XAT Geometry Questions

According to the question, we can draw the following table:

Here, EC refers to the shelf which is 7 feet and AB refers to the height of the robot. Here, AE reflects the hand of the robot. Since we need to find the minimum possible height, we need to maximize the length of the hands of the robot. The hands of the robot can be tilted to the maximum angle of 60, hence, we will assume that the hands of the robot tilt to exactly $$60^{\circ\ }$$.

Let the height of the robot i.e. AB be $$x$$ feet.

Now, we need to find the value of x. 

Since, BC = 1 foot, AD is also 1 foot. 

In triangle ADE, $$\tan\ 60^{\circ\ }=\dfrac{ED}{AD}$$

$$\tan\ 60^{\circ\ }=\dfrac{7-x}{1}$$

$$\sqrt{3}=\dfrac{7-x}{1}$$

$$x=7-\sqrt{3}$$

Hence, the minimum height of the robot from the ground to the shoulder for its arms to reach the topmost shelf is $$7-\sqrt{3}$$ feet.

Considering statement 1: We can only get the length of one of the sides of the land. But that is not sufficient to get the area.

Considering statement 2: 

We are given the diagonal length.

With this information and the perimeter, it is not possible to find the area of land.

Considering both 1 and 2 statements.

Now, we have two right-angle triangles. We know the hypotenuse of both triangles.

$$AB^2+AD^2=BD^2$$

$$110^2+AB^2= 255^2$$

AB$$\approx\ $$230

Let DC = x, BC = 700 - 110-230-x = 360-x.

$$x^2+(360-x)^2 = 255^2$$

Upon solving this we get x = 180$$\pm\frac{15}{\sqrt{2}}$$

360-x = 180$$\pm\frac{15}{\sqrt{2}}$$

Area of triangle ABD = $$\frac{1}{2}\cdot110\cdot230\ =\ 12650$$

Area of triangle BDC = $$\frac{1}{2}\left(180+\frac{15}{\sqrt{2}}\right)\left(180-\frac{15}{\sqrt{2}}\right)=16143.75$$

So, total area = 12650+16143.75 = 28793.75

Cost = 28793.75*400 = Rs. 11517500.

So, option C is the answer, as both statements 1 and 2 are required and sufficient to answer the question.

CD is on of the side of the rectangle. So, two of the remaining 3 sides will be perpendicular to CD.

 Length of CD is 4. We are told that area is 24.

We know area of rectangle is lb. 4*l = 24

l = 6

So, the other two points should be such that when jined with C and D respectively, should be perpendicular to CD and lie at a distance of 6.

There will be two pair of such points. One pair is (-2,6), (2,6) and the other pair is (-2,-6), (2,-6)

So, there are two possible equations of AB. One is y = 6 and the other is y = -6.

It is given that the perimeters of areas are equal for both of them. Let the equal perimeter be P.

Adu's piece is a square, and the perimeter of the square =  4s = P

Side of the square $$s=\ \dfrac{P}{4}$$

Amu's piece is a circle, and the perimeter of the circle = $$2\pi\ r=\ P$$

The radius of the circle $$r=\ \dfrac{P}{2\pi\ }$$

The ratio of areas is,

$$s^2\ :\ \pi\ r^2$$  $$=\left(\dfrac{P}{4}\right)^{^2}:\ \pi\ \left(\dfrac{P}{2\pi\ }\right)^{^2}$$ $$=\ \dfrac{P^2}{16}\ :\ \pi\times\dfrac{P^2}{4\pi^2\ }$$  $$=\ \dfrac{1}{4}\ :\ \dfrac{1}{\ \pi\ }\ =\ $$  $$\pi\ \ :\ \ 4$$

Hence, the correct answer is option D.

Given that the equation of line $$L_1$$ is $$y = k(x - 1)$$ and it intersects line $$L_2$$ at (5, 8).

So, the point (5, 8) must satisfy the equation of line $$L_1$$

We get $$8=k\left(5-1\right)$$

or $$k=2$$

Hence, the equation of line $$L_1$$ is $$y = 2(x - 1)$$ . . . (1)

Now, $$L_2$$ has a slope of 1 

Using slope form, the equation of line is $$y=mx+c$$, where $$m$$ is the slope and $$c$$ is the y-intercept

So, the equation of line $$L_2$$ is $$y=x+c$$

Now, as it passes through the point (5, 8), the equation must satisfy

Hence, $$8=5+c$$

or, $$c=3$$

Hence, the equation of line $$L_2$$ is $$y=x+3$$ . . . (2)

Plotting both lines on the graph, we get

From the graph we can clearly see that the distance between the y-intercepts of the two lines is 5 not 6. 

Hence, the statement the distance between the y-intercepts of the two lines is 6 is definitely false.

According to the question, we have the following figure:

According to the question, we have a frustum and a semisphere with a flat surface, and we need to find the curved surface area of the frustum along with the curved surface area of the hemisphere and the bottom flat area of the frustum. 

Curved surface area of the Frustrum = $$\pi\ \times\left(r_1+r_2\right)\times l$$

Here, r1 and r2 refer to the two different radii, and l refers to the slant height. 

$$l=\sqrt{h^2+\left(r_1-r_2\right)^2}$$

$$l=\sqrt{1600+100}=\sqrt{1700}=10\sqrt{17}$$

Hence, curved surface area of the frustrum = $$\pi\ \times\left(20+30\right)\times10\sqrt{17}$$  = $$\pi\ \times50\times10\sqrt{17}$$ = $$500\sqrt{17}\pi\ $$

Now, curved surface area of the hemisphere = $$2\pi r^2=2\times\pi\ \times\ \left(20\right)^2\ =\ 800\pi\ $$

And the flat bottom area of the frustrum = $$\pi r^2=\pi\ \times\left(30\right)^2=900\pi\ $$

Hence, the total surface area of the trophy =$$500\sqrt{17}\pi\ +\ 800\pi\ +\ 900\pi\ =500\sqrt{17}\pi\ +\ 1700\pi\ $$

The cost of gold coating is Rs. 40 per square cm.

Hence, the total cost of coating = $$40\times\ \left(500\sqrt{17}\pi\ +\ 1700\pi\ \right)$$

$$40\times\ \dfrac{22}{7}\times\ \left(500\sqrt{17}\ +\ 1700\ \right)$$

$$40\times\ \dfrac{22}{7}\times\ \left(2061.55\ +\ 1700\ \right)$$

$$40\times\ \dfrac{22}{7}\times\ 3761.55=Rs.\ 4,72,880.571\ \approx\ Rs.\ 4,73,000$$

It is given that ABC is a right-angled triangle, where $$\angle\ B\ =\ 90^{\circ\ }$$

Now, we can use Pythagoras theorem to calculate the value of AC, which is equal to $$\sqrt{\ 18^2+24^2}=\sqrt{\ 324+576}=\sqrt{\ 900}\ =\ 30$$ cm.

Here, we have drawn a common tangent MN, which is perpendicular to BC => Triangle MNC is a right-angled triangle.

It is also be concluded that traingle MNC is similar to traingle ABC

Hence, the ratio of the sides of triangle MNC is the same as the ratio of the sides of triangle ABC.

Thus, AB: BC: AC = MN: NC: MC = 18:24:30 = 3: 4: 5

Let the length of side MN be 3x => the length of NC = 4x, and the length of MC = 5x.

In triangle MNC, the circle inside it is the in circle of triangle MNC.

Thus, the in radius (r) of the circle in triangle MNC = (3x+4x-5x)/2 = x.

We know that BC = BN+NC, where BN is the diameter of the first circle, which is equal to 2r = 2x (since, r = x), and NC = 4x

=> 24 = 2x+4x 

=> 6x = 24 => x = 4 cm.

Hence, the radius of the circle (x = r) is 4 cm.

The correct option is B

The line AC is tangent to the circle. So, the line DE is perpendicular to AC and is radius to the circle.

Look at the $$\triangle\ $$ADC and  $$\triangle\ $$DEC. $$\angle\ $$DAC = $$\angle\ $$EDC, $$\angle\ $$C=$$\angle\ $$C. And both D and E are right angle triangles.

So, both the triangles are similar.

$$\triangle\ $$ ADC is a right-angled triangle.

$$AC^2$$=$$AD^2$$+$$DC^2$$

$$AC^2$$=$$200^2+150^2=250^2$$

AC=250, AD=200, DE=r, DC=150 

$$\frac{AD}{DE}=\frac{AC}{DC}$$

$$\frac{200}{r}=\frac{250}{150}$$

$$\Rightarrow$$ r=120.

Given half pond is inside the triangle, so he has to only buy half the area of the circle.

Cost for buying area of semi circle =$$\pi\ \frac{r^2}{2}$$*1400=$$\pi\ \frac{120^2}{2}$$*1400= 3,16,80,000

The question states that the path forms a right-angle triangle at Rivu. So, let's consider the distance from Raju to Rivu as x and that from Rivu to Ratan as y.

i)

Now, $$\triangle\ ABC$$ form a right-angled triangle. Using Pythagoras theorem

$$AB^2+BC^2=AC^2$$

=>$$x^2+y^2=80^2=6400$$ --->1

The area of $$\triangle\ ABC$$ is $$\frac{1}{2}\cdot b\cdot h$$=$$\frac{1}{2}\cdot x\cdot y$$ ---->2

If we consider the base of the triangle as AC of length 80(side of rectangle) and height as altitude from B of length 25 (half the breadth of rectangle), we can get the area as $$\frac{1}{2}\cdot b\cdot h$$=$$\frac{1}{2}\cdot 80\cdot 25=1000$$.

Now, use this in equation 2.

$$\frac{1}{2}\cdot x\cdot y$$=1000

xy=2000--->3

$$\left(x+y\right)^2=x^2+y^2+2xy$$=6400+4000=10400

x+y=20$$\sqrt{\ 26}$$

This is the distance the ball travels at 10m/s. The time taken is 2$$\sqrt{\ 26}$$+5 sec for holding.

So, 1 alone is sufficient.

ii) Given the area of the triangle is 1000

$$\frac{1}{2}\cdot x\cdot y$$=1000

=> xy=2000

With this alone, we can't find the travel time.

$$Let\ \sin^215^{\circ}\ =a\ and\ \sin^275^{\circ}\ =\ b\ $$

Then the given equation becomes,

$$\ \frac{\ a^3+b^3+6ab}{a^2+b^2+5ab}$$ 

=$$\ \frac{\ \left(a+b\right)^3-3ab\left(a+b\right)+6ab}{\left(a+b\right)^2-2ab+5ab}$$     $$\ \therefore\ \ a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)$$ , $$\ \therefore\ \ a^2+b^2=\left(a+b\right)^2-2ab$$

Using $$\sin\left(\theta\right)\ =\cos\left(90-\theta\ \ \right)$$

we get, $$\sin^275^{\circ\ }=\cos^215^{\circ\ }$$

now, a+b = $$\sin^215^{\circ\ }+\sin^275^{\circ\ }$$ = $$\sin^215^{\circ\ }+\cos^215^{\circ\ }$$ = 1.

The equation becomes, $$=\ \frac{\ 1-3ab+6ab}{1-2ab+5ab}\ =\ \frac{\ 1+3ab}{1+3ab}\ =1\ $$

Given that, $$AD\bot\ BC$$

                   $$BE=3$$

                   $$BC=5$$

Using Pythagoras' theorem,

    $$AD^2+BD^2=AB^2$$ .....(1)

    $$AD^2+DE^2=AE^2$$......(2)

    $$AD^2+DC^2=AC^2$$......(3)

(1) - (2) gives

$$BD^2+DE^2=AB^2-AE^2$$

$$x^2-\left(3-x\right)^2=AB^2-AE^2$$

$$AB^2-AE^2=6x-9$$

$$AB^2-AE^2+6CD=6x-9+6\left(5-x\right)$$       ($$CD=\left(5-x\right)$$)

$$AB^2-AE^2+6CD=6x-9+30-6x$$

$$AB^2-AE^2+6CD=21$$

The length of AB = $$\left(b+4c\right)-\left(b-2c\right)=6c$$  (X-coordinates of A&B are same).

The altitude of triangle ABC, CD = $$a-(-2a)=3a$$.

Area of triangle ABC = $$\ \dfrac{\ AB\ \times\ CD}{2}$$ = $$\ \dfrac{6c\ \times\ 3a\ }{2}=9ac$$

Option (D) is correct.

The shortest route ant takes to travel from the bottom corner to the diagonally opposite top corner is  shown below.

The route goes through lateral faces (vertical faces) of the cubical room.

The length of route, r = $$\sqrt{\ \left(2+2\right)^2+2^2}$$ = $$\sqrt{\ 20}$$ = $$2\sqrt{5}$$

Given that, CPD is an equilateral triangle

$$\angle\ CPD=\angle\ PDC=\angle\ DCP=60^{\circ\ }$$

As AQ is parallel to PC $$\angle\ CPD=\angle\ QAP=60^{\circ\ }$$

As BC is parallel to AD $$\angle\ QAP=\angle\ AQB=60^{\circ\ }$$ ( alternate interior angles).

Given, Area of equilateral triangle CPD is $$4\sqrt{\ 3}$$

This implies, $$\ \frac{\ \sqrt{3}a^2}{4}=4\sqrt{3}$$

Solving, we get $$a=\ 4$$.

From figure, length of $$AB$$ = Height of $$\triangle\ CPD=\frac{\sqrt{\ 3}a}{2}=2\sqrt{\ 3}$$.

From $$\triangle\ ABQ,\ BQ=\ \frac{\ AB}{\tan\ 60^{\circ\ }}=\frac{2\sqrt{\ 3}}{\sqrt{\ 3}}=2$$

Area of $$\triangle\ ABQ=\frac{1}{2}\times\ AB\times\ BQ=\ \ \frac{\ 2\times\ 2\sqrt{\ 3}}{2}=2\sqrt{\ 3}\ $$

Option A is correct.

Given: 

BD= 6 cm, AC= 9 cm, DC= 5 cm, BP=8 cm, and DP = 5 cm.

Since AD is the angular bisector applying the angular bisector theorem we have :

$$\frac{AB}{BD}=\ \frac{AC}{CD}$$

Hence : Considering AB = x cm.

$$\frac{9}{5}=\ \frac{x}{6}$$

x = 10.8 cm.

Now since BD is the angular bisector for angle PBA we have :

Applyinh the internal angle bisector theorem :

$$\frac{PB}{PD}=\ \frac{BA}{AD}$$

Considering AD = y cm.

$$\frac{8}{5}\ =\ \frac{10.8}{y}$$

y = 6.75 cm.

AP = AD + DP.

= 6.75 + 5 = 11.75 cm

Given the distances are :

AE = 4 meters , EB = 8 meters and EC = 16 meters.

Considering the length of ED = K.

Given the angles DAE and angle DCE  are complementary.

Hence the angles are A and 90 - A.

Tan(90-A) = Cot A

$$\ \tan\ DAE\ =\frac{k}{4}$$ and $$\ \tan\ DCE\ =\frac{1}{\tan\ DAE}=\ \frac{k}{16}$$

Hence $$\frac{k}{16}=\ \frac{4}{k}$$

k = 8 meters.

The angle DBE is given by 

$$Tan\ DBE\ =\ \frac{k}{8}=\ 1$$

Hence the angle is equal to 45 degrees.

Since A, B and C are adjacent faces. If we remove them, the resultant solid will also be a cube with side 5.

Hence total number of cubes unpainted = $$5^3$$ = 125

We will try to maximize the value of the angle ACD:

For a fixed triangle ABC, the angle ACD can be maximized when we take the median CD to be perpendicular to AB and the value of AC is as small as possible, so that the sine of angle ACD, and hence, the angle ACD itself if maximized, as the value of AD is fixed at half of AB at 0.5.

Now, the least possible value of AC is 2. The triangle will be of sides (1,2,2).

Value of sin(ACD)=$$\frac{0.5}{2}=0.25$$

$$\angle\ ACD=\sin^{-1}\left(0,25\right)=14.78\approx\ 15.$$

Let ED = $$\sqrt{\ 3}x$$
In triangle CDE, $$\tan\ 30\ =\ \frac{CD}{\sqrt{\ 3}x}$$ => CD = x

In triangle BDE, $$\tan\ 45\ =\ \frac{BD}{\sqrt{\ 3}x}$$ => BD = $$\sqrt{\ 3}x$$ => BC = $$\sqrt{\ 3}x\ -x$$

In triangle ADE, $$\tan\ 60\ =\ \frac{AD}{\sqrt{\ 3}x}$$ => AD = 3x => AB = $$3x\ -\ \sqrt{\ 3}x$$

AB : BC : CD = $$(3 - \sqrt 3) : (\sqrt 3 - 1) : 1$$

We know that radius of circle P is 2 cm 

Length of MO = 1 cm

radius of circle R = AO = $$\sqrt{\ 2^2-1^2}$$ = $$\sqrt{\ 3}$$

Area of overlap between the circle R and P(shaded region) = semi circle area of R + area of segment ANBOA

Area of segment ANBOA = area of sector ANBM - area of triangle AMB

                                        = $$\frac{120}{360}\pi\ \left(2\right)^2$$ - $$\frac{1}{2}\left(1\right)\left(2\sqrt{\ 3}\right)$$

                                        = $$\frac{4\pi}{3}-\sqrt{\ 3}$$

Area of overlap between circle R and P = $$\frac{\pi\ \left(\sqrt{\ 3}\right)^2}{2}$$+ $$\frac{4\pi}{3}-\sqrt{\ 3}$$

                                                               = $$\ \frac{\ 17\pi\ }{6}-\sqrt{\ 3}$$

Answer is option E.

Based on the information, the following figure can be obtained.

Given, CE=0.5, BC = 1.3 and ED=2.5

Triangle CEB is a right-angled triangle => EB = 1.2

$$\triangle\ $$ ECB is similar to $$\triangle\ $$ EDA

EB/EC = AE/ED  => AE = 6

Hence total distance travelled = AB + BC + CD = 7.2 + 1.3 + 3 = 11.5km

Triangle ACC' is similar to triangle ABB'

Considering AC = a, BC = b, CC' = h, AA' is given as 60, BB' is given to be 40.

AC/AB = CC'/BB' = h/40.

$$\left(\frac{a}{a+b}\right)\ =\ \frac{h}{40}$$  (1)

Similarly triangle BCC' is similar to BAA'.

BC/AB = CC'/AA' = h/60.

$$\left(\frac{b}{a+b}\right)\ =\ \frac{h}{60}$$  (2)

Adding (1) and (2).

$$\frac{h}{40}+\frac{h}{60\ }=\ 1$$

1/h = $$\left(\frac{1}{40}+\frac{1}{60}\right)$$

h = 24

Using crossed ladder theorem 

$$\frac{1}{CC'}=\frac{1}{AA'}+\frac{1}{BB'}$$=1/60 +1/40 =5/120 =24.  

The volume of cuboid will be 50*25*1=1250

The volume of the portion below cuboid will be Area of triangle*width =$$\frac{1}{2}\cdot3\cdot50\cdot25=1875$$

.'. Total volume = 1875+1250= 3125.

In triangle OAP, since AP=6, OA will be 12.

Area of AQBMA=Area of Triangle ABQ- (Area of minor arc AMB-Area of OAB)

Length of PQ=MQ+PM = 12+(12-$$6\sqrt{\ 3}$$)=24-$$6\sqrt{\ 3}$$

Area of Triangle ABQ=$$\frac{1}{2}\cdot12.\left(24-6\sqrt{\ 3}\right)=6\left(24-6\sqrt{\ 3}\right)$$.=81.64

Area of minor arc AMB-Area of OAB=$$\frac{60}{360}\cdot\pi\ \cdot144-\frac{1}{2}\cdot12\cdot6\sqrt{\ 3}$$ = $$24\pi\ -36\sqrt{\ 3}$$.=13.07

.'.Area of AQBMA=68.57$$\approx\ 69$$

Each circle on the end of the diagonal will touch sides of the rectangular field

Using Pythagoras' theorem, the distance between the vertex of the rectangle and center of the first circle drawn on the diagonal = $$1.25\sqrt{\ 2}$$

Distance between the vertex of the rectangle and circumference of the first circle drawn on the diagonal = $$1.25\sqrt{\ 2}$$ - 1.25 = 0.51 meters

Space that cannot be used to draw circle otherwise they will go outside rectangle on every diagonal = 0.51 * 2 = 1.02 meters

Space that can be used to draw circles = length of diagonal - unused space = 50 - 1.02 = 48.98 meters

On every diagonal, maximum number of such circles = usable length/diameter of each circle = 48.98/2.5 = 19

Or, on every diagonal, one circle will be at the center (intersection of diagonals) and 9 circles will be on each half of the diagonal

Therefore the circle in center will be common for both diagonals, and 9 circles can be drawn on each half of the diagonal. So total circles = 9*4 + 1 = 37

If we join YP and PZ, XYPZ will become a cyclic quadrilateral. YZ and XP will be diagonals of this quadrilateral.

Ptolemy's theorem states that the product of the length of the diagonals is equal to the sum of products of measures of the pairs of opposite sides.

As per this theorem for the quadrilateral :

XY*PZ +XZ*PY = YZ*XP.

For the equilateral triangle.

Now, in equilateral triangle XYZ, YZ = XZ = XY.

Hence equation 1 becomes, XP = YP + PZ.

The two circles are symmetric about the diagonal.

NC1=$$\frac{2\sqrt{\ 2}}{3}$$ = $$\sqrt{\ \left(\frac{2}{3}\right)^{^2}+\left(\frac{2}{3}\right)^2}=\ \frac{\left(2\sqrt{\ 2}\right)}{3}$$

The lengths FC1, EC1 are the radius of the circle which is 2km/3.

The length C1P is the radius of the circle.

Because of symmetry C1O = C2O and C1N = C2B.

