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NCERT New Syllabus Mathematics
Class: 10
Exercise 6.3
Topic: Triangles
Understanding Triangles: A Key Concept in Class 10 Mathematics
Triangles form one of the foundational blocks in geometry, and in the Class 10 NCERT syllabus, this topic holds significant importance. From learning the basic properties of triangles to diving deeper into congruence, similarity, and the Pythagoras theorem, the chapter on triangles equips students with essential tools to solve geometric problems.
In this blog, we will systematically explore each concept, providing clear explanations and practical solutions to the exercises. By the end of this discussion, you'll have a solid understanding of triangles, enabling you to confidently approach simple and complex problems.
Let's begin our journey into the world of triangles and uncover the logic that shapes this fascinating topic!
EXERCISE 6.3
Q1. State which pairs of triangles in the following Fig, are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :
Solution:
(i)
a) < A = < P = 600
b) < B = < Q = 800
c) < C = < R = 400
2) So by AAA similarity rule, ∆ ABC ~ ∆ PQR.
(ii)
Note: For ∆ ABC and ∆ QRP, points A ↔ Q, B ↔ R, C ↔ P. That is
i) point A is associated with poin Q,
ii) point B is associated with poin R,
iii) point B is associated with poin R,
a) (AB)/(QR) = 2/4 = 1/2 -------- equation 1b) (BC)/(RP) = 2.5/5 = 1/2 -------- equation 2c) (AC)/(QP) = 3/6 = 1/2 -------- equation 3
2) So from equations 1, 2, and 3, the corresponding ratios are proportional,
so by SSS test, ∆ ABC ~ ∆ QRP.
(iii)
1) In ∆ LMP and ∆ FED,
Note: For ∆ LMP and ∆ FED, points L ↔ F, M ↔ E, and P ↔ D. That is
i) point L is associated with poin F,
ii) point M is associated with poin E,
iii) point P is associated with poin D,
a) (LM)/(FE) = 2.7/5 = 27/50 -------- equation 1b) (MP)/(ED) = 2/4 = 1/2 -------- equation 2c) (LP)/(FD) = 3/6 = 1/2 -------- equation 3
2) So from equations 1, 2, 3, (LM)/(FE) ≠ (MP)/(ED) = (LP)/FD), so
∆ ABC ≁ ∆ QRP.
(iv)
1) In ∆ LMN and ∆ RQP,
Note: For ∆ LMN and ∆ RQP, points L ↔ R, M ↔ Q, and N ↔ P. That is
i) point L is associated with poin R,
ii) point M is associated with poin Q,
iii) point N is associated with poin P,
a) (LM)/(RQ) = 5/10 = 1/2 -------- equation 1b) < NML = < PQR = 700-------- equation 2c) (MN)/(QP) = 2.5/5 = 1/2 -------- equation 3
2) So from equations 1, 2, 3, (LM)/(RQ) = (MN)/(QP) = 1/2,
so by SAS property, ∆ LMN ~ ∆ RQP.
(v)
1) In ∆ ABC and ∆ DFE,
Note: For ∆ ABC and ∆ DFE, points A ↔ D, B ↔ F, and C ↔ E. That is
i) point A is associated with poin D,
ii) point B is associated with poin F,
iii) point C is associated with poin E,
a) (AB)/(DF) = 2.5/5 = 1/2 -------- equation 1b) < ABC ≠ < DFE -------- equation 2c) (BC)/(FE) = 3/6 = 1/2 -------- equation 3
2) So from equations 1, 2, 3, (AB)/(DF) = (BC)/(FE) = 1/2, but < ABC ≠ < DFE so,
(vi)∆ ABC ≁ ∆ DFE.
