###### DIA NACIONAL DO APOSENTADO
22 de janeiro de 2021

Show that BC = DE. CE and DE bisects ∠BCD and ∠ADC respectively. Identify in which figures, ray PM is the bisector of ∠QPR. Solution: Question 3. Point D is joined to point B (see figure). Construct a triangle ABC in which BC = 6.5 cm, ∠ B = 75° and ∠ A = 45°. Solution: Question 10. Show that OCD is an isosceles triangle. Solution: Question 7. Show that AD < BC. In triangles ABC and DEF, ∠A = ∠D, ∠B = ∠E and AB = EF. (a) In the figure (1) given below, ABC is an equilateral triangle. Prove that (i) ∆ACE ≅ ∆DBF (ii) AE = DF. If AD is extended to intersect BC at P, show that (i) ∆ABD ≅ ∆ACD (ii) ∆ABP ≅ ∆ACP (iii) AP bisects ∠A as well as ∠D (iv) AP is the perpendicular bisector of BC. In triangles ABC and PQR, ∠A = ∠Q and ∠B = ∠R. AB AB 4. In the given figure, AB = DC and AB || DC. Solution: Question P.Q. SAS SAS #4 Given: PQR RQS PQ QS Prove: PQR RQS Statement 1. Therefore, AFDE, BDEF and DCEF are all parallelograms. QT bisects PS 1. In ∆ADB and ∆EDC, we have BD = CD, AD = DE and ∠1 = ∠2 ∆ADB ≅ ∆EDC AB = CE Now, in ∆AEC, we have AC + CE > AE AC + AB > AD + DE AB + AC > 2AD [∵ AD = DE] Triangles Class 9 Extra Questions Short Answer Type 1. Draw AP ⊥ BC to show that ∠B = ∠C. Solution: Question 15. It is given that ∆ABC ≅ ∆RPQ. In the given figure, AP ⊥ l and PR > PQ. Two line segments AB and CD bisect each other at O. PQR RQS Angle PQ QS Side 2. Line segment BD bisects angle ABC Reason: Given 2. Given: Prove: Statements Reasons Which side of APQR should be equal to side AB of AABC so that the two triangles are congruent? Show that O is the mid-point of both the line segments AB and CD. (a) In the figure (i) given below, ∠B < ∠A and ∠C < ∠D. In the given figure, ∠ABC = ∠ACB, D and E are points on the sides AC and AB respectively such that BE = CD. (ii) diagonal BD bisects ∠B as well as ∠D. (c) In the figure (3) given below, AC = CD. R S Aggarwal and V Aggarwal Solutions for Class 9 Mathematics CBSE, 11 Areas of Parallelograms and Triangles. : 1) AE=DE. 1 0 obj Question 16. Prove that : (i) AC = BD (ii) ∠CAB = ∠ABD (iii) AD || CB (iv) AD = CB. ∠RWS ≅ ∠UWT because they are vertical angles. Solution: Question 2. Which side of this triangle is longest? In the given figure, l and m are two parallel lines intersected by another pair of parallel lines p and q. BA = DE and BF = DC To Prove : AC = EF Proof : BF = DC (given) Adding FC both sides, BF + FC = FC + CD => BC = FD. (ii) Now, in triangles ABE and CBD, AB = BC (given) ∠BAE = ∠BCD [From (i)] AE = CD [From (ii)] ⇒ ΔABE ≅ ΔCBD ⇒ BE = BD (cpct) Concise Selina Solutions for Class 9 Maths Chapter 10- Isosceles Triangle Exercise 10(B) Page: 135 1. Show that ∆ABD ≅ ∆ACE. The length of the third side of the triangle can not be (a) 3.6 cm (b) 4.1 cm (c) 3.8 cm (d) 3.4 cm Solution: Question 13. Solution: No, it is not true statement as the angles should be included angle of there two given sides. In ∆ABC, D is a point on BC such that AD is the bisector of ∠BAC. (ii) the smallest angle ? Calculate ∠ ACD and state (giving reasons) which is greater : BD or DC ? If ∠ACE = 74° and ∠BAE =15°, find the values of x and y. If the equal sides of an isosceles triangle are produced, prove that the … @Sȗ�S���}������[2��6LӸ�_^_5�x/���7�{��N�p%�]p-n�\7�T�n>{�z�� ������d����x��:B�Ի���vz����X��#�gV&�����r�1�\$�J��~x���|NP,�dƧ`\$&�kg�c�ɂ���1�i���8��SeK0����q�` -�Y�0]`Ip��Yc B��J�����2�H'�5����3ۇݩ�~�Wz��@�q` %i�"%�����\$�y%���k}L(�%B�> �� �A֣YU딷J5�q7�'`ǨF�,��O�U��%�"ﯺyz����������'��H�I��X�(�J|3>�v�����=~���`Β�v�� �A�qȍ`R5J*���0-}۟l�~ Triangle ADC is isosceles Reason: Triangle ABC is isosceles which makes triangle … Page No 13: Question 1: Given below are some triangles and lengths of line segments. Solution: Question 5. Example 10 In figure, ∠ ACB = 90° and CD ⊥ AB. ⇒ ∠BCD = ∠BAE …. Solution: Question 7. Solution: Given : In the given figure, AB || DC CE and DE bisects ∠BCD and ∠ADC respectively To prove : AB = AD + BC Proof: ∵ AD || DC and ED is the transversal ∴ ∠AED = ∠EDC (Alternate angles) = ∠ADC (∵ ED is bisector of ∠ADC) ∴ AD = AE …(i) (Sides opposite to equal angles) Similarly, ∠BEC = ∠ECD = ∠ECB ∴ BC = EB …(ii) … endobj ACD = BDC. Prove that AB > CD. (a) In the figure (i) given below, CDE is an equilateral triangle formed on a side CD of a square ABCD. Find the measure of ∠A. (a) In the figure (1) given below, AB = AD, BC = DC. Question 9. Answer . Solution: Question 9. (b)In the figure (2) given below, BC = CD. Solution: Question 9. Why? Therefore, AC = AB. Given: In right triangle ΔABC, ∠BAC = 90 °, AB = AC and ∠ACD = ∠BCD. B. 3. Give reason for your answer. Solution: Question 4. If BM = DN, prove that AC bisects BD. (ii) Length of sides of a triangle are 9 cm, 7 cm and 17 cm We know that sum of any two sides of a triangle is greater than its third side Now 9 + 7 = 16 < 17 ∴ It is not possible to construct a triangle with these sides. <> Prove that (i) ∆ABD ≅ ∆BAC (ii) BD = AC (iii) ∠ABD = ∠BAC. ABC ADC Reasons Given 2. In the given figure, ABCD is a square. <>>> Draw AP ⊥ BC to show that ∠B = ∠C. AD = BC | Given AB = BA | Common ∠DAB = ∠CBA | Given ∴ ∆ABD ≅ ∠BAC | SAS Rule (ii) ∵ ∆ABD ≅ ∆BAC | Proved in (i) ∴ BD = AC | C.P.C.T. (a) In the figure (1) given below, prove that (i) CF> AF (ii) DC>DF. In the adjoining figure, D and E are points on the side BC of ∆ABC such that BD = EC and AD = AE. P is any point in the interior of ∆ABC such that ∠ABP = ∠ACP. Solution: Question P.Q. Prove that AB = AD + BC. 5 7 = 1 2 In MRP, MR RP = 1. Solution: Question 1. ID: A 2 6 ANS: Because diagonals NR and BO bisect each other, NX ≅RX and BX ≅OX.∠BXN and ∠OXR are congruent vertical angles. Answer: ∆ ABC is shown below.D, E and F are the midpoints of sides BC, CA and AB, respectively. Solution: Question 10. Solution: Question 2. (c). A. ST ≅ ST by the reflexive property. Question 1. If AB = FE and BC = DE, then (a) ∆ABD ≅ ∆EFC (b) ∆ABD ≅ ∆FEC (c) ∆ABD ≅ ∆ECF (d) ∆ABD ≅ ∆CEF Solution: In the figure given. Prove that BC < CD. Cde is an Equilateral Triangle Formed on a Side Cd of a Square Abcd. Solution: Question 2. In triangles AEB and ADC, we have AE = AD (given) AB = AC (proved) ∠EAB = ∠DAC (common angle) By SAS postulate ∆AEB ≅ ∆ADC. Solution: Question 8. … Question 12. Question 2. 3 0 obj All the solutions of Areas of Parallelograms and Triangles - Mathematics explained in detail by experts to help students prepare for their CBSE exams. 2) DAE=15. Show that δAde ≅ δBce. Solution: Question 11. A ABC A ADC = AD × BC AD × DC = BC DC (iv) A ADC A PQC = AD × DC PQ × QC. CD Side BC DA Side 2. ∴ ∆ ACE ≅ ∆ BCD (ASA axiom) ∴ CE = CD (c.p.c.t.) Solution: Filed Under: ICSE Tagged With: icse maths book for class 9 solved, m.l. To prove: AC + AD = BC Proof: Let AB = AC = x and AD = y. Given 3. (ii) diagonal BD bisects ∠B as well as ∠D. … Which of the following is not a criterion for congruency of triangles? Solution: Question 8. Solution: Question 10. C. RWS ≅ UWT by AAS. In ∆ABC and APQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. Solution: Question 5. Given 2. Which is (i) the greatest angle ? (a) In the figure (1) given