2*(C1N+ C1O) = $$2\sqrt{\ 2}$$ the length of the diagonal of the square.

C1N+C1O = $$\sqrt{\ 2}$$

C1O = $$\sqrt{\ 2}-\frac{2\sqrt{\ 2}}{3}=\ \frac{\sqrt{\ 2}}{3}$$

OP = $$\frac{\left(2-\sqrt{\ 2}\right)}{3}$$

The diagonal perpendicularly bisects the line GH. Hence C1OH is 90 degrees.$$C1H^2\ =\ C1O^2+OH^2$$

OH = $$\frac{\sqrt{\ 2}}{3}$$. Similarly OG = $$\frac{\sqrt{\ 2}}{3}$$

HC1, C1G both are equal to $$\frac{2}{3}$$ each. H1G is $$\frac{2\sqrt{\ 2}}{3}$$. HC1G is a raight angled traingle with angle HC1G is 90 degrees.

Area of the region required is 2*(Area of segment OGH) 

.'. Area of required region=$$2\left(\frac{90}{360}\cdot\frac{4\pi\ }{9}-\frac{1}{2}\cdot\frac{2\sqrt{\ 2}}{3}\cdot\frac{\sqrt{\ 2}}{3}\right)$$=$$\frac{2(\pi - 2)}{9}$$

Let the radius of the cylinder be $$r$$ and height be $$h$$ (as shown).

Let the dimensions of the cone be $$r_c$$ and $$h_c$$ (as shown).

It is given that $$h_c = \frac{h}{2}$$

Area of base of cone $$A_{c}=\pi \times r_{c}^{2}$$

Area of base of cylinder $$A=\pi \times r^{2}$$

Given, $$A_{c}=\frac{1}{5} \times A $$

$$ \Rightarrow \pi \times r_{c}^{2} =\frac{1}{5} \times \pi \times r^{2} $$

$$  \Rightarrow r_{c}^{2} = \frac{1}{5} \times r^{2} $$

$$  \Rightarrow \frac{r^{2}}{r_{c}^{2}} = 5 $$

Now we know that the volume of cylinder = total volume of cones 

Let the number of cones be n. 

So, volume of cylinder = n x volume of  each cone

$$  \Rightarrow V = n \times V_{c} $$

$$  \Rightarrow \pi \times r^{2} \times h = n \times \frac{1}{3} \times \pi \times r_{c}^{2} \times h_c $$

$$  \Rightarrow r^{2} \times h = n \times \frac{1}{3} \times r_{c}^{2} \times h_c $$

$$  \Rightarrow \frac{r^{2}}{r_{c}^{2}} \times \frac{h}{h_c} \times 3 = n $$

$$  \Rightarrow 5 \times 2 \times 3 = n $$

$$  \Rightarrow 30 = n $$

Thus, 30 conical ingots can be made.

On solving for x and y from the equations 

3x + 4y = 2018 and 5x + 3y = 1

we get Q(-550,917)

Let, P(x,y)

So, $$\frac{y - 917}{x + 550}$$ = 2

=> y - 2x = 2017 ....(1)

Considering the equations

3x + 4y = 2a ........(2)

7x + 2y = 2018 .....(3)

On subtracting equation (2) from (3) we have,

4x - 2y = 2018 - 2a

=> 2x - y = 1009 - a

=> y - 2x = a -1009 .....(4)

From equation (1) and (4)

2017 = a - 1009

=> a = 3026

Hence, option C.

As mentioned in the question,

Lines L1 and L2 intersect at point P as shown in the figure.

On solving for x and y from equations

x + 2y - 18 = 0

2x - y - 1 = 0

We get x = 4 and y =7.

Given, radius = $$\sqrt{20}$$

Using the equation of a circle, we have

$$(4-a)^{2}$$ + $$(7-b)^{2}$$ = 20....(1)

The center of the circle will lie on the line: 2x - y - 1 = 0

a,b will satisfy this equation.

So 2a-b-1=0

b=2a-1

From equation 1...

$$(4-a)^{2}$$ + $$(7-b)^{2}$$ = 20

$$(4-a)^{2}$$ + $$(8-2a)^{2}$$ = 20

5$$(4-a)^{2}$$ = 20

a=6 or a=2

b=11 or b=3

The sum of the coordinates possible=6+11  or 2+3 

i.e. 17 or 5

Option B is one of the solution.

Construct a perpendicular in the trapezium. Let its height be 'h'

Now, $$sin\angle BAC = \dfrac{h}{AC}$$

and $$sin\angle BAD = \dfrac{h}{AD}$$

Thus, $$\dfrac{sin\angle BAC}{sin\angle BAD} = \dfrac{\frac{h}{AC}}{\frac{h}{AD}} = \dfrac{AD}{AC}$$ 

Let the sides of the smaller square be $$x$$ units and the sides of the larger square be $$y$$ units.

Construct diagonal HF on the larger square. Let the point of intersection of AD and diagonal be I and point of intersection of BC and diagonal be J.  

We know that since EFGH is a square, the diagonal will bisect the angle. Therefore $$\angle EHF=45^{o}$$

Observe $$\triangle AEH$$ and $$\triangle DHG$$ : 

AE=AH=DH=DG = $$x$$ units and EH=HG=$$y$$ units. 

Thus by SSS property, $$\triangle AEH \cong \triangle DHG$$

Thus, we know that $$\angle$$AHE=$$\angle$$DHG

Also, $$\angle$$AHE+$$\angle$$DHG+$$\angle$$AHD= $$90^{o}$$ (angle of a square)

$$ \Rightarrow (2 \times \angle AHE)+ 60^{o} = 90^{o}$$ (angle of an equilateral triangle)

$$ \Rightarrow (2 \times \angle AHE) = 30^{o}$$

$$ \Rightarrow \angle AHE = 15^{o}$$

Since $$\angle AHE = 15^{o}, \angle AHI = 45^{o}-15^{o} = 30^{o} $$

Since $$30^{o}= \dfrac{60^{o}}{2}$$ we can say that HF is the angle bisector of $$\angle$$AHD 

Since $$\triangle$$AHD is an equilateral triangle, we know that the angle bisector, median and altitude will all be the same line ie HI in this case. 

By symmetry, we know that JF will also be the angle bisector, median and altitude. 

Also, length of altitude of equilateral triangle ie HI and JF = $$\dfrac{\sqrt{3} \times x}{2} $$

Length of diagonal HF = $$\sqrt{2} \times y$$

From the figure, we can express the length of diagonal HF as given : 

$$ \sqrt{2} \times y = \dfrac{\sqrt{3} \times x}{2} + x + \dfrac{\sqrt{3} \times x}{2} $$

$$ \Rightarrow \sqrt{2} \times y = (\sqrt{3} \times x) + x $$

$$ \Rightarrow \sqrt{2} \times y = x \times (\sqrt{3} + 1) $$

$$ \Rightarrow y = x \times (\dfrac{\sqrt{3} + 1}{\sqrt{2}}) $$

Ratio of areas asked = $$(\dfrac{y}{x})^{2} = (\dfrac{\sqrt{3} + 1}{\sqrt{2}})^{2} = \dfrac{4+2\sqrt{3}}{2} = 2+\sqrt{3}$$

Alternate Explanation: 

Let AB=BC=CD=DA=x, then the area the area of square ABCD will be $$x^2$$

Since AHD, DGC, CFB, and BAE are equilateral triangles with side x, the sum of area of these traingles will be $$4\times\ \frac{\sqrt{\ 3}}{4}\times\ x^2=\sqrt{\ 3}x^2$$

We know that angle HAD=60, angle DAB=90, and angle BAE=60, thus angle HAE=360-(60+90+60)=150

Area of triangles HAE+HDG+GCF+FBE=4* area of triangle HAE

In HAE, AH=AE=x, and angle HAE=150, thus area will be $$\frac{1}{2}\times\ x\times\ x\times\ \sin\ 150$$

sin 150=sin(180-150)=sin 30=0.5

Thus area of 4 triangles will be 4*0.5*x*x*0.5=$$x^2$$

Thus, total area= $$x^2+\sqrt{\ 3\ }\ x^2+x^2$$=$$x^2\left(2+\sqrt{\ 3}\right)$$

Thus area of EFGH/area of ABCD=$$2+\sqrt{\ 3}$$

In order to accommodate maximum number of points on the circular disk, all the points should be placed on the boundary of the disk as any point if placed inside the boundary would be less than 1 metre apart from the centre of the disk. Also, all the points have to be exactly 1 metre apart from each other to place maximum number of  points. Hence, a maximum of 6 points can be placed on the boundary of the disk (each adjacent points placed 60 degrees apart) and 1 point at the centre. 

Hence, a total of 7 points can be placed on the disk.

The figure can be redrawn as shown above. 

$$\because $$ AB and OP are parallel, POAC is a rectangle. 

In right $$\triangle$$QCB, QB = $$\sqrt{5}$$

$$\triangle$$QBO is an isosceles right angled triangle with $$\angle $$OQB = $$ 90°$$ (By Pythagoras theorem)

$$\therefore $$ $$\angle $$QOB =  $$45°$$

$$\triangle$$ ABC can be represented as shown above where AD $$\bot$$ BC

Given, Sin b = $$\frac{3}{5}$$

Hence, AD = $$\frac{3x}{5}$$ and BD = $$\frac{4x}{5}$$

Now, BC = 2 BD = $$\frac{8x}{5}$$

$$\therefore$$ area of $$\triangle$$ ABC = $$\frac{1}{2}$$ * BC * AD = $$\frac{1}{2}$$ * $$\frac{8x}{5}$$ * $$\frac{3x}{5}$$ = $$\frac{12x^{2}}{25}$$

Only option E satisfies.

Hence, option E.

It is given that the length of the diagonals are in 3:4. Let '3x', and '4x' be the lengths of semi-diagonals as shown in the figure. We know that diagonals of a rhombus intersect each other perpendicularly.

In right angle triangle AOB,
$$AB^2=AO^2+BO^2$$

$$\Rightarrow$$ $$AB= \sqrt{AO^2+BO^2}$$

$$\Rightarrow$$ $$AB= \sqrt{(3x)^2+(4x)^2}$$

$$\Rightarrow$$ $$15= 5x$$

$$\Rightarrow$$ $$x= 3$$cm.

Therefore, we can say that the length of diagonals = 6x and 8x or 18 and 24 cm.

Hence, the area of the rhombus = $$\dfrac{1}{2}*18*24$$ = 216 cm$$^2$$. Therefore, option E is the correct answer. 

2 litres are required to paint the surface of a solid sphere. 
Surface area of the solid sphere = $$4\pi*r^2$$
Now, the solid sphere is cut into 4 identical parts. This is possible only when the sphere is cut into 4 quarter spheres. 
After making the first cut, 2 hemispheres will be formed. 2 circles of area $$\pi*r^2$$ will get exposed in addition to the surface area of the sphere (the base surface of the bottom hemisphere and the base of the top hemisphere).
Now, another perpendicular cut will be made along the diameter of the sphere. 2 additional surfaces will get exposed again (one on left hemisphere and the another on the right hemisphere).

