Geometry
As you have learned basics of geometry in Study Plan for Beginners, so we’ll not revise that basics here, rather we will move on to advance level topics and scenarios that are tested in exams like GAT.
Lines & Angles:
First of all, you need to remember some rules here as follows:
Rule 1:
When a line cross two parallel lines, then the opposite angles are always equal as shown in the figure bellow:
Here the green colored angles are always equal provided that the tree lines do intersect in such a way that two lines are parallel and third line intersect the two parallel lines.
Similarly,
The above two green angles would also be the same in such scenario.
Also,
The sky-blue angles would also be the same for being opposite angles to each others.
And similarly,
The above two sky-blue angles would also be the same in such scenario.
Also remember that the sum of one green angle and one blue angle would always be equals to 180° as shown bellow:
In short, angle formed by a line at one side is always 180, as shown above line l1 is sum of one green angle and one sky-blue angle, which is always 180° by rule.
If in any question a figure like show bellow is given, then never suppose the green nor the sky-blue angle to be the half of 180° i.e, 90°. Always remember that figure would not drawn to scale unless it’s written that ‘figure drawn to scale‘.
Triangles:
A triangle is as we know a figure which has three sides, three vertex (the corner of triangle where the two sides intersect is called vertex) and three angles.
Area of Triangle:
As mentioned in Study plan for Beginners the area of any triangle is: {1⁄2} × Base × Perpendicular height. This equation for calculating area of triangle is for any kinds of triangle.
Area of Right-angle Triangle vs area of Non right-angle Triangles:
Suppose a 100 meter wire is used to form a right-angle triangle and another 100 meter wire is used to form a non right-angle triangle. Would the areas of both triangle equal? if not which triangle would have greater area?
To solve such things, always suppose base of both triangles equal. Now, as we know area of triangle is {1⁄2} × Base × Perpendicular height, so only height is different for both triangles; while the other two things (i.e 1⁄2 and Base) are same. If you analyze it, you will always see that the height of right-angle triangle would be the highest, making the area of right-angle triangle to be greatest as compared to any triangle with the same perimeter (perimeter is used for sum of sides or total boundary length of any two-dimensional figure except circle), which we assumed 100. You can see this in drawing form as follows:
Pythagoras Thorium:
You are studying this term since secondary school education, if you are studying Math continuously. So it’s not unfamiliar to you.
First thing about usage of Pythagoras Thorium is that it’s used only on right-angle triangle. If the triangle is not a right-angle triangle, then Pythagoras Thorium can not be used.
You have already read Pythagoras Thorium in Study Plan for Beginners, so we will skip this here. It’s a basic concept.
Sum of angles:
As we read that the sum of any triangle is always 180°. So if any two angles are given, or the sum of any two angles of a triangle is given, then the third angle would be found by subtracting the sum of the two given angles from 180°. Because Sum of angles equals 180°, but when one of the angle is need to be find, then the other two angles would go to the other side of the equation with –ve sign, i.e 180° – (sum of two given angles) and don’t forget to place brackets around sum of two angles, which is general rule that you’ll learn in detail later in Arithmetic topic.
30°-60°-90° Triangle: (V.V.V Important Concept)
Remember that this is the only triangle, whose sides ratios you have to remember and it’s the most important topic of the whole geometry that is almost always tested in tough exams like Local GRE Commonwealth General Scholarship & employment Tests, and some times in GAT as well.
See the following 30°-60°-90° triangle:
Did you notice that if any one side is given you can find all the other sides and hence area of this triangle. For instance, if the side in front of angle 30° is given as 10, so this implies that the side in front of 90° (I call it as side of 90°), which is twice the side of 30°, would be 2(10) = 20; while the side of 60°, which is √3 times the side of 30°, would be 10√3. So in this way you can find all other sides of such triangle. You’ll get plenty of practice at end of this exercise as well as in weekly assignments.
Similar Triangles vs Congruent Triangles:
As the words indicated; similar means almost same but slightly differ, while congruent means exactly equal.
Similar Triangles
Two or more triangles are said to be similar triangles when their all three individual angles are equal, but sides triangles are not equal; i.e size of triangles are differ but angles are same. For instance Triangle ABC and Triangle DEF are similar triangles as follows:
And similarly,
When you see two triangles as similar triangles, then you may use the following two equations to find values of sides or area:
S1⁄S2 = L1⁄L2
Where S1 and S2 are the sides of smaller triangle, while L1 and L2 are the same sides(same sides mean if S1 is side made by angle a, then L1 would also be the side made by angle a of larger triangle.
Also
A1⁄A2 = {(S1)2⁄(L1)2} = {(S2)2⁄(L2)2} = {(P1)2⁄(P2)2}
Where A1 is the area of smaller triangle, while A2 is the area of larger triangle. P1 is perimeter of small triangle and P2 is perimeter of larger triangle.
These equations are use to find the missing thing, if other things are given. You’ll see practice exercise at end of this chapter.
Congruent Triangles:
Two triangles are said to be congruent when their angles and sides are same; in short two same and equal triangle (just copy & paste) are congruent triangles.
