10 AP = PB AP2 = PB2 1 ( + 4)2 + ( - 3)2 = 2 +( + 2)2 2 + 8 + 16 + 2 - 6 + 9 = 2 + 2 + 4 + 4 1 8 - 6 - 4 + 25 – 4 = 0 8 - 10 + 21 = 0 1 11 From P draw PQ AB AQ = QB = 3cm Join PB. Since arc APC is described by a circle with center B, so BA = BP=BC =6cm. In PQB Cos = = = 60o Area of sector BPA = (62) = 18.84cm 1 Area of BPQ = (QB) (PQ) = (3)( 6 Sin 60) = 7.794Sq.cm Area of portion APQ = Area of sector BPA Area of BPQ – = 18.84 7.794 = 11.046 Sq.cm – Area of shaded portion = 2 x Area of Quadrant ABC 2 Area APQ – = [2 x (6)2 - 2 x 11.046] =34.428 Sq.cm 1
10 AP = PB AP2 = PB2 ( + 4)2 + ( - 3)2 = 2 +( + 2)2 2 + 8 + 16 + 2 - 6 + 9 = 2 + 2 + 4 + 4 8 - 6 - 4 + 25 – 4 = 0 8 - 10 + 21 = 0 1 1 1 11 From P draw PQ AB AQ = QB = 3cm Join PB. Since arc APC is described by a circle with center B, so BA = BP=BC =6cm. In PQB Cos = = = 60o Area of sector BPA = (62) = 18.84cm Area of BPQ = (QB) (PQ) = (3)( 6 Sin 60) = 7.794Sq.cm Area of portion APQ = Area of sector BPA – Area of BPQ = 18.84 – 7.794 = 11.046 Sq.cm Area of shaded portion = 2 x Area of Quadrant ABC – 2 Area APQ = [2 x (6)2 - 2 x 11.046] =34.428 Sq.cm 1 1
453216800class_x_sample_papers.docx