– ( + ) = If are zeroes of the cubic polynomial then: * * * Division algorithm for polynomials: If then we can find polynomials ( Level - 1 ) If the two zeroes of the polynomial 4 and 7 find he polynomial Ans ; x2 -11x +28 If are the zeroes of then find Ans. (-1) Find a quadratic polynomial whose zeroes are Ans. If then find its sum and product of zeroes. Ans. Sum=15, Product = If the sum of zeroes of a given polynomial is 6. Find the value of K. Ans. k = 2 Find the zero of polynomial Ans. -4/3 Write the degree of zero polynomial. Ans: not defined Ans:(D) ( Level - 2 ) Form a cubic polynomial with zeroes 3, -2 and -4. Ans: x3+3x2-10x-24 Hints: Find the zeroes of the quadratic polynomial 10x2 +3x -4 and verify the relationship between the zeroes and the coefficients. Ans. Zeroes are -4/5,1/2. For what value of k, (-4) is a zero of polynomial Ans. k=9 Give an example of polynomials Ans. Find the zeroes of Ans. 0, -2 Find a quadratic polynomial, whose the sum and product of its zeroes are Ans. ( Level - 3 ) Find the zeroes of polynomial Ans. -1, 1, 2 If the zeroes of the polynomial are . Find Ans. Divide by Ans. Quotient=; Remainder Check whether the polynomial is a factor of polynomial
In the given figure, two line segments AC and BD intersect each other at the point P such that PS = 6 cm, PB = 3 cm, Pc = 2.5 cm, ∠APB = 50° and ∠CDP = 30°. Then, ∠PBA is equal to (A) 50° (B) 30° (C) 60° (D) 100° (Ans. D) If in two triangles DEF and PQR, ∠D=∠Q and ∠R=∠E, then which of the following is not true? (A) (B) (C) (D) (Ans. B) In Triangles ABC and DEF, ∠B = ∠E, ∠F = ∠C and AB= 3 DE. Then, the two Triangles are (A) Congruent but not similar (B) similar but not congruent (C) Neither congruent nor similar (D) congruent as well as similar (Ans. B) It is given that ∆ ABC ~ ∆ PQR, with , Then, is equal to (A) 9 (B) 3 (C) (D) (Ans. D) It is given that ∆ ABC ~ ∆ DEF, ∠A = 30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF = 7.5 cm. The, the following is true: (A) DE = 12 cm, ∠F = 50° (B) DE = 12 cm, ∠F = 100° (C) EF = 12 cm, ∠D = 100° (D) EF = 12 cm, ∠D = 30° (Ans. A) If in triangle ABC and DEF, , then they will be similar, when (A) ∠B = ∠E (B) ∠A = ∠D (C) ∠B = ∠D (D) ∠A = ∠F (Ans. C)
OBJECTIVE:- To verify the above statement through activity. PRE-REQUISITE KNOWLEDGE:- The concept of similar triangles. Division of a line segment into equal parts. Construction of lines parallel to given line. MATERIAL REQUIRED:- White paper sheet Scale /Rubber Paint box Black ball point pen or pencil TO PERFORM THE ACTIVITY:- STEPS:- On the poster paper sheet, draw two similar triangle ABC and DEF. We have the ratio of their corresponding sides same and let as have AB: DE= BC: EF=CA: FD=5:3 ie , AB/DE=5/3 , BC/EF=5/3 , CA/FD =5/3, ie DE =3/5 AB, EF=3/5 BC,FD=3/5 CA Divide each side of ∆ABC into 5 equal parts and those of ∆DEF into 3 equal parts as shown in Fig (i) and (ii). By drawing parallel lines as shown in Fig (i) and (ii)., we have partition ∆ABC into 25 smaller triangle of same size and also each smaller triangle in fig (i) has same size and as that of the smaller triangle fig (ii). Paint the smaller triangle as shown in Fig (i) and (ii).. OBSERVATION:- Area of ∆ABC= Area of 25 smaller triangle in fig (i)=25 square unit Where area of one smaller triangle in fig (i)=P (square unit ) Area of ∆DEF=Area of a smaller triangle in Fig (ii)=9p where area of one smaller triangle in fig (ii)=P square units.
girls On three different sheets of paper find mean height on the sheet of paper , median height on the second and the modal height on the third sheet of paper. Let us find mean by step deviation method:- Class of heights (in cm) Frequency p Class mark xi U1= a= 147.5,h=5 Fi x ui 135-140 140-145 145-150 150-155 155-160 160-165 4 7 18 11 6 5 137.5 142.5 147.5 152.5 157.5 162.5 -2 -1 0 1 2 3 -8 -7 0 11 12 13 Mean=a+h x =147.5+5 x 23/51 =147.5+115/51 =(147.5+2.255)cm=149.755cm Let us find median of the data:- Class of height (in cm) Frequency number of girls Cumulative frequency 135-140 140-145 145-150 150-155 155-160 160-165 4 7 18=f 11 6 5 4 11=cf 29 40 46 51 Total n median=l+ x h=145 + x5 =145+14.5 x5 18 =145+4.028 =149.028cm Let us find mode of the data:- Class of heights (in cm) FREQUENCY (No of Girls) 135-140 140-145 145-150 150-155 155-160 160-165 4 7=f1 18=fm 11=f2 6 5 Total 51 (Modal class) Modal class is 145-150 Thus l=145, h=5, fm=18, f1=7, f2=11 Mode=H xh=145 + x 5