SNAP Geometry Questions

Let AB represent the rock cliff with its height be h

 D represents the initial point and C is the point at which he observers the boat after Ten minutes less than half of an hour (20 minutes).

BD = $$\frac{h}{\tan\ \left(30\right)}\ =\ \sqrt{\ 3}h$$

BC = $$\frac{h}{\tan\ \left(60\right)}\ =\ \frac{h}{\sqrt{\ 3}}$$

In 20 minutes boat covered CD length which is = $$\sqrt{\ 3}h\ -\ \frac{h}{\sqrt{\ 3}}=\ \frac{2h}{\sqrt{\ 3}}$$

Hence remaining lenght $$\frac{h}{\sqrt{\ 3}}$$(BC) will be covered in 10 minutes

The above area can be obtained by the following operation: Area of Big Square - Area of the Bigger circle + $$4\times\ $$ Area of Smaller Circle + $$8\times\ $$ Overlapping Area of smaller circles

i) Area of Big Square = $$24\times\ 24=576$$

ii) Area of Big circle = $$\pi\ 12^2\ =\ 144\pi\ $$

iii) Area of smaller circle = $$\pi\ 6^2\ =\ 36\pi\ $$

iv) For Area of smaller common area = 2b from below figure

Area b = Area of circular arc(2b+a) - Area of triangle (a+b)

= $$\frac{1}{4}\left(36\pi\ \right)-\ \frac{1}{2}\left(6\times\ 6\right)$$  =$$9\pi\ -18$$

Area of common part of smaller circle = 2b = $$18\pi\ -36$$

Therefore required area = $$576-144\pi+\left(4\times\ 36\pi\right)-8\times\ \left(18\pi\ -36\right)$$

= $$288+144\pi\ $$    

None of the above are correct

Let the point between A and C be D

Triangle ABD and Triangle BCD are congruent using ASA congruency 

Therefore 2x=3y and 4x=5y+10

On solving X=15 and Y=10

Midpoint of P(x1,y1) , Q(x2,y2) = C($$\frac{x1+x2}{2},\ \frac{y1+y2}{2}$$)

So $$y=\frac{4-2}{2},\ -1=\ \frac{x+4}{2}$$ 

On solving x=-6 and y=1

AC= $$\sqrt{\ AB^2\ +BC^2}=\ \sqrt{\ 6^2+8^2\ }=10$$

In-radius of right angled triangle = $$\frac{\left(a+b-h\right)}{2}$$

Inradius = $$\frac{\left(6+8-10\right)}{2}=2$$

Since AB is the diameter $$\angle\ ADB\ =\ 90^0$$

Which gives $$\angle\ DAB\ =\ 50^0$$

ABCD is cyclic quadrilateral so $$\angle\ A+\angle\ C\ =\ 180^0$$

Thus $$\angle\ DCB\ =\ 130^0$$

In$$\bigtriangleup\ $$ BCD, 2x+130 = 150

$$x\ =\ 25^0$$

Total surface area of cube = $$6\times\ 8^2\ =384$$

Amount of paint required = $$\ \frac{\ 384}{16}=24\ kgs$$

Cost of 24kgs paint = $$24\times\ 36.5\ =876$$

$$m^2-n^2=\ \left(m+n\right)\left(m-n\right)=\left(2\tan\theta\ \right)\left(2\sin\theta\ \right)=4\tan\theta\ \sin\theta\ $$

Going by the options you have only mn or m/n

Case 1 : mn

mn= $$\left(\tan\theta\ +\sin\theta\right)\ \left(\tan\theta\ -\sin\theta\ \right)$$ = $$\tan^2\theta\ -\sin^2\theta$$ = $$\frac{\sin^2\theta}{\cos^2\theta\ }\ -\sin^2\theta\ =\sin^2\theta\ \left(\frac{1}{\cos^2\theta\ }-1\right)=\frac{\sin^2\theta}{\cos^2\theta\ }\ \left(1-\cos^2\theta\ \right)=\tan^{^2}\theta\ \cdot\sin^2\theta\ $$

So $$4\sqrt{\ mn}\ =\ 4\tan\theta\ \cdot\sin\theta\ $$

TPO is a right-angled triangle.

