#dailyquizadda #geometry #sscAdvancemaths
Answer : (B)
Answer : (B)
Note : Such questions are rare in geometry and in most of the questions you will have to pick up your pen, but still I shared these questions just to unleash the jugaad within you. Moreover, if you are able to solve even a single question with such approach, you will save at least 1 crucial minute in the examination hall.
(2) In geometry too, my beloved concept of 'symmetry' plays an important role. If in any question you find that some symmetrical expressions/equations are given, you can assume the triangle to be equilateral.
In the above question you can assume that the triangle is equilateral. Then angles A, B and C will be 60. Hence sin2A + sin2B + sin2C = (√3/2)2 + (√3/2)2 + (√3/2)2 = 9/4
Answer : (B)
Note : If in any question, the sides of a triangle are given (like a, b and c), then you can assume a = b = c, but make sure any additional detail is not given. Like in the above question that I solved, some additional equations were given, but then too I supposed the triangle to be equilateral only because the equation was symmetrical. Had the equation been unsymmetrical, I could not have been able to assume a =b =c
Now put a = b = c in all the options and check which one will give 3√3a2
A) √3a
Moving on...
I have found that SSC has some real love with 'Area of a triangle' and it keeps on asking it again and again under different contexts. Like in Tier 2 (2014) around 5 questions asked area of triangles.
Therefore it is very important that you memorize all the possible formulas to calculate it. This will save you a lot of time.
Note : All the below questions are taken from a single paper [Tier-2, 2014].
Formula 1 : (Applicable only for Right-Angled Triangle)
In the above question you may struggle to calculate the area. You can
try this question yourself (with a timer), to see how much time you are
taking...
There is a direct formula for such questions -
Apply this formula, area = (100^2 * sin30)/4 = 1250
Answer : (D)
Note : You can choose any of the angles of the triangle (except the 90 degree one) and you will get the same result. Like in the above question, the 3 angles of the triangle are 15, 75 and 90.
sin2*15 = sin30
sin2*75 = sin150
And we know that sin150 = sin30
Hence it doesn't matter which angle you take. But to avoid any confusion, always take the smaller angle (15 in this case).
Next question -
Again apply the same formula
Answer : (C)
Formula 2 : ½*b*c*sinϴ
This formula is only applicable when ϴ lies between the sides 'b' and 'c'.
Apply the formula, area = 1/2 * 10 * 10 * sin45
Answer : (D)
Formula 3 : Area of a triangle = r * S
where r = inradius
S = semi-perimeter
Given, perimeter = 50, hence semi-perimeter = 25
Area = 6 * 25 = 150
Answer : (C)
Now when you know this formula, in Q. 2 above, where we had to establish relation between 's' and 'a', you can establish it by applying this formula too.
Area of an equilateral triangle = (√3/4)s2
Area of a triangle = inradius * semi-perimeter = a * (3s)/2
Orthocentre, Incentre, Circumcentre and Centroid :
∠AHB + ∠ACB = 180 [Whenever you see the word 'orthocentre' in the paper, you should immediately recall this formula]
Over the years SSC has asked many questions based on this simple formula. Like -
Answer : (A)
Incentre
While solving questions on circumcentre, you should imagine the above figure. Here two things are quite useful -
∠AOB = 2∠ACB
OA = OB = OC [radius of the circle]. And hence,
∠OAB = ∠OBA
∠OCA = ∠OAC
∠OBC = ∠OCB
Each angle of an equilateral triangle is 60 and we have seen earlier that the angle at the centre is twice that of the triangle.
Hence ∠AOC = 120
Answer : (C)
Q. 3
Answer : (D)
Centroid
Remember one more property related to centroid,
Area(ΔOBC) = 1/3Area(ΔABC)
We know DG = 1/2AG
Hence DG = 2cm
Answer : (A)
Some additional things to remember :
1) The orthocentre, incentre, circumcentre and centroid of an equilateral traingle coincide, i.e., a single point acts as all the centres.
