Mathematics · Quantitative Aptitude

Triangle Properties

234 Questions

Triangle properties encompass the rules governing the sides, angles, and area of different types of triangles. Key areas include the Pythagorean theorem, similar triangles, and centroid calculations. These concepts form a foundational part of geometry in various competitive examinations.

Similar trianglesArea and perimeterPythagorean theoremTriangle inequalityEquilateral properties

Triangle Properties Questions

Multiple choice maths bearing and drawings mapping mapping space around us changing scale

State true or false:
On a map drawn to a scale of 1: 25,0000; a triangular plot of land has the following measurements AB=3 cm, BC=4 cm, and angle ABC=$\displaystyle 90^{\circ}.$
The area of the plot in sq. km is 37.5 sq. km.

  1. True

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

Scaling factor = $\dfrac{Length \quad of \quad model}{Length \quad of \quad building}$
$\dfrac{1}{250000} = \dfrac{AB}{Length _{AB}}$
$Length _{AB} = 250000 \times 3$
$Length _{AB} = 750000$
$Length _{AB} = 0.75 $ km

$\dfrac{1}{250000} = \dfrac{BC}{Length _{BC}}$
$Length _{BC} = 250000 \times 4$
$Length _{BC} = 1000000$
$Length _{BC} = 1 $ km

Area of plot = $\dfrac{1}{2} \times Length _{AB} \times Length _{BC}$ (Area of triangle =$ \dfrac{1}{2} base \times height$)
Area of plot = $\dfrac{1}{2} \times (0.75) \times 1$
Area of plot = $0.375$ $km^2$

Multiple choice maths bearing and drawings mapping mapping space around us changing scale

The perimeter of two similar triangles $ABC$ and $PQR$ are $36\ cm$ and $24\ cm$ respectively. If $PQ = 10\ cm$ then the length of $AB$ is

  1. $18\ cm$
  2. $12\ cm$
  3. $15\ cm$
  4. $30\ cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\triangle ABC = \triangle PQR$, then
$\dfrac {\text {Perimeter of}\triangle ABC}{\text {Perimeter of}\triangle PQR} = \dfrac {AB}{PQ} = \dfrac {BC}{QR} = \dfrac {AC}{PR}$
$\dfrac {36}{24} = \dfrac {AB}{10}; AB = \dfrac {36}{24}\times 10 = 15\ cm$.

Multiple choice maths mixture types of ratios ratios in proportion mathematical logic

The sides of a triangle are $10$ cm,$10$ cm and $12$ cm. If each of the two equal sides is increased in the ratio $5\colon4$ but the third side remains unchanged, in what ratio has its perimeter been increased?

  1. $\;5\colon32$
  2. $\;5\colon37$
  3. $\;37\colon32$
  4. $\;32\colon37$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, sides of the triangle $=10cm, 10cm, 12cm$


$\therefore $ perimeter of the given triangle $=10+10+12=32  cm$

As the two equal sides each $10 cm$ increased in the ratio $5:4$

So, the corresponding sides of the new triangle$=10\times \displaystyle \frac{5}{4}=12.5   cm$ 

Hence the 3 sides of the new triangles are $12.5 cm,12.5 cm,12 cm$.

Then the perimeter of the new triangle $=12.5+12.5+12=37   cm$

$\therefore$The ratio in which the perimeter of the new triangle is increased

$=\displaystyle \frac{perimeter \ of \   the \ new \  triangle}{perimeter \  of given \  triangle} =\frac{37}{32}$

Hence the ratio $=37:32$
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

A guy wire attached to a vertical pole of height $18m$ is $24m$ long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

  1. $15.87m$
  2. $16.8m$
  3. $15$
  4. $15.67$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Consider the wire and pole system as a right angled triangle such that length of pole is the perpendicular, length of wire the hypotenuse and distance between the base of pole and the stack is the base, such that, H = 24 m, P = 18 m
Now, from Pythagoras theorem,
$H^2 = P^2 +B^2$
$24^2 = 18^2 + B^2$
$576 = 324 + B^2$
$B^2 = 252$
$B = 15.87$ m
Thus, distance between the base of the pole and the stack is 15.87 m

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

If the Pythagorean triples of one member is $22$, find the other two members.