Isosceles Triangle:
A triangle, whose any of the two sides are equal, while third side is different in length, it’s known as isosceles triangle. Remember that when the two sides of any triangle are equal, it’s two correspondence angles, that make those two sides) would also be equal; given bellow are few examples of isosceles triangles:
Last triangle is known as isosceles right-angle triangle or 45-45-90 triangle.
Equilateral Triangle:
A triangle in which all three sides are equal as well as all three angles are equal (60°) is known as equilateral triangle as shown below:
Area of Equilateral Triangle: (V.V.V Important Concept)
Area of equilateral triangle is the same as area of triangle. But alternatively area of equilateral triangle can be find from an important equation below:
If any side of an Equilateral triangle is given as x as shown in figure above for Equilateral triangle, then we can find the area of an Equilateral Triangle as follows:
Area of Equilateral Triangle (if only one side is given) = x2 {√3⁄4}
Area of Equilateral Triangle (if only height is given) = h2 {√3⁄3}
Where h is the height of equilateral triangle.
And both of these are equal, i.e
x2 {√3⁄4} = h2 {√3⁄3}
So if side of an equilateral triangle, we can find its height, and vice versa through the given equation above.
Quadrilateral:
1. Square: As you have read little about square in study plan for beginners. Here we’ll discuss in detail.
In square, all sides are equal, and all angles are also equal (90° ). And we know area of square is length × width.
Perimeter of square = a + a + a + a = 4a
Diagonal length is the length between two farthest vertex as shown bellow:
Diagonal length of a square can be found by using Pythagoras thorium as follows, because diagonal of a square makes two isosceles right-angle triangle:
d2 = a2 + a2 = 2a2
⇒ d = √2a2
⇒ Diagonal length of square = d = a√2
Where a is a side of a square.
How many congruent triangles can be made in a square?
This question many times asked in exams like Local GRE (Commonwealth Scholarship).
If we draw this, we’ll get some sequence as follows:
So answer must be: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024 …
In short, the answer must be 2n, where n is a positive integer.
So you must select correct answer that must be any integer from the above sequence.
Rectangle:
As we have discussed little about rectangle as well in study plan for beginners. Here we’ll discuss this in detail.
Area of rectangle is length × width.
Diagonal length of rectangle can also be find through Pythagoras thorium in same say as in square as follows:
d2 = a2 + b2
⇒ d = √a2 + b2
How many congruent triangles can be made in a rectangle?
This question many times asked in exams like Local GRE (Commonwealth Scholarship).
If we draw this, we’ll get some sequence as follows:
If the radius of the circle is 28, what is the arc length AB?
Solution:
In order to find the arc length AB, we need two variables, 1) radius of the circle and 2) angle formed by the two radius or arc from the center of the circle. Fortunately we have both, so let’s find the arc length AB:
Remember that
Arc length AB = {angle⁄360} × 2πr
⇒ = {45⁄360} × 2π(28)
⇒ = {1⁄8} × 56π
⇒ = 7π
Or more specifically, as we know that π = 22⁄7
⇒ Arc length AB = 7 × {22⁄7}
⇒ Arc length AB = 22 Answer
Now, suppose the similar circle with center O as shown bellow:
Now, if the radius of the circle is 6, what is the area of sector OAB (shaded region)?
Solution:
In order to find the area of sector OAB, we need two variables, 1) radius of the circle and 2) angle formed by the two radius or arc from the center of the circle. Fortunately we have both, so let’s find the area of sector OAB:
Remember,
Area of sector OAB = {angle⁄360} × πr²
⇒ = {45⁄360} × π(6)²
⇒ = {45⁄360} × 36π
⇒ = {45⁄10} × π
⇒ = {9⁄2} × π
⇒ = {9π⁄2}
or more specifically, as we know that π = 22⁄7
⇒ Area of sector OAB = {9⁄2} {22⁄7}
⇒ Area of sector OAB = 9 {11⁄7}
⇒ Area of sector OAB = 99⁄7 Answer
Now let’s suppose another scenario and concept in circle,
Suppose a diameter is drawn from a circle such that the end points of diameter AB makes a triangle with the circumference inside the circle, then the angle formed at the circumference would always be 90°, just as at point C in the figure bellow:
Even, you may draw any triangle i.e that cut the circumference of the circle at any point except points A and B, that point at the circumference would always form an angle of 90°.
In another scenario, where points A and B are not at the end point of diameter, rather these points are at other part of the circumference of the circle as shown bellow:
When the two points A and B on the circumference form an angle 70° with the center of circle, these two points would always form half of this angle 70° with the circumference of the circle anywhere just as at point C in the figure above.
And from this principle, we can also further drive another rule as follows:
As in previous figure, it’s said that the two points A nd B would always form half of central angle with anywhere at the circumference, which means all the angles form at anywhere at the circumference of the circle (e.g at points C, D and E) would be equal.
Finally, any line that is tangent to the circumference of the circle (i.e a line which cut at the circumference of the circle such that it will not cross the circumference but just cut at one point at circumference of the circle), would always form right-angle (90° with the radius of the circle from that tangential point of the line), as shown bellow:
