswerNo, the two figures are not congruent: their arm lengths match, but the angles between the arms (about 90° and about 71°) do not.

Watch this explained “Congruent, and the test”, 3:59 into Same shape and same size: superimposition as the test · हिंदी में देखें

Open NCERT p. 3Checked by computer

1. Compare each pair by imagining one shape sliding onto the other.
2. The two teardrops are the same size and shape, just turned to a different angle.
3. The two clouds are not the same size -- the left one is visibly bigger than the right one.
4. The two starbursts are not the same size either -- one is clearly smaller.
5. The two leaf-clusters are the same size and shape, just mirrored.

AnswerThe teardrops and the leaves appear congruent; the clouds and the starbursts do not.

Watch this explained “Congruent, and the test”, 3:59 into Same shape and same size: superimposition as the test · हिंदी में देखें

Open NCERT p. 3One way to think about it

(a) Circle

1. A circle's size is completely fixed by one measurement.
2. Take the radius (or the diameter) of the circle.

In shortTake the radius; two circles are congruent exactly when their radii are equal.

(b) Rectangle

1. A rectangle needs two measurements to fix its size and shape.
2. Take the length and the breadth of the rectangle.

In shortTake the length and the breadth; two rectangles are congruent exactly when both their lengths match and both their breadths match.

Watch this explained “A circle takes one number”, 6:29 into Same shape and same size: superimposition as the test · हिंदी में देखें

Open NCERT p. 4Checked by computer

1. Each figure is three arms meeting at one point: two arms lie along one straight line, and the third arm branches off it.
2. To check two such figures, measure the three arm lengths and the angles between neighbouring arms. Two of the angles are enough, because the angles all the way round the point add up to 360°. If all of these match, one figure can be moved (turned or flipped if needed) to lie exactly on the other.
3. First pair: the two figures have the same three arm lengths and the same angles. The second figure is the first one slid across the page, without turning.
4. Second pair: again the three arm lengths and the angles are the same in both figures, so they also fit exactly on each other.

AnswerFirst pair: yes, congruent. Second pair: yes, congruent.

Watch this explained “A third arm”, 8:02 into Same shape and same size: superimposition as the test · हिंदी में देखें

Open NCERT p. 8Checked by computer

1. ΔHEN ≅ ΔBIG pairs the vertices in order: H with B, E with I, N with G.
2. Any other correct way must keep these same three pairs. So write the letters of ΔHEN in a new order, and write the letters of ΔBIG in exactly the same new order.
3. The letters H, E, N can be arranged in 6 orders. One of them, HEN, is the statement we were given, which leaves 5 others.
4. (Writing the two triangles the other way round, such as ΔBIG ≅ ΔHEN, keeps the same pairs too; the list below keeps ΔHEN's letters on the left, as the chapter does.)

AnswerΔHNE ≅ ΔBGI, ΔEHN ≅ ΔIBG, ΔENH ≅ ΔIGB, ΔNHE ≅ ΔGBI, ΔNEH ≅ ΔGIB.

Watch this explained “Six ways to say one thing”, 3:53 into Naming congruent figures so the correspondence is unambiguous · हिंदी में देखें

Open NCERT p. 8Checked by computer

1. List the three sides of each triangle: RE=3.5 cm, ED=5 cm, DR=6 cm and JM=6 cm, MA=5 cm, AJ=3.5 cm.
2. Both triangles have the same three side lengths, {3.5, 5, 6}, so by SSS they are congruent.
3. Match the vertices by which sides meet there: E joins the 3.5 and 5 sides, matching A; D joins the 5 and 6 sides, matching M; R joins the 6 and 3.5 sides, matching J.

AnswerYes, congruent by SSS: ΔRED ≅ ΔJAM.

Watch this explained “The result, and its name”, 6:12 into SSS: three sidelengths fix a triangle completely · हिंदी में देखें

Open NCERT p. 9One way to think about it

1. AC is a diagonal shared by ΔABC and ΔADC.
2. AB=AD and CB=CD are given, and AC=AC is common to both triangles.
3. All three sides of ΔABC match all three sides of ΔADC, so by SSS the triangles are congruent.
4. By CPCT, ∠BAC=∠DAC and ∠BCA=∠DCA -- and these are exactly the two halves AC cuts ∠BAD and ∠BCD into.

In shortYes -- ΔABC ≅ ΔADC by SSS. Because of this, AC does divide ∠BAD into ∠BAC=∠DAC and ∠BCD into ∠BCA=∠DCA, so both angles are cut into two equal parts.

