Hi SSCians,

In the last post about SUPERHACKS we promised to bring you SUPERHACKS FOR GEOMETRY and here we are with them!

Geometry is a very important topic since it covers the major part of the question paper as a topic with almost more than 5 questions every time in SSC CGL. We understand that Geometry has always given you the pains with lengthy theorems and questions but we also know that we will be short on time if we continue this approach. Therefore for you, Team SSCHacks have compiled these SUPERHACKS for you to make you love GEOMETRY. Spend your 5 minutes now to save this time while attempting the examination. 

Triangles are commonly tested in SSC CGL every year because questions on triangles involve many shortcuts for the savvy test taker. The Commission knows that the average student will rely on formulas and calculations, losing valuable time and possibly making careless mistakes. However, students who consistently report high scores know the following tips and tricks for saving time and attacking triangles.  

  1. When a square is inscribed in a circle, the diameter of the circle is also the diagonal of the square.

    This property is useful for finding the radius of the circle:


    The radius is one-half of the diameter: ½ of 4sqrt{2} is 2sqrt{2}. Once the radius is determined, you can find the area and the circumference of the circle, as in the following question:



    Inscribed Square: (C)

    Draw the diagonal of the square to find the diameter of the circle. This creates two 45: 45: 90 triangles, so you can use the ratio s: s: ssqrt{2} .


    Since the diameter is 12sqrt{2}, the radius is one-half of 12sqrt{2} or 6sqrt{2}. Use the radius to find the area of the circle:

    A = pr^2

    A = p(6)^2

    A = p(36)(2) = 72p

  2. Similar triangles are often hidden.

    Similar triangles have the exact same shape but different area. The corresponding angle measurements of similar triangles are equal, and the corresponding side lengths are proportionate:


    There are three ways the SSC may try to disguise similar triangles. The first involves a triangle within a triangle:


    If line BE is parallel to line CD, then triangle ABE is similar to triangle ACD.

    The second way ETS attempts to hide a set of similar triangles involves two parallel lines that are intersected by a third and fourth line:


    As you can see, the two intersecting lines create two triangles that share vertical and corresponding angles.

    The final disguise is probably the most camouflaged of all three:


    Both triangles have a right angle and share angle a. Therefore, the remaining angle, b, must have the same measurement in each triangle.

    Try solving a problem with hidden similar triangles:



    Similar Triangles: (C)

    The two triangles are similar. You know this because they both contain a right angle and they share angle W. Therefore, angle Y and angle XZW must have the same measurement.

    The sides of similar triangles are proportional. The shortest side of the small triangle is 3 units long. The shortest side of the large triangle is 6 units long. So to find the lengths of the sides of the large triangle, simply multiply the corresponding sides of the small triangle by 2.

    The longest side of the small triangle is 6. So to find WY, the longest side of the large triangle, multiply 6 by 2. The correct answer is 12, (C).

  3. The height of a triangle may be found outside of the triangle itself.

    When trying to compute the area of a triangle (½ base * height), many students get confused when the triangle has an angle greater than 90º (obtuse triangles). The height of a triangle is any perpendicular line segment from a base to a vertex. It is easy to locate the height of an acute triangle:


    But what about an obtuse triangle in which one of the shorter sides is the base?


    Sample Question:

    In this situation, the height can be found outside of the triangle. This knowledge is essential to solve the following question:


    Triangle Height: (C)

    The area of a triangle is ½ base * height. The base of triangle WYZ is 10. The height is WX, or 6. Many students miss this because they assume the height must be inside the triangle.

    A = ½bh

    A = ½(10)(6) = 30

  4. Certain integers that satisfy the Pythagorean Theorem are often used as the lengths of triangle legs.

    In a right triangle, the side opposite the right angle is called the hypotenuse, and usually labeled c for the convenience of formulas. The other two sides, called legs, are labeled a and b.

