Geometry

3D Geometry:

Cube:
When a square is made three dimensional, it’s become cube, as shown bellow:

A cube always have six faces of equal area. Each face is a square.

Now, let’s learn about Volume, Surface Area, and Diagonal length of a cube.

Let’s suppose x to be the each side of the cube as shown bellow:

Volume of cube = x × x × x = x ³

Now, as there are six faces of a cube, that faces are square of area x ²

⇒ Surface Area of cube = x ² + x ² + x ² + x ² + x ² + x ²

⇒ Surface Area of cube = 6x ²

Diagonal length of a cube is the distance between farthest vertexes of the cube as shown bellow with orange line:

In square, we find diagonal by using 2 dimensions, but here we’ll find diagonal by using 3 dimensions, so

⇒ Diagonal length of a cube = √x ² + x ² + x ²

⇒ Diagonal length of a cube = √3x ²

⇒ Diagonal length of a cube = x√3

Rectangular Solid:
When a rectangle is made three dimensional, it becomes rectangular solid, as shown bellow:

Volume of rectangular solid = l × w × h = lwh

Now, as there are six faces of a rectangular solid with two faces (back & front) of all faces have same area, i.e there are three different faces of areas (l × w), (w × h), and (l × h) and each face are of two size i.e each face is used in every back & front face of the rectangular solid. If you still not understood, then just see the rectangle bellow and you’ll see that bottom face has same area as top face; left face has same area as right face; and front face has same area as back face:

A rectangular solid always have six faces of different areas, but back and front same faces and back and front same areas.

Now, let’s learn about Volume, Surface Area, and Diagonal length of a rectangular solid.

Let’s suppose l, w and h being length, width and height respectively of a rectangular solid as shown bellow:

In above rectangular solid, there are two green faces, two pink faces and two blue faces and each same color are back and front of each others.

⇒ Surface Area of rectangular solid = 2lw + 2wh + 2lh

Diagonal length of a rectangular solid is also the distance between farthest vertexes of the rectangular solid.

In rectangle, we find diagonal by using 2 dimensions, but here we’ll find diagonal by using 3 dimensions, so

⇒ Diagonal length of a rectangular solid= √l ² + w ² + h ²

Now, let’s discuss two different scenarios of rectangular solid questions that mostly asked in exams.

Suppose a water tank of 20m width and 15m length. If the cost of painting on the external part of the tank is Rs 10400 and rate of painting on a square meter is Rs 8, what is the height of the water tank?

Solution:
Firstly, where you must start to solve such question? definitely, as painting on the external part suggests it’s talking about surface area of the whole tank. So let’s start from it.

⇒ Surface area of water tank = 2(20)(15) + 2(15)h + 2(20)h

⇒ Surface area of water tank = 600 + 30h + 40h

⇒ Surface area of water tank = 600 + 70h —————————— (eq. 1)

As we know that,

Total cost of painting on the water tank surface = Total surface area in square meter × cost of painting per square meter —————————-(eq. 2)

(Don’t confuse yourself with the unit meter and square meter; when meter multiplies with meter, it results in square meter, so unit of area is always in square units i.e square meter, square inches or square feet etc.)

By putting value of (eq. 1) and other values in (eq. 2), we’ll get,

⇒ 10400 = (600 + 70h) × 8

⇒ 10400 = 4800 + 560h

⇒ h = 10m Answer

Now, suppose an empty box of length 22m, width 20m and height 25m is needed to be filled with boxes of length 2m, width 3m and height 5m, such that minimum empty space would left in big box. How many maximum possible smaller boxes can be placed inside big box?

Solution:
This question is among the hard difficulty level question. But it’s really easy trust me.

99% students do this question wrong. And majority of them do mistake by placing the length of small box to the length of large box, width of small box to the width of large box and height of small box to height of large box, and happy with their calculation with feel like it’s so easy question and do the wrong way as bellow:

22m length of large box may have (²²⁄₂ =) 11 boxes on length wise; (²⁰⁄₃ =) 6 boxes on width wise; and (²⁵⁄₅ =) 5 boxes on height wise. Therefore, all in all you can place 11 × 6 × 5 = 330 boxes
(Few students may confuse, why we did multiply these?, well you’ll get answer later on while solving it on wright way.)

Or few students may do wrong way as follows:

2m side of smaller box along the 20m side of larger box, thereby placing (²⁰⁄₂ =) 10 boxes along 20m side of larger box. And placed 5m side of smaller box along the 25m side of larger box, thereby placing (²⁵⁄₅ =) 5 boxes along 25m side of larger box. And placed the remaining 3m side of smaller box along the remaining 22m side of larger box, thereby placing (²²⁄₃ =) 7 boxes along 22m side of larger box. Therefore, all in all it can place 10 × 7 × 5 = 350 boxes

Even if you take any combinations of dimensions, you would get different results each time.