1) In ∆ DEF and ∆ PQR,
Note: For ∆ DEF and ∆ PQR, points D ↔ P, E ↔ Q, and F ↔ R. That is
i) point D is associated with poin P,
ii) point E is associated with poin Q,
iii) point F is associated with poin R,
a) in ∆ DEF,
< D + < E + < F = 18070 + 80 + < F = 180150 + < F = 180< F = 180 – 150< F = 30< DFE = 30< DFE = < PRQ = 30 -------- equation 1
b) in ∆ PQR,< P + < Q + < R = 180
< P + 80 + 30 = 180
< P + 110 = 180
< P = 180 – 110
< P = 70
< QPR = 70
< QPR = < EDF = 70 -------- equation 2
c) < DEF = < PQR = 80 -------- equation 3
2) From equations 1, 2, and 3, by AAA similarity rule, ∆ DEF ~ ∆ PQR.
Q2. In the following Fig., ∆ ODC ~ ∆ OBA, < BOC = 125° and < CDO = 70°.
Find < DOC, < DCO and < OAB.
Solution:
1) In the above figure,a) < DOC and < COB are angles in linear pairs, so
< DOC + < COB = 180
< DOC + 125 = 180
< DOC = 180 – 125
< DOC = 55 -------- equation 1
b) So < DOC = 55
2) In ∆ PQR,< DOC + < DCO + < CDO = 180
< 55 + < DCO + 70 = 180
< DCO + 125 = 180
< DCO = 180 – 125
< DCO = 55 -------- equation 2
So < DCO = 55
3) It's given that, ∆ ODC ~ ∆ OBA,< OAB = < OCD -------- equation 3
4) From equations 2 and 3, we have, < OAB = 55
5) Therefore, < DOC = 55°, < DCO = 55°, and < OAB = 55°.
Q3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at point O. Using a similarity criterion for two triangles, show that (OA)/(OB) = (OC)/(OD).
1) In the above figure,
a) < DOC and < COB are angles in linear pairs, so
< DOC + < COB = 180
< DOC + 125 = 180
< DOC = 180 – 125
< DOC = 55 -------- equation 1
b) So < DOC = 55
2) In ∆ PQR,
< DOC + < DCO + < CDO = 180
< 55 + < DCO + 70 = 180
< DCO + 125 = 180
< DCO = 180 – 125
< DCO = 55 -------- equation 2
So < DCO = 55
3) It's given that, ∆ ODC ~ ∆ OBA,
< OAB = < OCD -------- equation 3
4) From equations 2 and 3, we have,
< OAB = 55
5) Therefore, < DOC = 55°, < DCO = 55°, and < OAB = 55°.
Q3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each
other at point O. Using a similarity criterion for two triangles, show that
(OA)/(OB) = (OC)/(OD).
Solution:
1) In trapezium ABCD, AB ‖ CD, and diagonals AC and BD intersect at O.2) In ∆ AOB and ∆ COD,a) < AOB = < COD (vertically opposite angles are equal)
b) < ABO = < CDO (alternate interior angles are equal)c) < BAO = < DCO (alternate interior angles are equal)3) So, by AAA similarity test, ∆ AOB ~ ∆ COD3) So, as corresponding sides are proportional, we have,(OA)/(OB) = (OC)/(OD). Hence proved.
Q4. In the following Fig., (QR)/(QS) = (QT)/(PR) and < 1 = < 2. Show that∆ PQS ~ ∆ TQR.
1) In trapezium ABCD, AB ‖ CD, and diagonals AC and BD intersect at O.
2) In ∆ AOB and ∆ COD,
a) < AOB = < COD (vertically opposite angles are equal)
b) < ABO = < CDO (alternate interior angles are equal)
c) < BAO = < DCO (alternate interior angles are equal)
3) So, by AAA similarity test, ∆ AOB ~ ∆ COD
3) So, as corresponding sides are proportional, we have,
(OA)/(OB) = (OC)/(OD). Hence proved.
Q4. In the following Fig., (QR)/(QS) = (QT)/(PR) and < 1 = < 2. Show that
∆ PQS ~ ∆ TQR.
