MAH MBA CET Geometry Questions

Let the breadth of the rectangle be ‘x’. So its length would be 2x.
Thus the area of the rectangle would be $$2x^{2}$$
Now the new length would be 2.4x and new breadth would be .6x
Hence the new area would be $$2.4x*.6x = 1.44x^{2}$$.
Hence we can see that the area has got reduced.
Percentage reduction = $$\frac{56}{200}*100 = 28\%$$
Hence the correct answer is option D.

Let the base and height be b and h respectively. Hence, the original area is = 1/2*b*h. After changing the base and height, area= 1/2 * (0.9b) * (1.1h) = 0.99 (1/2*b*h) = 198 sq-cm
Hence, (1/2*b*h) = 198/0.99 = 200 sq-cm

Given, the diagonal is 13x and the side is 5x.

Let the other side be b.

$$(13x)^2 = (5x)^2 + b^2$$
b = 12x

68 = 2 (12x + 5x)
x = 2m

Hence, the sides are 10m and 24m

Area = 240 sq. m

Volume of the initial sphere = $$\frac{4}{3}*\pi *r^3$$ = $$\frac{4}{3}*\pi *6^3$$
Volume of each of the smaller spheres = $$\frac{4}{3}*\pi *r^3$$
==> 3*$$\frac{4}{3}*\pi *r^3$$ = $$\frac{4}{3}*\pi *6^3$$ ==> $$r^3 = 6^3/3$$ ==> $$r^3 = 72$$ ==> $$r = 2*\sqrt[3]{9}$$
So the correct option to choose is E.

The area of square = 10*10 = 100$$cm^2$$
Now the side of the square is equal to the diameter of the square.
Hence the diameter of the circle is 10cm
So the radius of the circle will be 5 cm.
Hence the area of the circle = $$\pi*r^2$$ = $$\pi*5^2$$ = 78.5$$cm^2$$
Thus, the remaining area = 100 - 78.5 = 21.5 $$cm^2$$

We can use the Angle Bisector Theorem:

In triangle ABC, if AD bisects ∠A, then $$\dfrac{AB}{AC}=\dfrac{BD}{DC}$$

We know: BC=10 and DC=6

So, BD = BC − DC = 10 − 6 = 4

Applying the theorem

$$\dfrac{AB}{AC}=\dfrac{BD}{DC}$$ 

Given
AC=4.2, so:$$AB=\dfrac{2}{3}\times4.2=2.8\ cm$$

Volume remains conserved during melting.
Volume before melting = Volume after melting
Before melting, the wire was in cylindrical shape and after melting, it became sphere.
So, Vol. before = $$\pi r^2l$$ = Vol. after = $$\ \frac{\ 4}{3}\pi\ R^3$$, where R is the radius of sphere formed.
$$\left(2\times\ 10^{-3}\right)^2\times\ 243\ =\ \frac{\ 4}{3}R^3$$ 
$$R^3=729\times\ 10^{-6}$$
R=9cm.
So, diameter of the sphere = 18cm.
Option C is the answer.

Let the radius of sphere A be 'r' and sphere B be 'R'
Surface area of A = $$4\pi r^2$$
Surface area of B = $$4\pi R^2$$ = (1+300%) of A = 4 * $$4\pi r^2$$
R=2r
Volume of A = $$\frac{4}{3}\pi r^3$$ 
Volume of B = $$\frac{4}{3}\pi R^3\ =\ \ \frac{\ 4}{3}\pi\ \left(2r\right)^3\ =\ \ \frac{\ 32}{3}\pi\ r^3$$
Required percent = $$\ \frac{\ \left(\frac{32}{3}\pi\ r^3\ -\ \frac{4}{3}\pi\ r^3\right)}{\ \frac{32}{3}\pi\ r^3}\times\ 100\ =87.5\ \%\ $$
Option C is the answer.

Let the sides of the isosceles triangle be (a, a, b), where a and b are integers.

Case 1: The sum of the two equal sides is 12

a + a = 12 ⇒ a = 6

a + a = 12 ⇒ a = 6

Triangle inequality: 6 + 6 > b ⇒ b < 12

Also b > 0 and integer.

So, b = 1, 2, 3, …, 11

So, Total = 11 triangles

Case 2: One equal side + base = 12 

a + b = 12 ⇒ b = 12 − a

Triangle inequality: a + a > b ⇒ 2a > 12 − a ⇒ 3a > 12 ⇒ a > 4

Also: b > 0 ⇒ 12 − a > 0 ⇒ a < 12

So, a = 5,6,7,8,9,10,11

Hence, Total = 7 triangles

But when a=6, the triangle is (6,6,6), which was already counted in Case 1. So subtract 1 duplicate.

So final count: 11 + 7 - 1 = 17

The bottom part of the pool has a length of 12 m and the width of 6 m . So the area of the bottom ground = 12*6 = 72 m square.

Now, the pool has 2 opposite rectangular walls with the length of 12 m and the depth/width = 2 m so the area = 2*12*2 = 48 m square.

Further, it has another 2 opposite rectangular walls with length of 6 m and depth/width of 2 m so the area = 2*6*2 = 24 m square.

So, the total area to be painted = 72 + 48 + 24 = 144 m square.

Cost of painting = 144*25 = Rs. 3600

Number of diagonals can be found by using: $$^nC_2-n$$
$$\frac{n\left(n-1\right)}{2}-n=35$$
$$n^2-3n-70=0$$
$$\left(n-10\right)\left(n+7\right)$$

N cannot be negative so n=10

Internal angle can be calculated using, 
$$\frac{180\left(n-2\right)}{n}=\frac{180\left(8\right)}{10}=144$$

Hence the answer is 144. 

The area of a triangle formed by the medians of an equilateral triangle would be (3/4) of the area of the original triangle. 

The area of the original equilateral triangle can be calculated using the formula $$\frac{\sqrt{\ 3}}{4}a^2$$, where $$a$$ is the side length, giving the area of the original equilateral triangle to be $$\frac{\sqrt{\ 3}}{4}\times\ 12\times\ 12$$

This would give the area of the triangle formed by the medians to be $$\frac{\sqrt{\ 3}}{4}\times\ 12\times\ 12\times\ \frac{3}{4}=27\sqrt{\ 3}cm^2$$

Therefore, Option C is the correct answer. 

Volume of cone = $$\ \frac{\ 1}{3}\pi\ r^2h$$ 
Let initial height be 'h' and radius be 'r' of the cone.
So, Initial volume =  $$\ \frac{\ 1}{3}\pi\ r^2h$$
New height H = h(1+200%) = h(1+$$\ \frac{\ 200}{100}$$) = 3h
New radius R = r(1-50%) = r(1-0.5) = $$\ \frac{\ r}{2}$$
New volume = $$\ \ \frac{\ 1}{3}\pi\ \times\ \left(\frac{r}{2}\right)^2\times\ 3h$$ = $$\ \frac{\ \pi\ r^2h}{4}$$
Change in volume = $$\frac{\left(\frac{\ \pi\ r^2h}{4}-\frac{\ \pi\ r^2h}{3}\right)}{\ \frac{\ \pi\ r^2h}{3}}\times\ 100$$ = -25%
So, volume decreased by 25%.
Option B is the answer.

Let the common factor be K

∴ Length = 4K ; Breadth = 5K and Height = 7K

Tip:

Total Surface area of cuboid = 2(LB + BH + LH)

L = Length; B = Breadth/width; H = Height

Whole surface area of the rectangular plank = 2(4K x 5K + 5K x 7K + 7K x 4K)

∴ 20086 = 83*2*K^2

∴ K = 11

∴ Length = 4K = 44 sq.m