Mathematics

Properties of Triangles

30 Questions

Understanding the properties of triangles is essential for solving geometry problems in competitive exams. These questions specifically address the properties of medians, right-angled triangles, and area ratios. Mastering these rules builds a strong foundation for advanced mathematics.

Triangle mediansSimilar triangles areaRight-angled trianglesApollonius theorem

Properties of Triangles Questions

Multiple choice

On a graph paper, an insect moves from origin O and covers a distance of 4√3 in the direction of 60° east of north and reaches point A. Then, it turns and moves due south and reach point B on x-axis and again it turns and reaches to origin. What is the equation of median through A?

  1. √3y + 2x - 6 = 0

  2. √3y - 2x - 6 = 0

  3. √3y - 2x + 6 = 0

  4. √3y + 2x + 6 = 0

  5. 2y - √3x + 3√3 = 0
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Coordinates of point A (6,2√3).   Coordinates of point at x-axis through which median through A passes are (3,0). (As median divides the line joining O and B into two equal parts) Equation of line y-y1=m(x-x1), where m = (y2-y1)/(x2-x1) y2=2√3, y1=0, x2=6, x1=3 m=(2/√3) Putting m in the equation, we get y=(2/√3) (x-3) √3y-2x+6=0 (Correct Answer)

Multiple choice

Directions: Choose the correct answer from the options given below:

Name the point of intersection of the medians of triangle?

  1. Centroid

  2. Apical diagonal

  3. Lateral diagonal

  4. Ventral point of intersection
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The centroid is the geometric center of a triangle where all three medians intersect. A median is a line segment from a vertex to the midpoint of the opposite side. The centroid divides each median in a 2:1 ratio. The other options are not valid geometric terms.

Multiple choice business maths pair of straight lines condition for perpendicular and coincident lines and bisectors of angles pair of straight lines through origin analytical geometry

The triangle ABC has medians AD, BE, CF. AD lies along the line $y = x + 3$ , BE lies along the line $y = 2x + 4$, AB has length $60$ and angle $C = 90$, then the area of ABC is

  1. $400$
  2. $200$
  3. $100$
  4. $50$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given line 
$y=x+3-----(1)$
$y=2x+4-----(2)$
By shiftiing centroid to origin 
Thus now equation of median are $y=x$ and $y=2x$ 
Now the coordinates of A and B can be taken as $(a,a)$ and $B(b,2b)$
Using centroif formula $C(-a-b,-a-2b)$
Length of hypotenuse $AB=60$
$AB^2=(a-b)^2+(a-2b)^2$
$3600=a^2+b^2-2ab+a^2+4b^2-4ab$
$2a^2+5b^2-6ab=3600---(3)$
Slope of line AC from point $A,C$ is $m _{AC}=\dfrac{-a-2b-a}{-a-b-a}$
$m _{AC}=\dfrac{2a+2b}{2a+b}$
Slope of line BC from point $B,C$ is $m _{BC}=\dfrac{-a-2b-2b}{-a-b-b}$
$m _{BC}=\dfrac{a+3b}{a+2b}$
Angle $C=90$ means line AC and BC are perpendicular 
Hence $m _{AC}m _{BC}=-1$
$\dfrac{2a+2b}{2a+b}\times \dfrac{a+3b}{a+2b}=-1$
$2a^2+6ab+2ab+6b^2=-2a^2-4ab-ab-2b^2$
$4a^2+8b^2+13ab=0--------(4)$
Solving both $(3)$ and $(4)$ equations we get 
$ab=-\dfrac{800}{3}$
$Area of traingle=\dfrac{3}{2}(ab)=400$
Multiple choice maths triangles relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

If $AD$ and $PM$ are medians of triangles $ABC$ and $PQR$, respectivetly where $\triangle ABC \sim \triangle PQR$, then  $\dfrac {AB}{PR}=\dfrac {AC}{PM}$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For similar triangles, the ratio of corresponding medians is equal to the ratio of corresponding sides (AB/PQ = AD/PM). The provided ratio AB/PR = AC/PM is not a standard property of similar triangles.

Multiple choice maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

The ratio of the areas of two similar triangles is equal to the

  1. ratio ofcorresponding medians

  2. ratio ofcorresponding sides

  3. ratio of the squares ofcorresponding sides

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The ratio of the areas of two similar triangles is equal to the square of ratio of their corresponding sides.

