### Geometry Class 8th Mathematics Term 1 Tamilnadu Board Solution

##### Question 1.Choose the correct answer:In the isosceles ∆XYZ, given XY = YZ then which of the following angles are equal?A. ∠X and ∠YB. ∠Y and ∠ZC. ∠Z and ∠XD. ∠X, ∠Y and ∠ZAnswer:Reason: Angles opposite XY and YZ are angle X and Z respectively.[T3] Question 2.Choose the correct answer:In Δ ABC and ∆DEF, ∠B = ∠E, AB = DE, BC = EF. The two triangles are congruent under _____ axiomA. SSSB. AAAC. SASD. ASAAnswer:option C.If any two sides and the included angle of a triangle are respectively equal to any two sides and the included angles of another triangle then the two triangles are congruent. Question 3.Choose the correct answer:Two plane figures are said to be congruent if they haveA. the same sizeB. the same shapeC. the same size and the same shapeD. the same size but not same shapeAnswer:Reason: If two geometrical figures are identical in shape and size then they are said to be congruent.Question 4.Choose the correct answer:In a triangle ABC, ∠A = 40o and AB = AC, then ABC is _____ triangle.A. a right angledB. an equilateralC. an isoscelesD. a scaleneAnswer:In a triangle, When two sides are equal then triangle are said to be an isosceles triangle.Question 5.Choose the correct answer:In the triangle ABC, when ∠A = 90o the hypotenuse is ______A. ABB. BCC. CAD. None of theseAnswer:option B[T4] Question 6.Choose the correct answer:In the Δ PQR the angle included by the sides PQ and PR isA. ∠PB. ∠QC. ∠RD. None of theseAnswer:[T5] Question 7.Choose the correct answer:In the figure, the value of x° is_______ A. 80oB. 100oC. 120oD. 200oAnswer:In Δ ABD and Δ CBDAB = CB = 2cmAD = CD = 3cmBD = BD = common∴ Δ ABD ≅ Δ CBD∴ ∠ A = ∠ C⇒ 100° = x°Hence, option B is correct.Question 8.In the figure, ABC is a triangle in which AB = AC. Find x° and y°. Answer:Theorem 2:If a side of a triangle is produced, the exterior angle so formed, is equal to the sum of the two interior angle.∠ ACE + ∠ ACB = 180°⇒ x + 48° + x° = 180°⇒ 2x + 48° = 180°⇒ 2x = 180° - 48°⇒ 2x = 132°⇒ ⇒ x = 66°Ext. ∠ A = ∠ B + ∠ C⇒ y° = x° + x°⇒ y° = 2x°⇒ y° = 2 × 66°⇒ y° = 132°Question 9.In the figure, Find x°. Answer:Theorem 1: Sum of all the angles of the triangles is 180°In Δ COBOD = DC∴ ∠ COD = ∠DCO = 40°In Δ AOBOA = OB (Given)∴ ∠ OAB = ∠ OBA = x°∠ OAB + ∠ OBA + ∠ AOB = 180° (Angle Sum Property)⇒ x° + x° + ∠ AOB = 180°⇒ 2x° + ∠ AOB = 180°⇒ ∠ AOB = 180° - 2x°∠ AOB = ∠ COD (opposite angles are equal)180 – 2x° = 40°⇒ 180° = 40 + 2x°⇒ 180° - 40° = 2x°⇒ 140° = 2x°⇒ ⇒ x = 70°Question 10.In the figure ∆PQR and ∆SQR are isosceles triangles. Find x°. Answer:Theorem 1: Sum of all the angles of triangles is 180°Let ∠ PQR be yIn Δ SQR,SQ = SR∴ ∠ SQR = ∠ SRQ∠ SQR + ∠ SRQ + ∠ QSR = 180° (Angle Sum Property)⇒ 2 ∠ SQR + 70° = 180°⇒ 2 ∠ SQR = 180° - 70°⇒ 2 ∠ SQR = 110°⇒ ⇒ ∠ SQR = 55° = ∠ SRQIn Δ PQR,∠ PQR + ∠ PRQ + ∠ RPQ = 180°⇒ y + y + 40° = 180°⇒ 2y + 40° = 180°⇒ 2y = 180° - 40°⇒ 2y = 140°⇒ ⇒ y = 70°∠ PQR = ∠ PQS + ∠ SQR⇒ 70° = x° + 55°⇒ x° = 70° - 55°⇒ x° = 15°Question 11.In the figure, it is given that BR = PC and ∠ACB= ∠QRP and AB ∥ PQ. Prove thatAC = QR. Answer:Given: BR = PC and ∠ ACB = ∠ QRP , AB || PQTo Prove: AC = QRProof:In Δ ABC, we haveBC = BR + RCIn Δ PQRPR = PC + RCBut , BR = PC [Given]So, BC = PC + RC and PR = BR + RC⇒ BC = PRSo, in Δ ABC and Δ PQR, we have∠ ACB = ∠ QRP [Given]BC = PR [Proved Above]∠ ABC = ∠ QPR [AB || PQ, alternate interior angles]Thus, Δ ABC ≅ Δ PQR [Angle – Side – Angle]∴ AC = QR [C. P. C. T]Question 12.In the figure, AB = BC = CD, ∠A = xo. Prove that ∠DCF = 3∠A. Answer:Given: AB = BC = CD and ∠ A = x°To Prove: ∠ DCF = 3 ∠ AProof:In Δ ABCAB = BC [Given]∴ ∠ A = ∠ C = x°Now,∴ ext. ∠ B = ∠ A + ∠ C⇒ Ext. ∠ B = x° + x°⇒ Ext. ∠ B = 2x°In Δ CBDBC = CD [Given]∴ ∠ B = ∠ D = 2x°Now,In Δ ADC,Ext. ∠ DCF = ∠ CDA + ∠ CAD⇒ ∠ DCF = 2x + x⇒ ∠ DCF = 3x⇒ ∠ DCF = 3 ∠ A [ ∠ A = x°, Given]Hence Proved.Question 13.Find x°, y°, z° from the figure, where AB = BD, BC = DC and ∠DAC 30o. Answer:Theorem 1: Sum of all the angles in the triangle is 180°.In Δ ABD,We know that, AB = BD∴ ∠ A = ∠ D⇒ 30° = x°Hence, ∠ A + ∠ B + ∠ C = 180°⇒ 30° + ∠ B + 30° = 180°⇒ ∠ B + 60° = 180°⇒ ∠ B = 180° - 60°⇒ ∠ B = 120°∠ DBA + ∠ DBC = 180° (Sum of adjacent angles is 180°)⇒ 120° + y° = 180°⇒ y° = 180° - 120°⇒ y° = 60°In Δ DBCWe know that, BC = DC∴ ∠ B = ∠ D⇒ y° = z°⇒ 60° = z°Question 14.In the figure, ABCD is a parallelogram. AB is produced to E such that AB = BE. AD produced to F such that AD = DF. Show that ∆FDC ≡ ∆CBE. Answer:Given: Parallelogram ABCD and AB = BE and AD = FDTo prove: Δ FDC ≡ ΔCBEConstruction: Join DB Proof:We know that,AB = DC [ opposite sides of parallelogram]BE = DC [AB = BE, because B is the midpoint of AE]Similarly,AD = BC [ opposite sides of parallelogram]DF = BC [ AD = DF, because B is the midpoint of AE]Now, AD||BC and AB∠ A = ∠ B [corresponding angles] …(1)Now, AB||CD and AD∠ A = ∠ D [corresponding angles] …(2)∴ ∠ B = ∠ D (From 1 and 2)In Δ FDC and Δ CBEFD = CB [Proved Above]DC = BE [Proved Above]∠ D = ∠ B [Proved Above]Thus, Δ FDC ≡ Δ CBEHence Proved.Question 15.In figure, BO bisects ∠ABC of ∆ABC. P is any point on BO. Prove that the perpendicular drawn from P to BA and BC are equal. Answer:Given: A Δ ABC in which BO is bisector of ∠ABCAlso, we have PD ⊥ AB and PE ⊥ BCTo Prove: PD = PEProof:In Δ PBD and Δ PBEPB = PB [common]∠ PBD = ∠ PBE [ given]∠ PDB = ∠ PEB = 90° [Given]Thus, Δ PBD ≅ Δ PBE [Angle – Angle – Side]∴ PD = PEHence Proved.Question 16.The Indian Navy flights fly in a formation that can be viewed as two triangles with common side. Prove that ∆SRT ≡ ∆QRT, if T is the midpoint of SQ and SR = RQ. Answer:Given: T is the mid-point of SQ and SR = RQTo Prove: Δ SRT ≅ Δ QRTProofIn Δ SRT and QRTRT = RT [common]ST = QT [T is the mid-point of SQ]SR = RQ [Given]Thus, Δ SRT ≅ Δ QRT [Side – Side – Side]Hence Proved.

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