Area exposed after making 4 identical pieces = $$4*\pi*r^2$$ + $$2*\pi*r^2$$ + $$2*\pi*r^2$$
                                                                   = $$8*\pi*r^2$$
2 litres of paint is required to paint an area of $$4\pi*r^2$$
=> 4 litres of paint will be required to paint an area of $$8*\pi*r^2$$.

Therefore, option D is the right answer.

The ratio of the curved surface area of the upper cone to the lower frustum is 1:2.
=> the ratio of the curved surface area of the upper cone to the total cone = 1:3.
Curved surface area (CSA) of a cone = $$\pi*r*l$$
For the given cone, the slant height, $$l=12$$cm

CSA of the cone = $$48*\pi$$
CSA of the smaller cone = $$16*\pi$$
Both the slant height and the radius would have been reduced by the same ratio. Let that ratio be $$x$$.
$$x^2*48*\pi$$=$$16\pi$$
=>$$x^2=\frac{1}{3}$$
$$x=\frac{1}{\sqrt{3}}$$
Slant height of the smaller cone = $$\frac{12}{\sqrt{3}}$$
Slant height of the frustum = $$12-\frac{12}{\sqrt{3}}$$
= $$12*\frac{\sqrt{3}-1}{\sqrt{3}}$$
=$$12*\frac{(3-\sqrt{3})}{3}$$
=$$12-4\sqrt{3}$$
Therefore, option D is the right answer.

Let us draw the diagram according to the given info,

We can see that AD = AO*cos60° = 2R*$$\dfrac{1}{2}$$ = R

In triangle, ACD

$$\Rightarrow$$ $$sinACD=\dfrac{AC}{AC}$$

$$\Rightarrow$$ $$sinACD=\dfrac{R}{\sqrt{2}*R}$$ = $$\dfrac{1}{\sqrt{2}}$$

$$\Rightarrow$$ $$\angle$$ ACD = 45°

By symmetry we can say that $$\angle$$ BCD = 45°

Therefore we can say that $$\angle$$ ACB = 90°

Hence, the area colored by green color = $$\dfrac{270°}{360°}*\pi*(\sqrt{2}R)^2$$ = $$\dfrac{3}{2}*\pi*R^2$$ ... (1)

Area of triangle ACB = $$\dfrac{1}{2}*R*2R$$ = $$R^2$$ ... (2) 

Area shown in blue color = $$\dfrac{60°}{360°}*\pi*(2R)^2-\dfrac{\sqrt{3}}{4}*(2R)^2$$ = $$\dfrac{2}{3}*\pi*R^2-\sqrt{3}*R^2$$ ... (3)

By adding (1) + (2) + (3) 

Therefore, the area of the common region between two circles = $$\dfrac{3}{2}*\pi*R^2$$ + $$R^2$$ + $$\dfrac{2}{3}*\pi*R^2-\sqrt{3}*R^2$$

$$\Rightarrow$$ $$(\dfrac{13\pi}{6}+1-\sqrt{3})R^2$$

Hence, option C is the correct answer. 

Let us draw the diagram using the given conditions.

AB = 24 cm and P is the mid-point of AB. Therefore, AP=PB=12 cm. 
MN is perpendicular to AB and passes through P. 
PM < PN. Therefore, M should be closer to A and B than N.
MN and AB are 2 perpendicular chords intersecting at P. 
Therefore, according to the intersecting chords theorem, AP*PB = PM*PN
12*12=8*PN
=> PN = 18 cm.
Therefore, option B is the right answer. 

For smaller values of 'x', cos 'x' will be close to 1 and sin x will be close to 0. 
Sin 30° = 1/2 = 0.5
As we move closer to 0, the value of cos x increases and sin x decreases. Therefore, when we move from a larger value of x to a smaller value, the value of the expression cos x - sin x will increase. 

Let us evaluate the value of the expression at x= 30° (since we know the values of cos x° and sin x° at this point) and then logically deduce the range of the value of the expression. 

At x = 30°, cos x = $$\frac{\sqrt{3}}{2}$$ and sin x = $$\frac{1}{2}$$
Therefore, cos x° - sin x° would have been 1.732/2 - 0.5 = 0.866 - 0.5 = 0.366

sin 30° + cos x - sin x will be 0.866

At x = 0, sin 30 + cos x - sin x will be 0.5 + 1 - 0 = 1.5.
At x = 5°, the value of the expression sin 30 + cos x - sin x  will be slightly less than 1.5. 
Therefore, we can infer that the upper limit of the expression will be greater than 1. None of the given limits include values greater than 1. Therefore, option E is the right answer. 

Alternate solution:

The value of sin 15° and cos 15° can be found out using the identities sin (A-B) = sin A cos B - cos A sin B and cos (A-B) = cos A cos B + sin A sin B.
Substituting A = 45° and B = 15° in these expressions, we get, 
sin 15° = 0.2588
cos 15° = 0.9659

sin 30° + cos 15° - sin 15° = 0.5 + 0.9659 - 0.2588 = 1.2071.
None of the given options capture this value in the range. Therefore, option E is the right answer.

Volume of the pyramid = $$200$$ cubic cm. 
The volume of  a pyramid is usually a third of the volume of a cuboid of the same height. 
Therefore, a square cuboid of the height of the pyramid will have a volume of $$600$$ cubic cm.

We know that the height is $$13$$ cm.
Area of the base square* height = $$600$$ cm.
=> Area of the base square = $$\frac{600}{13}$$ cm$$^2$$.
Side of the base square = $$\sqrt{\frac{600}{13}}$$ cm.

Length of diagonal of the base square = $$\sqrt{\frac{600}{13}}*\sqrt{2}$$

Now, the height of slant edge of the pyramid can be found out by using the Pythagoras theorem. 
Length of half the diagonal of the base square will form one of the sides and the height of the pyramid will form the other side. The slant height of the pyramid will be the hypotenuse of the right-angled triangle. 

Height of slant edge = $$\sqrt{13^2 + \frac{1}{4}*2*\frac{600}{13}}$$
                   = $$\sqrt{169 + \frac{600}{26}}$$
                   = $$\sqrt{\frac{4394+600}{26}}$$
                   = $$\sqrt{ \frac{4994}{26}}$$
                   =$$\sqrt{192.07}$$
                   =$$13.85$$ cm

The nearest integer is $$14$$. Therefore, option C is the right answer.

There are 3 parallel lines AB, CD, and EF, in that order. 
Let the distance between CD and AB be 2x.
It has been given that the distance between CD and EF is x.

A quadrilateral PQRS is formed such that P is on AB, Q and S are on CD, and R is on EF. 
Also, the length of SR is the least possible value it can take. Therefore, SR must be perpendicular to the parallel lines. 

Area of quadrilateral PQRS = Area of triangle PQS + Area of triangle SRQ = 30 square cm. 

Area of triangle PQS = 2*area of triangle SRQ (Since they rest on the same base and height of SRQ is half the height of PQS)
=> Area of triangle SRQ = 10 square cm. 
Let the length of SQ be $$b$$. We know that SR= $$x$$
$$0.5*x*s = 10$$
=> $$xs=20$$
$$s=20/x$$
We do not have any other detail to evaluate the value of the expression. But, we have been given that d1 and d2 are integers. Therefore, the least value that 'x' can take is 1. 
The least value that S can take is 1. 
By Pythagoras theorem, $$QR =\sqrt{s^2 + x^2}$$
$$QR=\sqrt{20^2+1}$$
$$QR=\sqrt{401}$$
Therefore, the value of QR will be slightly greater than 20. Therefore, option E is the right answer. 

First, let us construct the rectangle using the given information.

Let the length of the rectangle be 'l' and the breadth of the rectangle be 'b'.
Area of rectangle ABCD = lb. 
We have to find the area of the triangle APS.

We consider length as 4cm and breadth as 2cm. ( Answer is irrespective of length and breadth) 

In triangle QRD

RS/RQ=XS/DQ

1/4=XS/2cm

XS= 0.5 cm

SY= 4cm-0.5cm= 3.5cm

We can now find the areas of all the sections.

area (RDQ)= 1/2 x 1cm x 2cm= $$1cm^2$$

area (ABP) =1/2 x 1cm x 4cm = $$2cm^2$$

area (ARS) =1/2 x 1cm x 0.5cm = $$0.25cm^2$$

area (PSQC) = Area (PSZ)-Area(QCZ)= 1/2 x 2cm x 3.5cm  - 1/2 x 1cm x 2cm =

                       =   $$2.5cm^2$$

Area (ABCD)= 4cm x 2cm = $$8cm^2$$

Area (APS) = 8 -1 -2 - 0.25 - 2.5 = $$2.25cm^2$$ 

Area (APS) / Area (ABCD) = 9/32  =  36/128

Hence answer (A).

AB = AC = CD, => $$\angle CAD = \angle CDA = 20^{\circ}$$

and $$\angle ABC = \angle ACB$$

In $$\triangle$$ ACD

=> $$\angle ACD + \angle CAD + \angle CDA = 180^{\circ}$$

=> $$\angle ACD = 180^{\circ} - 20^{\circ} - 20^{\circ} = 140^{\circ}$$

=> $$\angle ACB = 180^{\circ} - 140^{\circ} = 40^{\circ} = \angle ABC$$

Similarly, In $$\triangle$$ ABC

=> $$\angle BAC = 180^{\circ} - 40^{\circ} - 40^{\circ} = 100^{\circ}$$

$$\therefore \angle BAD = 100^{\circ} + 20^{\circ} = 120^{\circ}$$

AB = 54 cm and $$\triangle$$ANM , $$\triangle$$OCP , $$\triangle$$OPX are equilateral triangles.

=> MN = MR = NO = OP = PQ = QR = $$\frac{54}{3} = 18$$ cm

Thus,  MNOPQRM is a regular hexagon with side 18 cm

$$\therefore$$ Area of  MNOPQRM = $$\frac{3 \sqrt{3}}{2} (side)^2$$

= $$\frac{3 \sqrt{3}}{2} \times 18^2 = 486 \sqrt{3} cm^2$$

Area of rectangular floor = $$9.5 \times 11.5$$

= $$109.25 m^2$$

Now, cost of covering $$109 m^2$$ (with 1x1 tiles) = $$109 \times 100$$ = Rs. 10,900

Cost of covering $$0.25 m^2$$ (with 0.5 m square tile) = Rs. 30

$$\therefore$$ Total cost = $$10,900 + 30 = Rs. 10,930$$

When a square sheet is folded in half, its area is also halved. Now, an isosceles right triangle is formed after folded. For 2nd & 3rd folds, again areas halved each time and isosceles right triangle is formed

Equal sides = 10 cm

Area of last such triangle = $$\frac{1}{2} \times 10 \times 10 = 50 cm^2$$

$$\therefore$$ Area of original square = $$50 \times 2 \times 2 \times 2$$

= $$400 cm^2$$

Let radius of incircle = $$r$$, => Radius of circumcircle = $$2r$$

Difference in area = $$\pi [(2r)^2 - (r)^2] = 2156$$

=> $$3 \times \frac{22}{7} \times r^2 = 2156$$

=> $$r^2 = \frac{2156 \times 7}{66}$$

=> $$r = \sqrt{\frac{686}{3}}$$

Now, height of equilateral triangle = $$3 r = \frac{\sqrt{3}}{2} a$$    (where $$a$$ is side of triangle)

=> $$3 \times \sqrt{\frac{686}{3}} = \frac{\sqrt{3}}{2} a$$

=> $$a = 2 \sqrt{686}$$

$$\therefore$$ Area of triangle = $$\frac{\sqrt{3}}{4} a^2$$

= $$\frac{\sqrt{3}}{4} \times 4 \times 686 = 686 \sqrt{3} cm^2$$

Let P and Q be the centres of the circles with arcs x and y respectively.

