REVISION TEST FOR IGCSE EXAM

GEO TUITION CENTRE , KOTTAYAM , KERALA STATEREVISION TESTIGCSE FORM 4[1] The diagram shows a square of side 8 cm and four congruent triangles of height 7cm (a) (i) calculate the area of one triangle. (ii) the area of the whole shape (b) The shape is the net of a solid. Write down the special name of this solid[2] A straight road between P and Q is shown in the diagram. R is the point south of P and east of Q.PR = 8.3 km and QR = 4.8 kmCalculate(a) the length of the road PQ (b) the bearing of Q from P [3] In the diagram the lines AB and CD are parallel. The lines AD and BC intersect at XAngle XDC = 350 and angle CXD = 1200(a) (i) Write down the size of angle BAX (ii) Write down the size of angle ABX(b) Complete the following statement: Triangle AXB is ...................................to triangle DXC(c) AB = 8.3 cm , BX = 5.5 cm and CD = 16.6 cm.Calculate the length of CX.[4] In the diagram AB is the diameter of a circle, centre O. The length of AB is 12 cm.(a) Write down the the size of angle APB(b) Given angle PAB = 40o.Calculate the length of PB(c) calculate the area of the circleREVISION TEST 1 / ANSWERS 1] (a) (i) Area of triangle, A = ½ bh = ½ x 8 x 7 = 28 cm2 (ii) Total area = (4 x 28) + base area = 112 + (8x8) = 176 cm2 (b) Pyramid2] (a) PQ = ( to 3 s.f) (b) In the triangle , PQR , tan L QPR = 4.8 / 8.3 = 0.5783 ………. So angle QPR = tan-1 0.5783….. = 30.040 So the required bearing is angle NPQ = (180 + 30.04 ) = 2100 ( to 3 s. f)3] (a) (i) L BAX = 350 ( alternate angles) (ii) L AXB = 1200 ( vertically opposite angles) So angle ABX = 180 - (120 + 35) = 250 (b) SIMILAR triangles (c) Therefore CX = 5.5 x 2 = 11 cm4] (a) L APB = 900 ( all angles in a semicircle are 900 each) (b) sin 40 = So PB = 12 x sin 40 =12 x 0.64278………. = 7.713 …….. = 7.71 cm( 3 s.f) (c) Area of the circle = r2 = 3.142 x 6 x 6 = 113.112 = 113 cm2 (3 s . f)