Exercise 6.3 answers: Triangles

Open NCERT p. 94Matches NCERT’s answer

(i)

1. In ΔABC: ∠A = 60°, ∠B = 80°, ∠C = 40°. In ΔPQR: ∠P = 60°, ∠Q = 80°, ∠R = 40°.
2. So ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R. All three pairs of corresponding angles are equal.

AnswerSimilar, by AAA (two equal angles, AA, would already be enough); ΔABC ~ ΔPQR.

(ii)

1. Sides: AB = 2, BC = 2.5, CA = 3 and QR = 4, RP = 5, PQ = 6.
2. Match the shortest with the shortest and the longest with the longest: AB/QR = 2/4 = 1/2, BC/RP = 2.5/5 = 1/2, CA/PQ = 3/6 = 1/2. All three ratios are equal.
3. So A matches Q, B matches R and C matches P.

AnswerSimilar, by SSS; ΔABC ~ ΔQRP.

(iii)

1. Sides in increasing order: MP = 2, LM = 2.7, LP = 3 and DE = 4, EF = 5, FD = 6.
2. Any similarity must match the smallest side with the smallest and the largest with the largest, so the ratios would have to be 2/4, 2.7/5 and 3/6.
3. 2/4 = 0.5, 2.7/5 = 0.54, 3/6 = 0.5. They are not all equal, so no matching of the sides works.

AnswerNot similar.

(iv)

1. In ΔMNL, ∠M = 70° lies between MN = 2.5 and ML = 5. In ΔPQR, ∠Q = 70° lies between QP = 5 and QR = 10.
2. MN/QP = 2.5/5 = 1/2 and ML/QR = 5/10 = 1/2, and the angles between these sides are equal: ∠M = ∠Q.
3. So M matches Q, N matches P and L matches R.

AnswerSimilar, by SAS; ΔMNL ~ ΔQPR.

(v)

1. In ΔDEF, ∠F = 80° lies between DF = 5 and FE = 6.
2. In ΔABC, ∠A = 80°, but the given sides are AB = 2.5 and BC = 3, and ∠A is not between them (it lies between AB and AC).
3. The ratios 2.5/5 = 3/6 = 1/2 look right, but SAS needs the equal angle to be the one between the two sides, so SAS does not apply, and no other criterion fits the data.
4. They are in fact not similar: working out the remaining angles (with trigonometry, which comes later) gives about 44.8° and 55.2° in ΔABC but about 43.8° and 56.2° in ΔDEF.

AnswerNot similar.

(vi)

1. In ΔDEF: ∠D = 70°, ∠E = 80°, so ∠F = 180° − 70° − 80° = 30°.
2. In ΔPQR: ∠Q = 80°, ∠R = 30°, so ∠P = 180° − 80° − 30° = 70°.
3. So ∠D = ∠P, ∠E = ∠Q and ∠F = ∠R.

AnswerSimilar, by AA; ΔDEF ~ ΔPQR.

Watch this explained “Where it fails: the mark in the wrong place”, 7:00 into SAS: one angle plus the two sides that enclose it

Open NCERT p. 95Matches NCERT’s answer

1. B, O, D are collinear (they lie on diagonal BD), so ∠DOC and ∠BOC form a linear pair: ∠DOC = 180° − 125° = 55°.
2. In triangle ODC, the three angles add to 180°: ∠DCO = 180° − ∠DOC − ∠CDO = 180° − 55° − 70° = 55°.
3. Since ΔODC ~ ΔOBA, matching angles are equal; the angle at C in ΔODC (∠DCO) matches the angle at A in ΔOBA (∠OAB), so ∠OAB = ∠DCO = 55°.

Answer∠DOC = 55°, ∠DCO = 55°, ∠OAB = 55°.

Watch this explained “What the order of the letters commits you to”, 1:04 into AAA and AA: equal angles are enough on their own

Open NCERT p. 95One way to think about it

1. In triangles OAB and OCD: ∠AOB = ∠COD (vertically opposite angles at O).
2. Since AB || DC and AC is a transversal, ∠OAB = ∠OCD (alternate angles).
3. By the AA similarity criterion, ΔOAB ~ ΔOCD.
4. Corresponding sides of similar triangles are in one ratio: OA/OC = OB/OD (= AB/DC).

In shortHence OA/OC = OB/OD.

