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abcd is a square of side 8 , m is the centre of the circle taking ad as diameter , E is a point on the side ab such that Ce is tangent to the circle . find the area of the triangle cbe it's urgent no spam please give right answer I will mark brianlist |
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Answer» Step-by-step explanation: (i)AB, BC and AC are tangents to the circle at E, D and F. BD=30cm,DC=7cm,∠BAC=90 From the theorem stated, BE=BD=30cm Also FC=DC=7cm Let AE=AF=x …. (1) Then AB=BE+AE=(30+x) AC=AF+FC=(7+x) BC=BD+DC=30+7=37cm Consider right TRIANLGE ABC, by Pythagoras theorem we have BC 2 =AB 2 +AC 2
(37) 2 =(30+x) 2 +(7+x) 2
1369=900+60x+x 2 +49+14x+x 2
2x 2 +74x+949–1369=0 2x 2 +74x–420=0 x 2 +37x–210=0 x 2 +42x–5x–210=0 x(x+42)–5(x+42)=0 (x–5)(x+42)=0 (x–5)=0or(x+42)=0 x=5orx=–42 x=5[SINCE x cannot be negative] ∴AF=5cm[From (1)] ThereforeAB=30+x=30+5=35cm (II) AC=7+x=7+5=12cm Let ‘O’ be the centre of the circle and ‘r’ the radius of the circle. Join point O, F; points O, D and points O, E. From the figure, 2 1 ×AC×AB= 2 1 ×AB×OE+ 2 1 ×BC×OD+ 2 ×AC×OC AC×AB=AB×OE+BC×OD+AC×OC 12×35=35×r+37×r+12×r 420=84r ∴r=5 Thus the radius of the circle is 5 cm. |
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