  1. $124$ and $122$
  2. $123$ and $122$
  3. $121$ and $122$
  4. $120$ and $122$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As we know $2m$, $m^{2}+1, m^{2}-1$ form a Pythagorean triplet for any number $m > 1$.


Let us assume $2m = 22$

$\Rightarrow m = 11$

Therefore, $m^{2}+1$ $=$ $11^{2}+1$

$\Rightarrow 121 + 1 = 122$

and $m^{2}-1$ $=$ $11^{2}-1$

$\Rightarrow 121 - 1 = 120$

Hence, the other two members are $120$ and $122$.

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

The sides of a triangle are $25 m$, $39 m$ and $56 m$ respectively. Find the length of perpendicular from the opposite angle on the greatest sides.

  1. $56 m$
  2. $60 m$
  3. $15 m$
  4. $12 m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $a=25 \ m, \ b=39 \ m, \ c=56 \ m$


We have, semi-perimeter, $s=\dfrac{a+b+c}{2}$

$s=\dfrac{25+39+56}{2}$

$s=\dfrac{120}{2}=60 \ m$

$Area \ of \ triangle = \sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{60(60-25)(60-39)(60-56)}$

$=\sqrt{60(35)(21)(4)}=\sqrt{176400}$

$=420 \ m^2$


$Area \ of \ triangle = \dfrac{1}{2}bh$,    where b-base and h-height

$base=greatest side=56 \ m$

$420=\dfrac{1}{2}(56)h$

$h=\dfrac{840}{56}$

$h=15 \ m$

Multiple choice maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

An altitude of a triangle is five-third the length of its corresponding base. If the altitude is increased by $4 cm$ and the base is decreased by $2 cm$, the area of the triangle remains same. Find the base and the altitude of the triangle.

  1. The base of the triangle is $12 cm$ and altitude is $20 cm$.
  2. The base of the triangle is $4 cm$ and altitude is $34 cm$.
  3. The base of the triangle is $16 cm$ and altitude is $12 cm$.
  4. The base of the triangle is $8 cm$ and altitude is $32 cm$.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let the base of the triangle be $x$ cm. 
Then, the altitude of the triangle $=\cfrac { 5x }{ 3 } $
So, area of the triangle $=\cfrac { 1 }{ 2 } \times base\times altitude$
$=\cfrac { 1 }{ 2 } \times x\times \cfrac { 5 }{ 3 } x\\ =\cfrac { 5 }{ 6 } { x }^{ 2 }$        ...(1)
On increasing the altitude by $4 cm$ and the decreasing base by $2 cm$, the area remains the same.
Therefore, $\cfrac { 1 }{ 2 } \times (x-2)\times \left( \cfrac { 5x }{ 3 } +4 \right) =\cfrac { 5 }{ 6 } { x }^{ 2 }$       ...[using (1)]
$\Longrightarrow \cfrac { 1 }{ 2 } \times (x-2)(\cfrac { 5x+12 }{ 3 } )=\cfrac { 5 }{ 6 } { x }^{ 2 }$
$ \Longrightarrow (x-2)(5x+12)=5{ x }^{ 2 }$
$ \Longrightarrow 5{ x }^{ 2 }-10x+12x-24=5{ x }^{ 2 }$
$ \Longrightarrow 2x-24=0$
$ \Longrightarrow 2x=24$ or $x = 12$.
 Now, altitude of the triangle $=\cfrac { 5x }{ 3 } =\cfrac { 5\times 12 }{ 3 } =20 cm$
Hence, the base of the triangle is $12 cm$ and altitude is $20 cm$.
Multiple choice maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

If A is the area of a triangle in em", whose sides are 9 em, 10 cm and 11 em, then which one of the following is correct?

  1. $A < 40\:cm^2$
  2. $40\:cm^2 < A < 45\:cm^2$
  3. $45\:cm^2 < A < 50\:cm^2$
  4. $A>50\:cm^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle s=\frac{1}{2}(9+10+11):cm=15:cm$

$\therefore\Delta=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{15\times6\times5\times4}:cm^2$

$=30\sqrt{2}=30\times1.4=42:cm^2$
which lies between $40:cm^2$ and $45:cm^2$.