Multiple choice maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

If the lengths of the medians $AD, BE$ and $CF$ of the triangle $ABC$, are $6,8,10$ respectively, then

  1. $AD$ and $BE$ are perpendicular
  2. $BE$ and $CF$ are perpendicular
  3. area of $\Delta ABC=32$
  4. area of $\Delta DEF=8$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\begin{array}{l}AD = \sqrt {2A{B^2} + 2A{C^2} - B{C^2}}  = 6\2A{B^2} + 2A{C^2} - B{C^2} = 36\BE = \sqrt {2A{B^2} + 2B{C^2} - A{C^2}}  = 8\2A{B^2} + 2B{C^2} - A{C^2} = 64\CF = \sqrt {2A{C^2} + 2B{C^2} - A{B^2}}  = 10\2A{C^2} + 2B{C^2} - A{B^2} = 100\A{B^2} = x\A{C^2} = y\B{C^2} = z\2x + 2y - z = 36\2x + 2z - y = 64\2y + 2z - x = 100\x = A{B^2} = \frac{{100}}{9}\y = A{C^2} = \frac{{208}}{9}\z = B{C^2} = \frac{{292}}{9}\AD = 6,BE = 8,CF = 10\in,\Delta ABE\AD \bot BE\area,\Delta BEC = 16\area,\Delta ABE = 16 + 16 = 32\\frac{{area,\Delta ABE}}{{area,\Delta DEF}} = 4\\frac{{32}}{{area,\Delta DEF}} = 4\area,\Delta DEF = 8\end{array}$

Multiple choice maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

If $m {a},\ m _{b},\ m _{c}$ are lengths of medians through the vertices $A,B, C$ of $\triangle ABC$ respectively, then length of side $b=$___ 

  1. $\sqrt { { 2m } _{ a }^{ 2 }+{ 2m } _{ c }^{ 2 }-{ 2m } _{ b }^{ 2 } } $
  2. $\dfrac { 1 }{ 3 } \sqrt { { 2m } _{ a }^{ 2 }+{ 2m } _{ c }^{ 2 }-{ 2m } _{ b }^{ 2 } }$
  3. $\dfrac { 2 }{ 3 } \sqrt { { 2m } _{ a }^{ 2 }+{ 2m } _{ c }^{ 2 }-{ 2m } _{ b }^{ 2 } }$
  4. $\dfrac { 3 }{ 4 } \sqrt { { 2m } _{ a }^{ 2 }+{ 2m } _{ c }^{ 2 }-{ 2m } _{ b }^{ 2 } }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The length of a median m_b is given by m_b = 1/2 * sqrt(2a^2 + 2c^2 - b^2). Rearranging this formula for side b gives b = 2/3 * sqrt(2m_a^2 + 2m_c^2 - m_b^2).

Multiple choice maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity

If the medians of two equilateral triangles are in the ratio $3:2,$ then what is ratio of the sides$: ?$

  1. $1:1$
  2. $2:3$
  3. $3:2$
  4. $\sqrt{3}:\sqrt{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Equilateral triangles are similar triangles.
In similar triangles, the ratio of their corresponding sides is the same as the ratio of their medians.
Hence, ratio of sides = $3: 2$

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

In $\Delta$ABC, $\angle B = 90^{o}, AB = 8 \ cm$ and $BC = 6 \ cm.$ The length of the median $BM$ is:

  1. $3 \ cm$
  2. $5 \ cm$
  3. $4 \ cm$
  4. $7 \ cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$AC^2= AB^2+ BC^2$        (because $\angle B = 90^o$)
$= 64+36= 100$
$\therefore AC = 10$
In a right triangle, the median from the right angle to the hypotenuse is half the length of the hypotenuse. 
So, $\displaystyle BM = \frac{1}{2} AC = \frac{10}{2} = 5 \ cm.$
Multiple choice maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

The perimeter of a triangle is $.........$ than the sum of its medians.

  1. Greater

  2. Lesser

  3. Equal

  4. May be greater or lesser
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: $\triangle ABC$, AD, BE and CF are medians from A, B and C respectively on the corresponding sides.

We know that sum of any two sides of the triangle is greater than twice the median bisecting the third side 
Hence, $AB + AC > 2 AD$ (1) 
$AB + BC > 2 BE$ (2)
$BC + AC > 2 CF$ (3)
Adding the three equations:
$2 (AB + BC + AC) > 2 (AD + BE + CF)$
$AB + BC + AC > AD + BE + CF$
Hence, the perimeter of the triangle is greater than the sum of the medians.