Watch this explained “Two parallels, two joins, two similar triangles”, 12:11 into AAA and AA: equal angles are enough on their own

Open NCERT p. 96One way to think about it

1. ∠1 = ∠PQS = ∠PQR (S lies on QR) and ∠2 = ∠PRS = ∠PRQ (S lies on QR), so the given ∠1 = ∠2 means ∠PQR = ∠PRQ in triangle PQR.
2. Since ∠PQR = ∠PRQ, the sides opposite them are equal: PR = PQ — triangle PQR is isosceles.
3. Substitute PR = PQ into the given QR/QS = QT/PR to get QR/QS = QT/QP, i.e. QT/QP = QR/QS.
4. Since T lies on ray QP (Q, P, T are collinear) and S lies on QR, ∠TQR and ∠PQS are the very same angle at Q.
5. In ΔTQR and ΔPQS, QT/QP = QR/QS (sides) and the included angle at Q is common — by the SAS similarity criterion, ΔTQR ~ ΔPQS.

In shortHence ΔPQS ~ ΔTQR.

Watch this explained “The rule, and the word it hangs on”, 1:53 into SAS: one angle plus the two sides that enclose it

Open NCERT p. 96One way to think about it

1. ∠R is common to both triangles: since S lies on PR and T lies on QR, ∠PRQ (in ΔRPQ) and ∠SRT (in ΔRTS) are the same angle.
2. It is given that ∠P (i.e. ∠RPQ) = ∠RTS.
3. Two angles of ΔRPQ equal two angles of ΔRTS, so by the AA similarity criterion, ΔRPQ ~ ΔRTS.

In shortHence ΔRPQ ~ ΔRTS.

Watch this explained “Two angles were enough all along”, 10:54 into AAA and AA: equal angles are enough on their own

Open NCERT p. 96One way to think about it

1. Since ΔABE ≅ ΔACD (given), matching sides are equal: AB = AC and AE = AD (matching A↔A, B↔C, E↔D).
2. From AD = AE and AB = AC, dividing equals by equals gives AD/AB = AE/AC.
3. ∠A is common to both ΔADE and ΔABC — it is the same angle ∠DAE = ∠BAC.
4. In ΔADE and ΔABC, AD/AB = AE/AC (sides) with the included angle at A common — by the SAS similarity criterion, ΔADE ~ ΔABC.

In shortHence ΔADE ~ ΔABC.

Watch this explained “The rule, and the word it hangs on”, 1:53 into SAS: one angle plus the two sides that enclose it

Open NCERT p. 96One way to think about it

(i) Δ AEP ~ Δ CDP

1. AD ⊥ BC and CE ⊥ AB, and P is the intersection of AD and CE.
2. ∠AEP = 90° (part of the right angle ∠AEC made by CE ⊥ AB) and ∠CDP = 90° (part of the right angle ∠CDA made by AD ⊥ BC).
3. ∠APE = ∠CPD (vertically opposite angles, since A, P, D are collinear and C, P, E are collinear).
4. By AA, ΔAEP ~ ΔCDP.

In shortΔAEP ~ ΔCDP (AA).

(ii) Δ ABD ~ Δ CBE

1. ∠ADB = 90° (AD ⊥ BC) and ∠CEB = 90° (CE ⊥ AB).
2. ∠ABD and ∠CBE are the same angle ∠B, since D lies on BC and E lies on AB.
3. By AA, ΔABD ~ ΔCBE.

In shortΔABD ~ ΔCBE (AA).

(iii) Δ AEP ~ Δ ADB

1. ∠AEP = 90° and ∠ADB = 90° (both shown in part (i) and (ii)).
2. ∠EAP and ∠DAB are the same angle at A, since E lies on AB (ray AE = ray AB) and P lies on AD (ray AP = ray AD).
3. By AA, ΔAEP ~ ΔADB.

In shortΔAEP ~ ΔADB (AA).

(iv) Δ PDC ~ Δ BEC

1. ∠PDC = 90° (AD ⊥ BC) and ∠BEC = 90° (CE ⊥ AB).
2. ∠PCD and ∠BCE are the same angle at C, since P lies on CE (ray CP = ray CE) and D lies on BC (ray CD = ray CB).
3. By AA, ΔPDC ~ ΔBEC.

In shortΔPDC ~ ΔBEC (AA).

Watch this explained “Two angles were enough all along”, 10:54 into AAA and AA: equal angles are enough on their own

Open NCERT p. 96One way to think about it

1. ∠BAE = ∠BAD (E lies on ray AD produced) and ∠BCF = ∠BCD (F lies on CD); these are opposite angles of the parallelogram ABCD, so they are equal: ∠BAE = ∠BCF.
2. Since AD || BC (opposite sides of the parallelogram) and BE is a transversal cutting them at E and B, ∠AEB = ∠EBC (alternate angles); and ∠EBC = ∠FBC since F lies on BE.
3. So ∠AEB = ∠FBC.
4. In ΔABE and ΔCFB, ∠BAE = ∠BCF and ∠AEB = ∠FBC — two angles equal, so by AA, ΔABE ~ ΔCFB.

