CAT Mensuration Questions

Since BC is the diameter of the circle, which implies angle BAC is 90 degrees. Let AB = a cm, which implies AC = b cm. Hence, $$BC\ =\ \sqrt{\ a^2+b^2}$$, which is diameter of the circle (2r).

Hence, $$2r\ =\sqrt{\ a^2+b^2}$$

=> $$4r^2=a^2+b^2$$

The area of the triangle is $$\frac{1}{2}\times\ a\times\ b$$, which can be written as 

=> $$\ \frac{\ a\cdot b}{2\left(a^2+b^2\right)}\times\ \left(a^2+b^2\right)$$

=> $$\frac{\ a\cdot b}{2\left(a^2+b^2\right)}\times\ 4r^2$$

=> $$\frac{\ a\cdot b}{a^2+b^2}\times\ 2r^2$$

The correct option is B

Let us assume the length of the rectangle is 'l' and breadth of the rectangle is 'b'.

The radius, l/2 and b in the above diagram form a right-angled triangle.

=> $$\left(\frac{l}{2}\right)^2+b^2=2^2$$

We know that the area of the rectangle is l*b, which can be obtained by considering 2 times the geometric mean of $$\left(\frac{l}{2}\right)^2$$ and $$b^2$$.

Therefore, for the maximum area, the equality condition of AM-GM inequality should be satisfied

=> $$\left(\frac{l}{2}\right)^2=b^2$$ => l = 2b.
=> l/b = 2/1.

It is given that AB = 9 cm, BC = 6 cm.

It is also known that the areas of the figures ABP, APQ, and AQCD are in geometric progression.

Hence, the area of the ABP, APQ, and AQCD are k, 2k, and 4k respectively.

The ratio of BP, PQ, QC will be the ratio of the respective triangles. Hence, we can draw a line from point A to point C.

Let the area of triangle AQC be x, which implies the area of triangle ADC = ADQC - AQC = 4k -x, which is equal to the sum of the area of triangle APB, AQP, and ACQ, respectively.

Therefore, 4k-x = 3k+x 

=> x = k/2

Hence the ratio of BP: PQ: CQ = k:2k: k/2 =2:4:1

The sum of the interior angles of a polygon of 'n' sides is given by $$\left(2n-4\right)\times\ 90$$, and the sum of the exterior angles of a polygon is 360 degrees.

So, the difference between them will be 120 * n

=> $$\left(2n-4\right)90-360=120n$$

=> 60n = 720 => n = 12.

We know that the number of diagonals of a regular polygon is nC2 - n = 12C2 - 12 = 66 - 12 = 54.

CD = $$\sqrt{\ y^2+25}$$

$$11+y+\sqrt{y^2+25}=36$$

$$\sqrt{y^2+25}=25-y$$

$$y^2+25=25^2+y^2-50y$$

2y = 24

y = 12

Area of trapezium = $$3y+\frac{5y}{2}=\frac{11y}{2}=\frac{11}{2}\left(12\right)=66$$

$$A=lb$$

$$l^2+b^2=4r^2$$

$$P\ =2\left(l+b\right)$$

$$\frac{P}{2}=l+b$$

Squaring on both the sides, we get

$$\frac{P^2}{4}=l^2+b^2+2lb$$

$$\frac{P^2}{4}=4r^2+2lb$$

$$\frac{P^2}{8}-2r^2=lb$$

The answer is option B.

Area of a regular hexagon = $$\frac{3\sqrt{3}}{2}x^2$$

Area of an equilateral triangle = $$\frac{\sqrt{3}}{4}\left(a\right)^2$$ ; where a = side of the triangle

Since the area of the two figures are equal, we can equate them as folllows: $$\frac{3\sqrt{3}}{2}x^2=\frac{\sqrt{3}}{4}\left(12\right)^2$$

On simplifying: $$x^2=24\ $$

$$\therefore\ x=2\sqrt{6}$$

All the sides of the rhombus are equal.