Watch this explained “Two more figures with a shared side”, 8:48 into SSS: three sidelengths fix a triangle completely · हिंदी में देखें

Open NCERT p. 9Checked by computer

1. The two triangles share the side DE.
2. Match the equal sides: DF = DG (given), FE = GE (given) and DE = DE (common side).
3. So D goes with D, F goes with G, and E goes with E. All three pairs of sides are equal, so the triangles are congruent by SSS.
4. Write the letters in this matching order: ΔDFE ≅ ΔDGE. ΔGED is the same triangle, but writing ΔDFE ≅ ΔGED would wrongly pair D with G.

AnswerYes, ΔDFE and ΔGED are congruent by SSS, and the congruence is written ΔDFE ≅ ΔDGE.

Watch this explained “A dart whose letters mislead”, 8:36 into Naming congruent figures so the correspondence is unambiguous · हिंदी में देखें

Open NCERT p. 13Checked by computer

1. In ΔABC: AB=7 cm, ∠B=47°, BC=5 cm -- two sides with the angle between them.
2. In ΔXYZ: XZ=7 cm, ∠Z=47°, ZY=5 cm -- the same two side lengths with the same angle between them.
3. By SAS, the triangles are congruent, matching B (where the known sides meet) to Z, so A matches X and C matches Y.

AnswerYes, congruent by SAS: ΔABC ≅ ΔXZY.

Watch this explained “Naming it: side, angle, side”, 3:03 into SAS, and why the angle has to be the included one · हिंदी में देखें

Open NCERT p. 13Checked by computer

1. The segments AC and BD cross at O, making two triangles: ΔAOB and ΔCOD.
2. AB ∥ CD and AC cuts both lines, so the alternate angles are equal: ∠OAB = ∠OCD.
3. BD also cuts both lines, so ∠OBA = ∠ODC (alternate angles).
4. ∠AOB = ∠COD, because they are vertically opposite angles.
5. In ΔAOB and ΔCOD: ∠OAB = ∠OCD, AB = CD (given) and ∠OBA = ∠ODC. The given side lies between the two angles, so the triangles are congruent by ASA, with A ↔ C, O ↔ O, B ↔ D.
6. Corresponding sides of congruent triangles are equal, so OA = OC and OB = OD. In other words, O is the midpoint of both AC and BD.

AnswerThe other equal parts are ∠OAB = ∠OCD, ∠OBA = ∠ODC, ∠AOB = ∠COD, OA = OC and OB = OD. Yes, the triangles are congruent by ASA: ΔAOB ≅ ΔCOD.

Watch this explained “Naming it: angle, side, angle”, 1:27 into ASA and AAS: two angles are enough to fix the third · हिंदी में देखें

Open NCERT p. 14One way to think about it

1. BC is a side shared by ΔABC and ΔDBC.
2. ∠ABC=∠DBC and ∠ACB=∠DCB are given -- two angles and the included side BC match in both triangles.
3. By ASA, ΔABC ≅ ΔDBC.
4. By CPCT, the remaining angles match too: ∠BAC=∠BDC.

In shortYes, the triangles are congruent (ΔABC ≅ ΔDBC, by ASA), and this is exactly why ∠BAC=∠BDC.

Watch this explained “Naming it: angle, side, angle”, 1:27 into ASA and AAS: two angles are enough to fix the third · हिंदी में देखें

Open NCERT p. 14Checked by computer

1. The diagonal BD splits ∠ABC into ∠ABD and ∠DBC; the diagonal CA splits ∠DCB into ∠DCA and ∠ACB.
2. Adding the two given equal pairs, ∠ABD+∠DBC = ∠DCA+∠ACB, which is exactly ∠ABC = ∠DCB.
3. In ΔABC and ΔDCB, BC=CB is a shared side lying between ∠ABC,∠ACB and ∠DCB,∠DBC -- two angles and the side between them match, so by ASA the triangles are congruent.
4. By CPCT: AB=DC, AC=DB, and ∠BAC=∠CDB.

Answer∠ABC = ∠DCB, and ΔABC ≅ ΔDCB (ASA), giving AB = DC, AC = DB and ∠BAC = ∠CDB.

Watch this explained “Naming it: angle, side, angle”, 1:27 into ASA and AAS: two angles are enough to fix the third · हिंदी में देखें

Open NCERT p. 20Checked by computer

1. ΔAIR ≅ ΔFLY is written so the letters line up in order: A↔F, I↔L, R↔Y.
2. Each side is named by the two vertices at its ends, so the matching sides follow the same correspondence: AI↔FL, IR↔LY, RA↔YF.
3. Each angle is named by its own vertex, so the matching angles follow too: ∠A↔∠F, ∠I↔∠L, ∠R↔∠Y.

AnswerVertices: A↔F, I↔L, R↔Y. Sides: AI↔FL, IR↔LY, RA↔YF. Angles: ∠A↔∠F, ∠I↔∠L, ∠R↔∠Y.

