Free Geometry Set II - 02 Practice Test - CAT 

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= $\frac{2}{3}$(height of the triangle xyz)

Height=$\left(\frac{\sqrt{3}}{2}\right)$(2R) =$\sqrt{3}$R

Therefore R+r = $\frac{2}{3}$($\sqrt{3}$R)  $\Rightarrow$ $\frac{r}{R}$ = $\frac{(2-\sqrt{3})}{\sqrt{3}}$ 

Shortcut:- Using the approximation technique used in class, the radius of the bigger circle: smaller circle is close to 0.2.

 In a 45-45-90 triangle, sides are in the ratio 1:1:$\sqrt{2}$

Thus OB= 150

In a 30-60-90 triangle, sides are in the ration 1:$\sqrt{3}$:2

Thus OA = $\frac{50}{\sqrt{3}}$

Distance travelled in 2 mins = AB = 150- 50$\sqrt{3}$= 63.4 m

speed = $\left(\frac{63.4}{2}\right)$ $\times$ $\left(\frac{60}{1000}\right)$ = 1.9 kmph

option (d)

From the figure, 0.338xy = O.2ba $\therefore$ ab = 1.69xy ..... (1)Also, since ¨ABCD and ¨PQRS are similar,$\frac{b}{a}-\frac{y}{x}\therefore \frac{b}{y}-\frac{a}{x}$..... (2)From (1) and (2), we get $\frac{a}{x}$ = 1.69 $\frac{x}{a}$a${}^2$ = 1.69x${}^2$       a = 1.3x $\frac{a}{x}$= 1.3          answer = 1.3.

Sides are in the ratio 2:4:6

Volume of cuboid= 2x .4x . 6x = 48 x${}^3$

Volume of cube= 2.2.2=8

Number of cubes= 6x${}^3$

Length of all edges of the cuboid= 144 =4(2x+4x+6x) $\Rightarrow$  x=3cm

Ratio = 2$\frac{[72+108+216]}{6\left(27\right)}$ x 6 x 4

= $\frac{11}{54}$

Number of cubes = $\frac{2a\times 3a\times 9a}{3\times 3\times 3}$ = 2a${}^3$
Diagonal of a cube = $\sqrt{4a^2+9a^2+81a^2}$
Sum of the diagonals of cubes = 2a${}^3$ $\times$ 3$\sqrt{3}$
Ratio = $\frac{\sqrt{4a^2+9a^2+81a^2}}{2a^3\times 3\sqrt{3}}$

Let M be the end point of the altitude on the hypotenuse. Since, we are dealing with right angle
triangles, $\Delta$MAC ~ $\Delta$ABC, so AM = $\frac{12}{5}$. Let N be the endpoint he reaches on side AC. $\Delta$
MAC ~ $\Delta$ NAM, So, $\frac{MN}{AM}$ =$\frac{4}{5}$. This means that each altitude that he walks gets shorter
by a factor of $\frac{4}{5}$. The total distance is thus an infinite G.P. =$\left[\frac{a}{1-r}\right]$ = 12

Graphical Division: - 

Assume the square to have a side 2a. Hence, area of square = 4a${}^2$

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 $\frac{2}{8}$${}^{th}$ 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

Altitude of the traingular faces = $\sqrt{\frac{a^2}{4}+9}$
Area of faces =2 x a x $\sqrt{\frac{a^2}{4}+9}$
The total surface area = area of base + lateral surface area
                                    = a${}^2$ + 2 x a x $\sqrt{\frac{a^2}{4}+9}$
Lets us put a=8 (smallest value whch will yield an integer area, Area = 64 + 2 $\times$ 8 $\times$ 5 =144)

= 12${}^2$

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${}^{\circ }$ is twice subtended at the circle i.e. 130${}^{\circ }$)

In triangle DAC,  $\angle$DAC = $\angle$DCA = 40${}^{\circ }$.Let $\angle$ADB = 2x $\Rightarrow$ $\angle$ACB = x, and let $\angle$BDC = $\angle$CBD = y $\Rightarrow$ 2x+y = 100${}^{\circ }$, and 2y+x = 140${}^{\circ }$.

