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Question 1. In the quadrilateral ABCD, AD = DC = CB, and ADC = 100, ABC = 130. Then the measure of ACB is   
  1.    20           
  2.    30  
  3.    50  
  4.    cannot be determined
Answer: Option A
: A

Ans. (a) A, B, C will lie on a circle with centre at D (as the angle subtended by the arc at the centre i.e. 260 is twice subtended at the circle i.e. 130) In triangle DAC, DAC = DCA = 40.Let ADB = 2x ACB = x, and let BDC = CBD = y 2x+y = 100, and 2y+x = 140. Hence (a) is the right answer

Question 2. Four identical circles are drawn taking the vertices of a square as centers. The circles are tangential to one another. Another circle is drawn so that it is tangential to all the circles and lies within the square. Find the ratio of the sum of the areas of the four circles lying within the square to that of the smaller circle?

  1.    2(1−√2)
  2.    2(√2−1)2
  3.    1(√2−1)2
  4.    1(√2+1)2
Answer: Option C
: C

Let the radius of the identical circles be r. Hence, Areas of the four circles lying within the square = 4 xπ x r24 =π x r2 Radius of the smaller circle = (diagonalofsquare2r)2 = (22r2r)2 = r (2-1) Required ratio = πxr2πx[x2(21)2] = 1(21)2 Shortcut : Lateral Thinking The question is basically the ratio areas of larger to smaller circle because sum of areas within the square of the 4 circles add up to a full large circle. Approximately compare the radii of small and large circles; the ratio is nearly 3.5/1, square of which is approx. 6/1. Only option c) works. This is a good approach which can be used for a lot of geometry questions.

Question 3. Two rectangles, ¨ABCD and ¨PQRS overlap each other as shown in the figure below. Also, the overlapped area (shaded region) is 20% ¨ABCD and 33.8% of ¨PQRS. If the ratio of corresponding sides of the two rectangles is same is then ratio AD : PR equals
  1.    1.2
  2.    1.5
  3.    1.6
  4.    none of these
Answer: Option D
: D

option (d) From the figure, 0.338xy = O.2ba ab = 1.69xy ..... (1)Also, since ¨ABCD and ¨PQRS are similar,bayxbyax..... (2)From (1) and (2), we get ax = 1.69 xaa2 = 1.69x2 a = 1.3x ax= 1.3 answer = 1.3.

Question 4. Two travelers start walking from the same point at an angle of 1500 with each other at the rate of 4 kmph and 3 kmph. Find the distance between them after 2 hours
  1.    2√25+8√3
  2.    28-48√3
  3.    2√25+12√3
  4.    none of these
Answer: Option C
: C

option c let OA and OB be the paths traveled by the two travelers in 2 hours. Let BCâÂ?´ AO at C. then BOC= 180-150= 30 In right angled triangle OCB, BC= 62=3 KM and OC= 33 km In right angled triangle ACB, AB2= AC2 + BC2 = (8+3√3)2+32= 100+483 AB= 225+123 Shortcut a2+b22abcosα; α is the angle between the paths a and b between the 2 people. 100+48is the answer

Question 5. Find the distance between the incentre and the circumcentre of a triangle with circumradius 6 and inradius 2?
  1.    3√2
  2.    2√3
  3.    √10
  4.    None of these
Answer: Option B
: B

Use Eulers triangle theorem which states that the distance,d between the incentre and circumcentre of a trinagle is given by d2 = R(R-2r) where R = circum radius, r= inradius d2= R(R-2r) = 6(6 - 2 x 2) = 12 d=23

Question 6. A metal cuboid with sides in the ratio of 2:4:6 is melted to form small cubes of side 2 cm. If the sum of all the edges of the cuboid is 144 cm, then find the ratio of the total surface area of the cuboid to the total surface area of the cubes.
  1.    2:1
  2.    13:4
  3.    11:54
  4.    4:1
Answer: Option C
: C

Sides are in the ratio 2:4:6 Volume of cuboid= 2x .4x . 6x = 48 x3 Volume of cube= 2.2.2=8 Number of cubes= 6x3 Length of all edges of the cuboid= 144 =4(2x+4x+6x) x=3cm Ratio = 2[72+108+216]6(27) x 6 x 4 = 1154