The area of a rhombus is $$12\ cm^2$$

Considering d1 to be the length of the longer diagonal, d2 to be the length of the shorter diagonal.

The area of a rhombus is $$\left(\frac{1}{2}\right)\left(d1\right)\cdot\left(d2\right)\ =\ 12$$

d1*d2 = 24.

The length of the side of a rhombus is given by $$\frac{\sqrt{\ d1^2+d2^2}}{2}$$. This is because the two diagonals and a side from a right-angled triangle with sides d1/2, d2/2 and the side length.

$$\frac{\sqrt{\ d1^2+d2^2}}{2}=\ 5$$

Hence $$\sqrt{\ d1^2+d2^2}\ =\ 10$$

$$d1^2+d2^2\ =\ 100$$

Using d1*d2 = 24, 2*d1*d2 = 48.

$$d1^2+d2^2\ +2\cdot d1\cdot d2=\ 100+48\ =\ 148$$

$$d1^2+d2^2\ -2\cdot d1\cdot d2=\ 100-48\ =\ 52$$

$$d1+d2\ =\ \sqrt{\ 148}$$  (1)

d1-d2 = $$\sqrt{52}$$   (2)

(1) + (2)= 2*(d1) = 2*($$\sqrt{\ 37}+\sqrt{\ 13}$$)

d1 = $$\sqrt{\ 37}+\sqrt{\ 13}$$

or 

In a rhombus the area of a Rhombus is given by :

The diagonals perpendicularly bisect each other. Considering the length of the diagonal to be 2a, 2b.

The area of a Rhombus is : $$\left(\frac{1}{2}\right)\cdot\left(2a\right)\cdot\left(2b\right)\ =\ 12$$

ab =6.

The length of each side is : $$\sqrt{\ a^2+b^2}$$ = 5, $$a^2+b^{2\ }=\ 25,\ $$

$$\left(a+b\right)^2=\ 37,\ \left(a+b\right)\ =\ \sqrt{\ 37}$$

($$\left(a-b\right)^2=\ 13,\ a-b\ =\ \sqrt{\ 13}$$

$$2a\ =\ \left(\sqrt{\ 37}+\sqrt{\ 13}\right)$$,  $$2b\ =\ \left(\sqrt{\ 37}-\sqrt{\ 13}\right)$$.

2a is longer diagonal which is equal to $$\ \left(\sqrt{\ 37}+\sqrt{\ 13}\right)$$

Let us draw the rectangle.

Now, definitely, three sides should be fenced at Rs 100/ft, and one side should be fenced at Rs 200/ft.

In this question, we are going to assume that the L is greater than B.

Hence, the one side painted at Rs 200/ft should be B to minimise costs.

Hence, the total cost = 200B + 100B + 100L + 100L = 300B + 200L

Now, L x B = 60000

B = 60000/L

Hence, total cost = 300B + 200L = 18000000/L + 200L

To minimise this cost, we can use AM>=GM,

$$\frac{\frac{18000000}{L}+200L}{2}\ge\sqrt{\ \frac{18000000}{L}\times\ 200L}$$

$$\frac{18000000}{L}+200L\ge2\sqrt{\ 18000000\times\ 200}$$

$$\frac{18000000}{L}+200L\ge2\times\ 60000$$

Hence, minimum cost = Rs 120000.

We can say 40m is the perimeter of the park 
so side of rhombus = 10 
Now $$\frac{1}{2}\times\ d_1\times\ d_2\ =\ 96$$
so we get $$\ d_1\times\ d_2\ =\ 192$$   (1)
And as we know diagonals of a rhombus are perpendicular bisectors of each other :
so $$\ \frac{d_1^2}{4}+\ \frac{d_2^2}{4}=\ 100$$
so we get $$\ d_1^2+\ d_2^2=\ 400$$    (2)
Solving (1) and (2)
We get $$d_1=12$$ and$$d_2=16$$
Now the cost, in INR, of laying electric wires along its two diagonals, at the rate of ₹125 per m, is= (12+16)(125) =3500

