A semi-circle of diameter 14 cm has three chords of equal length connecting the two end points of the diameter so as to form a trapezoid inscribed within the semi-circle. What is the value of the area enclosed by the trapezoid ?
Answer: C Draw a entire circle and the same trapezoid in another semicircle. Thus it becomes a regular hexagon. ∠BAD = 60° AB = AO = 7 cm Area of regular hexagon ABCDEF = 6 *√3/4 * (side)2 Area of trapezoid = 1/2 * 6 * √3/4 *
side)2 =1/2 * 6 * √3/4 * 7 * 7 = 147√3/4 sq. cm.
Q. No. 26:
A filter paper of the form of a right circular cone of base radius 20 cm and altitude 40 cm is placed with its axis vertical and the vertex downwards. Water flows out at the rate of 22.5 cc. The rate at which the level of the water falls when the depth of the water is 30 cm is :
Read the following information and answer the questions that follow ::
A bucket is in the shape of an inverted truncated right-circular cone with a base radius of 20cm and height 35 cm. The base angles, of a vertical cross sections through the centre of the base, are 1300 each. It contains water whose height is 10 cm. A solid iron ball of radius 5*(74)1/3 cm is dropped into the bucket.
Answer: A Length = 6m + 24 cm = 624 cm Width = 4m + 80 cm = 480 cm HCF of 624 and 480 = 48 cm. Hence, minimum number of identical squares = (486*624)/(48*48) = 130.
Q. No. 29:
Each side of a regular hexagon measures a cm. By joining the mid points of each side, another hexagon is formed inside it. The ratio of the areas of the outer and inner hexagons would be
Answer: A Area of outer hexagon : Area of inner hexagon 6*√3 /2 *a2 : 6 *√3 /2 *(√3 a/2)2 = 4 : 3.
Q. No. 30:
In a triangle ABC the length of side BC is 295. If the length of side AB is a perfect square, then the length of side AC is a power of 2 and the length of side AC is twice the length of side AB. Determine the perimeter of the triangle.
Answer: C Let the length of side AB = 2k-1 Then, side AC = 2k As, side BC = 295, the powers of 2 satisfy the given conditions will be 256 and 512 respectively. Thus the perimeter = 256+295+512 = 1063.