Thus, $$\angle APB = 90$$ and $$\angle AQB = 60$$

Also, length of arc $$y = 10 \pi$$ cm

=> $$\frac{\theta}{360} \times 2 \pi r = 10 \pi$$

=> $$\frac{1}{6} \times 2 \times r = 10$$

=> $$r = AQ = 10 \times 3 = 30$$ cm

=> AB = 30 ($$\because \triangle$$ AQB is equilateral triangle)

Also, $$\triangle$$ APB is right isosceles triangle, => $$AP = \frac{30}{\sqrt{2}}$$ 

$$\therefore$$ Arc length = $$x = \frac{90}{360} \times 2 \pi \times \frac{30}{\sqrt{2}}$$

= $$\frac{15 \pi}{\sqrt{2}}$$

Let $$DQ = x$$, => $$BP = 2x$$

Acc. to ques,

=> $$ar (\triangle BPD) + ar (\triangle BQD) \leq ar (PBQD)$$

=> $$(\frac{1}{2} \times AD \times BP) + (\frac{1}{2} \times BC \times QD) \leq 150$$

=> $$(\frac{1}{2} \times 9 \times 2x) + (\frac{1}{2} \times 13 \times x) \leq 150$$

=> $$31x \leq 300$$ => $$x \leq \frac{300}{31}$$

=> $$x \leq 9.68$$

Thus, for $$x$$ to be an integer and positive, 9 different values (1 to 9) are possible. 

Volume of Cylinder = $$\pi r^2 h = \pi \times 7^2 \times 10 = 490\pi$$

Now, The solid metal cylinder is re-cast into two cones in the proportion 3 : 4 i.e. the volumes of cone 1 and cone 2 is

$$210 \pi$$ and $$280 \pi$$ respectively.

So, flat Surface area of cylinder before melting = $$2 \pi r^2 = 2 \pi \times 7^2 = 98 \pi$$

Volume of cone 1 = $$\frac{1}{3} \pi r_1^2 h = 210 \pi$$

=> $$r_1^2 = \frac{210 \times 3}{10} = 63$$

Volume of cone 2 = $$\frac{1}{3} \pi r_2^2 h = 280 \pi$$

=> $$r_2^2 = \frac{280 \times 3}{10} = 84$$

Flat surface area of cones = $$\pi r_1^2 + \pi r_2^2$$

= $$\pi (63 + 84) = 147 \pi$$

$$\therefore$$ Percentage change in surface area = $$\frac{147 \pi - 98 \pi}{98 \pi} \times 100$$

= $$\frac{1}{2} \times 100 = 50 \%$$

Area of given figure = 144 sq meter

It is given that BCE becomes square when we will unfold it, so to find the complete area of the figure shown as dotted after unfolding we need to add the area of triangle BCE.

Thus, BC = CE = 6 m

=> Area of $$\triangle$$ BCE = $$\frac{1}{2} \times 6 \times 6 = 18$$ sq meter

$$\therefore$$ Final area of whole figure = 144 + 18 = 162 square meter.

Perimeter of square ABCD = 200 ft

=> AB = $$\frac{200}{4} = 50$$ ft

=> $$DB = \sqrt{50^2 + 50^2} = 50 \sqrt{2}$$ ft

=> $$BO = r = \frac{50 \sqrt{2}}{2} = 25 \sqrt{2}$$ ft

Width of the road = BX = $$7 \sqrt{2}$$ ft

=> $$BX = R = 25 \sqrt{2} + 7 \sqrt{2} = 32 \sqrt{2}$$

Area of bigger circle = $$\pi R^2 = \pi (32 \sqrt{2})^2 = 2048 \pi$$ sq. ft

Area of smaller circle = $$\pi r^2 = \pi (25 \sqrt{2})^2 = 1250 \pi$$ sq. ft

=> Area of road = $$2048 \pi - 1250 \pi = 798 \times \frac{22}{7} = 2508$$ sq. ft

But we have to calculate cost of construction of 50% road.

Required Construction = $$\frac{2508}{2} = 1254$$ sq. ft

$$\therefore$$ Cost of 1254 ft = $$1254 \times 100 = Rs. 1,25,400$$

In $$\triangle$$ ABF

=> $$tan 30 = \frac{BF}{AB}$$

=> $$\frac{1}{\sqrt{3}} = \frac{10}{AB}$$

=> $$AB = 10 \sqrt{3}$$

Similarly, $$ED = 10 \sqrt{3}$$

Also, $$\angle ECD = \angle BCF = 60$$ (Vertically opposite angles)

In $$\triangle$$ BCF

=> $$tan 60 = \frac{BF}{BC}$$

=> $$\sqrt{3} = \frac{10}{BC}$$

=> $$BC = \frac{10}{\sqrt{3}}$$

=> Height = $$AD = AB + BC + CD = 10 \sqrt{3} + \frac{10}{\sqrt{3}} + 10 = \frac{40 + 10 \sqrt{3}}{\sqrt{3}}$$

$$\therefore area (\triangle AED) = \frac{1}{2} \times AD \times ED$$

= $$\frac{1}{2} \times \frac{40 + 10 \sqrt{3}}{\sqrt{3}} \times 10 \sqrt{3}$$

= $$50 (\sqrt{3} + 4)$$

Using distance formula,

$$CX = \sqrt{(17.5 - 5.5)^2 + (23.5 - 7.5)^2} = \sqrt{12^2 + 16^2}$$

= $$\sqrt{144 + 256} = \sqrt{400} = 20$$

=> $$AC = 2 \times CX = 40$$

$$BX = \sqrt{(17.5 - 13.5)^2 + (23.5 - 16)^2} = \sqrt{4^2 + 7.5^2}$$

= $$\sqrt{16 + 56.25} = \sqrt{72.25} = 8.5$$

=> $$BD = 2 \times BX = 17$$

AB + BC + CD + AD = 1120 ------------Eqn(I)

PB + BC + CD + PD = 1000 -------------Eqn(II)

Subtracting eqn(II) from (I), we get :

=> AB - PB + (AD - PD) = 120

=> AB - PB + AP = 120

=> AB + AP = 120 + PB

Now, if AB = PB, => AP = 120

=> AD = 600 and BC = 480, then AB + PB + CD = 40, which is not possible (We know that BC = PD. If BC = PD = 480, then BC+PD = 960. PB + BC + CD + PD = 1000.
=> PB+CD = 40. Therefore, AB + PB+CD should be greater than 40).

Similarly, AB = AP is also not possible. Thus $$AP = BP$$

=> $$\angle ABC = x + (180 - 2x) = (180 - x)$$

=> $$sin \angle ABC = sin (180 - x) = sin x$$

Also, perimeter of PBCD = $$10y = 1000$$ => $$y = 100$$

and perimeter of ABCD = $$AB + 10y = 1120$$ => $$AB = 120$$

Applying cosine rule in $$\triangle$$ ABP

=> $$cos x = \frac{(AB)^2 + (AP)^2 - (BP)^2}{2 AB AP}$$

=> $$cos x = \frac{(120)^2 + (100)^2 - (100)^2}{2 \times 120 \times 100}$$

=> $$cos x = \frac{120}{200} = \frac{3}{5}$$

$$\therefore sin x = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}}$$

= $$\sqrt{\frac{16}{25}} = \frac{4}{5}$$

If a point is equidistant from all 3 vertices, it has to be the circumcentre. The given circle with centre S is concentric and touches two sides.

As S is equidistant from 2 of the sides (say AB and AC), => It lies on angle bisector of $$\angle A$$.

=> $$\triangle ABC$$ is isosceles with AB = AC

Radius of the circle = RS = SQ = 175 cm and SA = SB = SC = 625 cm

=> $$AR = \sqrt{625^2 - 175^2} = 600$$

Let SP = x

=> $$(BP)^2 = (BA)^2 - (AP)^2 = (BS)^2 - (SP)^2$$

=> $$1200^2 - (625 + x)^2 = 625^2 - x^2$$

=>$$1200^2 - 625^2 - x^2 - 2*625x = 625^2 - x^2$$

=> $$1200^2 - 2 * 625^2 = 1250x$$

=> $$x = \frac{658750}{1250} = 527$$

=> $$BP = \sqrt{625^2 - 527^2} = 336$$

$$\therefore$$ ar $$(\triangle ABC)$$ = $$\triangle ASB + \triangle ASC + \triangle SBC$$

= $$(600 \times 175) + (600 \times 175) + (527 \times 336)$$

= $$105000 + 105000 + 177072 = 387072$$

Let O be the centre of the clock. Let the person's eye be at A and the tip of minute hand at 5.00 p.m. is at P and at 5.10 p.m. at Q

AM = 1800 m and OP = OQ = $$200\sqrt{3}$$ m

In $$\triangle$$ APM

=> $$tan 30 = \frac{PM}{AM}$$

=> $$\frac{1}{\sqrt{3}} = \frac{PM}{1800}$$

=> $$PM = \frac{1800}{\sqrt{3}} = 600 \sqrt{3}$$

=> $$OM = PM - OP = 600 \sqrt{3} - 200 \sqrt{3} = 400 \sqrt{3}$$

In $$\triangle$$ OBM

=> $$tan 60 = \frac{OM}{BM}$$

=> $$\sqrt{3} = \frac{400 \sqrt{3}}{BM}$$

=> $$BM = 400$$ m

=> $$AB = AM - BM = 1800 - 400 = 1400$$ m

Time taken to reach B from A = 10 minutes = 600 sec

$$\therefore$$ Speed of the person = $$\frac{1400}{600} = \frac{7}{3}$$ m/s

= $$(\frac{7}{3} \times \frac{18}{5})$$ km/hr = $$8.4$$ km/hr

In a triangle , sum of two sides is greater than the third side and difference of two sides is less than third side.

In $$\triangle$$ PQS

=> $$QS < 9 + 5$$ => $$QS < 14$$ ------Eqn(I)

In $$\triangle$$ QRS

=> $$QS > 17 - 5$$ => $$QS > 12$$ ------Eqn(II)

From eqn(I) & (II)

=> $$12 < QS < 14$$

Given : Length of ladder = PA = P'B = 30 m and PQ = 26 m

$$\angle$$ PAQ = 2 $$\angle$$ P'BQ

To find : AB = ?