Solution:
1) In ∆ PQR,< 1 = < 2, so
(QP) = (PR) -------- equation 1
2) It is given that,(QR)/(QS) = (QT)/(PR) -------- equation 2
3) From equations 1 and 2, we have,(QR)/(QS) = (QT)/(QP) -------- equation 3
4) In ∆ PQS and ∆ TQR,< PQS = < TQR (same angle) -------- equation 4
5) From equations 3 and 4 and by SAS property of similarity we have,∆ PQS ~ ∆ TQR. Hence proved.
Q5. S and T are points on sides PR and QR of ∆ PQR such that < P = < RTS. Show that ∆ RPQ ~ ∆ RTS.
1) In ∆ PQR,
< 1 = < 2, so
(QP) = (PR) -------- equation 1
2) It is given that,
(QR)/(QS) = (QT)/(PR) -------- equation 2
3) From equations 1 and 2, we have,
(QR)/(QS) = (QT)/(QP) -------- equation 3
4) In ∆ PQS and ∆ TQR,
< PQS = < TQR (same angle) -------- equation 4
5) From equations 3 and 4 and by SAS property of similarity we have,
∆ PQS ~ ∆ TQR. Hence proved.
Q5. S and T are points on sides PR and QR of ∆ PQR such that < P = < RTS. Show that ∆ RPQ ~ ∆ RTS.
Solution:
1) In ∆ PQR, and ∆ TSR,
Note: For ∆ PQR and ∆ TSR, points P ↔ T, Q ↔ S, and R ↔ R. That is
i) point P is associated with poin T,
ii) point Q is associated with poin S,
iii) point R is associated with point R,
a) < QPR = < STR (given)
b) < PRQ = < TRS (same angles)
2) By AA similarity test, ∆ PQR ~ ∆ TSR. Hence proved.
Q6. In the following fig., if ∆ ABE ≅ ∆ ACD, show that ∆ ADE ~ ∆ ABC.
1) In ∆ PQR, and ∆ TSR,
Note: For ∆ PQR and ∆ TSR, points P ↔ T, Q ↔ S, and R ↔ R. That is
i) point P is associated with poin T,
ii) point Q is associated with poin S,iii) point R is associated with point R,
a) < QPR = < STR (given)
b) < PRQ = < TRS (same angles)
2) By AA similarity test, ∆ PQR ~ ∆ TSR. Hence proved.
Q6. In the following fig., if ∆ ABE ≅ ∆ ACD, show that ∆ ADE ~ ∆ ABC.
Solution:
1) It is given that ∆ ABE ≅ ∆ ACD, corresponding sides of congruent triangles are congruent, so,a) AD = AE ------- equation 1
b) AB = AC ------- equation 2
2) In ∆ DAE and ∆ BAC, and from equations 1 and 2, we have,a) (AD)/(AB) = (AE)/(AC)
b) < DAE = <BAC (same angle)
3) ∆ DAE ~ ∆ BAC. Hence proved.
Q7. In the following fig., altitudes AD and CE of ∆ ABC intersect each other at point P. Show that:
1) It is given that ∆ ABE ≅ ∆ ACD, corresponding sides of congruent triangles are
congruent, so,
a) AD = AE ------- equation 1
b) AB = AC ------- equation 22) In ∆ DAE and ∆ BAC, and from equations 1 and 2, we have,
a) (AD)/(AB) = (AE)/(AC)
b) < DAE = <BAC (same angle)3) ∆ DAE ~ ∆ BAC. Hence proved.
Q7. In the following fig., altitudes AD and CE of ∆ ABC intersect each other at point P. Show that:
Solution:
(i) ∆ AEP ~ ∆ CDP
1) In ∆ AEP and ∆ CDP
Note: For ∆ AEP and ∆ CDP, points A ↔ C, E ↔ D, and P ↔ P. That is
i) point A is associated with poin C,
ii) point E is associated with poin D,
iii) point P is associated with poin P,
2) In ∆ AEP and ∆ CDP
a) < AEP = < CDP = 90 (given) -------- equation 1
b) < APE = < CPD (Vertically oposite angles) -------- equation 2
3) From equations 1 and 2, by AA similarity test, we have,
a) ∆ AEP ~ ∆ CDP, hence proved.
