### Geometry Class 9th Mathematics Term 1 Tamilnadu Board Solution

##### Question 1.If an angle is equal to one third of its supplement, its measure is equal toA. 40°B. 50°C. 45°D. 55°Answer:Let the required angle be x and the supplement of x is y.As we know if two angles are supplement of each other, then sum of those two angles is 180°.⇒ x + y = 180⇒ y = 180 – xNow, according to question;Required angle is equal to one third of its supplement;i.e. ⇒ y = 3xNow putting the value of y in this equation,⇒ 180 – x = 3x⇒ ⇒ ⇒ Hence, required angle is 45°.Hence, correct option is (c).Question 2.In the given figure, OP bisect ∠ BOC and OQ bisect ∠ AOC. Then ∠ POQ is equal to A. 90°B. 120°C. 60°D. 100°Answer:Suppose that BOC = 2x and ∠ AOC = 2yNow, by question; OP and OQ is the bisect or of ∠ BOC and ∠ AOC respectively.⇒ Similarly, Now,∠ POQ = ∠ POC + ∠ COQ = x + ySo, we have to find ∠ POQ i.e.x + y Now, ∠ BOC + ∠ AOC = 180⇒ 2x + 2y = 180⇒ 2(x + y) = 180⇒ x + y = 180/2⇒ x + y = 90°Hence, ∠ POQ = 90°.Question 3.The complement of an angle exceeds the angle by 60°. Then the angle is equal toA. 25°B. 30°C. 15°D. 35°Answer:Let the required angle be x and it’s complement is y.So, ⇒ y = 90 – xNow, according to question;Complement of the angle exceeds the angle by 60° i.e. y exceeds the x by 60°.⇒ y – 60 = xNow, putting the value of your in this equation,⇒ (90 – x) – 60 = x⇒ 30 – x = x⇒ 2x = 30⇒ x = 15°Hence, required angle is 15°.Question 4.Find the measure of an angle, if six times of its complement is 12° less than twice of its supplement.A. 48°B. 96°C. 24°D. 58°Answer:Let the required angle be x and it’s complement is y and it’s supplement is z.⇒ x + y = 90i.e. y = 90 – xAnd, x + z = 180i.e. z = 180 – xNow, according to question,Six times of complement of x is 12°less than twice of its supplement.i.e. 6y = 2z – 12Putting the value of y and z in this equation,⇒ 6 (90 – x) = 2 (180 – x) – 12⇒ 540 – 6x = 360 – 2x – 12⇒ 6x – 2x = 540 – 360 + 12⇒ 4x = 192⇒ ⇒ x = 48°Hence required angle is 48°.Question 5.In the given figure, ∠ B:∠ C = 2:3, find ∠ B A. 120°B. 52°C. 78°D. 130°Answer:Let ∠ B and ∠ C be 2x and 3x respectively.Now, As we know sum of two interior angles of a triangle is equal to third exterior angle;⇒ ∠ B + ∠ C = 130°⇒ 2x + 3x = 130°⇒ 5x = 130°⇒ x = 26°So, ∠ B = 2x = 2×26° = 52°Hence, required angle is 52°.Question 6.ABCD is a parallelogram, E is the mid – point of AB and CE bisects ∠ BCD. Then ∠ DEC isA. 60°B. 90°C. 100°D. 120°Answer:Given: ABCD is a parallelogram and E is the mid – point of AB such that CE bisects ∠ BCD.construction: Join DE. The figure is given below: Let ∠ BCD = 2x⇒ ∠ BCE = ∠ ECD = ∠ BCD/2 Now, let ∠ ADC = 2yAnd as we joined DE it will also bisect ∠ ADC.Therefore, ∠ EDC = Now, as we know sum of adjacent angles of parallelogram is equal to 180°.⇒ ∠ BCD + ∠ ADC = 180°⇒ 2x + 2y = 180°⇒ 2 (x + y) = 180°⇒ x + y = 90°Now, In triangle CED,∠ CED + ∠ EDC + ∠ DCE = 180°⇒ ∠ CED + x + y = 180°⇒ ∠ CED + 90° = 180°⇒ ∠ CED = 180° – 90°⇒ ∠ CED = 90°Hence required angle is 90°.

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