From a point O in the interior of a DABC if perpendiculars OD, OE and OF are drawn to the sides BC, CA and AB respectively, then which of the -Maths 9th

From a point O in the interior of a DABC if perpendiculars OD, OE and OF are drawn to the sides BC, CA and AB respectively, then which of the -Maths 9th
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 (i) In Δ O C E  ,D C 2 = D E 2 + E C 2 Δ O B D , D B 2 = O D 2 + B D 2 Δ O A F , O A 2 = O F 2 + A F 2 Adding we get O A 2 + O B 2 + O C 2 = O F 2 + O D 2 + O F 2 + E C 2 + B D 2 + A F 2 A F 2 + B D 2 + C E 2 = O A <span style="font-size:
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From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. -Maths 9th

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In Fig. 8.53,ABCD is a parallelogram and E is the mid - point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L.Prove that (i) AF = 2DC (ii) DF = 2DL -Maths 9th

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If a+b+c= 5 and ab+bc+ca =10, then prove that a3 +b3 +c3 – 3abc = -25. -Maths 9th

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If AB = QR, BC = PR and CA = PQ, then -Maths 9th

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If (log x)/(a^2+ab+b^2) = (log y)/(b^2+bc+c^2) = (log z)/(c^2+ca+a^2), then x^(a-b). y^(b-c). z^(c-a) = -Maths 9th

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Last Answer : a+b+c=9 and a2+b2+c2=35 Using formula, (a+b+c)2=a2+b2+c2+2(ab+bc+ca) 92=35+2(ab+bc+ca) 2(ab+bc+ca)=81−35=46 (ab+bc+ca)=23 using formula, (a3+b3+c3)−3abc=(a2+b2+c2−ab−bc−ca)(a+b+c) a3+b3+c3−3abc=(35−23)×9=9×12=108

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In figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to -Maths 9th

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Last Answer : (a) We know that, the perpendicular from the centre of a circle to a chord bisects the chord. AC = CB = 1/2 AB = 1/2 x 8 = 4 cm given OA = 5 cm AO2 = AC2 + OC2 (5)2 = (4)2 + OC2 25 = 16 + OC2 ... length is always positive] OA = OD [same radius of a circle] OD = 5 cm CD = OD - OC = 5 - 3 = 2 cm

In figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to -Maths 9th

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If a + b + c = 9 and ab + bc + ca = 26, find a2 + b2 +c2. -Maths 9th

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A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB, BC and CA are concurrent. -Maths 9th

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If a+b+c=5 and ab+bc+ca=10 find the value of a^3+b^3+c^3-3abc -Maths 9th

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Last Answer : We know , a³ + b³ + c³ -3abc = (a + b + c )(a² + b² + c² -ab -bc-ca) now , a + b + c = 5 ab + bc + ca = 10 (a + b + c)² = a² + b² + c² +2(ab + bc+ca) (5)² -2 10 = a² + b² + c² a² + b² + c² = ... )(a² + b² + c² -ab- bc-ca) =( 5)( 5 - 10) = 5 (-5) = -25 Hope this will help u..... by :- RAXTAR.....