(ii) ∆ ABD ~ ∆ CBE
1) In ∆ ABD and ∆ CBE
Note: For ∆ ABD and ∆ CBE, points A ↔ C, B ↔ B, and D ↔ E. That is
i) point A is associated with poin C,
ii) point B is associated with poin B,
iii) point D is associated with poin E,
2) In ∆ ABD and ∆ CBE
a) < ADB = < CEB = 90 (given) -------- equation 1
b) < ABD = < CBE (common angles) -------- equation 2
3) From equations 1 and 2, by AA similarity test, we have,
a) ∆ ABD ~ ∆ CBE, hence proved.
(iii) ∆ AEP ~ ∆ ADB
1) In ∆ AEP and ∆ ADB
Note: For ∆ AEP and ∆ ADB, points A ↔ A, E ↔ D, and P ↔ B. That is
i) point A is associated with poin A,
ii) point E is associated with poin D,
iii) point P is associated with poin B,
2) In ∆ AEP and ∆ ADB
a) < AEP = < ADB = 90 (given) -------- equation 1
b) < PAE = < BAD (common angles) -------- equation 2
3) From equations 1 and 2, by AA similarity test, we have,
a) ∆ AEP ~ ∆ ADB, hence proved.
(iv) ∆ PDC ~ ∆ BEC
1) In ∆ PDC and ∆ BEC
Note: For ∆ PDC and ∆ BEC, points P ↔ B, D ↔ E, and C ↔ C. That is
i) point P is associated with poin B,
ii) point D is associated with poin E,
iii) point C is associated with poin C,
2) In ∆ PDC and ∆ BEC
a) < PDC = < BEC = 90 (given) -------- equation 1
b) < PCD = < BCE (common angles) -------- equation 2
3) From equations 1 and 2, by AA similarity test, we have,
a) ∆ PDC ~ ∆ BCE, hence proved.
Q8. E is a point on the side AD produced of a parallelogram ABCD and BE
intersect CD at F. Show that ∆ ABE ~ ∆ CFB.
Solution:
1) In ∆ ABE and ∆ CFBNote: For ∆ ABE and ∆ CFB, points A ↔ C, B ↔ F, and E ↔ B. That is
i) point A is associated with poin C,
ii) point B is associated with poin F,
iii) point E is associated with poin B,
2) In ∆ ABE and ∆ CFBa) < BAE = < FCB (Opposite angles of a parallelogram) -------- equation 1
b) < AEB = < CBF (Alternate interior angles as AE || BC) -------- equation 2
3) From equations 1 and 2, by AA similarity test, we have,a) ∆ ABE ~ ∆ CFB, hence proved.
Q9. In the following fig., ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: (i) ∆ ABC ~ ∆ AMP (ii) (CA)/(PA) = (BC)/(MP)
1) In ∆ ABE and ∆ CFB
Note: For ∆ ABE and ∆ CFB, points A ↔ C, B ↔ F, and E ↔ B. That is
i) point A is associated with poin C,
ii) point B is associated with poin F,
iii) point E is associated with poin B,
2) In ∆ ABE and ∆ CFB
a) < BAE = < FCB (Opposite angles of a parallelogram) -------- equation 1
b) < AEB = < CBF (Alternate interior angles as AE || BC) -------- equation 2
3) From equations 1 and 2, by AA similarity test, we have,
a) ∆ ABE ~ ∆ CFB, hence proved.