2) For an equilateral triangle-
Here is a question based on this fact-
Radius of the circumcircle = 15, hence diameter = 30cm
AC = 30cm
We have to find AB and BC.
AB = BC [as ∠ACB = ∠ABC = 45]
AB = sin45 * AC = 30/√2 = 15√2
Answer : (C)
Interior and Exterior Angles
Another important topic of geometry is "Internal and External Angles" of polygon.
Now remember few formulae for such questions-
1) Sum of a interior angle and its corresponding exterior angle adds up to 180 degrees.
2) Sum of all exterior angles of a regular polygon is always 360.
3) Sum of interior angles = (n-2)*180
4) Each interior angle of a regular polygon is equal to [(n-2)*180]/n
5) Each exterior angle of a regular polygon is equal to 360/n
Mug all these formulae well and you will be able to solve all questions on interior and exterior angles easily.
Now let's solve some CGL questions
(n - 2)*180 = 2*360
So n - 2 = 4
or n = 6
Answer : (B)
Note : Here the difference between the "angles" of two polygons is given. You might be wondering that they haven't mentioned the type of angle, whether interior or exterior. Well, it doesn't matter! Because as far as two polygons are concerned-
Difference between their interior angles = Difference between their exterior angles
But I will take the angles to be exterior, because that will make our calculations simple
Let the sides be 5x and 4x.
Measure of each exterior angle of Polygon 1 = 360/5x = 72/x
Measure of each exterior angle of Polygon 1 = 360/4x = 90/x
Given, 90/x - 72/x = 6
so x = 3
The two angles are 5x and 4x
5x = 5*3 = 15
4x = 4*3 = 12
Had I taken the angles to be interior -
We have seen earlier that the measure of each angle of a polygon is [(n - 2)*180]/n.
Given [(5x - 2)*180]/5x - [(4x - 2)*180]/4x = 6
Solve it and you will get x = 3
So the angles are 15 and 12
Answer : (D)
Note : Please note that each and every formula that I have mentioned in this post is extremely important and I can guarantee that after mugging all these formulae you will be able to solve all the questions on orthocentre, incentre, circumcentre, centroid and interior-exterior angles...
Disclaimer: Use this method only when you are unable to figure out how to solve a question.
Q. 1) If BE and CF are two medians of a triangle ABC and G is the centroid and EF and AG intersects each other at O then find - AO:OG
(A) 1:1 (B) 1:2 (C) 2:1 (D) 3:1
Since no information about the type of triangle is given, hence we will assume it to be 'Equilateral'.
Step 1: Draw an Equilateral triangle of any size(but it should be big enough for the purpose). Take 6 cm.
How to draw an equilateral triangle? Just draw a line AB = 6 cm, and make ∠A = 60 and ∠B = 60, the point where the two angles will intersect, will be point C.
Step 2: Now draw the medians BE and CF. Make sure that E and F lie exactly in the middle of AC and AB respectively.
Step 3: Complete the remaining figure.
Step 4: Measure AO and OG with a ruler.
You will find that AO:OG = 3:1
Answer: (D)
Again, a simple question.
Step 1: Draw a circle of any radius with centre O and diameter AB.
Step 2: Put protractor at A and draw ∠CAB = 34.
Step 3: Join BC
Step 4: Measure ∠CBA with protractor.
Answer: (C)
This may appear tricky if you go on solving it with conventional methods. But it is way too easy when solved through construction.
Step 1: Draw a triangle ABC with AC = 6 cm. Here we are using the concept of scaling. In the exam you can't draw a line 12 cm long. Hence we are reducing AC to 1/2. Hence, when we will measure AE, we will have to multiply it by 2 to get the exact answer.
Step 2: Draw internal bisector of ∠ABC. Now, be very cautious, a slight mis-measurement give you a wrong answer. Use protractor for accurate measurements (For simplicity, you can take ∠A = 60 and hence use protractor to make ∠CBD = 30)
Step 3: Again with the help protractor, make ∠BDA = 90
Step 4: Draw DE parallel to BC
Step 5: Measure AE with a scale. You will get AE = 3 cm.
