Updated On : August 1, 2023
Entrance exams give a reasonably good amount of weightage to quantitative aptitude. In the IPMAT exam, 40 questions are being asked in the MCQ form and 20 questions in short answer type from the quant section. In the DUJAT exam, quantitative ability comprises a total of 25 questions for 100 marks. Therefore, preparing well for topics under quantitative ability becomes important for securing good marks.
Geometry is one of the important sections in the quant section from which questions are asked every year. Questions from Lines, Angles and Triangles are asked in Geometry. This article provides a blueprint about how to go on preparing triangle based questions, properties of triangles, explained answers, etc.
Triangle is a sub-topic under Geometry with a very wide domain. Before attempting questions on Triangle, your basics of the properties of triangle should be crystal clear.
A triangle is a plane figure bounded by three lines and joins three non-collinear points. Triangles come in a variety of shapes and sizes, each with its own set of characteristics. Some features, however, apply to all triangles. One such feature is that the total of any two triangle sides is always bigger than the triangle's third side. It is depicted as follows:
Here, x + y > z, x + z > y, and y + z > x.
Triangles are divided into three sorts based on their sides:
The angles opposite the congruent sides are congruent in an isosceles triangle. In addition, if two triangle angles are identical, the sides opposite them are likewise equal.
See the images given below:
In the above right angled triangle, (AB)2 + (AC) 2 = (BC)2.
Example: if AB = 6 and AC= 8, then BC= 10, because 62 + 82 = 36 + 64 = 100 = (BC) 2
There are some Pythagorean triplets in the questions that are regularly used. It is preferable to learn these triplets by heart.
Any multiple of these Pythagorean triplets will also be a Pythagorean triplet, i.e. if we say 5,12,13 is a triplet, we may multiply all of these numbers by 4 and get 20, 48, 52 which is also a Pythagorean triplet.
Special features of Right Angled Triangle
In ∆ABC, ∠ABC + ∠BAC + ∠ACB = 180°
∆ DBA is similar to ∆ DCB which in turn is similar to ∆ BCA.
The geometric mean of the segments into which the hypotenuse is divided is the altitude from the vertex of the right angle to the hypotenuse.
i.e. (DB)2 = AD * DC
Triangle congruence: Two triangles are said to be congruent if and only if the sides and angles of the first triangle are equal to the corresponding sides and angles of the other triangle.
Triangle similarity: Two triangles are considered to be similar if they are similar in shape. These triangles' corresponding angles are equal, but their corresponding sides are merely proportionate.
Congruent triangles are all similar, however not all similar triangles are congruent.
Note: Where, s = (m1+m2+m3)/2 ; m1,m2,m3 are three median of the Triangle.
Do not try to attempt all the questions as the cut off is never 100%. Smartly pickup the low hanging fruits. Attempt questions in which you are confident first. Rather than attempting all the questions, focus on the accuracy of the questions which you attempt.
Here is the list of books recommended by toppers-
You can learn some of the super tricks to solve Triangle-based questions for DUJAT and IPM on the youtube channel of Supergrads. Click here to brush up your concepts and solve super questions based on Triangles.
Some relevant questions with explanation
1. The area of a triangle is 615 cm2. Find length of the perpendicular that is dropped on that particular side from the opposite vertex if one of its sides is 123 centimetres?
A. 10 cm B. 50 cm C. 0.16 cm D. 0.2 cm
Area = 615cm2. Base = 123 cm Perpendicular = ?
Also A = 1/2 * B * H ∴ 615 = 1/2 * 123 * H ⇒ H = 10 cm.
2. Two sides of an isosceles triangle equal 12.5 cm each and the third side is 20 cm. Find the area of the triangle?
A. 25 cm2 B. 75 cm2 C. 12 cm2 D. 10 cm2
Answer : Lengths of the sides are 12.5 cm, 12.5 cm and 20 cm. ∴ s = (12.5 + 12.5 + 20) / 2 ⇒ s = 22.5. Area = 75 cm2
3. One of the sides of a triangle is 3 cm and the perimeter is 8 cm. If the area of the triangle is maximum then the other two sides are:
A. (3,3) B. (7,5) C.(2/3,2) D.(5/2,5/2)
Some questions of geometry borrow concepts from Algebra.
Let a, b and c be three sides of the triangle. Perimeter= 8 cm, so a+b+c = 8 cm, a+b+3 = 8 cm, a+b= 5 cm. If we know the angle between two sides and the angle between two sides, then we can use the formula A= ½ ab sin θ. Now the angle between two sides is constant, we cannot change that. But when product of a and b is maximum then area will be maximum. Product of a and b is maximum when a=b. Now we know that a+b = 5. This implies that a=b= 5/2 .
Option D. (5/2,5/2) is the answer.
4. There are obtuse angle triangles with sides as 8 cm, 15 cm, and the third side is unknown. If the length of the third side is an integer, how many such triangles exist?
To solve this question, one should know the general properties of triangles and specific properties of obtuse angled triangles. Two properties of obtuse angled triangle will be used here:
Let the third side be x cm. Either 15 cm will be the largest side of x. Let's consider one by one
Let x be the largest side : 8 + 15 > x cm or 23 > x cm or x< 23 cm. Because x is an integer, it can be at most 22. x2 > 8*8 + 15*15 x2 > 64 + 225 x2 > 289 x > 17 . Now we have 17 < x < 23. So, 18, 19, 20, 21 and 22 are the 5 possible values.
Let the largest side be 15 cm. 8 + x > 15 cm or x > 7 cm. c² > a² + b² => 15*15 > 8*8 + x*x => 225 > 64 + x2 or x2 < 161 or x < 12.56 cm . So, 7 cm < x < 12.56 cm; so, 8,9,10,11,12 are possible values of x.
Total number of possible values = 5 + 5 = 10. Option B is the right answer.
5. A circle is put inside an equilateral triangle with side 24 cm, touching its sides. Find the area of the remaining portion of the triangle?
A. 144√3 - 48π cm2
B. 122√3 - 36π cm2
C. 147√3 - 36π cm2
D. 130√3 - 47π cm2
The radius of the circle is one third of the height of the triangle, when it is inscribed in an equilateral triangle.
Height of an equilateral triangle(h) = √3/2 * a, (a is the length of side of triangle)
h = √3/2 * 24 = 12 √3.
Radius of Circle (r) = 1/3 * h = 12 √3/3 = 4 √3.
Area of an equilateral triangle (A) = √3/4 * a2, (a is the length of a side of triangle)
A = √3/4 * (24)2 = √3 * 144.
Area of the Circle (a) = π r2., ( r is the radius of the circle)
a = π (4 √3)2 = 48 π.
Area of rest of the portion of the triangle= A – a = √3 * 144 - 48 π = 144 √3 - 48 π. Option A is the correct answer.
To practice more such questions, visit the youtube channel of Supergrads click here.