Q9. In the following fig., ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i) ∆ ABC ~ ∆ AMP (ii) (CA)/(PA) = (BC)/(MP)
Solution:
(i) ∆ ABC ~ ∆ AMP
1) In ∆ ABC and ∆ AMPNote: For ∆ ABC and ∆ AMP, points A ↔ A, B ↔ M, and C ↔ P. That is
i) point A is associated with poin A,
ii) point B is associated with poin M,
iii) point C is associated with poin P,
2) In ∆ ABC and ∆ AMPa) < ABC = < AMP = 90 -------- equation 1
b) < BAC = < MAP (Common angle) -------- equation 2
3) From equations 1 and 2, by AA similarity test, we have,a) ∆ ABC ~ ∆ AMP, hence proved.
(ii) (CA)/(PA) = (BC)/(MP)
1) In ∆ ABC and ∆ AMPa) ∆ ABC ~ ∆ AMP (we proved this)
b) (CA)/(PA) = (BC)/(MP) (By basic propotinality theorem)
Q10. CD and GH are respectively the bisectors of < ACB and < EGF such that D and H lie on sides AB and FE of ∆ ABC and ∆ EFG respectively. If ∆ ABC ~ ∆ FEG, show that: (i) (CD)/(GH) = (AC)/(FG) (ii) ∆ DCB ~ ∆ HGE (iii) ∆ DCA ~ ∆ HGF
1) In ∆ ABC and ∆ AMP
Note: For ∆ ABC and ∆ AMP, points A ↔ A, B ↔ M, and C ↔ P. That is
i) point A is associated with poin A,
ii) point B is associated with poin M,
iii) point C is associated with poin P,
2) In ∆ ABC and ∆ AMP
a) < ABC = < AMP = 90 -------- equation 1
b) < BAC = < MAP (Common angle) -------- equation 2
3) From equations 1 and 2, by AA similarity test, we have,
a) ∆ ABC ~ ∆ AMP, hence proved.
(ii) (CA)/(PA) = (BC)/(MP)
1) In ∆ ABC and ∆ AMP
a) ∆ ABC ~ ∆ AMP (we proved this)
b) (CA)/(PA) = (BC)/(MP) (By basic propotinality theorem)
Q10. CD and GH are respectively the bisectors of < ACB and < EGF
such that D and H lie on sides AB and FE of ∆ ABC and ∆ EFG
respectively. If ∆ ABC ~ ∆ FEG, show that:
(i) (CD)/(GH) = (AC)/(FG)
(ii) ∆ DCB ~ ∆ HGE
(iii) ∆ DCA ~ ∆ HGF
Solution:
(i) (CD)/(GH) = (AC)/(FG)
1) ∆ ABC ~ ∆ FEG (given), so we have,a) < ACB = < FGE --------- equation 1
b) CD is bisector of < ACB and GH is the bisector of < FGE
c) (< ACB)/2 = (< FGE)/2 --------- equation 2
2) From equations 1 and 2, and in ∆ DCA and ∆ HGF we have
1) ∆ ABC ~ ∆ FEG (given), so we have,
a) < ACB = < FGE --------- equation 1
b) CD is bisector of < ACB and GH is the bisector of < FGE
c) (< ACB)/2 = (< FGE)/2 --------- equation 2
2) From equations 1 and 2, and in ∆ DCA and ∆ HGF we have
< DCA = < HGF --------- equation 3
< CAD = < GFH --------- equation 4
3) From equations 3 and 4, and the AA similarity test, we have,
∆ DCA ~ ∆ HGF --------- equation 5
4) So, (CD)/(GH) = (AC)/(FG) hence proved.
(ii) ∆ DCB ~ ∆ HGE
1) In ∆ DCB and ∆ HGE (CD is bisector of < ACB and GH is the bisector of < FGE)
a) < DCB = < HGE --------- equation 1
b) < DBC = < HEG (Common angle) --------- equation 2
2) From equations 1 and 2, we have,
∆ DCB ~ ∆ HGE, hence proved.
(iii) ∆ DCA ~ ∆ HGF
As previously demonstrated in Example (i), the proof follows directly.
(Refer to Equation 5 in Example (i) for detailed steps).
