SECTION A1. The number 0.47 in the form, q 0 and p, q z isa.b.c.d.2. If x 1 is a factor of √ then the value of k isa.b. √c. √d. √3. The equation y = 4x 7 hasa. infinitely many solutionsb. exactly 2 solutionsc. a unique solutiond. no solution4. Euclid stated that all right angles are equal to each other in the form ofa. an axiomb. a definitionc. a postulated. a proof5. In the given figure, if OP ‖ RS, OPQ = 110 , QRS = 130 then PQR is equal toa. 40b. 60c. 50d. 706. The angles of a triangle ate in the ratio 5:3:7. The triangle isa. an acute angled triangleb. an obtuse angled trianglec. a right triangled. an isosceles triangle7. In two triangles ABC and PQR, A = 30 , B = 70 , P = 70 and Q = 80 and AB = RPthen,a. △ ABC △ PQRb. △ ABC △ QRPc. △ ABC △ RQPd. △ ABC △ RPQ8. In triangles ABC and PQR, AB = AC, C = P and B = Q. The triangles area. isosceles and congruentb. isosceles but not congruentc. congruent but not Isoscelesd. neither congruent nor isoscelesSECTION B9. Represent √ on a number line.10. Factorise11. In the given figure, X and Y are the mid-points of AB and BC respectively. AX = CY. Showthat AB = BC12. If the angles and are complementary angles, find x13. Which of the points A (0, 3), B (1, 0), C (0, -1), D (-5, 0), E (1, 2) and F (-2, -3) do not lie on Xaxis?14. The base of a right triangle is 16 cm and its hypotenuse is 34 cm. Find the area of the triangle..SECTION C15. Writein decimal form.16. Determine the rational numbers a and b if √√√√√17. a) Factorise using suitable identity.b) Write the expansion ofc) Without actually calculating cubes, find the value of18. If (x 2) is a factor of polynomial then find the value of.19. If is a solution of the equation , find the value of . Also findtwo more solutions of the equation.20. In given figure, XOY is a straight line and OQ ⊥ XY at O. Show that 2 QOP = YOP -XOP21. In the given figure, EF ‖ DQ and AB ‖ CD. If FEB = 64 , PDC = 27 then, find PDQ,AED and DEF22. In the given figure AC = AE, AD = AB and BAD = CAE. Prove that BC = DELearnhive Education Pvt. Ltd.www.learnhive.comCopyright © 2013 Learnhive Education Pvt. Ltd.23. In triangle PQR, PQ = QR. S is any point on side PR. Prove that SR SQ.24. Two triangular walls of a flyover have been used for advertisement. The sides of each wall are100 m, 80 m and 40 m. The advertisements yield an earning of Rs. 1200/ per year. Find theamount of revenue earned in one year (Take √SECTION - D25. Find the value ofa)b)26. Rationalise the denominator √ √√27. The polynomials and , whendivided by leave remainders and respectively. If , find the value of k.28. Find zeroes of polynomial , if is a factor of the polynomial.29. Draw the graph of the equation . Find the co-ordinates of the point where thegraph cuts axis.30. Plot the points A (-2, 3), B (-2, 0), C (2, 0) and D (2, 6) on the graph paper. Join themconsecutively. Find the lengths of AC and AD31. In the given figure AD divides angle BAC in the ration 1 : 3 and AD = DB. CAE = 108 .Determine the value of CLearnhive Education Pvt. Ltd.www.learnhive.comCopyright © 2013 Learnhive Education Pvt. Ltd.32. In the given figure, BC ‖ AD and AB ‖ DC. If, ADB = 36finda) xb) zc) ABCd) BAD33. Show that the difference of any two sides of a triangle is less than the third side.34. ABC is an isosceles triangle, with AB = AC. Side BA is produced to a point D such that AB =AD. Prove that BCD is a right angle.

Hi,
33)

Let ABC be a triangle having sides AB, BC and CA.

We know that the sum of any two sides of a triangle is greater than the third side.

∴ AB + BC > CA

⇒AB > CA – BC

This shows that difference of two sides is less than the third side of the triangle.

Similarly, we can  prove it for other sides as well.

Regards

• 1

In ΔABC,

AB = AC (Given)

⇒ ACB = ABC (Angles opposite to equal sides of a triangle are also equal)

In ΔACD,

⇒ ADC = ACD (Angles opposite to equal sides of a triangle are also equal)

In ΔBCD,

ABC + BCD + ADC = 180º (Angle sum property of a triangle)

⇒ ACB + ACB +ACD + ACD = 180º

⇒ 2(ACB + ACD) = 180º

⇒ 2(BCD) = 180º

⇒ BCD = 90º

• 5

thanks

• 4

In ΔABC,

AB = AC (Given)

⇒ ACB = ABC (Angles opposite to equal sides of a triangle are also equal)

In ΔACD,

⇒ ADC = ACD (Angles opposite to equal sides of a triangle are also equal)

In ΔBCD,

ABC + BCD + ADC = 180º (Angle sum property of a triangle)

⇒ ACB + ACB +ACD + ACD = 180º

⇒ 2(ACB + ACD) = 180º

⇒ 2(BCD) = 180º

⇒ BCD = 90º

• 3
AB = AC (Given)

⇒ ACB = ABC (Angles opposite to equal sides of a triangle are also equal)

In ΔACD,

⇒ ADC = ACD (Angles opposite to equal sides of a triangle are also equal)

In ΔBCD,

ABC + BCD + ADC = 180º (Angle sum property of a triangle)

⇒ ACB + ACB +ACD + ACD = 180º

⇒ 2(ACB + ACD) = 180º

⇒ 2(BCD) = 180º

⇒ BCD = 90º
• 1
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