Hence actual length of AE = 2*3 = 6 cm
Answer: (B)
Method:
AD extended meets BC at F.
∠ADB = ∠BDF = 90
∠ADB = ∠FDB (BD is the angle bisector)
∠BAD = ∠BFD
Triangles ABD and FBD are congruent. So AD = DF
Triangles ADE and AFC are similar
AE/AC = AD/AF = 1/2
AE = 1/2 * 12 = 6 cm
Q. 4)
Although this is a very simple question and you would get 60 degrees as your answer when you solve it. But, if by any chance, if you are not able to figure out the method, solve it using construction.
Method: The median of an isosceles triangle cuts the opposite side at right angles. Hence ∠AOB = 90.
Similarly you can solve many geometry questions with proper measurements. Try solving some questions on your own...
#dailyquizadda #SSC #Advance Maths
[Important Formulae]
(1) Be it algebra or geometry, such questions are always there that don't deserve your rough space.
Q 1.
Like the above questions asks you the value of x. Now value of x can't
be negative in this context. Because CO = x - 3 and any negative value
of 'x' will make CO negative, like if x = -8, then CO = -11. We know
that side can never be negative. Hence all the options that have
negative value for x are wrong.Answer : (B)
Q 2.
In the above question a certain area is asked. Now see the options and
observe that in options A, C and D if we take r=7, 2 and 1 respectively,
then the area will become zero, and area can never be zero in this
case, although value of 'r' can be 7, 2 or 1.Answer : (B)
Note : Such questions are rare in geometry and in most of the questions you will have to pick up your pen, but still I shared these questions just to unleash the jugaad within you. Moreover, if you are able to solve even a single question with such approach, you will save at least 1 crucial minute in the examination hall.
(2) In geometry too, my beloved concept of 'symmetry' plays an important role. If in any question you find that some symmetrical expressions/equations are given, you can assume the triangle to be equilateral.
Q. 3 
In the above question you can assume that the triangle is equilateral. Then angles A, B and C will be 60. Hence sin2A + sin2B + sin2C = (√3/2)2 + (√3/2)2 + (√3/2)2 = 9/4
Note : If in any question, the sides of a triangle are given (like a, b and c), then you can assume a = b = c, but make sure any additional detail is not given. Like in the above question that I solved, some additional equations were given, but then too I supposed the triangle to be equilateral only because the equation was symmetrical. Had the equation been unsymmetrical, I could not have been able to assume a =b =c
Q. 4
Here the lengths of perpendiculars are given to be a, b and c. Note that
no additional information is given, hence it is safe to assume a = b =
c. Let the side of the triangle be 's'. The figure will look like this
-
We have to establish a relation between 'a' and 's'. In an equilateral
triangle, the incentre, orthocentre, circumcentre and centroid, all
coincide. So you can calculate 'a', by which ever method you like.
a = inradius = s/2√3 [The inradius of an equilateral triangle is s/2√3 and the circumradius is s/√3]
Hence s = 2√3a
We know the formula for calculating the area of an equilateral triangle = (√3/4)s2
= (√3/4) (2√3a)2
= 3√3a2
Now put a = b = c in all the options and check which one will give 3√3a2
A) √3a
B) 3√2a2
C) √2a
D) 3√3a2
Answer : (D)
I have found that SSC has some real love with 'Area of a triangle' and it keeps on asking it again and again under different contexts. Like in Tier 2 (2014) around 5 questions asked area of triangles.
Therefore it is very important that you memorize all the possible formulas to calculate it. This will save you a lot of time.
Note : All the below questions are taken from a single paper [Tier-2, 2014].
Formula 1 : (Applicable only for Right-Angled Triangle)
Q. 5
There is a direct formula for such questions -
Answer : (D)
Note : You can choose any of the angles of the triangle (except the 90 degree one) and you will get the same result. Like in the above question, the 3 angles of the triangle are 15, 75 and 90.
sin2*15 = sin30
sin2*75 = sin150
And we know that sin150 = sin30
Hence it doesn't matter which angle you take. But to avoid any confusion, always take the smaller angle (15 in this case).
Next question -
Q. 6
Answer : (C)
Formula 2 : ½*b*c*sinϴ
This formula is only applicable when ϴ lies between the sides 'b' and 'c'.
Q. 7
Apply the formula, area = 1/2 * 10 * 10 * sin45
Answer : (D)
Formula 3 : Area of a triangle = r * S
where r = inradius
S = semi-perimeter
Q. 8
Given, perimeter = 50, hence semi-perimeter = 25
Area = 6 * 25 = 150
Answer : (C)
Now when you know this formula, in Q. 2 above, where we had to establish relation between 's' and 'a', you can establish it by applying this formula too.
Area of an equilateral triangle = (√3/4)s2
Area of a triangle = inradius * semi-perimeter = a * (3s)/2
Now,
a * (3s)/2 = (√3/4)s2
or s = 2√3a [Same result]
[Centres of a Triangle]
Orthocentre, Incentre, Circumcentre and Centroid :
This Topic is very important from SSC point of view. You all must be knowing the theory of these terms, but I will reiterate -
Orthocentre - Point of intersection of altitudes of a triangle
Incentre - Point of intersection of angle bisectors
Circumcentre - Point of intersection of perpendicular bisectors
Centroid - Point of intersection of medians
The above information, although important, is not sufficient enough to
solve the questions. Each of these terms require different approach to
solve the questions based on them.
Orthocentre
∠AHB + ∠ACB = 180 [Whenever you see the word 'orthocentre' in the paper, you should immediately recall this formula]
Over the years SSC has asked many questions based on this simple formula. Like -
Q. 1
∠AIC = 90 + ∠ABC/2 [Whenever you see the word 'incentre' in the paper, you should immediately recall this formula]
Circumcentre
While solving questions on circumcentre, you should imagine the above figure. Here two things are quite useful -
OA = OB = OC [radius of the circle]. And hence,
∠OAB = ∠OBA
∠OCA = ∠OAC
∠OBC = ∠OCB
Q. 2
Hence ∠AOC = 120
Answer : (C)
Q. 3
Answer : (D)
Centroid divides the medians in the ratio 2:1. Hence for the median AE,
OA = 2/3AE
OE = 1/3AE
OA = 2OE
Similarly for the medians CD and BF.
Remember one more property related to centroid,
Area(ΔOBC) = 1/3Area(ΔABC)
Q. 4
We know DG = 1/2AG
Hence DG = 2cm
Answer : (A)
Some additional things to remember :
1) The orthocentre, incentre, circumcentre and centroid of an equilateral traingle coincide, i.e., a single point acts as all the centres.
2) For an equilateral triangle-
- Inradius = h/3 = a/2√3
- Circumradius or outer-radius= 2h/3 = a/√3
- Height(h) = (√3/2)a
3) For a right-angled triangle
- Orthocentre is at the right angle vertex
- Circumcentre is the midpoint of the hypotenuse
Here is a question based on this fact-
Q. 5
Two angles are 45 degrees and hence the third angle is 90 degrees. The figure will look like this-
AC = 30cm
We have to find AB and BC.
AB = BC [as ∠ACB = ∠ABC = 45]
AB = sin45 * AC = 30/√2 = 15√2
Answer : (C)
Interior and Exterior Angles
Another important topic of geometry is "Internal and External Angles" of polygon.
1) Sum of a interior angle and its corresponding exterior angle adds up to 180 degrees.
2) Sum of all exterior angles of a regular polygon is always 360.
3) Sum of interior angles = (n-2)*180
4) Each interior angle of a regular polygon is equal to [(n-2)*180]/n
5) Each exterior angle of a regular polygon is equal to 360/n
Mug all these formulae well and you will be able to solve all questions on interior and exterior angles easily.
Now let's solve some CGL questions
Q. 6
Given(n - 2)*180 = 2*360
So n - 2 = 4
or n = 6
Answer : (B)
Q. 7
Note : Here the difference between the "angles" of two polygons is given. You might be wondering that they haven't mentioned the type of angle, whether interior or exterior. Well, it doesn't matter! Because as far as two polygons are concerned-
Difference between their interior angles = Difference between their exterior angles
But I will take the angles to be exterior, because that will make our calculations simple
Let the sides be 5x and 4x.
Measure of each exterior angle of Polygon 1 = 360/5x = 72/x
Measure of each exterior angle of Polygon 1 = 360/4x = 90/x
Given, 90/x - 72/x = 6
so x = 3
The two angles are 5x and 4x
5x = 5*3 = 15
4x = 4*3 = 12
Had I taken the angles to be interior -
We have seen earlier that the measure of each angle of a polygon is [(n - 2)*180]/n.
Given [(5x - 2)*180]/5x - [(4x - 2)*180]/4x = 6
Solve it and you will get x = 3
So the angles are 15 and 12
Answer : (D)
Note : Please note that each and every formula that I have mentioned in this post is extremely important and I can guarantee that after mugging all these formulae you will be able to solve all the questions on orthocentre, incentre, circumcentre, centroid and interior-exterior angles...
[Construction]
You can tackle many tricky geometry questions with a single word 'Construction'. Yes, just precisely draw the diagram as asked in the question and manually measure the unknown side/angle that has been asked...Disclaimer: Use this method only when you are unable to figure out how to solve a question.
Q. 1) If BE and CF are two medians of a triangle ABC and G is the centroid and EF and AG intersects each other at O then find - AO:OG
(A) 1:1 (B) 1:2 (C) 2:1 (D) 3:1
Since no information about the type of triangle is given, hence we will assume it to be 'Equilateral'.
Step 1: Draw an Equilateral triangle of any size(but it should be big enough for the purpose). Take 6 cm.
How to draw an equilateral triangle? Just draw a line AB = 6 cm, and make ∠A = 60 and ∠B = 60, the point where the two angles will intersect, will be point C.
Step 2: Now draw the medians BE and CF. Make sure that E and F lie exactly in the middle of AC and AB respectively.
Step 3: Complete the remaining figure.
Step 4: Measure AO and OG with a ruler.
You will find that AO:OG = 3:1
Answer: (D)
Q. 2)
Step 1: Draw a circle of any radius with centre O and diameter AB.
Step 2: Put protractor at A and draw ∠CAB = 34.
Step 3: Join BC
Step 4: Measure ∠CBA with protractor.
Answer: (C)
Q. 3)
This may appear tricky if you go on solving it with conventional methods. But it is way too easy when solved through construction.
Step 1: Draw a triangle ABC with AC = 6 cm. Here we are using the concept of scaling. In the exam you can't draw a line 12 cm long. Hence we are reducing AC to 1/2. Hence, when we will measure AE, we will have to multiply it by 2 to get the exact answer.
Step 2: Draw internal bisector of ∠ABC. Now, be very cautious, a slight mis-measurement give you a wrong answer. Use protractor for accurate measurements (For simplicity, you can take ∠A = 60 and hence use protractor to make ∠CBD = 30)
Step 3: Again with the help protractor, make ∠BDA = 90
Step 4: Draw DE parallel to BC
Step 5: Measure AE with a scale. You will get AE = 3 cm.
Hence actual length of AE = 2*3 = 6 cm
Answer: (B)
Method:
∠ADB = ∠BDF = 90
∠ADB = ∠FDB (BD is the angle bisector)
∠BAD = ∠BFD
Triangles ABD and FBD are congruent. So AD = DF
Triangles ADE and AFC are similar
AE/AC = AD/AF = 1/2
AE = 1/2 * 12 = 6 cm
Q. 4)
Although this is a very simple question and you would get 60 degrees as your answer when you solve it. But, if by any chance, if you are not able to figure out the method, solve it using construction.
Method: The median of an isosceles triangle cuts the opposite side at right angles. Hence ∠AOB = 90.
Similarly you can solve many geometry questions with proper measurements. Try solving some questions on your own...
#dailyquizadda #SSC #Advance Maths
No comments:
Post a Comment