Solution : In $$\triangle$$ PAQ

=> $$(QA)^2 = (PA)^2 - (PQ)^2$$

=> $$(QA)^2 = 30^2 - 26^2 = 900 - 676$$

=> $$QA = \sqrt{224} \approx 15$$

Also, $$cos 2 \theta = \frac{QA}{PA}$$

=> $$2 \theta = cos^{-1}(\frac{15}{30})$$

=> $$2 \theta = 60$$ => $$\theta = \frac{60}{2} = 30$$

In $$\triangle$$ P'QB

=> $$cos 30 = \frac{QB}{P'B}$$

=> $$QB = \frac{\sqrt{3}}{2} \times 30$$

=> $$QB = 15 \times 1.732 = 25.98$$

$$\therefore$$ AB = QB - QA = 25.98 - 15 = 10.98 m

ABC is right angled triangle. D and E are mid points of AB and BC respectively.

As it is not given that which angle is 90°. So we need statement (III) to find the value of AC.

Whereas using statement (I) & (II) alone, we cannot find the value of AC.

But using all the three statements. We can find value of $$(AB)^2 + (BC)^2$$.

Hence, the value of $$AC = \sqrt{(AB)^2 + (BC)^2}$$ using Pythagoras Theorem.

=> Ans - (E) : All statements are required.

Given : $$OP_1 = 3 , O'P_2 = 8 , OX = 5$$ units

To find : $$P_1P_2 = ?$$

Solution : In $$\triangle OP_1X$$

=> $$(P_1X)^2 = (OP_1)^2 - (OX)^2$$

=> $$(P_1X)^2 = 5^2 - 3^2 = 25 - 9$$

=> $$P_1X = \sqrt{16} = 4$$

In $$\triangle OP_1X$$ and $$\triangle O'P_2X$$

$$\angle OXP_1 = O'XP_2$$   (Vertically opposite angles)

$$\angle OP_1X = O'P_2X = 90$$

=> $$\triangle OP_1X \sim \triangle O'P_2X$$

=> $$\frac{XP_1}{XP_2} = \frac{OP_1}{O'P_2}$$

=> $$XP_2 = 4 \times \frac{8}{3} = 10.66$$

$$\therefore P_1P_2 = P_1X + XP_2 = 4 + 10.66 = 14.66$$ units

Radius = 3 units, => Diameter = PQ = 6 units

y-coordinates of P and Q are same as PQ is parallel to x-axis

x-coordinates of P and Q will have a difference of 6 units.

Equating y-coordinate of P and Q

=> $$a^x = 2 a^{x + 6}$$

=> $$\frac{1}{2} = \frac{a^{x + 6}}{a^x}$$

=> $$a^{x + 6 - x} = \frac{1}{2}$$

=> $$a = \frac{1}{\sqrt[6]{2}}$$

Let us draw the figure according to the available information, 

In $$\triangle$$ AMP and $$\triangle$$ CMR

$$\angle$$ MAP = $$\angle$$ MCR

$$\angle$$ AMP = $$\angle$$ CMR

Therefore, we can say that $$\triangle$$ AMP $$\sim$$ $$\triangle$$ CMR

Hence, we can say that $$\dfrac{AP}{CR}=\dfrac{MP}{MR}$$

$$\Rightarrow$$ $$\dfrac{AB/3}{AB/2}=\dfrac{MP}{MR}$$

$$\Rightarrow$$ $$MR=\dfrac{3}{2}*MP$$

Therefore, we can say that $$\Rightarrow$$ $$MR=\dfrac{3}{5}*RP$$   ... (1)

Similarly, in $$\triangle$$ ANQ and $$\triangle$$ CNR

$$\angle$$ NAQ = $$\angle$$ NCR

$$\angle$$ ANQ = $$\angle$$ CNR

Therefore, we can say that $$\triangle$$ ANQ $$\sim$$ $$\triangle$$ CNR

Hence, we can say that $$\dfrac{AQ}{CR}=\dfrac{NQ}{NR}$$

$$\Rightarrow$$ $$\dfrac{2AB/3}{AB/2}=\dfrac{NQ}{NR}$$

$$\Rightarrow$$ $$NR=\dfrac{3}{4}*NQ$$

Therefore, we can say that $$\Rightarrow$$ $$NR=\dfrac{3}{7}*RQ$$  ... (2)

In $$\triangle$$ RMN and $$\triangle$$ RPQ 

$$\dfrac{\text{Area of triangle RMN}}{\text{Area of triangle RPQ}} = \dfrac{0.5*RM*RN*sinMRN}{0.5*RP*RQ*sinPRQ}$$

$$\text{Area of triangle RMN}=\dfrac{3}{5}*\dfrac{3}{7}*\text{Area of triangle RPQ}$$

We know that, Area of triangle RPQ = 1/6*Area of rectangle ABCD = 1/6*90 = 15 sq. units

$$\Rightarrow$$ $$\text{Area of triangle RMN}=\dfrac{3}{5}*\dfrac{3}{7}*15$$ = $$\dfrac{27}{7}$$ sq. units

Hence, the area of the quadrilateral PQNM = 15 - $$\dfrac{27}{7}$$ = $$\dfrac{78}{7}=11\dfrac{1}{7}$$ sq. units. Therefore, option D is the correct answer. 

For every 2.6 m that one walks along the slanting part of the pool, there is a height of 1 m that is gained.

=> $$\frac{AC}{BC} = \frac{2.6}{1}$$

=> $$AC = 2.6 \times BC$$

Also, dimensions of cuboidal part = $$48 \times 20 \times 1$$

 In $$\triangle$$ ABC

=> $$(AC)^2 = (AB)^2 + (BC)^2$$

=> $$(2.6 \times BC)^2 = (48)^2 + (BC)^2$$

=> $$6.76 (BC)^2 - (BC)^2 = 2304$$

=> $$(BC)^2 = \frac{2304}{5.76} = 400$$

=> $$BC = \sqrt{400} = 20$$ m

$$\therefore$$ Volume of water in the pool = Volume of cuboid + Volume of triangle

= $$(l \times b \times h) + (\frac{1}{2} \times AB \times BC) \times height$$

= $$(48 \times 20 \times 1) + (\frac{1}{2} \times 48 \times 20 \times 20)$$

= $$960 + 9600 = 10560 m^3$$

Area of $$S_1 = 8$$ sq. units

=> Side of $$S_1 = PS = \sqrt{8} = 2 \sqrt{2}$$ units

Similarly, Side of $$S_2 = CD = \sqrt{9} = 3$$ units

=> $$a + b = 3$$

In $$\triangle$$ PDS

=> $$b^2 + a^2 = 8$$

=> $$b^2 + (3 - b)^2 = 8$$

=> $$b^2 + 9 + b^2 - 6b = 8$$

=> $$2b^2 - 6b + 1 = 0$$

=> $$b = \frac{6 \pm \sqrt{36 - 8}}{4} = \frac{6 \pm \sqrt{28}}{4}$$

=> $$b = \frac{3 + \sqrt{7}}{2}$$     $$(\because b > a)$$

=> $$a = 3 - \frac{3 + \sqrt{7}}{2} = \frac{3 - \sqrt{7}}{2}$$

$$\therefore \frac{b}{a} = \frac{\frac{3 + \sqrt{7}}{2}}{\frac{3 - \sqrt{7}}{2}}$$

= $$\frac{3 + \sqrt{7}}{3 - \sqrt{7}} \approx 15.9$$

Radius of cone = $$BP = 13$$ cm

In $$\triangle$$ ABC

=> $$(AB)^2 = (BC)^2 - (AC)^2$$

=> $$(AB)^2 = 17^2 - 15^2 = 289 - 225$$

=> $$AB = \sqrt{64} = 8$$ cm

=> AP = OC = $$BP - AB = 13 - 8 = 5$$ cm

Let time elapsed before water stops coming out of the pipe = $$T$$ min

Volume of frustum = Volume discharged

=> $$\frac{1}{3} \pi h (R^2 + r^2 + R r) = \pi  (0.5)^2 \times f \times T$$    (where f is the flowrate given by 10m/s = 1000cm/s)

(Total volume = Cross sectional area of pipe $$\times$$ flowrate $$\times$$ Total time)

=> $$\frac{1}{3} \times 15 \times (13^2 + 5^2 + 13 * 5) = 0.25 \times 1000 \times T$$

=> $$5 \times 259 = 250 \times T$$

=> $$T = \frac{259}{50} = 5.18$$ mi

The fish travels North for 300 m and on hitting the edge, travels East for 400 m. The directions North and East are perpendicular to each other. On connecting the initial and final positions of the fish in the tank, we get a right-angled triangle. 
The line AC subtends an angle of 90 degree on the circumference. Therefore, AC must be the diameter of the pond. 
Applying Pythagoras theorem, we get, 
$$AB^2 + BC^2$$ = $$AC^2$$
$$AC^2 = 300^2 + 400^2$$
$$AC = 500$$ m.
=> Radius of the pond = $$250$$ m.
Area of the pond = $$\pi*r^2$$
=> Area = $$62500\pi m^2$$.
Therefore, option A is the right answer. 

PQ = QR = RS = SP = 6 cm
PA = 2AS and PA + AS = 6cm
=> PA = 4 cm and AS = 2 cm
Similarly, RB = 2BS and RB + BS = 6 cm
=> RB = 4 cm and BS = 2cm

Area of △ ABQ = area of PQRS - area of △APQ - area of △RBQ - area of △ASB
 = (36 - 12 - 12 - 2) sq. cm
= 10 sq. cm

Hence, option C is the correct answer.

Let us draw the diagram according to the information given in the questions. 

Let us assume that 'T' is the width of the road as shown in the diagram. 

Total area covered by road = $$30*T+40*T-T^2$$

Also it is given that the mayor wants that the area of the two roads to be equal to the remaining area of the park.

$$\Rightarrow$$ $$30*T+40*T-T^2$$ = $$\dfrac{30*40}{2}$$

$$\Rightarrow$$ $$T^2-70T+600$$ = $$0$$

$$\Rightarrow$$ $$(T-10)(T-60)$$ = $$0$$

$$\Rightarrow$$ $$T$$ = $$10$$ or $$60$$ m. T $$\neq$$ 60 m as the width of park is 30 m only. 
Therefore, we can say that T = 10 meters. Hence, option A is the correct answer.