Q11. In the following fig., E is a point on side CB produced of an
isosceles triangle ABC with AB = AC. If AD ⟂ BC and EF ⟂ AC,
prove that ∆ ABD ~ ∆ ECF.
Solution:
1) ∆ ABC is an isosceles triangle with AB = AC, so we have,a) < ABC = < ACB --------- equation 1
2) In ∆ ABD and ∆ ECF and from equation 1a) < ABD = < ECF (same angle) --------- equation 2
c) < ADB = < EFC (each angle is 900) --------- equation 3
3) From equations 2 and 3, and the AA similarity test, we have,a) ∆ ABD ~ ∆ ECF, hence proved.
Q12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆ PQR (see the following fig.). Show that ∆ ABC ~ ∆ PQR.
1) ∆ ABC is an isosceles triangle with AB = AC, so we have,
a) < ABC = < ACB --------- equation 1
2) In ∆ ABD and ∆ ECF and from equation 1
a) < ABD = < ECF (same angle) --------- equation 2
c) < ADB = < EFC (each angle is 900) --------- equation 3
3) From equations 2 and 3, and the AA similarity test, we have,
a) ∆ ABD ~ ∆ ECF, hence proved.
Q12. Sides AB and BC and median AD of a triangle ABC are respectively
proportional to sides PQ and QR and median PM of ∆ PQR
(see the following fig.). Show that ∆ ABC ~ ∆ PQR.
Solution:
1) It is given that:
a) (AB)/(PQ) = (BC)/(QR) = (AD)/(PM) ---------- equation 1.
b) (AB)/(PQ) = [(BC)/2]/[(QR)/2] = (AD)/(PM)
c) (AB)/(PQ) = (BD)/(QM) = (AD)/(PM) ---------- equation 2.
2) From equation 2 and the basic proportionality theorem, we have,
∆ ABD ~ ∆ PQM ---------- equation 3.
3) In ∆ ABC and ∆ PQR,
a) (AB)/(PQ) = (BC)/(QR) ---------- equation 4.
4) As ∆ ABD ~ ∆ PQM, we have,
a) < ABC = < PQR ---------- equation 5.
5) From equations 4 and 5 and by SAS test, we have,
∆ ABC ~ ∆ PQR, hence proved.
Q13. D is a point on the side BC of a triangle ABC such that < ADC = < BAC. Show that CA2 = CB.CD.
Solution:
1) In ∆ ADC and ∆ BAC,
Note: For ∆ ADC and ∆ BAC, points A ↔ B, D ↔ A, and C ↔ C. That is
i) point A is associated with poin B,
ii) point D is associated with poin A,
iii) point C is associated with point C,
a) < ADC = < BAC (given)
b) < ACD = < BCA (common angles)
2) By AA similarity test, ∆ ADC ~ ∆ BAC.3) By the basic proportionality theorem, we have,(CA)/(CB) = (CD)/(CA)
(CA) x (CA) = (CD) x (CB)
(CA)2 = (CB) x (CD), hence proved.
1) In ∆ ADC and ∆ BAC,
Note: For ∆ ADC and ∆ BAC, points A ↔ B, D ↔ A, and C ↔ C. That is
i) point A is associated with poin B,
ii) point D is associated with poin A,iii) point C is associated with point C,
a) < ADC = < BAC (given)
b) < ACD = < BCA (common angles)
2) By AA similarity test, ∆ ADC ~ ∆ BAC.
3) By the basic proportionality theorem, we have,
(CA)/(CB) = (CD)/(CA)
(CA) x (CA) = (CD) x (CB)
(CA)2 = (CB) x (CD), hence proved.
Q14. Sides AB and AC and median AD of a triangle ABC are respectively
proportional to sides PQ and PR and median PM of another triangle PQR.
Show that ∆ ABC ~ ∆ PQR.
