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 maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

A point taken on each median of a triangle divides the median in the ratio 1:3 reckoning from the vertex . then the ratio of the area of the triangle with vertices at these points  to that of the original triangle is :  

  1. 5 : 13

  2. 25 : 64

  3. 13 : 32

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

Using the properties of medians and area ratios, a point dividing a median in ratio 1:3 creates a triangle with vertices at these points that has an area ratio of 13/32 relative to the original triangle.

Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

The areas of two similar triangles are $121$ cm$^{2}$ and $64$ cm$^{2}$, respectively. If the median of the first triangle is $12.1$ cm, then the corresponding median of the other is:

  1. $6.4$ cm
  2. $10$ cm
  3. $8.8$ cm
  4. $3.2$ cm
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 ratio of the squares of the corresponding medians. Therefore,
$\displaystyle \frac{121}{64}=\frac{\left ( 12.1 \right )^{2}}{x^{2}},$ where $x$ is the median of the other $\triangle .$
$\Rightarrow $ $\displaystyle x^{2}=\frac{\left ( 12.1 \right )^{2}\times 64}{121}\Rightarrow x=\sqrt{\frac{121}{100}\times 64}$
   $\displaystyle =\frac{11}{10}\times 8=8.8$ cm.

Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

The ratio of areas of two similar triangles is $81 : 49$. If the median of the smaller triangle is $4.9\ cm$, what is the median of the other?

  1. $4.9\ cm$
  2. $6.3\ cm$
  3. $7\ cm$
  4. $9\ cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Area of $\triangle ABC= \cfrac 12 \times base \times height$

In similar triangles, $\cfrac {base 1}{base 2}=\cfrac {height 1}{height 2}=\cfrac {side 1}{side}$
$\therefore \cfrac {Area 1}{Area 2}= (\cfrac {Median 1}{Median 2})^2$
Ratio of Medians $=\sqrt{\cfrac {81}{49}}=\cfrac 97 >1$ 
$\therefore$ Altitude of smaller triangle $=4.9 \times \cfrac 97=6.3$

Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

The areas of two similar triangles are 100 $cm^2$ and 64 $cm^2$. If the median of greater side of first triangle is 13 cm, find the corresponding median of the other triangle.

  1. 20 cm

  2. 15 cm

  3. 10 cm

  4. 16 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given area of two similar triangles are $100$ sq cm and $64$ sq cm
The areas of two Similar-Triangles are in the ratio of the squares of the corresponding medians
The ratio of area of triangle $=\dfrac{100}{64}=\dfrac{25}{16}$
Median of greater triangle is $13$ cm and let other median is $x$ cm
$\therefore \dfrac{(13)^{2}}{(x)^{2}}=\dfrac{25}{16}$

$\Rightarrow \dfrac{169}{x^{2}}=\dfrac{25}{16}$

$\Rightarrow 25x^{2}=169\times 16$

$\Rightarrow x^{2}=\dfrac{2704}{25}=108.16$

$\Rightarrow x=10cm$

Multiple choice maths constructions mid-point formula midpoints division of a line segment

A triangle has vertices A(1,-1) B(2,4) and C(6,0) The length of the median from A is

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

Midpoint of BC = L = (4, 2)
$\displaystyle \therefore AL=\sqrt{\left ( 1-4 \right )^{2}+\left ( -1-2 \right )^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$

Multiple choice maths constructions mid-point formula midpoints division of a line segment

The length of the median from the vertex A of a triangle whose vertices are $A (-1, 3),$ B $(1, -1)$ and C$(5,1)$ is 

  1. $5$
  2. $4$
  3. $1$
  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Length of the median from the vertex $A$ of a triangle $\triangle{ABC}$
Let $AD$ be the median.

$\Rightarrow\,D$ is the midpoint of $BC$

Using midpoint formula,$D=\left(\dfrac{1+5}{2},\,\dfrac{-1+1}{2}\right)=\left(3,\,0\right)$

Length of median $=AD=\sqrt{{\left(-1-3\right)}^2{}+{\left(0-3\right)}^{2}}=\sqrt{16+9}=\sqrt{25}=5$units.
Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

$CM$ and $RN$ are respectively the medians of $\triangle {ABC}$ and $\triangle{PQR}$. If $\triangle {ABC}\sim \triangle{PQR}$, then
  $\cfrac{CM}{RN}=\cfrac{AB}{PQ}$

  1. True

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

In similar triangles, the ratio of corresponding medians is equal to the ratio of corresponding sides. Thus, CM/RN = AB/PQ is a true statement.

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

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

  1. ratio of corresponding medians

  2. ratio of corresponding sides

  3. ratio of the squares of corresponding sides

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

The area of triangle is proportional to the square of the side of the triangle.
ratio of areas of two similar triangles= ratio of the squares of corresponding sides 

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

Is the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians?