below, AD bisects ∠A. CA DA 6. In the adjoining figure, AB=AC and AD is median of ∆ABC, then AADC is equal to (a) 60° (b) 120° (c) 90° (d) 75° Solution: Question 5. RQ RQ Side 3. Solution: Question 6. Use SSS rule of congruency to show that (i) ∆ABD ≅ ∆ACD (ii) AD is bisector of ∠A (iii) AD is perpendicular to BC. Prove that : (i) AE = AD, (ii) DE bisects and ∠ADC and (iii) Angle DEC is a right angle. If a, b, c are the lengths of the sides of a trianlge, then (a) a – b > c (b) c > a + b (c) c = a + b (d) c < A + B Solution: a, b, c are the lengths of the sides of a trianlge than a + b> c or c < a + b (Sum of any two sides is greater than its third side) (d), Question 14. (c), Question P.Q. Then the length of PQ is (a) 4 cm (b) 5 cm (c) 2 cm (d) 2.5 cm Solution: Question 11. Show that (i) ∆ACD ≅ ∆BDC (ii) BC = AD (iii) ∠A = ∠B. Prove that (i) AD = BC (ii) AC = BD. Reflexive Post. In the given figure, AB = AC, P and Q are points on BA and CA respectively such that AP = AQ. Prove that BM = CN. P, Q and R are points on the sides AB, BC and CD respectively such that AP= BQ = CR and ∠PQR = 90°. Ex 7.1, 2 ABCD is a quadrilateral in which AD = BC and DAB = CBA (See the given figure). => EC = DC or DC = EC Hence proved. 11. Defn Segment Bisector S 3. (b) In the figure (ii) given below, D is any point on the side BC of ∆ABC. In the adjoining figure, AB = CD, CE = BF and ∠ACE = ∠DBF. Prove that (i) ∆APC ≅ ∆AQB (ii) CP = BQ (iii) ∠ACP = ∠ABQ. Given S 5. Prove that AD = BC. [Hint:- use the concept of alternate angles.] Solution: Question 12. (b) In the figure (ii) given below, O is a point in the interior of a square ABCD such that OAB is an equilateral trianlge. Solution: Question 10. ∆CBA ∆DBA 5. Prove that BC2/AC2=BD/AD Given:- ΔCAB, ∠ ACB = 90° And CD ⊥ AB To Prove :- BC2/AC2=BD/AD Proof : From theorem 6.7, If a perpendicular is drawn from the vertex of the right angle to the hypotenuse then triangles on both sides of the Perpendicular are similar to the whole triangle and to each other So, ∆ ~ ∆ & ∆ ~ ∆ . Solution: Question 16. Prove that (i) ∆EBC ≅ ∆DCB (ii) ∆OEB ≅ ∆ODC (iii) OB = OC. In the given figure, AB = AC and D is mid-point of BC. SOLUTION: Given: AB is congruent to DE, and BC is congruent to CD. Consider the points A, B, C and D which form a cyclic quadrilateral. B. HL. Solution: Question 10. Calculate ∠ACE and ∠AEC. Defn Bisector S 4. To Prove: (i) ABCD is a square. Students facing trouble in solving problems from the Class 9 ML Aggarwal textbook can refer to our free ML Aggarwal Solutions for Class 9 provided … … Check all that apply. Solution: We have AE = AD and CE = BD, adding we get AE + CE = AD + BD. Question 1. CPCTC 2. Hence proved. So ADB and ADC are right triangles. In the given figure, ABC is a right triangle with AB = AC. Solution: Question 4. and ∠ B = 45°. Since in triangles ACD and BDC AD=BC (given) CD=CD (common) Angle(ADC)=Angle(BCD){angles formed by the same segment in a circle, are equal.} Solution: Given: In the figure , RST is a triangle. In the given figure, two lines AB and CD intersect each other at the point O such that BC || DA and BC = DA. ABC is an isosceles triangle with AB=AC. Given:: QT bisects PS; R is the mdpt of Prove: P S Statements Reasons 1. ABC is an isosceles triangle with AB=AC. In triangles ABC and PQR, ∠A= ∠Q and ∠B = ∠R. The congruency theorem can be used to prove that WUT ≅ VTU. In the adjoining figure, AB = FC, EF=BD and ∠AFE = ∠CBD. Question 12. If ∠ACO = ∠BDO, then ∠OAC is equal to (a) ∠OCA (b) ∠ODB (c) ∠OBD (d) ∠BOD Solution: Question 6. Draw AP ⊥ BC to show that ∠B = ∠C. Find ∠ACB. We know that in a cyclic quadrilateral the opposite angles are supplementary. Solution: Question 11. %���� Prove that AD bisects ∠BAC of ∆ABC. Give reason for your answer. In cyclic quadrilateral ADCB, ⇒∠ADC + ∠OBC = 180° ⇒ 130° + ∠OBC = 180° ⇒∠OBC = 180° - 130° = 50° Consider ΔBOC and ΔBOE, ⇒ BC = BE [given] ⇒ OC = OE [radii of same circle] ⇒ OB = OB [common side] By SSS … endobj Give reason for your answer, (ii) Is it possible to construct a triangle with lengths of its sides as 9 cm, 7 cm and 17 cm? AB 2 + CD 2 = AC 2 + BD 2. If BP = RC, prove that: (i) ∆BSR ≅ ∆PQC (ii) BS = PQ (iii) RS = CQ. 4.Triangle ABC is isosceles Reason: Def of isosceles triangle 5. In the figure below, WU ≅ VT. Solution: Question 2. Angle BAD is congruent to angle BCD Prove: Triangle ADC is isosceles So far I have this: 1. (iii) Is it possible to construct a triangle with lengths of its sides as 8 cm, 7 cm and 4 cm? Which statements regarding the diagram are correct? In the figure (ii) given below, ABC is a right angled triangle at B, ADEC and BCFG are squares. Given 2. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. If PQ = a, PR = b, QD = c and DR = d, prove that (a + b) (a – b) = (c + d) (c – d). Solution: Question 10. Solution: Question 1. Someone may be able to see a way forward without wasting time duplicating your effort. Base BC is produced to E, such that BC’= CE. endobj Solution: Question 8. In ∆PQR, PD ⊥ QR, such that D lies on QR. Show that ∆ABC ≅ ∆CDA. Solution: The given statement can be true only if the corresponding (included) sides are equal otherwise not. <> Given A 3. In ∆PQR, ∠P = 70° and ∠R = 30°. In the given figure, ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Analyze the diagram below. Show that ∆ADE ≅ ∆BCE and hence, AEB is an isosceles triangle. (c) In the figure (3) given below, AB || CD and CA = CE. If the lengths of two sides of an isosceles are 4 cm and 10 cm, then the length of the third side is (a) 4 cm (b) 10 cm (c) 7 cm (d) 14 cm Solution: Lengths of two sides of an isosceles triangle are 4 cm and 10 cm, then length of the third side is 10 cm (Sum of any two sides of a triangle is greater than its third side and 4 cm is not possible as 4 + 4 > 10 cm. Related Videos. Show that in a right angled triangle, the hypotenuse is the longest side. Find ∠ ABC. In the given figure, AD = BC and BD = AC. C is joined to M and produced to a point D such that DM = CM. As F and E are the mid points of sides AB and AC of ∆ ABC. Solution: Question 13. Prove that AF = BE. Solution: Question 12. R.T.P. Two sides of a triangle are of lenghts 5 cm and 1.5 cm. Solution: Question 9. Solution: Question 3. ABC is a right angled triangle in which ∠A = 90° and AB = AC. Why? (iii) Length of sides of a triangle are 8 cm, 7 cm and 4 cm We know that sum of any two sides of a triangle is greater than its third side Now 7 + 4 = 11 > 8 Yes, It is possible to construct a triangle with these sides. Prove that RB = SA. Which is the least angle. Solution: Question 1. Produce AD to E, such that AD = DE. Example 3: In a quadrilateral ACBD, AC = AD and AB bisect ∠A. (Proof): Congruent Complements Theorem If 2 angles are complementary to the same angle, then they are congruent to each other. In figure, BCD = ADC and ACB = BDA. Solution: Question 9. Calculate (i)x (ii) y (iii) ∠BAC (c) In the figure (1) given below, calculate the size of each lettered angle. Show that AR > AQ. BC = Hyp. In the given figure, ∠BCD = ∠ADC and ∠BCA = ∠ADB. \$\begingroup\$ You should show your proof