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F4 math problems ( Circles )

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1.In the figure, TA, TB and PQ are tangents to circle and O is the centre of the circle. TA ⊥ TB, AP = 2cm and BQ = 3 cm.Fing the radius of the circle.Figure http://server7.pictiger.com/img/421430/picture-hosting/--.jpg2.In the figure, a circle is inscribed in trianglePQR, PR and PQ touch the circle at A and B... 顯示更多 1.In the figure, TA, TB and PQ are tangents to circle and O is the centre of the circle. TA ⊥ TB, AP = 2cm and BQ = 3 cm. Fing the radius of the circle. Figure http://server7.pictiger.com/img/421430/picture-hosting/--.jpg 2.In the figure, a circle is inscribed in trianglePQR, PR and PQ touch the circle at A and B respectively. IF AB//RQ, show that PQ = PR. Figure http://server7.pictiger.com/img/421456/picture-hosting/7.jpg

最佳解答:

( 1 ) By properties of tangents, AP = PR = 2cm , BQ = QR = 3cm and BT = AT Hence PQ = 5 cm Let PT = x and QT = y. Considering ΔPQT, x2 + y2 = 52 ( Pyth. Theorem ) --- ( 1 ) 3 + y = 2 + x ( since BT = AT ) x = y + 1 --- ( 2 ) Put ( 1 ) into ( 2 ). ( y + 1 )2 + y2 = 25 y2 + 2y + 1 + y2 = 25 y2 + y – 12 = 0 ( y – 3 )( y + 4 ) = 0 y = 3 or y = - 4 ( rejected ) So x = 3 + 1 = 4 Then BT = AT = 6cm Besides, TA ⊥ TB and the tangents are ⊥ radius with OB = OA, therefore OBTA is a square. Hence the radius of the circle is 6cm. ( 2 ) PA = PB ( properties of tangents ) To proveΔAPB ~ΔRPQ ∠APB = ∠RPQ ( common ∠ ) ∠PAB = ∠PRQ ( corr. ∠s , AB // RQ ) SoΔAPB ~ΔRPQ ( AA ) Hence, PQ = QR ( corr. sides, ~Δ )

其他解答:

1. Let r be the radius of the circle AP = PR = 2 [reason?] BQ = QR = 3 [reason?] PQ = PR + RQ = AP + BQ = 2 + 3 = 5 OA = OB = AT = BT = r [reason?] PT^2 + QT^2 = PQ^2 (path. th.) => (r - AP)^2 + (r - BQ)^2 = 5^2 => (r - 2)^2 + (r - 3)^2 = 25 => 2r^2 -10r -12 = 0 => r^2 - 5r -6 = 0 => (r - 6)(r + 1) = 0 => r = 6 (r = -1 rejected) So the radius of the circle is 6 2. PA = PB [reason ?] PA/PR = PB/PQ (since AB // RQ) => PQ = PR (since PA = PB)
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