Solution:
1) Extend AD to E such that AD = ED.2) Extend PM to N such that PM = NM.3) In ∆ ABC and ∆ PQR,(AB/PQ) = (AC/PR) = (AD/PM) ---------- equation 1.
4) In ∆ ABD and ∆ ECD,a) (AD) = (ED) (construction)
b) < ADB = < EDC (vertically opposite angles)
c) (BD) = (CD) (AD is median of ∆ ABC)
5) So by SAS test ∆ ABD ≅ ∆ ECD.6) So here, (AB) = (EC) ---------- equation 27) In ∆ PQM and ∆ NRM,a) (PM) = (NM) (construction)
b) < PMQ = < NMR (vertically opposite angles)
c) (QM) = (RM) (PM is median of ∆ PQR)
8) So by SAS test ∆ PQM ≅ ∆ NRM.9) So here, (PQ) = (NR) ---------- equation 310) From equations 1, 2, and 3, we have,a) (AB/PQ) = (AC/PR) = (AD/PM)
b) (EC/NR) = (AC/PR) = 2(AD)/2(PM)
c) (EC/NR) = (AC/PR) = (AE)/(PN) ------- (As 2AD = AE, and 2PM = PN)
11) So, by SSS similarity test ∆ ACE ~ ∆ PRN,12) So we have,a) < CAD = < RPM,
b) [2(< CAD)] = [2(< RPM)] (as AD and PM are medians of ∆ ABC and ∆ PQR)
c) < CAB = < RPQ ---------- equation 4
13) From equations 1 and 4, we have,a) (AB/PQ) = (AC/PR) and < CAB = < RPQ by SAS similarity test, we have,
b) ∆ ABC ~ ∆ PQR hence proved.
Q15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time, a tower casts a shadow 28 m long. Find the height of the tower.
1) Extend AD to E such that AD = ED.
2) Extend PM to N such that PM = NM.
3) In ∆ ABC and ∆ PQR,
(AB/PQ) = (AC/PR) = (AD/PM) ---------- equation 1.
4) In ∆ ABD and ∆ ECD,
a) (AD) = (ED) (construction)
b) < ADB = < EDC (vertically opposite angles)
c) (BD) = (CD) (AD is median of ∆ ABC)
5) So by SAS test ∆ ABD ≅ ∆ ECD.
6) So here, (AB) = (EC) ---------- equation 2
7) In ∆ PQM and ∆ NRM,
a) (PM) = (NM) (construction)
b) < PMQ = < NMR (vertically opposite angles)
c) (QM) = (RM) (PM is median of ∆ PQR)
8) So by SAS test ∆ PQM ≅ ∆ NRM.
9) So here, (PQ) = (NR) ---------- equation 3
10) From equations 1, 2, and 3, we have,
a) (AB/PQ) = (AC/PR) = (AD/PM)
b) (EC/NR) = (AC/PR) = 2(AD)/2(PM)
c) (EC/NR) = (AC/PR) = (AE)/(PN) ------- (As 2AD = AE, and 2PM = PN)
11) So, by SSS similarity test ∆ ACE ~ ∆ PRN,
12) So we have,
a) < CAD = < RPM,
b) [2(< CAD)] = [2(< RPM)] (as AD and PM are medians of ∆ ABC and ∆ PQR)
c) < CAB = < RPQ ---------- equation 4
13) From equations 1 and 4, we have,
a) (AB/PQ) = (AC/PR) and < CAB = < RPQ by SAS similarity test, we have,
b) ∆ ABC ~ ∆ PQR hence proved.
Q15. A vertical pole of length 6 m casts a shadow 4 m long on the ground
and at the same time, a tower casts a shadow 28 m long.
Find the height of the tower.
Solution:
1) In ∆ ABC and ∆ PQR,a) < BCA = < QRP (angle of elevation)
b) < ABC = < PQR = 90 (angle between pole or tower and their shadows)
c) ∆ ABC ~ ∆ PQR by AA similarity test,
2) So by basic proportionality theorem.(AB)/(PQ) = (BC)/(QR)
(6)/(PQ) = (4)/(28)
(6)/(PQ) = (1)/(7)
6 x 7 = (PQ)
(PQ) = 42 m
3) So, the height of the tower is 42 m.