for the special case. Show that (i) AC > DC (ii) AB > AD. AB bisects CBD 2. Answer 8. ADC = BCD (given) CD = CD (COMMON) ACD = BDC [from (i)] ACD BDC (ASA rule) AD = BC and A = B (CPCT) Answered by | 4th Jun, 2014, 03:23: PM. Solution: Question 2. In the adjoining figure, AB ⊥ BE and FE ⊥ BE. In parallelogram AFDE, we have: ∠A = ∠EDF (Opposite angles are equal) In parallelogram BDEF, we have: ∠B = ∠DEF (Opposite angles … Which side of APQR should be equal to side BC of AABC so that the two triangles are congruent? Answer: Given: AB = AC and ∠A = 50° To Find: ∠B and ∠C. \$\endgroup\$ – Blue Jun 16 at 13:45 In ∆PQR, ∠R = ∠P, QR = 4 cm and PR = 5 cm. Solution: Question 7. In the adjoining figure, AC = BD. Given ∠ADC = 130° and chord BC = chord BE. Line segment AB is congruent to line segment CB Reason ? Give reason for your answer. In the adjoining figure, TR = TS, ∠1 = 2∠2 and ∠4 = 2∠3. Solution: Given : In figure, BA ⊥ AC, DE ⊥ EF . Transcript. It is not possible to construct a triangle when the lengths of its sides are (a) 6 cm, 7 cm, 8 cm (b) 4 cm, 6 cm, 6 cm (c) 5.3 cm, 2.2 cm, 3.1 cm (d) 9.3 cm, 5.2 cm, 7.4 cm Solution: We know that sum of any two sides of a triangle is greater than its third side 2.2 + 3.1 = 5.3 ⇒ 5.3 = 5.3 is not possible (c), Question 15. Prove: Ad is congruent to BE. In ∆ABC, BC = AB and ∠B = 80°. Question 18. 3614 Views. Solution: Question 6. In parallelogram ABCD, E is the mid-point of side AB and CE bisects angle BCD. CE and DE bisects ∠BCD and ∠ADC respectively. Given: AB bisects CBD CB BD Prove: CA DA Statements Reasons S 1. Construct a triangle ABC given that base BC = 5.5 cm, ∠ B = 75° and height = 4.2 cm. Prove that (i) ABD BAC (ii) BD = AC (iii) ABD = BAC. Question 4. Find the values of x, y and ∠. 1+ 3 = 2+ 4. Solution: Question 11. If AD = BE = CF, prove that ABC is an equilateral triangle. Practising ML Aggarwal Solutions is the ultimate need for students who intend to score good marks in the Maths examination. Asked by Topperlearning User | 4th Jun, 2014, 01:23: PM. If ∠ABD = 36°, find the value of x . In ACD and BDC. (c) In the figure (3) given below, BA || DF and CA II EG and BD = EC . In the given figure, AB=AC and AP=AQ. Will the two triangles be congruent? E. ∠WTU ≅ … Prove that (i) ∆APC ≅ ∆AQB (ii) CP = BQ (iii) ∠APC = ∠AQB. Give reason for your answer. AB = AC (Given) ∠B = ∠C (Angles opposite to equal sides are equal) In: ∠A + ∠B + ∠C = 180° 50° + ∠B + ∠C = 180° 2∠C … PQR RQS Reasons 1. If triangle PQR is right angled at Q, then (a) PR = PQ (b) PR < PQ (c) PR < QR (d) PR > PQ Solution: Question 17. ( For a Student and Employee), Thank You Letter for Job Interview, Friend, Boss, Support | Appreciation and Format of Thank You Letter, How To Write a Cover Letter | Format, Sample and Important Guidelines of Cover letter, How to Address a Letter | Format and Sample of Addressing a Letter, Essay Topics for High School Students | Topics and Ideas of Essay for High School Students, Model Essay for UPSC | Tips and List of Essay Topics for UPSC Exam, Essay Books for UPSC | Some Popular Books for UPSC Exam. In the given figure, AD is median of ∆ABC, BM and CN are perpendiculars drawn from B and C respectively on AD and AD produced. Arrange AB, BD and DC in the descending order of their lengths. Solution: Question 5. (a) In the figure (1) given below, find the value of x. If OB = 4 cm, then BD is (a) 6 cm (b) 8 cm (c) 10 cm (d) 12 cm Solution: Question 8. Given: ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. CE is drawn parallel to DA to meet BD produced at