Unfortunately, it’s not the right way.
As the question said you must place the maximum number of smaller boxes, such that minimum space would left in larger box. So you must place dimensions of smaller box in such a way minimum space would left. That is possible only if you follow rule bellow:

Largest side/dimension of smaller box must be placed with smallest side of largest box, and smallest side of smaller box must be placed with largest side of largest box; but while doing so just try your best to have an no space left if possible. For instance

Place 5m side of smaller box along the 20m side of larger box, thereby you may place (²⁰⁄₅ =) 4 boxes along 20m side of larger box, this results to no space left. Now place 2m side/dimension of smaller box along the 25m dimension of larger box, this wouldn’t result to an integer clearly and showing some space left, so you must not place this dimension here, rather try to place 2m side of smaller box along the 22m side of larger box (remember you must not take 2m of smaller box with the 25m of larger box because result wouldn’t be an integer), thereby you may place (²²⁄₂ =) 11 boxes along 22m side of larger box. And you must place the remaining 3m side of smaller box along the remaining 25m side of larger box, thereby you may place (²⁵⁄₃ ≈) 8 boxes along 25m side of larger box; although here the result is still not an integer, but as you have only one combination left, so you must do so. Therefore, all in all you can place 4 × 11 × 8 = 352 boxes Answer

Now, we did multiply the number of boxes because, let’s suppose we need to place boxes in 2 dimensional plane (say xy-plane). If you have 4 boxes along x-plane and 5 boxes along y-plane; how many boxes can you place in total in that rectangular region (2-dimensional). Obviously there would be 5 rows having 4 boxes each; and hence 4 columns of 5 boxes each as follows:

All in all you would have 4 × 5 = 20 boxes in that xy-plane. That’s why we did multiply the number of boxes in each dimensions in order to find total number of boxes that could be placed according to the given condition.

Cylinder:
When the circumference of two circles joins straight away from every point such that the two circles are just opposite of each others, it form a cylinder as shown bellow:

The volume of cylinder can be found by multiplying area of circle (i.e base area) with the height of cylinder (which is similar to area of cube as base area multiplied by height) as follows:

Volume of cylinder = πr ² h

Where r is the radius of circle and h is the height of cylinder.

Surface area of cylinder can be found by addition areas of two circles and the circumferential height (which can be find by multiplying circumference of the circle with height of cylinder to make spring type area of surface of 3-D height) as follows:

Surface area of cylinder = 2πr ² + 2πrh

Now, let’s apply all the concepts in some scenario based practice so you’ll not face a big issue while solving quantitative questions from geometry.

If a circle is inscribed in a square of area 100 cm square, what is the area of the shaded region shown above?

Solution:

As we know the area of square is x², where x is a side of the square. Therefore,

x² = 100

⇒ x = 10

Now, if you analyze closely, you’ll come to know that the side of the square is actually equals to the diameter of the circle. That’s the main point of this type of question, and there’s how you may solve it easily.

So, diameter of circle = 10

⇒ Area of circle = πr² = π(5)² = 25π

Now,

Area of Shaded region = Area of Square – Area of circle

⇒ Area of Shaded region = 100 – 25π Answer

Similarly, a square inscribed in a circle or rectangle inscribed in a circle scenarios can be easily solved by analyzing the diagonal of square or rectangle with the diameter of circle, you’ll come to know these would be the same. There’s how you may lead towards the answer comfortably.

The below question of very high difficulty level scenario came in GRE exam of one of my student who attempted well in quantitative:

Two circles of equal radius 6cm overlap such that they intersect at center of each other as shown above. What is the area of shaded region.

Solution:

First of all, you need to create intelligent drawing as show below:

Here, we just analyzed that the lines OP, OX, OY, PX and PY are all same and radius of circle (i.e 6cm). Now,

Moreover, triangles OPX and OPY are equilateral triangles, as their lengths are same (and therefore angles i.e 60° each).

Therefore the areas of the two equilateral triangles can be find as you have a length of a side of equilateral triangle. So,

Area of green region in figure above = 2 × S²(^√3⁄₄)                                    (Where S is side of equilateral triangel)

⇒ Area of green region in figure above = 6²(^√3⁄₂)

⇒ Area of green region in figure above = 36(^√3⁄₂)

⇒ Area of green region in figure above = 18√3 ———————— (eq. 3)

Now, we need to find the area shown in blue color in figure above. For that purpose, we need to find area of sector OXP by using the method we’ve learned before.

Area of sector OXP = (^60°⁄_360°) × π(6)²                                   (As radius = 6cm)

⇒ Area of sector OXP = (¹⁄₆) × 36π

⇒ Area of sector OXP = 6π

Now, if we subtract area of one equilateral triangle (which is half the area of green region) from this area of sector OXP, we’ll get area of one blue region. And if we multiply one blue region by 4, we’ll get area of all 4 blue regions as follows:

⇒ Area of total blue region = 4 × (6π – 9√3)

⇒ Area of total blue region = 24π – 36√3) ——————— (eq. 4)

Now, by adding (eq. 3) and (eq. 4), we’ll get

⇒ Area of required shaded region = 18√3 + 24π – 36√3

⇒ Area of required shaded region = 24π – 18√3 Answer

You see that in high difficulty level question, more than one concept is tested. If you know all the concepts that we’ve discussed till now in geometry, you’ll able to solve any question by using only these concepts. In practice questions, you’ll get more cleared in which concept should be used in which scenario or question.