In shortHence ΔABE ~ ΔCFB.

Watch this explained “Two angles were enough all along”, 10:54 into AAA and AA: equal angles are enough on their own

Open NCERT p. 96One way to think about it

(i) Δ ABC ~ Δ AMP

1. ∠ABC = 90° (given, right angle at B) and ∠AMP = 90° (given, right angle at M).
2. ∠A is common to both triangles — it is the same angle ∠BAC = ∠MAP, since B and P lie on the same line through A, and M lies on AC.
3. By AA, ΔABC ~ ΔAMP.

In shortΔABC ~ ΔAMP (AA).

(ii)

1. To prove: CA/PA = BC/MP.
2. Since ΔABC ~ ΔAMP with correspondence A↔A, B↔M, C↔P, the corresponding sides are in one ratio.
3. Side CA (C to A) matches side PA (P to A), and side BC matches side MP.
4. So CA/PA = BC/MP.

In shortCA/PA = BC/MP.

Watch this explained “Two angles were enough all along”, 10:54 into AAA and AA: equal angles are enough on their own

Open NCERT p. 96One way to think about it

(i)

1. To prove: CD/GH = AC/FG.
2. Since ΔABC ~ ΔFEG, ∠ACB = ∠FGE and AC/FG = BC/EG = AB/FE.
3. CD bisects ∠ACB and GH bisects ∠FGE (=∠EGF), so ∠ACD = ½∠ACB and ∠FGH = ½∠FGE; since ∠ACB = ∠FGE, ∠ACD = ∠FGH.
4. Also ∠A = ∠F (from ΔABC ~ ΔFEG).
5. In ΔACD and ΔFGH, ∠A = ∠F and ∠ACD = ∠FGH, so by AA, ΔACD ~ ΔFGH.
6. So the corresponding sides are in one ratio: CD/GH = AC/FG.

In shortCD/GH = AC/FG.

(ii) Δ DCB ~ Δ HGE

1. From the bisectors, ∠DCB = ½∠ACB and ∠HGE = ½∠FGE; since ∠ACB = ∠FGE (ΔABC ~ ΔFEG), ∠DCB = ∠HGE.
2. Also ∠B = ∠E (from ΔABC ~ ΔFEG).
3. In ΔDCB and ΔHGE, ∠DCB = ∠HGE and ∠B = ∠E, so by AA, ΔDCB ~ ΔHGE.

In shortΔDCB ~ ΔHGE (AA).

(iii) Δ DCA ~ Δ HGF

1. This is the same pair of triangles found while proving part (i), ΔACD ~ ΔFGH, simply written in the vertex order ΔDCA ~ ΔHGF.

In shortΔDCA ~ ΔHGF, as already shown while proving part (i).

Watch this explained “Two angles were enough all along”, 10:54 into AAA and AA: equal angles are enough on their own

Open NCERT p. 97One way to think about it

1. Since AB = AC, triangle ABC is isosceles, so its base angles are equal: ∠ABC = ∠ACB.
2. ∠ABD = ∠ABC (D lies on BC) and ∠ECF = ∠ACB (E lies on line CB produced, so ray CE = ray CB, and F lies on AC, so ray CF = ray CA).
3. So ∠ABD = ∠ECF.
4. ∠ADB = 90° (AD ⊥ BC) and ∠EFC = 90° (EF ⊥ AC).
5. In ΔABD and ΔECF, ∠ADB = ∠EFC (both 90°) and ∠ABD = ∠ECF — by AA, ΔABD ~ ΔECF.

In shortHence ΔABD ~ ΔECF.

Watch this explained “Two angles were enough all along”, 10:54 into AAA and AA: equal angles are enough on their own

Open NCERT p. 97One way to think about it

1. Given: AB/PQ = BC/QR = AD/PM.
2. AD, PM are medians, so D is the midpoint of BC and M is the midpoint of QR: BD = BC/2 and QM = QR/2.
3. So BC/QR = BD/QM, and therefore AB/PQ = BD/QM = AD/PM — the three sides of triangle ABD are in the same ratio to the three sides of triangle PQM.
4. By the SSS similarity criterion, ΔABD ~ ΔPQM.
5. So ∠ABD = ∠PQM (corresponding angles), i.e. ∠ABC = ∠PQR, since D lies on BC and M lies on QR.
6. In ΔABC and ΔPQR, AB/PQ = BC/QR (given) and the included angle ∠ABC = ∠PQR (just shown) — by the SAS similarity criterion, ΔABC ~ ΔPQR.