    If you know the lengths of any two sides of a right triangle, you can find the length of the third side using the Pythagorean Theorem:

    a^2 + b^2  = c^2

    Pythagorean triples are groups of three certain positive integers that fit perfectly into the Pythagorean Theorem. The most common Pythagorean triple is (3, 4, 5):

    3^2 + 4^2 = 5^2

    9 + 16 = 25


    Therefore, given a right triangle with a leg length of 3 and a hypotenuse of 5, you know the remaining side length is 4 without having to run the numbers through the Pythagorean Theorem:

    Knowing combinations of Pythagorean triples can help you save time and potential errors by avoiding the Pythagorean Theorem. The following sets of Pythagorean triples, especially those with smaller integers, may appear on the SSC CGL:

    (3, 4, 5) (5, 12, 13) (8, 15, 17) (7, 24, 25)
    (9, 40, 41) (11, 60, 61) (12, 35, 37) (20, 21, 29)

    Note that the multiples of a triple also work in the Pythagorean Theorem. For example, multiply the (3, 4, 5) triple by 2 to get (6, 8, 10):

    6^2 + 8^2 = 10^2

    36 + 64 = 100

    You should also memorize some of the multiples of the first few Pythagorean triples:

    (6, 8, 10) (9, 12, 15) (12, 16, 20) (15, 20, 25)
    (18, 24, 30) (10, 24, 26) (16, 30, 34) (14, 48, 50)

    Try to solve the following question without using a pencil and paper or a calculator:



    Pythagorean Triples: (D)

    The triangle is a (5, 12, 13), so the missing side is 12. The area of a square is the length of a side squared, so 12^2 = 144.

  5. Any right triangle with a hypotenuse twice as long as one of the sides is a 30º: 60º: 90º triangle.

    Some triangles have special properties. The 30º: 60º: 90º triangle property states that in any triangle with angle measurements of 30º, 60º, and 90º, the length of the sides have a ratio of

    x: xsqrt{3}: 2x. Therefore, you can identify two side lengths given just one side length:


    Because this is a 30º: 60º: 90º triangle, you can determine that the hypotenuse is 8 and the base is 4sqrt{3}. This is a triangle trick in itself, but is pretty common knowledge among test takers. However, did you know that you can determine a right triangle is a 30º: 60º: 90º triangle when one side is half the hypotenuse?


    There are three facts in the diagram above:

    1. The triangle is a right triangle.
    2. One leg is 10 units long.
    3. The hypotenuse is 20 units long.

    This is enough information to complete the remaining angle measurements and side length. The fact that the hypotenuse is twice as long as the leg indicates a 30º: 60º: 90º triangle:


  6. An equilateral triangle has two internal 30º: 60º: 90º triangles.

    An equilateral triangle has three equal angles, all measuring 60º. But if you divide the triangle in half, there are two 30º: 60º: 90º triangles:


    You can use the properties of a 30º: 60º: 90º triangle to find the height of the equilateral triangle:


    Use this property to solve the following question:



    Equilateral Triangles: (C)

    If NO = 10, then PO = 10 and MP = 10. Therefore the base of MNO is 20. To find the height, you must divide the equilateral triangle, PNO, into two 30: 60: 90 triangles:


    Using the 30: 60: 90 triangle ratio, you can see that the height is 5sqrt{3}. Now solve for the area using the base and height:

    A = ½bh

    A = ½(20)( 5sqrt{3}) = 50sqrt{3}

  7. The length of a square’s diagonal is the side length multiplied by square root of 2.

    Like the 30º: 60º: 90º triangles, 45º: 45º: 90º triangles have a special property that states that the lengths of the sides have a ratio of s: s: ssqrt{2}. Therefore, you can identify the side lengths of two sides given just one side length:


    Because this is a 45º: 45º: 90º, you can determine that the hypotenuse is 5sqrt{2} and the base is 5.

    A square has four equal angles, all measuring 90º. But if you divide the triangle by a diagonal, there are two 45º: 45º: 90º triangles:


    You can use the properties of a 45º: 45º: 90º triangle to find the diagonal of the square:


    Use this property to solve a question on your own:



    Squares: (B)

    The diagonal of any square is its side length times sqrt{2} because the diagonal is the hypotenuse of two 45: 45: 90 triangles. Since the side of the square is 3, the diagonal is 3sqrt{2}.

Comment your queries below.

All the Best!

– Team SSCHacks

PS: Team SSCHacks recommends the following books to boost your preparation: