SKDG26U I, 2013 SUMMATIVE ASSESSMENT I, 2013 / MATHEMATICS IX / Class IX

Transcription

1 SKDG6U I, 0 SUMMATIVE ASSESSMENT I, 0 / MATHEMATICS IX / Class IX 90 Time Allowed : hours Maximum Marks : 90 General Instructions: All questions are compulsory. 6 0 The question paper consists of questions divided into four sections A, B, C and D. Section- A comprises of questions of mark each; Section-B comprises of 6 questions of marks each; Section-C comprises of 0 questions of marks each and Section-D comprises of questions of marks each. There is no overall choice in this question paper Use of calculator is not permitted. SECTION A Question numbers to carry mark each. (6) 0.5 Page of 8

2 Find the value of (6) 0.5 f(x)x x 8x x Find the value of the remainder of the polynomial f(x)x x 8x, when it is divided by x. ABCD x In given figure ABCD, find the value of x. 5 A point whose ordinate is 5 and abscissa is will lie in which quadrant? / SECTION B 5 0 Question numbers 5 to 0 carry marks each Simplify : x 7 x 5 Factorise : 6 x 7 x 5 Page of 8

3 7 ABBC BXBY AXCY In the given figure ABBC and BXBY. Show that AXCY. State the Eculid axiom used Prove that the sum of the four angles of a quadrilateral is cm cm cm Find the area of a triangle whose two sides are cm and cm and the perimeter is cm. 0 5 cm cm cm Find the area of a triangle of sides 5 cm, cm and cm. / SECTION C 0 Question numbers to 0 carry marks each p q p q q 0 Express 0 7. in the form p q, where p and q are integers and q p q p q q 0 Page of 8

4 Express in the form of p q, where p and q are integers and q 0. (x) (x ), ax 5xb, ab If (x) and (x ) are both factors of ax 5xb, show that ab. x ax x xb, x a b Find the values of a and b if the polynomial x ax x xb is exactly divisible by the polynomial x. 5 ABC ABD ADBC, BDAC In the figure ABC and ABD are such that ADBC, and. Prove that BDAC 6 DBC ECB BO CO BAC70 ABC0 BOC In the figure, BO and CO are bisectors of DBC and ECB respectively. If BAC70 and ABC0, find the measure of BOC. Page of 8

5 7 ABC BC D ADAC AB > AD D is a point on side BC of ABC (see figure), such that ADAC. AB > AD Show that 8 PR > PQ QPR PS PSR >PSQ In figure PR > PQ and PS bisects QPR. Prove that PSR >PSQ. 9 (, ), (5, ), (, ), (, ) State the quadrants in which the following points lie and also plot the points to verify your answer : (, ), (5, ), (, ), (, ) Page 5 of 8

6 0 8 cm 0 cm cm Find the area of a triangle, two sides of which are 8 cm and 0 cm and the perimeter is cm. Also find the length of the altitude corresponding to the third side. / SECTION D Question numbers to carry marks each Two classmates Salma and Anil simplified two different expressions during the revision hour and explained to each other their simplifications. Salma explains simplification of 5 and Anil explains simplifications of Write both the simplifications. What value does it depict? : Evaluate : : x x x Factorise : x x x y y y Using Factor Theorem, factorise the polynomial y y y. Page 6 of 8

7 5 (pq) 0 (pq) Factorise : (pq) 0 (pq)5 6 x x x0 x5 If x5 is a factor of x x x0, find its other factors. 7 ABAD, APAQ In figure ABAD, and. Prove that APAQ (x7), (x5) (x) x Prove that the sum of three angles of a triangle is 80. Using this result find the value of x and all the three angles if the angles are (x7), (x5) and (x) 9 If two parallel lines are intersected by a transversal, prove that the bisectors of two pairs of interior angles encloses a rectangle. 0 In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle. ABAC,CHCB HKBC CAX7 CHK Page 7 of 8

