Ex 9.4, 8 (Optional) - In Fig, ABC is a right triangle right angled at

Ex 9.4, 8 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 2
Ex 9.4, 8 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 3 Ex 9.4, 8 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 4 Ex 9.4, 8 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 5 Ex 9.4, 8 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 6 Ex 9.4, 8 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 7 Ex 9.4, 8 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 8 Ex 9.4, 8 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 9 Ex 9.4, 8 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 10 Ex 9.4, 8 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 11 Ex 9.4, 8 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 12 Ex 9.4, 8 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 13 Ex 9.4, 8 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 14 Ex 9.4, 8 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 15

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Question 8 In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that: ∆ MBC ≅ ∆ ABD Since all angles of square is 90° ∴ ∠ABM = 90° ∠DBC = 90° Now, ∠ABM = ∠DBC ∠ ABM + ∠ABC = ∠OBC + ∠ABC ∠MBC = ∠ABD In ∆MBC and ∆ABD, MB = AB ∠ MBC = ∠ABD BC = BD ∆MBC ≅ ∆ABD Hence proved (Sides of same square) (Proved above) (Sides of same square) (SAS Congruence rule ) Question 8 In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that: (ii) ar (BYXD) = 2 ar (MBC) Here, we first prove that BYXD is a parallelogram Given that AX ⊥ DE ∴ ∠YXE = 90° Also, ∠BDE = 90° (Angle of a square is 90°) For lines BD and YX, with transversal DX ∠YXE = ∠BDE So, corresponding angles are equal Thus, lines must be parallel ∴ BD ∥ YX Now, In quadrilateral BYXD, BD ∥ YX BY ∥ DX Since both pair of opposite sides are parallel ∴ BYXD is Parallelogram (As opposite sides of square are parallel) Now, Parallelogram BYXD and ∆ABD lie on the same base BD and between same parallel lines BD & YX ∴ ar (BYXD) = 2ar (ABD) From part (i), we proved that ∆MBC ≅ ∆ABD Thus, ar (MBC) = ar (ABD) From (1) and (2) ar (BYXD) = 2 ar (MBC) Hence proved Question 8 In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that: (iii) ar (BYXD) = ar (ABMN) Parallelogram ABMN and ∆MBC lie on the same base MB and between same parallel lines BM & NC ∴ ar (BYXD) = 2ar (MBD) From (2) and (3) ar (BYXD) = ar (ABMN) Question 8 In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that: (iv) ∆ FCB ≅ ∆ ACE Since all angles of square is 90° ∴ ∠ FCA = 90° ∠ ECB = 90° Now, ∠FCA = ∠ECB ∠ FCA+ ∠ACB = ∠ECB + ∠ACB ∠FCB = ∠ACE In ∆FCB and ∆ACE, FC = AC ∠FCB = ∠ACE CB = CE ∆ FCB ≅ ∆ACE Hence proved In ∆FCB and ∆ACE, FC = AC ∠FCB = ∠ACE CB = CE ∆ FCB ≅ ∆ACE Hence proved (Sides of same square) (Proved above) (Sides of same square) (SAS Congruence rule ) Question 8 In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that: (v) ar (CYXE) = 2 ar (FCB) Here, we first prove that CYXE is a parallelogram Given that AX ⊥ DE ∴ ∠YXD = 90° Also, ∠CED = 90° (Angle of a square is 90°) For lines CE and YX, with transversal EX ∠YXD = ∠CEX So, corresponding angles are equal Thus, lines must be parallel ∴ CE ∥ YX Now, In quadrilateral CYXE, CE ∥ YX CY ∥ EX Since both pair of opposite sides are parallel ∴ CYXE is Parallelogram Now, Parallelogram CYXE and ∆ACE lie on the same base CE and between same parallel lines CE& AX ∴ ar (CYXE) = 2ar (ACE) From part (iv), we proved that ∆ FCB ≅ ∆ACE Thus, ar (FCB) = ar (ACE) Thus, ar (CYXE) = 2 ar (FCB) Hence proved Question 8 In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that: (vi) ar (CYXE) = ar (ACFG) Parallelogram ACFG and ∆FCB lie on the same base FC and between same parallel lines FC & BG ∴ ar (ACFG) = 2ar (MBD) From (4) and (5) ar (CYXE) = ar (ACFG) Question 8 Show that (vii) ar (BCED) = ar (ABMN) + ar (ACFG) Note : Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X. From Proof of part (iii) & part (vi) ar (BYXD) = ar (ABMN) ar (CYXE) = ar (ACFG) Adding both equations ar (BYXD) + ar (CYXE) = ar (ABMN) + ar (ACFG) ar (BCED) = ar (ABMN) + ar (ACFG) Hence proved Note: The final result is also called Pythagoras Theorem For a right angled triangle Δ ABC with sides a, b, c Let’s draw square of side a on BC, square of side b on AC square of side c on AB Now, Since BCED, ACFG and ABNM are squares ar (BCED) = a2 ar (ACFG) = b2 ar (ABNM) = c2 And, we proved that ar (BCED) = ar (ACFG) + ar (ABMN) a2 = b2 + c2 i.e. Hypotenuse2 = Base2 + Height2 Thus, Pythagoras Theorem is proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.