Mensuration section in quantitative aptitude is one of the complex sections in any competitive exam. The reason for this is the variety of questions it asks. There is no structure for this section. Also, you need to remember many formulas to solve the questions from this section. Out of many types, the triangle is a very important topic. Because you only need to remember few properties to solve the questions of the triangle. It is also important that you remember the triangle formulae.
A polygon that has three sides is called a triangle. The sum of all the sides of a triangle is always 180°. Some of the common properties of the triangle are as follow:
- Sum of the length of any two sides in a triangle has to be greater than the third side.
- Difference between the lengths of any two sides of a triangle has to be always lesser than the third side.
- The side that is opposite to smallest angle is the smallest and similarly, the side opposite to greatest angle is the greatest.
- Two triangles are congruent if all the sides of one are equal to the corresponding sides of another. It also states that all the angles of one are equal to the corresponding angles of another.
Area of a triangle = ½ x base x height, here height is the perpendicular distance between the base and the vertex opposite to it.
Browse more Topics under Mensuration
- Volumes and Areas
- Results on Quadrilaterals
- Cylinder, Cone and Sphere
- Data Sufficiency
- Mensuration Practice Questions
Types of a Triangle
Based on the angles and sides of the triangle, there are six types of triangle.
- Equilateral triangle: In this type of triangle, all the sides are equal and each angle measures 60°.
- Isosceles triangle: In this triangle, any two sides are equal to each other. The angle opposite to this two sides is also equal.
- Scalene triangle: Triangle that has no side equal is called scalene triangle.
- Acute triangle: In this triangle, all the angles are less than 90°.
- Obtuse angle triangle: In this triangle, one of the three angles is more than 90°.
- Right angled triangle: In this triangle, one of the angles is of 90°.
Formulae and Properties
Area of an equilateral triangle: ½ × base × height = ½ × a × a√3/2 = √3/4 a²
h = a√3/2
In radius, r = h/3 = a/2√3
Circumradius, R = a/√3
Area of an isosceles triangle = b/4√(4a² – b²)
Right Angled Triangle
In right angled triangle it is important to know the Pythagoras theorem. For ∆ ABC given in the figure, a² = b² + c².
Area of a right triangle = 1/2 × product of two perpendicular sides.
Circumradius, R = hypotenuse/2
Some of the basic triplets that you need to remember for Pythagoras theorem and that might come handy for you in the solving the questions are, => 3,4,5 / 5,12,13 /8,15,17 / 7,24,25 / 9,40,41 / 11,60,61 / 12,35,37. These are some of the basic triplets there are many more triplets as well. Any multiple of the above triplets will also form a triplet. For example, as 3,4,5 is triplet 6,8,10 will also be a triplet.
Q. If the height of an equilateral triangle is 10cm, its area will be equal to:
A. 200√3/3 cm² B. 100/3 cm²
C. 100√3 cm² D. 100√3/3 cm²
In this type of question, you need to find the area of the equilateral triangle base on the height given in the question. As mentioned above the formula for the area of an equilateral triangle = 1/2 x base x height. We have height given in the question but we need to find the base. For that, h = a√3/2 => a = 10 × 2/√3 =20/√3cm.
Area, = 1/2 × 20/√3 10 = 100√3/3cm²
So, the correct answer is option D.
Q. If three sides a triangle are 7,24,25 then the circumradius of the triangle will be?
A. 3.5 B. 12 C. 12.5 D. 13
In this question, you can see that the sides of the triangle are actually a triplet. They follow the Pythagoras theorem. Thus, the triangle given in the question is the right-angled triangle. Circumradius for right-angled triangle = h/2.
Here, h = 25. So R = 25/2 = 12.5
So the correct answer is C.
Q. If AD, BE, CF are medians of ∆ABC and O is the centroid of ∆ABC. If the area of ∆AOF is 36 cm^2 then the area of ∆OFB + area of ∆OEC =?
A. 54 cm² B. 72 cm²
C. 36 cm² D. None of the above
Here, ‘O’ is the centroid of ∆ABC. Then are of ∆AOF = area of ∆OFB = area of ∆OEC. Area of ∆OFB + area of ∆OEC = 36 + 36 = 72 cm²
So the correct answer is option B.
Q. If we draw a ∆PQR inside a circle, (P, Q, R are on the circumference of a circle). Then the area of the ∆PQR is maximum when:
A. ∟QPR = 90°
B. ∆PQR is obtuse angle triangle
C. PQ = QR = PR
D. PQ = QR ≠ PR
The correct answer is C.
Q. Find the area of the triangle whose sides are 11, 60, 61.
A. 275 B. 330 C. 315 D. 210
The correct answer is B.
Q. The sides of the triangle are in the ratio of 4:5:6. The triangle is:
D. Either obtuse-angled or right-angled.
The correct answer is A.