8 In figure ABAC, CHCB and HKBC. If CAX7 then find CHK. ***** Page 8 of 8

9 MARKING SCHEME SKDG6U SUMMATIVE ASSESSMENT I, 0 MATHEMATICS Class IX SECTION A Question numbers to carry mark each (6) (C). IV Question numbers 5 to 0 carry marks each. SECTION B Given expression x 7 x x x 5 x 5 x x x 5 5 x x 5 7 AB BC AXXB BYYC AX CY ( Q BX BY) Axiom If equals are subtracted from equal the remainders are equal. 8 ABCD is a quadrilateral join AC. In ABC, B (i). ½ In ACD, D (ii). ½ Adding (i) and (ii) ()B()D60 ABCD60 Hence sum of four angles of a quadrilateral60. 9 Perimeter cm a cm, b cm c() Page of 6

10 0 ie. c5 cm a b c s cm Area ss as bs c s 5 5 Area cm 8 cm SECTION C Question numbers to 0 carry marks each. Let x x.7777.() 00x () () () 90x x 90 Let x x xx x.. x 99 x 990 Hence 0. 5 where pq are integers and q p()0 9ab5 p 0 a 9b 5 Solving ab Let f (x)x ax x x6 g (x)x (x) (x) f ()ab0 ab0 f ()ab ab 6 ab0 Page of 6

11 a, b 5, DAB CBA DAB CBA (SAS) BD AC (CPCT) 6 DBC 800 (Linear Pair) 0 7 CBO DBC 070 ACB 80(700)70 BCE 80700(Linear Pair) BCO 055 (Angle bisector) BOC 80( CBO BCO) 80(7055) 55 ADAC (see figure) In ADC, ADCACD. ½ Now ADC is ext. angle of ABD ADCABDBAD. ½ ADC >ABD. ½ Or ACD >ABD. ½ or ACB >ABC AB > AC But ACAD AB > AD (Proved). 8 PR > PQ PQS >PRS () QPSSPR P In PQS PPQSPSQ80 PSQ 80 P PQS () In PSR PSR 80 P PRS () (), () and () PSR >PSQ Page of 6

12 9 Stating the quadrants II, I, IV, III Plotting points in co-ordinate system 0 s cm third side cm 8 0 cm h h cm SECTION D Question numbers to carry marks each Value : cooperative learning among classmates without any gender and religious bias. 5 6 ( ) ( ) ( ) The factors of are, (x) is a factor of x x x x x xx x x xx (x) [x x] (x) (x) (x) p(y)y y y p()() () () 0 y is a factor y y yyy y y y(y)(y) p(y)(y)(y)(y) 5 On putting (pq) a in () we get (pq) 0(pq)5a 0a5 a 5a5a (5) (5) (a 5a)[5a(5) (5)] Page of 6

13 a(a5)5(a5) (a5)(a5) () Replacing a by (pq) on both sides of (), we get (pq) 0(pq)5(pq5) (pq5) 6 To get the other factors, divide x x x0 by x5. x5 x x x0 x x x 5x x x x 5x x0 x0 x x x0(x5) (x x) (x5) (x) (x) 0 7 BACDAC. In DAC and BAC ADAB (given) ACAC (common) DACBAC (proved above) DAC BAC (SAS). ADCABC (cpct).½ In ADQ &ABP ADAB (given) (given) ADQABC (proved above) ADQ ABP (ASA). AQAP (cpct).½ 8 Given, to prove, construction and correct figure correct proof x7 x5x80 6 x50 x5 The angles are, 50, 87 9 Here lm and n is a transversal E G, F G, F H and E H are the bisectors of the interior angles. Now AEFEFD Page 5 of 6

14 AEF EFD GEFEFH. But they are alternate angles. Thus EGFH. Similarly FGEH. EGFH is a parallelogram. Again AEFBEF80 (Linear pair) AEF BEF90 GEFHEF90 GEH90 EGFH is a rectangle. 0 Let ABC is isosceles triangle such that AB AC B C x (say) A (BC) (given) x In ABC A B C 80 xxx80 x0a0, B0, C0 XAKKAH80 (LP) KAH807. ABAC ABCACB ½ CHCB CBACHB68.5.½ HCB807. CHKHCB (alternate angles). Page 6 of 6