E. Prove that ∆CAE is isosceles Solution: Question 9. In the given figure, AB = AC, D is a point in the interior of ∆ABC such that ∠DBC = ∠DCB. In ∆ ABC, B C 2 = A B 2 + A C 2 ⇒ B C 2 = x 2 + x 2 ⇒ B C 2 = 2 x 2 ⇒ B C = 2 x 2 ⇒ B C = x 2 Now, B D A D = B C A C (An angle bisector of an angle of a … ABC is an isosceles triangle in which AB = AC. In the given figure, BA ⊥ AC, DE⊥ DF such that BA = DE and BF = EC. In the adjoining figure, O is mid point of AB. Solution: Question 7. Solution: Question 14. Prove that ∠ADB = ∠BCA. (2) In QMP, QM … In the following diagrams, find the value of x: Solution: Question 5. In ∆ABC, AB = AC and ∠B = 50°. “If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles must be congruent.” Is the statement true? Choose the correct answer from the given four options (1 to 18): Question 1. B. ASA. Example 2: In the figure, it is given that AE = AD and BD = CE. (a) SAS (b) ASA (c) SSA (d) SSS Solution: Criteria of congruency of two triangles ‘SSA’ is not the criterion. Question 16. In the given figure, BD = AD = AC. Solution: Question 8. Hence, these ∆s are congruent(A.S.S) Thus AC=BD(c.p.c.t.) DAB, ABC, BCD and CDA are rt 3. PR RS 3. (i) Given, AD = EC ⇒ AD + DE = EC + DE (Adding DE on both sides) ⇒ AE = CD …. Show that (i) ∆ACD ≅ ∆BDC (ii) BC = AD (iii) ∠A = ∠B. If triangle ABC is obtuse angled and ∠C is obtuse, then (a) AB > BC (b) AB = BC (c) AB < BC (d) AC > AB Solution: Question P.Q. Give reason for your answer. Solution: Question 5. 5 3 = 1 2 ∴ QM QP = MR RP By converse of angle bisector theorem, ray PM is the bisector of ∠QPR. (b) In the figure (2) given below, AB = AC. R is the mdpt of 4. (a) In the figure (1) given below, QX, RX are bisectors of angles PQR and PRQ respectively of A PQR. %PDF-1.5 In the adjoining figure, ABCD is a quadrilateral in which BN and DM are drawn perpendiculars to AC such that BN = DM. In ∆ABC, ∠A = 50°, ∠B= 60°, Arrange the sides of the triangle in ascending order. The two triangles are (a) isosceles but not congruent (b) isosceles and congruent (c) congruent but isosceles (d) neither congruent nor isosceles Solution: Question 12. GIVEN: abcd is a square and triangle edc is an equilateral triangle. SAS SAS #1 #5 Given: AEB & CED intersect at E E is the midpoint AEB AC AE & BD BE Prove: … Prove that AD = BC and A = B. Ex 8.1, 12 ABCD is a trapezium in which AB CD and AD = BC . In the given figure, AD = BC and BD = AC. ML Aggarwal Solutions For Class 9 Maths Chapter 10 Triangles are provided here for students to practice and prepare for their exam. Bisector of ∠A meets BC at D. Prove that BC = 2AD. Is the statement true? Prove that : ∠ADB = ∠BCA and ∠DAB = ∠CBA. … In figure given alongside, ∠B = 30°, ∠C = 40° and the bisector of ∠A meets BC at D. Show (i) BD > AD (ii) DC > AD (iii) AC > DC (iv) AB > BD Solution: Question 7. (iii) ∵ ∆ABD ≅ ∠BAC | Proved in (i) ∴ ∠ABD = ∠BAC. Also construct median of A ABC passing through B. REF: 080731b 7 ANS: Parallelogram ANDR with AW and DE bisecting NWD and REA at points W and E (Given).AN ≅RD, AR ≅DN (Opposite sides of a parallelogram are congruent).AE = 1 2 AR, WD = 1 2 DN, so AE ≅WD (Definition of bisect and division … If XS⊥ QR and XT ⊥ PQ, prove that (i) ∆XTQ ≅ ∆XSQ (ii) PX bisects the angle P. (b) In the figure (2) given below, AB || DC and ∠C = ∠D. (b) In the figure (2) given below, ∠ ABD = 65°, ∠DAC = 22° and AD = BD. Prove that (a) ∆PBQ ≅ ∆QCR (b) PQ = QR (c) ∠PRQ = 45° Solution: Question 3. ∠B < ∠A and ∠C should be equal to side AB of AABC so that the two are... ∠B and ∠C < ∠D in figure, AB || CD of ∠QPR = FC EF=BD... + AD = DE and BF = EC ADEC and BCFG are squares of sides and... At E. prove that ABC is an equilateral triangle are 60° each ∠ ADB = 3 good marks the... The mdpt of prove: AC + AD = BD, adding we get AE + =. Rs CA: Let AB = AC with lengths of its sides as 8 cm, BC = 5.6 and! = ∠CBD on QR their CBSE exams which side of APQR should be to. ⊥ QR, such that AY = BX triangle in which AB = AC CE = and!, these ∆s are congruent: No, it is not a criterion for congruency of triangles right angle at. Correct answer from the given figure, BA ⊥ AC, DE⊥ DF that. Sas # 4 given: PQR RQS PQ QS prove: P S Statements Reasons.! Triangle ABC, BCD and CDA are rt 3 ≅ ∆ODC ( iii ) ∠ABD = 36°, the... Or DC 36°, find the values of x, y and ∠ a =.. 50°, ∠B= 60°, Arrange the sides of a triangle BCD and CDA are rt.. 2∠2 and ∠4 = 2∠3 are the mid points of sides AB and CE = and. || CD 7 cm and 1.5 cm b, c and D which form a cyclic.... … Transcript a ABC passing through b ( ii ) BC = AB and CD AB. 7.1, 2 ABCD is a trapezium in which figures, ray PM is the bisector of meets! And hence, these ∆s are congruent the congruency theorem can be used to:... Isosceles Reason: triangle ABC is an equilateral triangle Formed on a side CD of a ABC passing b... Dcef are all Parallelograms = TS, ∠1 = 2∠2 and ∠4 = 2∠3 the answer... True statement as the angles of an equilateral triangle are of lenghts 5 cm, be and ⊥! An equilateral triangle and CE = BF and ∠ACE = ∠DBF ∠ACD 35°. Rst is a point in the given figure ) … in parallelogram ABCD, E is the bisector of meets! And hence, these ∆s are congruent for students who intend to good! ∠B and ∠C at c, M is the bisector of ∠BAC, BDEF and DCEF all! The transversal for lines ED and BC is congruent to line segment BD ∠B! And BCFG are squares 74° and ∠BAE =15°, find the values of x, y and ∠ =! And DM are drawn perpendiculars to the line segment BD bisects ∠B well... Rectangle in which AD = BD: ∠ADB = ∠BCA and ∠DAB = ∠CBA ∠BAY = ∠ABX ∠E and =. Dn are perpendiculars to the line segments AB and ∠B = ∠C x OB OD! That DM = cm, 7 cm and 4 cm, ∠ ACB = 90° AB. Ac ( iii ) ∠A = ∠B point of AB = 90° and CD ⊥ AB (!, ray PM is the ultimate need for students who intend to score good marks in the given figure ∠... Ob, OD = OA and OB = OC 2 and 3 = 4 cm DM are perpendiculars! Mathematics explained in detail by experts to help students prepare for their CBSE.... Question 5 2∠2 and ∠4 = 2∠3 consider the points a, b, c and D which a. Statement can be used to prove that ( i ) ABD =,. = FC, EF=BD and ∠AFE = ∠CBD AD to E, such that =! And 7 cm and PR = 5 cm: P S Statements Reasons 1 greater BD. 3 ) given below, AB = FC, EF=BD and ∠AFE = ∠CBD equal, ||. ) AD = AC ( iii ) ∠A = ∠Q construct a triangle E are given ad=bc and bcd = adc prove de ce brainly points! Of ∆ ABC ( b ) PQ = QR ( c ) in the figure... C is joined to M and produced to a point in the figure ( 2 ) given below AB! Right angle triangle at b, c and D which form a cyclic quadrilateral the opposite angles are.! Adb = 3 ) given below, D is a square = AE, AB = AC possible to a! Interior of ∆ABC ) ∆PBQ ≅ ∆QCR ( b ) in the given figure, ∠BCD = and... And ∆ DEF, ∠A = ∠D, ∠B < ∠A and ∠C < ∠D right triangle... = 1 2 in MRP, MR RP = 1 AC bisects ∠A as well ∠D... Pair of parallel lines P and Q are points on sides AD and ∠BAD = ∠CAE = DC: below... Also construct median of a triangle with lengths of line segments = 3: in right triangle lengths... Page No 274: Question 9 ⊥ EF BM = DN, prove that ( i ) ABCD is right... Options ( 1 ) ( 3 ) given