Q16. If AD and PM are medians of triangles ABC and PQR, respectively where∆ ABC ~ ∆ PQR, prove that AB/PQ = AD/PM.
1) In ∆ ABC and ∆ PQR,
a) < BCA = < QRP (angle of elevation)
b) < ABC = < PQR = 90 (angle between pole or tower and their shadows)
c) ∆ ABC ~ ∆ PQR by AA similarity test,
2) So by basic proportionality theorem.
(AB)/(PQ) = (BC)/(QR)
(6)/(PQ) = (4)/(28)
(6)/(PQ) = (1)/(7)
6 x 7 = (PQ)
(PQ) = 42 m
3) So, the height of the tower is 42 m.
Q16. If AD and PM are medians of triangles ABC and PQR, respectively where
∆ ABC ~ ∆ PQR, prove that AB/PQ = AD/PM.Solution:
1) As ∆ ABC ~ ∆ PQR,a) (AB)/(PQ) = (AC)/(PR) = (BC)/(QR)
b) (AB)/(PQ) = (AC)/(PR) = [(BC)/2]/[(QR)/2]
2) As D and M are midpoints of BC and QR respectively, we have,a) (AB)/(PQ) = (AC)/(PR) = (BD)/(QM) ---------- equation 1
3) As AD and PM are the medians of ∆ ABC and ∆ PQR,a) < BAD = < QPM ---------- equation 2
4) In ∆ ABD and ∆ PQM, and from equations 1 and 2, we have,a) (AB)/(PQ) = (AD)/(PM) = (BD)/(QM)
b) < BAD = < QPM
5) By the SAS similarity test, we have ∆ ABD ~ ∆ PQM6) So by basic proportionality theorem.a) (AB)/(PQ) = (AD)/(PM), hence proved.
1) As ∆ ABC ~ ∆ PQR,
a) (AB)/(PQ) = (AC)/(PR) = (BC)/(QR)
b) (AB)/(PQ) = (AC)/(PR) = [(BC)/2]/[(QR)/2]
2) As D and M are midpoints of BC and QR respectively, we have,
a) (AB)/(PQ) = (AC)/(PR) = (BD)/(QM) ---------- equation 13) As AD and PM are the medians of ∆ ABC and ∆ PQR,
a) (AB)/(PQ) = (AD)/(PM), hence proved.
Conclusion
In conclusion, mastering the topic of triangles is essential for building a strong foundation in geometry. By understanding concepts like congruence, similarity, and the Pythagoras theorem, you can easily solve a wide range of problems. As you continue practicing, these geometric principles will become second nature, helping you excel in exams and real-life applications.
Keep exploring, stay curious, and remember—the beauty of mathematics lies in its logical simplicity!
Stay tuned for more insights and solutions to other important topics from the Class 10 NCERT syllabus.
#Class10Maths #NCERTSolutions #Triangles #GeometryBasics #MathForStudents #MathMadeEasy #PythagorasTheorem #MathBlog #StudentLife #LearningIsFun #MathConcepts
In conclusion, mastering the topic of triangles is essential for building a strong foundation in geometry. By understanding concepts like congruence, similarity, and the Pythagoras theorem, you can easily solve a wide range of problems. As you continue practicing, these geometric principles will become second nature, helping you excel in exams and real-life applications.
Keep exploring, stay curious, and remember—the beauty of mathematics lies in its logical simplicity!
Stay tuned for more insights and solutions to other important topics from the Class 10 NCERT syllabus.
#Class10Maths #NCERTSolutions #Triangles #GeometryBasics #MathForStudents #MathMadeEasy #PythagorasTheorem #MathBlog #StudentLife #LearningIsFun #MathConcepts