Watch this explained “One pairing, three lists”, 1:37 into Naming congruent figures so the correspondence is unambiguous · हिंदी में देखें

Open NCERT p. 20Checked by computer

(a) AB = DE … CA = DF

1. All three sides of ΔABC (AB, BC, CA) match all three sides of ΔDEF (DE, EF, FD) in order.

AnswerCongruent by SSS: ΔABC ≅ ΔDEF.

(b) AB = EF ∠A = ∠E AC = ED

1. ∠A sits between sides AB and AC; ∠E sits between sides EF and ED -- the same two sides and the angle between them match.

AnswerCongruent by SAS: ΔABC ≅ ΔEFD.

(c) AB = DF … AC = FE

1. ∠B=∠D=90° are the right angles, AC and FE are the hypotenuses (opposite the right angles) and they're equal, and AB=DF is one matching leg -- right angle, hypotenuse, side all match.

AnswerCongruent by RHS: ΔABC ≅ ΔFDE.

(d) ∠A = ∠D ∠B = ∠E AC = DF

1. ∠A=∠D and ∠B=∠E are two matching angles, and AC=DF is a side that is not the one lying between those two angles (that side would be AB/DE) -- two angles and a non-included side match.

AnswerCongruent by AAS: ΔABC ≅ ΔDEF.

(e) AB = DF ∠B = ∠F AC = DE

1. AB and AC are two sides at vertex A, but the given angle ∠B is not the angle between them (that would be ∠A) -- this is side-side-angle, not one of the conditions that guarantees congruence.
2. The chapter's own construction earlier in this section showed two genuinely different triangles can share an SSA measurement, so this case cannot be concluded congruent.

AnswerCannot be concluded congruent -- the given measurements are SSA, which does not guarantee congruence.

Watch this explained “Five, and what is missing”, 8:19 into RHS: the right-angled special case · हिंदी में देखें

Open NCERT p. 20One way to think about it

1. AD and BC are two straight lines crossing at O, so ∠AOB=∠DOC (vertically opposite angles).
2. With OA=OD and OB=OC given, and ∠AOB=∠DOC, ΔAOB ≅ ΔDOC by SAS.
3. By CPCT, ∠OAB=∠ODC, that is, ∠DAB=∠ADC.
4. AD is a transversal cutting AB and DC, and ∠DAB and ∠ADC are the alternate angles it makes with them -- since these alternate angles are equal, AB is parallel to CD.

In shortΔAOB ≅ ΔDOC (SAS), so ∠DAB=∠ADC by CPCT; these are alternate angles for transversal AD, so AB ∥ CD.

Watch this explained “Which condition actually fires”, 3:50 into ASA and AAS: two angles are enough to fix the third · हिंदी में देखें

Open NCERT p. 21One way to think about it

1. In the square, AB = AD and CB = CD (all four sides of a square are equal), and AC is a side of both triangles. So ΔABC ≅ ΔADC by SSS, matching A ↔ A, B ↔ D, C ↔ C.
2. Now try the matching A ↔ C, B ↔ D, C ↔ A. Then AB goes with CD, BC goes with DA, and CA goes with AC. AB = CD and BC = DA (sides of the square), and CA = AC. So ΔABC ≅ ΔCDA by SSS as well.
3. So the same two triangles are congruent in two different ways. This happens because each triangle has two equal sides (AB = BC), so the two equal sides can be matched either way round.
4. More examples (there are many; here is one): any two congruent isosceles triangles. If PQ = PR, XY = XZ, PQ = XY and QR = YZ, then ΔPQR ≅ ΔXYZ and also ΔPQR ≅ ΔXZY.
5. Six different ways: two equilateral triangles with the same side length, say ΔPQR and ΔXYZ. All six sides are equal, so every matching works: ΔPQR ≅ ΔXYZ, ΔPQR ≅ ΔXZY, ΔPQR ≅ ΔYXZ, ΔPQR ≅ ΔYZX, ΔPQR ≅ ΔZXY and ΔPQR ≅ ΔZYX.

In shortYes: ΔABC ≅ ΔADC and also ΔABC ≅ ΔCDA, both by SSS. Two congruent isosceles triangles are congruent in two ways (one example of many), and two equilateral triangles with the same side are congruent in six ways.

Watch this explained “The other six is a different six”, 7:42 into Naming congruent figures so the correspondence is unambiguous · हिंदी में देखें

Open NCERT p. 21Checked by computer

1. AB and AC are both radii of the circle, so AB=AC.
2. In a triangle, the angles opposite equal sides are equal, so ∠B=∠C.
3. The three angles of ΔABC add to 180°: 120°+∠B+∠C=180°.
4. Since ∠B=∠C, each one is (180°-120°)÷2 = 30°.