Multiple choice maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

Two sides of a triangle have lengths $7$ and $9$. Which of the following could not be the length of the third side?

  1. $4$
  2. $5$
  3. $7$
  4. $11$
  5. $16$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

An important rule to remember about triangles is called the third side rule: the length of the third side of a triangle is less than the sum of the lengths of the other two sides and greater than the (positive) difference of the lengths of the other two sides. 

For this triangle, the length of the third side must be greater than $9-7=2$97=2 and less than $9+7=16$9+7=16. All the answers are possible except for answer E, which is equal to $16$16 but not less than $16$16.

Multiple choice maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

Which of the following sets of measurements can be used to construct a triangle?

  1. $4\ cm, 5\ cm, 6\ cm$
  2. $4\ cm, 3\ cm, 8\ cm$
  3. $5\ cm, 6\ cm, 12\ cm$
  4. $6\ cm, 3\ cm, 10\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Because the sum of the length of any $2$ sides of the triangle should be greater than the third side, which is only satisfied by option 1.

$(4+5)>6$
$(4+6)>5$
$(5+6)>4$

Multiple choice maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

The longest side of a triangle is three times the shortest side and the third side is $2$ cm shorter than the longest side. If the perimeter of the triangle is at least $61$ cm, find the minimum length of the shortest-side.

  1. $9$ cm
  2. $11$ cm
  3. $16$ cm
  4. $61$ cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the shortest side $= s$
Longest side $= 3s$
Third side $= 3s -2$
Now perimeter $= s + 3s +3s -2 = 7s -2 \ge 61$
$\therefore 7s \ge 63$
$\therefore s \ge 9$
Thus minimum length of the shortest side$= 9$ cm

Multiple choice maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

Triangle ABC has integral sides AB, BC measuring $2001$ unit and $1002$ units respectively. Then the number of such triangles, is?

  1. $3002$
  2. $2003$
  3. $1003$

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

Let the sides be a=2001, b=1002, and c be the third side. By the triangle inequality, |a-b| < c < a+b. So 999 < c < 3003. The number of possible integer values for c is 3003 - 999 - 1 = 2003.

Multiple choice maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

Which of the following will form the sides of a triangle?

  1. $23\,cm,\,17\,cm,\,8\,cm$
  2. $12\,cm,\,10\,cm,\,25\,cm$
  3. $6\,cm,\,7\,cm,\,16\,cm$
  4. $8\,cm,\,7\,cm,\,16\,cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Triangle Inequality Theorem, states that the sum of two side lengths of a triangle is always greater than the third side. 
$(a)17+8>23$ is  true
Hence the sides with the measure $23\ cm,17\ cm,8\ cm$ will  form the sides of a triangle.
$(b)10+12>25$ is not true
Hence the sides with the measure $10\ cm,12\ cm,25\ cm$ will not form the sides of a triangle.
$(c)6+7>16$ is not true
Hence the sides with the measure $6\ cm,7\ cm,16\ cm$ will not form the sides of a triangle.
$(d)8+7>16$ is not true
Hence the sides with the measure $8\ cm,7\ cm,16\ cm$ will not form the sides of a triangle.

Multiple choice maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

The length  of altitude through $A$ of the triangle $ABC$, where $A = (-3,\ 0);\ B = (4,\ -1);\ C = (5,\ 2).$

  1. $\dfrac{2}{\sqrt{10}}$
  2. $\dfrac{4}{\sqrt{10}}$
  3. $\dfrac{11}{\sqrt{10}}$
  4. $\dfrac{22}{\sqrt{10}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Equation of line BC 
$\frac { y-2 }{ x-5 } =\frac { 2-(-1) }{ 5-4 } \\ \Rightarrow { (y-2) }\times { (5-4) }={ (x-5) }\times { (2+1) }\\ \Rightarrow { 3x }-{ y }-{ 13 }={ 0 }$
Length of altitude from point A to the line BC
$\\ =\frac { \left| 3\times (-3)-1\times 0-13 \right|  }{ \sqrt { { 3 }^{ 2 }+{ (-1) }^{ 2 } }  } \\ =\frac { \left| -9-13 \right|  }{ \sqrt { 10 }  } \\ =\frac { 22 }{ \sqrt { 10 }  } $
Corrct answer is D