So $$\sin\left(30\right)=\frac{OP}{TO}=\frac{3}{TO}\ =>TO=6$$

Applying Pythagoras theorem in the triangle TPO, we get,

$$PT^2=TO^2-OP^2=36-9=27$$

PT= 3$$\sqrt{\ 3}$$ cm

We have MN || AB 
Now let area of CMN be A 
so area of ABMN will be 2A 
Now Area of triangle ABC will be 3A

Now since MN || AB 
so we can say CMN is similar to CAB
So we can say
(CM/CA)^2 = Area of CMN : Area of CAB
so we can say CM :CA =$$1:\sqrt{\ 3}$$
Now therefore CM :AM = $$1:\sqrt{\ 3}-1\ =\ \frac{\ 1\ \left(\sqrt{\ 3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{\ 3}+1\right)}=\ \sqrt{\ 3}+1:2$$

$$\angle\ DCB\ =90^0$$ (Since each angle in a square is 90)

$$\angle\ BCE\ =60^0$$ (Since each angle in an equilateral triangle is 60)

$$\angle\ DCE\ =90^0+ 60^0$$  = $$150^0$$

In $$\triangle\ DCE$$, the sides DC and CE are equal, so the angles opposite to them are equal.

$$\angle\ DCE$$ = $$\angle\ CDE$$ = $$\ \frac{\ 180-150}{2}$$

=$$15^0$$

A is the correct answer.

Let r be the radius of the circular plot

Perimeter of the plot = $$2\pi\ r$$

Cost of fencing the circular plot at Rs 15 per meter =  $$2\pi\ r$$ *15

$$2\pi\ r$$ *15 = 3300

r = 35m

Cost of flooring per sq m = Rs 100

Area  of the circular plot = $$\pi\ r^2$$ 

= 3850 sq m

Cost  of flooring the circular plot = 3850*100 = Rs.385000

A is the correct answer.

The intercept form of the line = $$\ \frac{\ x}{a}+\ \frac{\ y}{b}\ =\ 1$$ where a is x-intercept and b is y-intercept.

3x + 4y - 12 = 0

3x + 4y = 12 

$$\ \frac{3\ x}{12}+\ \frac{\ 4y}{12}\ =\ 1$$

$$\ \frac{\ x}{4}+\ \frac{\ y}{3}\ =\ 1$$

So the intercepts are 4 and 3.

B is the correct answer.

Let the radius of solid sphere be r.

The volume of sphere $$V_s$$= $$\ \frac{\ 4}{3}$$ πr³

It is melted and recast into a right circular cone of radius r and height h.

The volume of cone $$V_c$$= $$\ \frac{\ 1}{3}$$ πr²h

The same volume of material is used for recasting. Therefore,

$$V_s=V_c$$

$$\ \frac{\ 4}{3}$$πr³ =  $$\ \frac{\ 1}{3}$$ πr²h

h = 4r

$$\ \frac{\ h}{r}$$  = 4
D is the correct answer.

Area of the park = (60 x 40) m² = 2400 m².

Area of the lawn = 2109 m².

Area of the crossroads = (2400 - 2109) m² = 291 m².

Let the width of the road be x metres.

Then, 60x + 40x - x² = 291

x² - 100x + 291 = 0

(x - 97)(x - 3) = 0

x = 3. 

B is the correct answer

AD is parallel to BE  is parallel to CF 

$$\ \frac{\ AB}{BC}\ =\ \ \frac{\ DE}{EF}$$   [Basic proportionality theorem]

$$\ \frac{\ 2}{4}\ =\ \ \frac{\ 1.5}{EF}$$

EF = 3 cm

B is the correct answer.