Answer∠B = 30°, ∠C = 30°.

Watch this explained “Two radii, and the same argument”, 6:52 into Angles opposite equal sides are equal · हिंदी में देखें

Open NCERT p. 21Checked by computer

1. Name the four unlabelled dots: P is the dot just below R, Q is the dot with the 90° mark, S is the lower-left dot (joined to A and K), and T is the lower-right dot (joined to L, B and F). The figure is then made of 16 triangles.
2. Read the tick marks. Single: CR = RV = VD = CU = UA = VQ = US = UP. Double: TF = FB = TB. Triple: RQ = SA = KL = LB. Two facts do the work: angles facing equal sides are equal, and the angles of a triangle add up to 180° (angles along a straight line add up to 180°, and all the way round a point to 360°).
3. ΔCRU: CR = CU and ∠C = 90°, so ∠CRU = ∠CUR = (180° − 90°) ÷ 2 = 45°.
4. ΔRVQ: RV = VQ and ∠RVQ = 68°, so ∠VRQ = ∠VQR = (180° − 68°) ÷ 2 = 56°.
5. ΔVQD: ∠QVD = 180° − 68° = 112° (R, V, D are on one straight line). VQ = VD, so ∠VQD = ∠VDQ = (180° − 112°) ÷ 2 = 34°.
6. ΔQDF: corner D is a right angle, so ∠QDF = 90° − 34° = 56°. With ∠QFD = 98°, ∠DQF = 180° − 56° − 98° = 26°.
7. ΔTFB: all three sides are equal, so ∠TFB = ∠FBT = ∠BTF = 60°.
8. ΔTBL: corner B is a right angle, so ∠TBL = 90° − 60° = 30°. With ∠TLB = 90°, ∠BTL = 180° − 90° − 30° = 60°.
9. ΔTLK: ∠TLK = 180° − 90° = 90°. KL = LB and TL is common, so ΔTLK ≅ ΔTLB (SAS). So ∠TKL = ∠TBL = 30° and ∠KTL = ∠BTL = 60°.
10. ΔQFT: D, F, B are on one straight line, so ∠QFT = 180° − 98° − 60° = 22°. All the way round Q the angles make 360°, so ∠FQT = 360° − (56° + 44° + 46° + 90° + 26° + 34°) = 64°. Then ∠QTF = 180° − 22° − 64° = 94°.
11. ΔSQT: ∠QST = 180° − 90° − 56° = 34°.
12. ΔSKA: ∠SKA = 180° − 44° − 34° = 102°. A, K, L are on one straight line, so in ΔTKS, ∠TKS = 180° − 102° − 30° = 48°, and ∠KST = 180° − 30° − 48° = 102°.
13. ΔSAU: UA = US, so ∠USA = ∠UAS = 56° and ∠AUS = 180° − 56° − 56° = 68°.
14. ΔUSP: UP = US and ∠PUS = 34°, so ∠UPS = ∠USP = (180° − 34°) ÷ 2 = 73°.
15. ΔPSQ: all the way round S the angles make 360°, so ∠PSQ = 360° − (73° + 56° + 44° + 102° + 34°) = 51°. Then ∠SPQ = 180° − 51° − 46° = 83°.
16. ΔRPQ: ∠RPQ = 180° − 34° − 44° = 102°.
17. ΔRUP: C, U, A are on one straight line, so ∠RUP = 180° − 45° − 34° − 68° = 33°. All the way round P, ∠RPU = 360° − 73° − 83° − 102° = 102°. So ∠URP = 180° − 33° − 102° = 45°.
18. Check at R, on the straight line C, R, V: 45° + 45° + 34° + 56° = 180°, as it must be.

Answer∠CRU = ∠CUR = 45°; ∠VRQ = ∠VQR = 56°; ∠QVD = 112°, ∠VQD = ∠VDQ = 34°; ∠QDF = 56°, ∠DQF = 26°; ∠TFB = ∠FBT = ∠BTF = 60°; ∠TBL = 30°, ∠BTL = 60°; ∠TLK = 90°, ∠TKL = 30°, ∠KTL = 60°; ∠QFT = 22°, ∠FQT = 64°, ∠QTF = 94°; ∠QST = 34°; ∠SKA = 102°; ∠TKS = 48°, ∠KST = 102°; ∠USA = 56°, ∠AUS = 68°; ∠UPS = ∠USP = 73°; ∠PSQ = 51°, ∠SPQ = 83°; ∠RPQ = 102°; ∠RUP = 33°, ∠RPU = 102°, ∠URP = 45°.

Watch this explained “Put the eighty back: fifty each”, 6:12 into Angles opposite equal sides are equal · हिंदी में देखें