Applying cosine rule in triangle ACD, 

$$100+X^2-2\times\ 10\times\ X\cos30=400$$

$$X^2-10X\sqrt{\ 3}-300=0$$

Solving, we get X = $$\left(\frac{10\sqrt{\ 3}+10\sqrt{\ 15}}{2}\right)$$

Hence, area = 10Xsin 30 = $$\frac{\left(\frac{10\sqrt{\ 3}+10\sqrt{\ 15}}{2}\right)10}{2}$$

= $$25(\sqrt{3}+\sqrt{15})$$

Let the base radius be 3r.

Height of upper cone is 9 so, by symmetry radius of upper cone will be r.

Volume of frustum=$$\frac{\pi}{3}\left(9r^2\cdot27-r^2.9\right)$$

Volume of upper cone = $$\frac{\pi}{3}.r^2.9$$

Difference= $$\frac{\pi}{3}\cdot9\cdot r^2\cdot25=225$$ => $$\frac{\pi}{3}\cdot r^2=1$$

Volume of larger cone = $$\frac{\pi}{3}\cdot9r^2\cdot27=243$$

Let ABCD be the rectangle with length 2l and breadth 2b respectively.

Area of the circle i.e. area of painted region = $$\pi\ b^2$$.

Given, 4lb-$$\pi\ b^2$$=(2/3)$$\pi\ b^2$$.

=> 4lb=(5/3)$$\pi\ b^2$$.

=>l=$$\frac{5\pi}{12}b$$.

Given, 4lb=135 => 4*$$\frac{5\pi}{12}b^2$$=135 => b= $$\frac{9}{\sqrt{\ \pi\ }}$$

=> l=$$\frac{15}{4}\sqrt{\ \pi\ }$$
Perimeter of rectangle =4(l+b)=4($$\frac{15}{4}\sqrt{\ \pi\ }$$+$$\frac{9}{\sqrt{\ \pi\ }}$$ )=$$3\sqrt{\pi}(5+\frac{12}{\pi})$$.

Hence option A is correct. 

Let the sides of the rectangle be "a" and "3a" m. Hence the perimeter of the rectangle is 8a.

Let the side of the equilateral triangle be "m" cm. Hence the perimeter of the equilateral triangle is "3m" cm. Now we know that 8a+3m=90......(1)

Moreover area of the equilateral triangle is $$\frac{\sqrt{\ 3}}{4}m^2$$ and area of the rectangle is $$3a^2$$

According to the relation given $$\left(\frac{\sqrt{\ 3}}{4}m^2\right)^{^2}=\ 3a^2$$

$$\frac{3}{16}m^4=\ 3a^2\ or\ a^2=\frac{m^4}{16}$$

$$a=\frac{m^2}{4}$$  

Substituting this in (1) we get $$2m^2+3m-90\ =0$$ solving this we get m=6 (ignoring the negative value since side can't be negative)

Hence a=9 and the longer side of the rectangle will be 3a=27cm

Let the length of radius be 'r'.

From the above diagram,

$$x^2+r^2=6^2\ $$....(i)

$$\left(10-x\right)^2+r^2=8^2\ $$----(ii)

Subtracting (i) from (ii), we get: 

x=3.6 => $$r^2=36-\left(3.6\right)^2$$ ==> $$r^2=36-\left(3.6\right)^2\ =23.04$$.

Area of circle = $$\pi\ r^2=23.04\pi\ $$

Area of rhombus= 1/2*d1*d2=1/2*12*16=96.

.'. Ratio of areas = 23.04$$\pi\ $$/96=$$\frac{6\pi}{25}$$

Given, BC = DE = 4

CD = BE = 5

In triangle ADE, EAD=45^{0}$$

$$\tan\ 45\ =\ \frac{DE}{AE}$$ => AE = 4

Area of trapezium = Area of rectangle BCDE + Area of triangle AED

= 20 + 8 = 28