We know that ABC is a right angled triangle.
=> $$AC=\sqrt{6^2+2^2}$$
$$AC=\sqrt{40}$$
$$AC=2*\sqrt{10}$$
Let the coordinates of A be (0,0)
We know that the radius of the circle, OA = $$\sqrt{50}$$cm
Let OD be the height of the triangle AOC.
$$AC/2=*\sqrt{10}$$

By applying Pythagoras theorem, we get,
=>The height(y) of  the point  O $$= \sqrt{50-10}$$
The height(y) of  the point  O$$=\sqrt{40}$$cm
=> Coordinates of point O = $$(\sqrt{10},\sqrt{40})$$

Area of triangle ABC = $$0.5*6*2$$ = $$6$$ square units.
Let the height of triangle ABC be h.
0.5*h*AC=6 
h*AC = 12
h*$$2*\sqrt{10}$$ = $$12$$
h = $$\frac{6}{\sqrt{10}}$$
X-coordinate of point B =$$\sqrt{6^2-\frac{36}{10}}$$
                                    = $$\sqrt{32.4}$$ cm
Distance between points O and B = $$\sqrt{(\sqrt{10}-\sqrt{32.4})^2+({\sqrt{40}-\frac{6}{\sqrt{10}})^2}}$$
Expanding, we get,
Square of the distance between points O and B = $$26$$ cm.
Therefore, option A is the right answer.

We can draw the towns as shown in the diagram below.

For the minimum length the Government should diagonal roads. The length of road will be same as the length of a diagonal of a square whose side length is 10 km.

Length of one diagonal = $$\sqrt{2}*10$$ = 1.414*10 = 14.14 km

Therefore, total length of both the roads = 2*14.14 $$\approx$$ 28.30 km. 

Let the volume of watermelon = $$V$$

Total surface area = $$S$$

Thickness of the skin = $$t$$

=> Volume usable for juice = $$V - S t$$

Hence, if total surface area is minimum then usable volume of the watermelon will be max.

For equal volume, sphere has the least surface area.

Ans - (E)

Radius of larger sphere = $$R = 10$$ cm

Let radius of each of the smaller spheres = $$r$$ cm

=> $$\frac{4}{3} \pi R^3 = 1000 \times \frac{4}{3} \pi r^3$$

=> $$10^3 = 1000 r^3$$

=> $$r = \sqrt[3]{1} = 1$$ cm

Initial surface area of sphere = $$4 \pi R^2 = 4 \pi \times 100 = 400 \pi$$

Final surface area of 1000 spheres = $$1000 \times 4 \pi r^2 = 1000 \times 4 \pi = 4000 \pi$$

$$\therefore$$ Increase in surface area = $$4000 \pi - 400 \pi = 3600 \pi$$

=> $$\frac{3600 \pi}{400 \pi} = 9$$ times

To find : $$(BD)^2 + (BE)^2 + (BF)^2 = ?$$

AC = 80 m

AD = DE = EF = FC = 20

Let $$AB = a$$ and $$BC = b$$

In $$\triangle$$ ABC

$$(a)^2 + (b)^2 = (80)^2$$

Also, $$(BE) = 1/2 (AC) = 40 $$

=> $$BE = 40$$

Using Apollonius theorem in $$\triangle$$ ABE, as AD = DE

=> $$(AB)^2 + (BE)^2 = 2 [(BD)^2 + (AD)^2]$$

=> $$(BD)^2 + 20^2 = \frac{1}{2} (a^2 + 40^2)$$ --------Eqn(I)

Similarly, for $$\triangle$$ BEC, as EF = FC

=> $$(BE)^2 + (BC)^2 = 2 [(BF)^2 + (FC)^2]$$

=> $$(BF)^2 + 20^2 = \frac{1}{2} (b^2 + 40^2)$$ --------Eqn(II)

Adding eqns (I) & (II), we get :

=> $$(BD)^2 + (BF)^2 + 20^2 + 20^2$$ $$= \frac{1}{2} (a^2 + 40^2 + b^2 + 40^2)$$

=> $$(BD)^2 + (BF)^2 + 20^2 + 20^2$$ $$= \frac{1}{2} (80^2 + 40^2 + 40^2)$$

 => $$(BD)^2 + (BF)^2 = 4800 - 800 = 4000$$

$$\therefore$$ $$(BD)^2 + (BE)^2 + (BF)^2 = 4000 + 40^2$$

= $$4000 + 1600 = 5600$$

Height = 7 cm and Radius = 0.25 cm

Volume of cone = $$\frac{1}{3} \pi r^2 h$$

= $$\frac{1}{3} \times \frac{22}{7} \times (0.25)^2 \times 7 = 0.458 cm^3$$

$$\because$$ $$0.458 cm^3$$ can write $$330$$ words

=> $$1 cm^3$$ can write = $$\frac{330}{0.458} = 720.05$$ words

Now, $$1$$ litre = $$1000 cm^3$$

=> $$\frac{3}{5}$$ litre = $$\frac{3}{5} \times 1000 = 600 cm^3$$

$$\therefore$$ $$600 cm^3$$ can write = $$600 \times 720.05$$

$$\approx 4,32,000$$ words

It is given that the 3rd disk also has the maximum diameter. This is possible only when all 3 discs touch each other externally. 

Also the sum of the diameter of the two disks is 10+20 = 30 feets. Hence, we can say that centres of these two disk lie on a diameter of original plywood.The figure can be drawn as shown below,

Here, in the diagram O is the centre of original disk. A and C are the centres of disks having radius 10 feet and 5 feet respectively. D is the touching point of circles with centre A and C.

Let 'x' be the radius of the disk that is to be cut out from the remaining part.

We can say that AO = OD = CD = 5 feet. Also, AB = 10+x, BC = 5+X, OB = 15-x and AC = 15 feet. 

Also, let 'y' be the length of BD.

By applying apollonius theorem in triangle ABD,

$$AB^2+BD^2=2(OB^2+OD^2)$$

$$(10+x)^2+y^2=2((15-x)^2+5^2)$$

$$x^2+20x+100+y^2=500-60x+2x^2$$

$$y^2 - x^2+80x=400$$ ... (1) 

Similarly, by applying apollonius theorem in triangle OBC,

$$OB^2 + BC^2 = 2(BD^2+DC^2)$$

$$(15-x)^2 + (5+x)^2 = 2(y^2+5^2)$$

$$2x^2-20x+250=2y^2+50$$

$$2x^2-2y^2-20x+200=0$$ ... (2) 

By equation (1) and (2) we can say that, 

$$160x-20x+200=800$$

$$x=600/140 = 30/7$$

Therefore, the diameter of the largest disk that can be cut out from the remaining portion of the plywood piece = 2*30/7 = 8.57 feet. Hence, option C is the correct answer.

Let length = $$l$$ ft and breadth = $$b$$ ft

ABCD is original plot. First square AEFD is sold of side = DF = $$b$$, => FC = $$(l - b)$$

After that square EBHG is sold of side GH = $$(l - b)$$

=> HC = $$b - (l - b) = (2b - l)$$

Perimeter of remaining land GHCF = $$2 \times [(l - b) + (2b - l)]$$

= $$2b$$

Perimeter of original land ABCD = $$2 (l + b)$$

Acc to ques

=> $$\frac{2b}{2 (l + b)} = \frac{3}{8}$$

=> $$8b = 3l + 3b$$

=> $$5b = 3l$$

Only one combination is possible, i.e. $$l = 5$$ and $$b = 3$$     ($$\because$$ It is given that : $$l < 2b$$) 

=> Area of land = $$5 \times 3 = 15$$ sq. ft

=> Cost of land = $$15 \times 1000 = Rs. 15,000$$

For overall profit to be 10 %, S.P. = $$15,000 \times \frac{110}{100}$$

= $$Rs. 16,500$$

Side of AEFD = $$b = 3$$

=> S.P. of AEFD = $$3^2 \times 1200 = Rs. 10,800$$

Side of EBHG = $$(l - b) = 5 - 3 = 2$$

S.P. of EBHG = $$2^2 \times 1150 = Rs. 4,600$$

Length of remaining rectangular land GHCF = $$(l - b) = 5 - 3 = 2$$

Breadth = $$(2b - l) = 6 - 5 = 1$$

Let selling price per sq. ft of GHCF = $$Rs. x$$

S.P. of GHCF = $$2 \times 1 \times = Rs. 2x$$

$$\therefore$$ Total S.P.

=> $$10,800 + 4,600 + 2x = 16,500$$

=> $$2x = 16,500 - 15,400$$

=> $$x = \frac{1100}{2} = Rs. 550$$

Using properties of secant, $$PS \times ST = QS \times SR$$ -------------Eqn(I)

Also, for two numbers, $$PS$$ and $$ST$$, we know that harmonic mean is less than geometric mean.

=> $$\frac{2}{\frac{1}{PS} + \frac{1}{ST}} < \sqrt{PS \times ST}$$

=> $$\frac{1}{PS} + \frac{1}{ST} > \frac{2}{\sqrt{PS \times ST}}$$

Using Eqn(I)

=> $$\frac{1}{PS} + \frac{1}{ST} > \frac{2}{\sqrt{QS \times SR}}$$ ------Eqn(II)

Also, for two numbers, $$QS$$ and $$SR$$, geometric mean is less than arithmetic mean.

=> $$\sqrt{QS \times SR} < \frac{QS + SR}{2}$$

=> $$\frac{1}{\sqrt{QS \times SR}} > \frac{2}{QR}$$  $$(\because QS + SR = QR)$$

Multiplying both sides by $$2$$

=> $$\frac{2}{\sqrt{QS \times SR}} > \frac{4}{QR}$$ ----------Eqn(III)

From eqn(II) and (III)

$$\therefore \frac{1}{PS} + \frac{1}{ST} > \frac{4}{QR}$$

To find the minimum or maximum value, we need to find first derivative and put it equal to zero

Expression : $$f(x) = 21 sin x + 72 cos x$$

=> $$f'(x) = 21 cos x - 72 sin x = 0$$

=> $$21 cos x = 72 sin x$$

=> $$\frac{sin x}{cos x} = tan x = \frac{21}{72} = \frac{7}{24}$$

=> $$sin x = \frac{7}{\sqrt{7^2 + 24^2}} = \frac{7}{\sqrt{49 + 576}}$$

=> $$sin x = \frac{7}{\sqrt{625}} = \frac{7}{25}$$

Similarly, $$ cos x = \frac{24}{25}$$

Putting it in the original expression, we get :

=> $$f(x) = (21 \times \frac{7}{25}) + (72 \times \frac{24}{25})$$

= $$\frac{147}{25} + \frac{1728}{25} = \frac{1728 + 147}{25}$$

= $$\frac{1875}{25} = 75$$

Shortcut Method : Maximum value of $$a sin x + b cos x = \sqrt{a^2 + b^2}$$

and minimum value = $$- \sqrt{a^2 + b^2}$$

=> Max value of $$21 sin x + 72 cos x = \sqrt{(21)^2 + (72)^2}$$

= $$\sqrt{441 + 5184} = \sqrt{5625}$$

= $$75$$

Original position of the ladder = AC = $$25$$ ft

Base = BC = $$7$$ ft

=> In $$\triangle$$ ABC, using Pythagoras theorem

=> $$(AB)^2 = (AC)^2 - (BC)^2$$

=> $$(AB)^2 = (25)^2 - (7)^2$$

=> $$AB = \sqrt{625 - 49} = \sqrt{576}$$

=> $$AB = 24$$ ft

After drawing out the base, the new position of ladder = ED = $$25$$ ft

and $$AD = x$$ and $$CE = 2x$$

To find : CE = ?