In shortHence ΔABC ~ ΔPQR.

Watch this explained “Halving both terms of a ratio”, 10:48 into SAS: one angle plus the two sides that enclose it

Open NCERT p. 97One way to think about it

1. ∠C is common to ΔCAD and ΔCBA (D lies on BC, so it's the same angle ∠ACD = ∠BCA).
2. It is given that ∠ADC = ∠BAC.
3. In ΔCAD and ΔCBA, ∠C is common and ∠ADC = ∠BAC — by AA, ΔCAD ~ ΔCBA (matching C↔C, A↔B, D↔A).
4. Corresponding sides of similar triangles are in one ratio: CA/CB = CD/CA.
5. Cross-multiplying, CA² = CB × CD.

In shortHence CA² = CB.CD.

Watch this explained “Two angles were enough all along”, 10:54 into AAA and AA: equal angles are enough on their own

Open NCERT p. 97One way to think about it

1. Given: AB/PQ = AC/PR = AD/PM.
2. Produce AD to X so that AD = DX, and join BX and CX. Since diagonals AX and BC of quadrilateral ABXC bisect each other at D, ABXC is a parallelogram, so CX = AB and BX = AC (its two pairs of opposite sides).
3. Similarly produce PM to Y so that PM = MY, and join QY and RY. PQYR is a parallelogram, so RY = PQ and QY = PR.
4. Since AX = 2AD and PY = 2PM: AC/PR = AD/PM gives AC/PR = AX/PY, and since CX = AB, RY = PQ, also AB/PQ = CX/RY. So in ΔACX and ΔPRY, AC/PR = CX/RY = AX/PY — by SSS, ΔACX ~ ΔPRY, giving ∠DAC = ∠XAC = ∠YPR = ∠MPR.
5. Likewise AB/PQ = AD/PM gives AB/PQ = AX/PY, and since BX = AC, QY = PR, also AB/PQ = BX/QY. So in ΔABX and ΔPQY, AB/PQ = BX/QY = AX/PY — by SSS, ΔABX ~ ΔPQY, giving ∠DAB = ∠XAB = ∠YPQ = ∠MPQ.
6. Adding the two equal angles, ∠DAC + ∠DAB = ∠MPR + ∠MPQ, i.e. ∠BAC = ∠QPR.
7. In ΔABC and ΔPQR, AB/PQ = AC/PR (given) and the included angle ∠BAC = ∠QPR (just shown) — by SAS, ΔABC ~ ΔPQR.

In shortHence ΔABC ~ ΔPQR.

Watch this explained “The criterion, and the older rule with the same letters”, 5:43 into SSS: sides in proportion drag the angles into agreement

Open NCERT p. 97Matches NCERT’s answer

1. At the same moment, the sun's rays make the same angle with the ground everywhere, so the pole-with-its-shadow triangle and the tower-with-its-shadow triangle are similar (AA: both have a right angle where the object meets the ground, and the equal sun-angle).
2. So height/shadow is the same for both: pole height/pole shadow = tower height/tower shadow.
3. 6/4 = h/28, so h = 6 × 28 ÷ 4 = 42.

AnswerThe tower is 42 m tall.

Watch this explained “A tower, measured with a stick”, 14:56 into AAA and AA: equal angles are enough on their own

Open NCERT p. 97One way to think about it

1. Since ΔABC ~ ΔPQR, AB/PQ = BC/QR = CA/RP, and ∠ABC = ∠PQR (corresponding angles).
2. AD, PM are medians, so D, M are midpoints of BC, QR: BD = BC/2 and QM = QR/2.
3. So BD/QM = BC/QR = AB/PQ.
4. In ΔABD and ΔPQM, AB/PQ = BD/QM (shown) and the included angle ∠ABD = ∠PQM, since ∠ABD = ∠ABC, ∠PQM = ∠PQR, and ∠ABC = ∠PQR — by SAS, ΔABD ~ ΔPQM.
5. Corresponding sides of similar triangles are in one ratio, so AB/PQ = AD/PM.

In shortHence AB/PQ = AD/PM.

Watch this explained “Halving both terms of a ratio”, 10:48 into SAS: one angle plus the two sides that enclose it