Side of an equilateral triangle= 24 cm

The mid-points of its sides are joined to form another triangle whose mid-points are joined to form another triangle. This process continues indefinitely.

Side of 2nd equilateral triangle=12 cm [ Line segment joining midpoint of two sides of a triangle is parallel to the third side and half of it.]

Similarly, Side of third equilateral triangle=6 cm

Sides are 3, 1.5,3/4,....

Perimeter of 1st triangle=24+24+24=72 cm

Perimeter of 2nd triangle=12+12+12=36 cm

Perimeter of 3rd triangle=6+6+6=18 cm

..........

The sum of the perimeter of all the triangles=72+36+18+9+9/2+9/4+........

This is a geometric progression having a common ratio =36/72=1/2

Sum of an infinite G.P=

= 72 × 2

=144 which is the sum of the perimeter of all triangles.

A is the correct answer.

We can make the following figure from the information given in the question: 

Suppose the side of the square ABCD = 2a

Angle HAP= 30$$^{\circ\ }$$

HA= a

HP=$$\frac{a}{\sqrt{\ 3}}$$

Area HAP = $$\frac{a^2}{2\sqrt{\ 3}}$$

Remaining Area = 4* $$\frac{a^2}{2\sqrt{\ 3}}$$

Area of the shaded region= 4a$$^2$$ - 4* $$\frac{a^2}{2\sqrt{\ 3}}$$

Ratio= Shaded/ Remaining = 2$$\sqrt{\ 3}$$-1

Volume will be same 
Let length be h
we get :
$$66\ =\pi\ \times\ \frac{1}{20}\times\ \frac{1}{20}\times\ x$$
Cross multiplying
we get x=8400 cm or 84 metres

Let length be 3x, breadth be 2x, and height be x
Now we know that the area of four walls = 2h(l+b) = 2x(5x) = 10$$x^2$$
Now new length = 6x , new height = 0.5x and new breadth = x
So the area of four walls will now be: 2(0.5x)(7x)= 7$$x^2$$
Now therefore percentage change in area : $$\frac{7x^2-10x^2}{10x^2}\times\ 100\ =\ -30\%$$

Hence, the area decreases by 30%.

The diagonal of the first square is $$4\sqrt{2}$$
Now we know if side of a square is a
then diagonal is $$a\sqrt{\ 2}$$
Therefore here a = 4
Now area = 16
Now area of second square = 2(16) = 32
Now therefore side of 2nd square = $$\sqrt{\ 32}$$ = $$4\sqrt{\ 2}$$
Now therefore diagonal will be : $$4\sqrt{\ 2}\left(\sqrt{\ 2}\right)=\ 8$$

The maximum length of pencil will be along the longest diagonal of box
so we can say the maximum length will be : $$\sqrt{\ 4^2+6^2+8^2}=\sqrt{\ 104}\ =2\sqrt{\ 26}$$

We know that Circumference is dependent linearly on radius 
so if circumference is increased by 5% , we can conclude radius is increased by 5%
Now Area varies directly with square of radius 
Now if r is changed to 1.05 r 
so r^2 will change to (1.05r)(1.05r) = 1.1025 r^2
So we can conclude area will change by 10.25%

The cost of levelling and turfing a square field at Rs. 160 per hectare is Rs. 2624.40.

Hence, the area = $$\ \frac{\ 2624.40}{ 160}$$ = 16.4025 * 10000 square metres             (1 Hectare = 10000 square metres)

Assume the length of the side=a, the area, a$$^2$$ = 16.4025 * 10000 

=> a = 405 m

Then the perimeter = 4a = 405*4

The cost of surrounding = 405*4*25 paise = Rs 405

Assume the height of the tower = x and the shadow formed when the inclination = 60 is y. 