Hence (a) is the right answer

Use Eulers triangle theorem which states that the distance,d between the incentre and circumcentre of a trinagle is given by d${}^2$ = R(R-2r) where R = circum radius, r= inradius

d${}^2$ = R(R-2r) = 6(6 - 2 x 2) = 12

d=2$\sqrt{3}$

option c

let OA and OB be the paths traveled by the two travelers in 2 hours. Let BC� AO at C. then $\angle$BOC= 180-150= 30${}^{\circ }$

In right angled triangle OCB, BC= $\frac{6}{2}$=3 KM and OC= 3$\sqrt{3}$ km

In right angled triangle ACB, AB${}^2$= AC${}^2$ + BC${}^2$ = (8+3√3)${}^2$+3${}^2$= 100+48$\sqrt{3}$

AB=  2$\sqrt{25+12\sqrt{3}}$

Shortcut

$\sqrt{a^2+b^2-2abcos\alpha }$; α is the angle between the paths a and b between the 2 people. $\sqrt{100+48}$ is the answer

Option (b)

 O is the centre of the circle and the mid-point of BC. DO is parallel to AC. So, $\angle$DOB = 60°

 Area of Δ BDO =$\frac{3}{4}$ * 49

Area of sector OBD = $\frac{49}{6}$

Hence area of the shaded region

 = 2[$\frac{49}{6}$ – $\frac{3}{4}$*49]

= 49[$\frac{1}{3}$ – $\frac{3}{2}$]

 Shortcut

By graphical division, there are 3 equilateral triangles of areas of $\frac{\sqrt{3}}{4}$ * 49

Area of interest = (area of semi circle [$\frac{r^2}{2}$] - area of three triangles) = (49*$\frac{1}{2}$ - 3*3*$\frac{49}{4}$)*($\frac{2}{3}$) = 49*($\frac{1}{3}$ – $\frac{3}{2}$)

Let the radius of the identical circles be r. Hence, Areas of the four circles lying within the square = 4 x $\pi$ x $\frac{r^2}{4}$ = $\pi$ x $r^2$ 

Radius of the smaller circle = $\frac{(diagonalofsquare-2r)}{2}$ = $\frac{(2\sqrt{2}r-2r)}{2}$ = r ($\sqrt{2}$-1) 

Required ratio = $\frac{\pi x{r}^2}{\pi x\left[x^2\right(\sqrt{2}-1{)}^2]}$ = $\frac{1}{(\sqrt{2}-1{)}^2}$

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.

Interior angle = [(2n-4)x90]/n = 108

2x= 360-(108x2) = 144

angle P = 180-144= 36

Total = 36x5= 180

 Useful to remember: Sum of angles in a star= 180${}^{\circ }$

 Shortcut: Clearly the star point trisects the angle in a pentagon. Hence each angle = 108/3 = 36.

Sum of angles = 36*5 = 180

Alternatively:  Sum of angles of a n point star = (n – 4) x 180${}^{\circ }$ , where n = number of sides of a polygon.

Here, n = 5. Sum of angles of a n point star = (5 – 4) x 180${}^{\circ }$ = 180${}^{\circ }$.

AB = BC and $\angle$ABC=60°. Therefore, ΔABC is an equilateral triangle

Now see that ABOP is a rectangle.

And $\angle$BAN=60°, Therefore, $\angle$NAP=90${}^{\circ }$ − 60${}^{\circ }$ = 30${}^{\circ }$

And $\angle$ANP =$\frac{1}{2}$ * 90 = 45${}^{\circ }$

Now in ΔANP,

$\angle$NPA=180${}^{\circ }$−45${}^{\circ }$−30${}^{\circ }$=105${}^{\circ }$

And hence $\angle$NPO = $\angle$NPA - $\angle$OPA = 105${}^{\circ }$ - 90${}^{\circ }$ = 15${}^{\circ }$

Shortcut

ABC is an equilateral triangle and ABM is a 30 – 60 -90 triangle (M being the point of intersection of AN and the circle). OMN is also 30${}^{\circ }$. MOP = 90${}^{\circ }$, and MNP = 45${}^{\circ }$; MPO = PMO = 45${}^{\circ }$. NPO = 180${}^{\circ }$ – 75${}^{\circ }$ – 45${}^{\circ }$ – 45${}^{\circ }$ = 15${}^{\circ }$