  1. True

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

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

Multiple choice maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

The lengths of the medians through acute angles of a right-angled triangle are 3 and 4. Find the area of the triangle:

  1. $\displaystyle \frac{4}{3}\sqrt{11}$
  2. $\displaystyle \frac{2}{3}\sqrt{11}$
  3. $\displaystyle \frac{1}{3}\sqrt{11}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $AD=3,CE=4$
Using Appaloneaus theorem for median $AD$
We have $\displaystyle{ c }^{ 2 }+{ b }^{ 2 }=2\left( \frac { { a }^{ 2 } }{ 4 } +9 \right) $   ...(1)
Using Appaloneaus theorem for median $CE$
We have $\displaystyle{ b }^{ 2 }+{ a }^{ 2 }=2\left( \frac { { c }^{ 2 } }{ 4 } +10 \right) $   ...(2)
Also, ${ a }^{ 2 }+{ c }^{ 2 }={ b }^{ 2 }$
Adding (1) and (2)
$\displaystyle 3{ b }^{ 2 }=2\left( \frac { { b }^{ 2 } }{ 4 } +25 \right) \Rightarrow { b }^{ 2 }=20$
Solving (1) and (2) we get,
$\displaystyle c=\frac { 4 }{ \sqrt { 3 }  }$ and $\displaystyle a=2\frac { 4 }{ \sqrt { 3 }  } $
Hence, area of triangle
$\displaystyle = \frac{1}{2}\left ( \frac{4}{\sqrt{3}} \right )\left ( 2\sqrt{\frac{11}{3}} \right )= \frac{4}{3}\sqrt{11}$.

Multiple choice maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

If $AD,BE$ and $CF$ are the medians of a $\Delta ABC,$ then evaluate  $\displaystyle \left ( AD^{2}+BE^{2}+CF^{2} \right ):\left ( BC^{2}+CA^{2}+AB^{2} \right )=$

  1. $3:4$
  2. $4:3$
  3. $5:3$
  4. $4:1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $AD,BE$ and $CF$ are the medians of a $\Delta ABC$.
$\Rightarrow AB^2+AC^2=2(AD^2+BD^2)$
$\Rightarrow AB^2+AC^2=2AD^2+\displaystyle\frac{BC^2}{2}$
$\Rightarrow 2AD^2=AB^2+AC^2-\displaystyle\frac{BC^2}{2}$ -----(1)
Similarly,
$2BE^2=BC^2+BA^2-\displaystyle\frac{AC^2}{2}$ -----(2)
$2CF^2=CA^2+CB^2-\displaystyle\frac{AB^2}{2}$ -----(3)
Adding equation 1,2 and 3, we get
$2(AD^2+BE^2+CF^2)=\displaystyle\frac{3}{2}(AB^2+BC^2+CA^2)$
$\therefore (AD^2+BE^2+CF^2):(AB^2+BC^2+CA^2)=3:4$

Multiple choice maths pythagoras theorem similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

In $\triangle ABC$, AP is the median. If $AP=7$ and $AB^2+AC^2=260$, then find BC.

  1. $14$ cm
  2. $18$ cm
  3. $15$ cm
  4. $12$ cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By Apollonius Theorem (states that "the sum of the squares of any two sides of any triangle equals twice the square on half the third side, together with twice the square on the median bisecting the third side".) , we have


$BC^2=AB^2+AC^2+2AP^2$

$BC^2=260+2(7)^2$

$BC^2=260+98=358$

$BC=18.92$

Multiple choice maths pythagoras theorem similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

Find the length of median. If the sides of triangle are:
$a = 5, b = 6, c = 8$. and $m = 3, n = 2$.

  1. $\sqrt{\dfrac{206}{5}}$
  2. $\sqrt{206}$
  3. $\dfrac{\sqrt{206}}{5}$

  4. None
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
We have from Appollonius theorem,

$a(mn+p^2)=b^2m+c^2n$

$5(3\times2+p^2)=6^2\times3+8^2\times2$

$5(6+p^2)=36\times3+64\times2$

$30+5p^2=108+128$

$5p^2=236−30 \ \implies 5p^2=206$

$p^2=\dfrac{206}{5}$

$p=\sqrt{\dfrac{206}{5}}$

Option A.
Multiple choice maths pythagoras theorem similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

According to Apolloneous Theorem, if $\overline AD$ is a median of $\triangle ABC$, then $AB^{2}+AC^{2}=$

  1. $AD+BD$
  2. $AD-BD$
  3. $2(AD^{2}+BD^{2})$
  4. $2(AD^{2}-BD^{2})$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} Then,\, \, A{ B^{ 2 } }+A{ C^{ 2 } }=? &  \ AD\, \, is\, \, median &  \ \Rightarrow A{ B^{ 2 } }+A{ C^{ 2 } }=2\left| { \frac { { B{ C^{ 2 } } } }{ 4 }  } \right| +2A{ D^{ 2 } } & \left[ \begin{array}{l} BC=2BD \ or\, \, BC=DC \end{array} \right]  \ \Rightarrow A{ B^{ 2 } }+A{ C^{ 2 } }=2{ \left| { \frac { { B{ C^{  } } } }{ 2 }  } \right| ^{ 2 } }+2A{ D^{ 2 } } &  \ A{ B^{ 2 } }+A{ C^{ 2 } }=2B{ D^{ 2 } }+2A{ D^{ 2 } }=2\left( { A{ D^{ 2 } }+B{ D^{ 2 } } } \right)  &  \end{array}$