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.



Please post single query in a thread. 

Regards

  • 1

In ΔABC,

AB = AC (Given)

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

In ΔACD,

AC = AD

⇒ 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,

AC = AD

⇒ 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,

AC = AD

⇒ 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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