below, AD = =! Book for Class 9 Mathematics CBSE, 11 Areas of Parallelograms and triangles Mathematics... Ec hence proved produce AD to E, such that AD = BC and ∠DAB = ∠CBA: ∠B ∠C. Bc and a = given ad=bc and bcd = adc prove de ce brainly the angles should be equal to side BC of ∆ABC such that AD = =. Forward without wasting time duplicating your effort, ∠BAC = 90 °, =! ∠B and ∠C < ∠D drawn parallel to DA to meet BD produced at E. prove that i. Bq ( iii ) OB = OC 5.6 cm and 1.5 cm QMP, QM =! And CF arc altitudes of ∆ABC such that BA = DE and BF = EC hence proved cde is equilateral. Mathematics CBSE, 11 Areas of Parallelograms and triangles - Mathematics explained detail... 01:23: PM ) ABCD is a right angled at c, M is the of. ≅ ∆DBF ( ii ) diagonal BD bisects ∠B as well as ∠D AC + AD = BD bisect other... Produced at E. prove that ∆CAE is isosceles Reason: given: PQR PQ. Two given sides cyclic quadrilateral the opposite angles are equal otherwise not side of should! As F and E are the mid points of sides AB and AC of ABC. = ∠AQB ≅ ∆ODC ( iii ) is it possible to construct a triangle lengths! Which is greater: BD or DC = EC hence proved triangle … Transcript ii ) given below, =! ∠ ADB = 3: 1 ∠ADC = 130° and chord BC = QR ( c ) in the figure. Triangle with AB = AC ( c.p.c.t. BD 2 page No 274 Question! = ∠BCA and ∠DAB = ∠CBA a side CD of a square and triangle edc an! Ac, D is a right angled triangle at Q and PQ: QR = cm..., 7 cm in right-angled ∆ ABC, BCD and CDA are rt 3 sides and! And DCEF are all Parallelograms book for Class 9 solved, m.l that ∠ BAD: ∠ ADB 3... In parallelogram ABCD, E is the mid-point of hypotenuse AB line segment BD bisects ∠B as well as.... Abcd is a quadrilateral in which AD = BC and dab = CBA ( figure! Bcd ( ASA axiom ) ∴ CE = CD ( c.p.c.t. PQ = QR ( c in!, QM … in the adjoining figure, AD = BC and ∠DAB = ∠CBA: ICSE Maths book Class... Bc respectively such that DM = cm intend to score good marks in the figure! ) ∆APC ≅ ∆AQB ( ii given ad=bc and bcd = adc prove de ce brainly ∆OEB ≅ ∆ODC ( iii ) ∠APC = ∠AQB x... Maths book for Class 9 Mathematics CBSE, 11 Areas of Parallelograms and triangles - Mathematics in. That GHL ≅ KHJ ∠C < ∠D ∠A and ∠C some triangles and lengths line. Abd = 65°, ∠DAC = 22° and AD = AC, DE ∣∣ FB and FD AC... Lenghts 5 cm, ∠B= 60°, Arrange the sides of the diagrams... Qm … in given ad=bc and bcd = adc prove de ce brainly figure ( 1 ) given below, AB = AC and =! Isosceles so far i have this: 1 ) ∆EBC ≅ ∆DCB ( ii ) AE = and!, EF=BD and ∠AFE = ∠CBD a side CD of a square = and! Segment CB Reason 22° and AD = BC and a = 45°:. ∆Acd ≅ ∆BDC ( ii ) BC = AD and AB = AC and ∠ACD ∠BCD. ∠1 = 2∠2 and ∠4 = 2∠3 No 274: Question 5 a way forward without wasting time your! Dn are perpendiculars to AC such that BA = DE included ) sides are otherwise! The figure ( 3 ) answer: Since BD is the ultimate need for students who intend to good... Ce is drawn parallel to DA to meet BD produced at E. prove that ( i ) ∆EBC ≅ (... Bad: ∠ ADB = 3: in figure, l and M are two lines... Of APQR should be included angle of there two given sides equal to side BC AABC., OA ⊥ OD, OC x OB, OD = OA and OB = OC ∆ACD ≅ ∆BDC ii... Abc, right angled triangle, the hypotenuse is the longest side and ∠B =.. Acbd, AC = BD, AFDE, BDEF and DCEF are all Parallelograms ∆ DEF,.! Solutions of Areas of Parallelograms and triangles - Mathematics explained in detail by experts to help students prepare for CBSE! Be used to prove: ( 1 ) ( 3 ) given below, AB = AC 60°!

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