Exams > Cat > Quantitaitve Aptitude
GEOMETRY SET II MCQs
:
A
Option (a)
Let the centers of the large balls be x, y, z and radius R.
O is the centre of the smaller ball and radius r...
x, y, z form an equilateral triangle with side equal to 2R.
O is the centroid of this triangle.
Therefore ox=oy=oz=R+r= 23(height of the triangle xyz)
Height=(√32)(2R) =√3R
Therefore R+r = 23(√3R) ⇒ rR = (2−√3)√3
Shortcut:- Using the approximation technique used in class, the radius of the bigger circle: smaller circle is close to 0.2.
:
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
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B
Number of cubes = 2a×3a×9a3×3×3 = 2a3
Diagonal of a cube = √4a2+9a2+81a2
Sum of the diagonals of cubes = 2a3 × 3√3
Ratio = √4a2+9a2+81a22a3×3√3
Suresh is standing on vertex A of triangle ABC, with AB = 3, BC = 5, and CA = 4. Suresh walks according to the following plan: He moves along the altitude-to-the-hypotenuse until he reaches the hypotenuse. He has now cut the original triangle into two triangles; he now walks along the altitude to the hypotenuse of the larger one. He repeats this process forever. What is the total distance that Suresh walks?
:
C
Let M be the end point of the altitude on the hypotenuse. Since, we are dealing with right angle
triangles, ΔMAC ~ ΔABC, so AM = 125. Let N be the endpoint he reaches on side AC. Δ
MAC ~ Δ NAM, So, MNAM =45. This means that each altitude that he walks gets shorter
by a factor of 45. The total distance is thus an infinite G.P. =[a1−r] = 12
:
B
Altitude of the traingular faces = √a24+9
Area of faces =2 x a x √a24+9
The total surface area = area of base + lateral surface area
= a2 + 2 x a x √a24+9
Lets us put a=8 (smallest value whch will yield an integer area, Area = 64 + 2 × 8 × 5 =144)
= 122
:
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
:
C
Graphical Division: -
Assume the square to have a side 2a. Hence, area of square = 4a2
Using graphical division, we can divide the figure into 8 parts as follows. The triangle in question is the shaded part
Thus the triangle is 28th of the square area
Hence ratio = 1:4
Shortcut:- Assumption method. Assume a simple case as the isosceles triangle and square in this case. Just substitute and solve
:
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= 3√3 km
In right angled triangle ACB, AB2= AC2 + BC2 = (8+3√3)2+32= 100+48√3
AB= 2√25+12√3
Shortcut
√a2+b2−2abcosα; α is the angle between the paths a and b between the 2 people. √100+48 is the answer