In the given figure, tan60= $$\ \frac{\ x}{y}$$...............(1)

tan30=$$\ \frac{\ x}{y+60}$$...................(2)

Dividing 1 by 2, we get, $$\ \frac{\ y}{y+60}$$ = 3

=> y=30  =>x=y$$\sqrt{\ 3}$$ = 30$$\times\ $$1.732=51.96 m 

Assuming the sides of the cube = a, then surface painted blue = 6a$$^2$$

Now after cutting into half the blue surface = 3a$$^2$$ for each solid and a new colourless surface with area a$$^2$$ is generated.

Hence the total surface area  = 3a$$^2$$ + a$$^2$$ = 4a$$^2$$

% surface area not painted blue = $$\ \dfrac{\ a^2}{4a^2}\times\ 100$$ = 25

Assume the height of water in the cylinder = h

Then the volume of water on the roof = 90000 square cm * 0.01 cm = 900 cubic cm

For the cylinder, we have the volume as base of area*height= 900h cubic cm.

=> 900 = h*900

=> h = 1 cm

Let us draw the diagram of the ladder against the wall as given in the question.

Let the length of the ladder be t metres.

We can find BC, i.e the distance of the ladder from the foot of the wall using Pythagorus theorem

$$\ BC^2=\ t^2-5^2$$.,

=>$$\ BC^2=\ t^2-25$$

=> BC=$$\ BC^{ }=\ \sqrt{\ t^2-25}$$

Now, when the ladder is slips 2 m away from the wall, the ladder lies on the ground completely. This can be depicted as:

Now, BC+ BC'= t 

=> $$\ \ \sqrt{\ t^2-25}+2=t$$

=> $$\ \ \sqrt{\ t^2-25}=t-2$$

Squaring both sides,

$$\ \ t^2-25=\left(t-2\right)^2$$

=> $$\ \ t^2-25=\left(t^2-4t+4\right)$$

=>$$4t=29.\ And\ so,\ t=7.25$$

Lets assume the park is an equilateral triangle with side '2x' units, => the side of the flower bed = 'x' units

Area of park = $$\frac{\sqrt{\ 3}}{4}\cdot\left(2x\right)^2$$ =$$\frac{\sqrt{\ 3}}{4}\cdot4x^2$$

Area of flower bed = $$\frac{\sqrt{\ 3}}{4}\cdot x^2$$ 

Area of path = area of park -  area of flower bed = $$\frac{\sqrt{\ 3}}{4}\cdot3x^2$$

Ratio of area of path to area of flower bed = $$\frac{\left(\frac{\sqrt{\ 3}}{4}\cdot3x^2\right)}{\frac{\sqrt{\ 3}}{4}x^2}$$ = 3:1

Cubes are a 3-D figure, and so, the intersection of two 3-D figure must be a 2-D figure. 

So, Cube cannot be the intersection of two cubes.

Let the side of the square lake be of length x kms.

Using pythagorus theorem, we can find the length of the diagonal as $$\sqrt{x^2+x^2}=\sqrt{2x^2}= x\sqrt{2}$$

'.' Area of the square= 50 sq. kms,

$$x^2=50$$ or x= $$5\sqrt{\ 2}$$

Therefore length of the diagonal = $$5\sqrt{\ 2}\cdot\sqrt{\ 2}=5\cdot2=10$$ km

Let AB and CD be the vertical poles such that the height of one is double to that of the other

It is given that the angles of elevation are complimentary. Thus let us assume the angle made by smaller tower be $$\theta\ $$. Then angle made by DE is $$90-\theta\ $$

In $$\bigtriangleup\ $$ ABC,  $$\tan\left(\theta\ \right) = \frac{\ h}{20}$$

In $$\bigtriangleup\ $$ CDE, $$\tan\left(\theta\ \right) = \frac{\ 20}{2h}$$

Equating the values of$$\tan\left(\theta\right)$$ we get, $$\ \frac{\ h}{20}=\ \frac{\ 20}{2h}$$ 

$$h^2\ =\ 200$$ $$\Rightarrow$$ $$h=14.14$$

Hence height of poles are $$14.14$$ and $$28.28$$

Using pythagoras theorem, we can clearly see from above that the shortest distance between Sameera and Meera is 10km.