Solution : In $$\triangle$$ DBE

$$DB = (24 - x)$$ and $$BE = (7 + 2x)$$

=> $$(DB)^2 + (BE)^2 = (ED)^2$$

=> $$(24 - x)^2 + (7 + 2x)^2 = (25)^2$$

=> $$(576 - 48x + x^2) + (49 + 28x + 4x^2) = 625$$

=> $$625 - 20x + 5x^2 = 625$$

=> $$5x^2 = 20x$$

=> $$x = \frac{20}{5} = 4$$

$$\therefore$$ $$CE = 2 \times 4 = 8$$ ft

None of the options include 8 in the interval.

Given : PQ = 10 , QT = 8 , RS = 7

Now, TQ*TP=TR*TS (Secant from same external point)

8(PQ+QT)=TR(TR+RS)

8*18=TR(TR+7)

TR=9

TS=9+7=16

$$\therefore ar (\triangle PTS) = \frac{1}{2} \times PT \times TS \times sin60 $$

= $$\frac{1}{4} \times (16 \sqrt{3}) \times 18$$

= $$72 \sqrt{3}$$

Given : Side of square = CD = 60 cm

=> AB = CD = 60 cm , => Radius of circles centered at A and D have equal radius of 60 cm

To find : OP = $$r = ?$$

Solution : $$AO = 60 - r$$ and $$AQ = OP = r$$

In $$\triangle$$ AOQ

=> $$(OQ)^2 = (AO)^2 - (AQ)^2$$

=> $$(OQ)^2 = (60 - r)^2 - (r)^2$$

=> $$(OQ)^2 = 3600 - 120r + r^2 - r^2$$

=> $$(OQ)^2 = 3600 - 120r$$

Now, $$OD = 60 + r$$ and $$QD = 60 - r$$

In $$\triangle$$ DOQ

=> $$(OD)^2 = (QD)^2 + (OQ)^2$$

=> $$(60 + r)^2 = (60 - r)^2 + (3600 - 120r)$$

=> $$(3600 + 120r + r^2)$$ = $$(3600 - 120r + r^2) + (3600 - 120r)$$

=> $$120r + 120r + 120r = 3600$$

=> $$r = \frac{3600}{360} = 10 cm$$

$$(PL + PN)$$ will be minimum if P lies on LN, and $$(PM + PO)$$ will be minimum if P lies on OM.

=> P must be the intersection point of the diagonals of the quadrilateral.

$$\therefore$$ Min (PL + PM + PN + PO)

= $$LN + OM$$

= $$(\sqrt{(0 + 5)^2 + (5 - 0)^2}) + (\sqrt{(1 + 1)^2 + (-1 - 5)^2})$$

= $$(\sqrt{25 + 25}) + (\sqrt{4 + 36})$$

= $$\sqrt{50} + \sqrt{40} = 5 \sqrt{2} + 2 \sqrt{10}$$

Original area for grazing = $$\pi \times (25)^2$$

= $$625 \pi$$ sq. feet

After the shed (15 $$\times$$ 10 feet) is built, quarter of the area will be reduced.

=> New area = $$\frac{3}{4} \pi \times (25)^2 + \frac{1}{4}[\pi (10)^2 + \pi (15)^2]$$

= $$\frac{1875 \pi}{4} + \frac{325 \pi}{4}$$

= $$\frac{2200 \pi}{4} = 550 \pi$$

Now, original area corresponds to Rs. 1,000

=> $$625 \pi \equiv 1000$$

$$\therefore 550 \pi \equiv 1000 \times \frac{550}{625}$$

= $$Rs. 880$$

KL = lighthouse

BA = initial position man of man and BC = shadow

After moving 300 m east, DE = new position of man and EF = shadow

Given : AB = DE = 6 m

BC = 24 m and EF = 30 m and BE = 300 m

$$\triangle$$ LBE is right angled triangle (sea level).

To find : KL = ?

Solution : In $$\triangle$$ KLF and $$\triangle$$ DEF

=> $$\angle KLF = \angle DEF = 90$$

$$\angle KFL = \angle DFE$$ (common angle)

=> $$\triangle KLF \sim \triangle DEF$$

=> $$\frac{KL}{DE} = \frac{LF}{EF}$$ -----------Eqn(I)

Similarly, $$\triangle KLC \sim \triangle ABC$$

=> $$\frac{KL}{AB} = \frac{LC}{BC}$$ ----------Eqn(II)

From eqn (I) and (II), and using AB = DE

=> $$\frac{LC}{BC} = \frac{LF}{EF}$$

=> $$\frac{LC}{24} = \frac{LF}{30}$$

=> $$\frac{LC}{LF} = \frac{24}{30} = \frac{4}{5}$$

If, LC is 4 part $$\equiv$$ LF is 5 part

=> $$LB = 4x$$ and $$LE = 5x$$

$$\because$$ $$\triangle$$ LBE is right angled triangle

=> $$(LE)^2 - (LB)^2 = (BE)^2$$

=> $$25X^2 - 16X^2 = 90000$$

=> $$x^2 = \frac{90000}{9} = 10000$$

=> $$x = \sqrt{10000} = 100$$

=> $$LB = 400$$ and $$LE = 500$$

=> $$LC = LB + BC = 400 + 24 = 424$$

Now, using Eqn (II), we get : 

=> $$KL = \frac{LC}{BC} \times AB$$

= $$\frac{424}{24} \times 6 = \frac{424}{4}$$

= $$106 m$$

KL = lighthouse

BA = initial position man of man and BC = shadow

After moving 300 m east, DE = new position of man and EF = shadow

Given : AB = DE = 6 m

BC = 24 m and EF = 30 m and BE = 300 m

$$\triangle$$ LBE is right angled triangle (sea level).

To find : LE = ?

Solution : In $$\triangle$$ KLF and $$\triangle$$ DEF

=> $$\angle KLF = \angle DEF = 90$$

$$\angle KFL = \angle DFE$$ (common angle)

=> $$\triangle KLF \sim \triangle DEF$$

=> $$\frac{KL}{DE} = \frac{LF}{EF}$$ -----------Eqn(I)

Similarly, $$\triangle KLC \sim \triangle ABC$$

=> $$\frac{KL}{AB} = \frac{LC}{BC}$$ ----------Eqn(II)

From eqn (I) and (II), and using AB = DE

=> $$\frac{LC}{BC} = \frac{LF}{EF}$$

=> $$\frac{LC}{24} = \frac{LF}{30}$$

=> $$\frac{LC}{LF} = \frac{24}{30} = \frac{4}{5}$$

If, LC is 4 part $$\equiv$$ LF is 5 part

=> $$LB = 4x$$ and $$LE = 5x$$

$$\because$$ $$\triangle$$ LBE is right angled triangle

=> $$(LE)^2 - (LB)^2 = (BE)^2$$

=> $$25^2 - 16X^2 = 90000$$

=> $$x^2 = \frac{90000}{9} = 10000$$

=> $$x = \sqrt{10000} = 100$$

$$\therefore LE = 5 \times 100 = 500 m$$

XADY = XA + AD + DY = 2/2 + (2 * 3.14 * 2)/4 + (30/360) * (2 * 3.14 * 2) = 5.19
XOBY = XO + OB + BY = 2/2 + 2 + (60/360) * (2 * 3.14 * 2) = 5.09
XODY = XO + OD + DY = 2/2 + 2 + (30/360) * (2 * 3.14 * 2) =  4.04
Hence, option D is the correct answer.

Given : AB = AC = BC = 3  cm and BD = $$\frac{1}{2}$$ CD

AE is median.

To find : $$AD = ?$$

Solution : BD + CD = 3

=> $$BD + 2BD = 3BD = 3$$

=> $$BD = \frac{3}{3} = 1$$ cm

Also, since AE is media => $$BE = CE = \frac{3}{2}$$ cm

=> $$DE = BE - DE = \frac{3}{2} - 1 = \frac{1}{2}$$ cm

Also, AE = $$\frac{\sqrt{3}}{2} a = \frac{3 \sqrt{3}}{2}$$ cm

In $$\triangle$$ ADE

=> $$(AD)^2 = (AE)^2 + (DE)^2$$

=> $$(AD)^2 = (\frac{3 \sqrt{3}}{2})^2 + (\frac{1}{2})^2$$

=> $$(AD)^2 = \frac{27}{4} + \frac{1}{4} = \frac{28}{4}$$

=> $$AD = \sqrt{7}$$ cm

To find : $$EF = x = ?$$

Solution : In $$\triangle$$ ABC and $$\triangle$$ EFC

$$\angle ACB = \angle ECF$$  (common)

$$\angle ABC = \angle EFC = 90$$

=> $$\triangle ABC \sim \triangle EFC$$

=> $$\frac{x}{2} = \frac{5 - d}{5}$$ ----------Eqn(I)

Similarly, $$\triangle BCD \sim \triangle BFE$$

=> $$\frac{x}{3} = \frac{d}{5}$$ --------Eqn(II)

Adding Eqns (I) & (II), we get :

=> $$\frac{x}{2} + \frac{x}{3} = \frac{5 - d}{5} + \frac{d}{5}$$

=> $$\frac{3x + 2x}{6} = \frac{5}{5}$$

=> $$5x = 6 \times 1 = 6$$ 

=> $$x = \frac{6}{5} = 1.2 m$$

Area of parallelogram = $$ab sin 60 = 15\frac{\sqrt{3}}{2}$$

=> $$\frac{\sqrt{3}}{2} ab = 15 \frac{\sqrt{3}}{2}$$

=> $$ab = 15$$

Using cosine rule in $$\triangle$$ ABD

=> $$cos 120 = \frac{a^2 + b^2 - 7^2}{2 ab}$$

=> $$\frac{-1}{2} = \frac{a^2 + b^2 - 49}{30}$$

=> $$a^2 + b^2 = 49 - 15 = 34$$

Also, $$(a + b)^2 = a^2 + b^2 + 2ab$$

=> $$(a + b)^2 = 34 + 2(15) = 64$$

=> $$(a + b) = \sqrt{64} = 8$$

$$\therefore$$ Perimeter of parallelogram = $$2 (a + b) = 2 \times 8 = 16$$ cm

Volume of tank = $$150 \times 120 \times 100 = 18,00,000 cm^3$$

Volume of water in the tank = $$12,81,600 cm^3$$

Volume to be filled in the tank = $$18,00,000 - 12,81,600 = 5,18,400 cm^3$$

Let the number of bricks to be placed in the tank = $$x$$

Volume of $$x$$ bricks = $$x \times 20 \times 6 \times 4 = 480x cm^3$$

Each brick absorbs $$(\frac{1}{10})^{th}$$ of its volume in water

=> $$x$$ bricks will absorb = $$\frac{480 x}{10} = 48x cm^3$$

$$\therefore$$ $$5,18,400 + 48x = 480x$$

=> $$480x - 48x = 432x = 5,18,400$$

=> $$x = \frac{518400}{432} = 1200$$