Congruence- Extension Sheet - Math with JM - home

Question 5In the given figure, a quadrilateral PQRS is shown.Arrange all the line segments (shown in the figure) indescending order of their lengths.Question 6In ΔABC, AD is the bisector of ∠BAC and is perpendicular to theside BC. Find the values of x and y.Question 7ΔPQR is an isosceles triangle with PR = PQ. An exterior angle isdrawn by extending the side PR of ΔPQR.A line XY is drawn such that it passes through R and bisects theexterior angle QRS. Prove that XY || PQ.

Question 8In the given figure, an isosceles triangle is shown with AC = BC.Also, the line EF is parallel to the line AB.Find the measure of ∠BCA.Question 9In the given figure, an isosceles ΔXYZ is shown with XY = YZ. Aline ZA is drawn intersecting XY at A such that ZA = XZ and ZA= AY. Find the measure of ∠XYZ.Question 10ΔABC is right-angled at B. CX is the bisector of ∠BCA and BY isperpendicular to side AC. Prove that BX = BO

Congruence- Extension SheetSolutionsQuestion 1In the given figure, an isosceles triangle PQR with PQ = QR is shown. S is a point on side PR.Prove that QR > QS.Solution:It is given that in ΔPQR, PQ = QR⇒∠RPQ = ∠QRP … (1)Now, ∠QSR is an exterior angle of ΔPQS.∴ ∠QSR = ∠SPQ + ∠PQS⇒∠QSR > ∠SPQi.e., ∠QSR > ∠RPQ⇒ ∠QSR > ∠QRP [Using (1)]We know that in a triangle, the side opposite to the larger angle is greater.∴QR > QSHence, provedQuestion 2In ΔPQR, PS is the bisector of ∠RPQ.Prove that PU = PVSolution:In ΔPUS and ΔPVS,

∠UPS = ∠VPS (PS is the bisector of ∠RPQ)PS = PS (Common)∠PUS = ∠PVS (Each equal to 90°)∴ ΔPUS ≅ ΔPVS (By ASA congruency criterion)∴ By CPCT, PU = PVHence, provedQuestion 3Prove that the angles opposite to equal sides of an isosceles triangle are equal.Solution:Let PQR be an isosceles triangle with PQ = PRLet PS be the bisector of ∠RPQ.To prove: ∠PQR = ∠QRPIn ΔPQS and ΔPRS,PQ = PR (Given)PS = PS (Common)∠QPS = ∠RPS (By construction)∴ΔPQS ≅ ΔPRS (By SAS congruency criterion)We know that corresponding parts of congruent triangles are equal.Therefore, we have ∠PQR = ∠PRQHence, the angles opposite to equal sides of an isosceles triangle are equal.Question 4In ΔLMN, LM = 3.6 cm, MN = 6.1 cm, and NL = 5.5 cm

Arrange the angles x, y, z in ascending order of their measures.Solution:We know that in a triangle, the angle opposite to the longer side is larger (or greater). It isgiven that LM = 3.6 cm and MN = 6.1 cm∴LM < MNThus, ∠MNL < ∠NLM, i.e., z < x … (1)Also, LM = 3.6 cm and NL = 5.5 cm∴LM < NLThus, ∠MNL < ∠LMN, i.e., z < y … (2)Also, MN = 6.1 cm and NL = 5.5 cm∴ NL < MNThus, ∠LMN < ∠NLM, i.e., y < x … (3)Therefore, from (1), (2), and (3), we have z < y < x ( z < y and y < x)Question 5In the given figure, a quadrilateral PQRS is shown.Arrange all the line segments (shown in the figure) in descending order of their lengths.Solution:In ΔPQS,∠PQS + ∠QSP + ∠SPQ = 180° (By angle sum property)⇒ 47° + ∠QSP + 91° = 180°⇒ ∠QSP = 180° − 138° = 42°Thus, in ΔPQS,∠PQS = 47°, ∠QSP = 42°, and ∠SPQ = 91°We know that in a triangle, the side opposite to the larger angle is longer.Thus, we have SQ > PS > PQ … (1)In ΔQRS,

AB = AC⇒ 2x = 5y − 2⇒ 5y − 2 = 8 ( x = 4)⇒ 5y = 10⇒ y = 2Thus, the respective values of x and y are 4 and 2.Question 7ΔPQR is an isosceles triangle with PR = PQ. An exterior angle is drawn by extending the sidePR of ΔPQR.A line XY is drawn such that it passes through R and bisects the exterior angle QRS.Prove that XY || PQ.Solution:It is given that ΔPQR is an isosceles triangle with PR = RQ.We know that angles opposite to equal sides are equal.∴ ∠PQR = ∠RPQ … (1)Also, it is given that XY bisects ∠QRS.∴ ∠QRY = ∠YRS … (2)We know that the measure of an exterior angle of a triangle is equal to the sum of themeasures of its two opposite interior angles.∴∠QRS = ∠RPQ + ∠PQR⇒ ∠QRS = 2∠PQR [Using (1)]⇒ ∠QRY + ∠YRS = 2∠PQR (∠QRS = ∠QRY + ∠YRS)⇒ 2∠QRY = 2 ∠PQR⇒ ∠QRY = ∠PQRThus, the alternate interior angles made by the lines XY and PQ, when cut by the transversalRQ, are equal.

Therefore, XY||PQHence, provedQuestion 8In the given figure, an isosceles triangle is shown with AC = BC.Also, the line EF is parallel to the line AB.Find the measure of ∠BCA.Solution:It is given that in ΔABC, AC = BC∴ ∠ABC = ∠CAB … (1)(Since angles opposite to equal sides of a triangle are equal)Also, EF||AB and these lines are intersected by a transversal AC.∴∠CAB = ∠FCD (Corresponding angles are equal)⇒∠CAB = 66°∴ ∠ABC = ∠CAB = 66° [From (1)]In ΔABC,∠ABC + ∠BCA + ∠CAB = 180° (By angle sum property)⇒ 66° + ∠BCA + 66° = 180°⇒ ∠BCA = 180° − 132°⇒ ∠BCA = 48°Question 9In the given figure, an isosceles ΔXYZ is shown with XY = YZ. A line ZA is drawn intersectingXY at A such that ZA = XZ and ZA = AY. Find the measure of ∠XYZ.

Solution:It is given that XY = YZWe know that angles opposite to equal sides of an isosceles triangle are equal.∴ ∠ZXY = ∠YZX i.e., ∠1 = ∠2 + ∠3 … (1)( ∠YZX = ∠2 + ∠3)Also, ZA = XZ (Given)∴ ∠1 = ∠4 … (2)Also, ZA = AY∴ ∠6 = ∠3 … (3)Now, ∠4 = ∠6 + ∠3 (Using exterior angle property)⇒ ∠4 = ∠6 + ∠6 [Using (3)]⇒ ∠1 = 2∠6 [Using (2)]Also, using (1),∠2 + ∠3 = ∠YZX = 2∠6By applying angle sum property in ΔXYZ, we obtain∠1 + ∠6 + (∠2 + ∠3) = 180°⇒ 2∠6 + ∠6 + 2∠6 = 180°⇒5∠6 = 180°⇒∠6 =⇒∠6 = 36°Thus, ∠XYZ = 36°Question 10ΔABC is right-angled at B. CX is the bisector of ∠BCA and BY is perpendicular to side AC.Prove that BX = BO