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 congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

If two sides of a triangle are 8 cm and 13 cm, then the length of the third side is between a cm and b cm. Find the values of a and b such  that a is less than b.

  1. a = 2 cm and b =9 cm

  2. a = 7 cm and b = 12 cm

  3. a = 5 cm and b = 21 cm

  4. a = 8 cm and b = 24 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$If  \ two \  sides  \ of \  a \  triangle \  are \  8 cm \  and \  13 cm.$
$Then  \ the  \ length \  of  \  third \  side  \  is$
$more \  than \  13-8 \ =  5cm$
$and  \ less \  than \  13+8  =  21cm$
$a = 5cm , b = 21cm$

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

Two sides of a triangle are $7$ and $10$ units, which of the following length can be the length of the third side?

  1. $19$ units
  2. $17$ units
  3. $13$ units
  4. $3$ units
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
The sum of any two sides is greater than the third side.
and difference between two sides should be lesser than the third side.
For option $A,$ $7+10\ngtr 19$ 
For option $B,$ $7+10\ngtr 17$ 
For option $C,$ $7+10 > 13$ 
Also, $ |10-7| < 13$
For option $D,$ $7+10 > 3$ 
but $|7-10| \nless 3$
Only, $C$ is correct.
Multiple choice maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

Find the length of the third side of the triangle inequality, if sides of a triangle have $a = 4$ and $b = 8$.

  1. $c = 10$
  2. $c = 2$
  3. $c = 3$
  4. $c = 4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The Triangle Inequality theorem states that the sum of any $2$ sides of a triangle must be greater than the measure of the third side.
Take option A: $a = 4, b = 8, c = 10$
$4 + 8 > 10 (a + b > c)$
$8 + 10 > 4 (b + c > a)$
$4 + 10 < 8 (a + c > b)  $
Therefore, the third side, $c = 10$ will satisfy the triangle inequality
Similarly check for option 2: $c = 2$
$4 + 8 > 2 (a + b > c)$
$8 + 2 > 10 (b + c = a)$
$4 + 2 < 8 (a + c < b)  $
Hence, the second option will not satisfying the triangle inequality.
Similarly check for option $3$ and $4$.
Here, the third side of the triangle inequality, $c = 10$.

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

Two sides of a triangle have lengths $5$ and $8$, and the length of the third side is an integer. What is the greatest possible value of the perimeter of the triangle?

  1. $22$
  2. $24$
  3. $25$
  4. $26$
  5. $27$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The third side rule says that the length of the third side of the triangle in this case must be less than $5+8=13$
Since the length of the third side is an integer, the length must be $12$. 

The biggest possible perimeter is then $5+8+12=25$
As we know, perimeter of a triangle is $=1^{st}$ side $+2^{nd}$ side $+ 3^{rd}$ side.

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

In a triangle $ABC$, $(a+b+c)(b+c-a)=k$$bc$ if:

  1. $k< 0$
  2. $k> 6$
  3. $0< k< 4$
  4. $k> 4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The expression (a+b+c)(b+c-a) = (b+c)^2 - a^2 = b^2 + c^2 + 2bc - a^2. By the Law of Cosines, a^2 = b^2 + c^2 - 2bc cos(A). Substituting this, we get 2bc + 2bc cos(A) = 2bc(1+cos(A)). Since 0 < 1+cos(A) < 2, the value k must be between 0 and 4.

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The area of a triangle is computed using the formula $S=\dfrac {1}{2}$ bc sin A. If the relative errors made in measuring b, c and calculating S are respectively $0.02$, $0.01$ and $0.13$ the approximate error in A when $A=\pi /6$ is

  1. $0.05$ radians
  2. $0.01$ radians
  3. $0.05$ degree
  4. $0.01$ degree
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Error formula for given equation is
$\dfrac{\Delta s}{s}=\dfrac{\Delta b}{b}+\dfrac{\Delta c}{c}+\dfrac{\Delta sinx}{sinx}$
$0.13=0.02+0.01+\dfrac{\Delta sinx}{1/2}$
$\Delta sinx=0.05$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

In a $\triangle ABC, C=90^{o}$. Then $\tan A$ and $\tan B$ are the roots of the equation

  1. $abx^{2}-c^{2}x+1=0$
  2. $abx^{2}-(a^{2}+b^{2})x+ab=0$
  3. $c^{2}x^{2}-abx+c^{2}=0$
  4. $ax^{2}-bx+a=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\tan A$ & $\tan B$ are roots 
Sum $=(\tan A+\tan B)$
Product $=\tan A\tan B$
$\angle C=90^o$
$\angle A+\angle B=90^o$
$\tan A \tan B=1$
$\tan B=1/\tan A$
$x^2-(\tan A+\tan B)x+1=0$
$x^2\dfrac {2}{\sin 2A}x+1=0$
$\Rightarrow \ x^2-\dfrac {c^2}{ab}x+1=0 \Rightarrow \ x^2 (ab)-c^2 (x)+1=0$
$\tan A+\tan B=\tan A+\dfrac {1}{\tan A}$
$=\dfrac {2\tan ^2 A+1}{2\tan A}=\dfrac {2}{\sin 2A}$
$\sin 2A=\dfrac {2\tan A}{1+\tan^2 A}$


Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

If $B$ is reflection of $A(a,5)$ about line $4x-3y=0$, then area of triangle $ABC$ is equal to

  1. $\dfrac{253}{50}$
  2. $\dfrac{506}{25}$
  3. $\dfrac{253}{25}$
  4. $\dfrac{506}{50}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The reflection B of A(a, 5) across 4x-3y=0 is found using the reflection formula. The area of triangle ABC (where C is the origin or a fixed point) is calculated using the coordinates of A, B, and the intersection point.

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

If $\triangle ABC\sim \triangle DEF$ and $AB:DE=3:4$, then the ratio of area of triangles taken in order is 

  1. $\dfrac{9}{16}$
  2. $\dfrac{16}{9}$
  3. $\dfrac{15}{9}$
  4. $\dfrac{9}{15}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
 In similar triangle,
         Rratio of areas of triangle = square of ratio of corresponding sides
       Ratio of areas of triangle=${\dfrac {3^2}{4^2}}$
                                                 =$\dfrac{9}{16}$
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 $16cm^2$ and $36cm^2$ respectively. If the altitude of the first triangle is $3cm$, then the corresponding altitude of the other triangle is:

  1. $4cm$
  2. $6.5cm$
  3. $4.5cm$
  4. $6cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let ${A} _{1}$ and ${A} _{2}$ be the areas of the similar triangles.

$\Rightarrow \dfrac{{A} _{1}}{{A} _{2}}=\dfrac{{s} _{1}^{2}}{{s} _{2}^{2}}$

$\Rightarrow \dfrac{16}{36}=\dfrac{{\left(3\right)}^{2}}{{s} _{2}^{2}}$ given $({s} _{1}=3 \ cm )$

$\Rightarrow {s} _{2}^{2}=\dfrac{36\times 9}{16}$

$\Rightarrow {s} _{2}=\dfrac{6\times 3}{4}=4.5 \ cm$  

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 $12$ ${cm}^{2}$ and $48$ ${cm}^{2}$. If the height of the smaller one is $2.1$ $cm$, then the corresponding height of the bigger one is:

  1. $4.41$ $cm$
  2. $8.4$ $cm$
  3. $4.2$ $cm$
  4. $0.525$ $cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Areas of two similar triangles are $12 cm^2$ and $48 cm^2$
For similar triangles the ratio of areas is equal to the ratio of square of corresponding heights
Hence, $\dfrac{A _1}{A _2} = \dfrac{(h _1)^2}{(h _2)^2}$

$\dfrac{12}{48} = \dfrac{(2.1)^2}{(h _2)^2}$

$(h _2)^2= 4 \times (2.1)^2$

$h _2 = 2 \times 2.1$

$h _2 = 4.2 cm$

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

The corresponding sides of two similar triangles are in the ratio $2$ to $3$. If the area of the smaller triangle is $12$ the area of the larger is

  1. $24$
  2. $27$
  3. $18$
  4. $8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Area of similar triangles are in the ratio of square of the corresponding sides.
Hence, $\dfrac{\text{Area of smaller triangle}}{\text{Area of larger triangle}} = \frac{2^2}{3^2}$
$\Rightarrow \dfrac{12}{\text{Area of larger triangle}} = \dfrac{4}{9}$
$\text{Area of larger triangle}  = 27$

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

If $\Delta ABC\sim \Delta DEF$ such that area of $\Delta ABC$ is $9 cm^2$ and area of $\Delta DEF$ is $16 cm^2$ and $BC=1.8 cm$, then EF is

  1. 2.4 cm

  2. 1.35 cm

  3. 2.1 cm

  4. 3.2 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$ar(\triangle ABC)=9cm^2,\,ar(\triangle DEF)=16cm^2$ and $BC=1.8cm$

$\triangle ABC\sim\triangle DEF$                [ Given ]

$\Rightarrow$  $\dfrac{ar(\triangle ABC)}{ar(\triangle DEF)}=\dfrac{(BC)^2}{(EF)^2}$                    [ Area of similar triangle theorem ]

$\Rightarrow$  $\dfrac{9}{16}=\dfrac{(1.8)^2}{(EF)^2}$
Taking square root on both sides,

$\Rightarrow$  $\dfrac{3}{4}=\dfrac{1.8}{EF}$

$\Rightarrow$  $EF=\dfrac{1.8\times 4}{3}$

$\Rightarrow$  $EF=\dfrac{7.2}{3}$

$\Rightarrow$  $EF=2.4\,cm$


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

The sides of two similar triangles are in the ratio $4:9$ Areas of these triangles are in the ratio

  1. $3 : 5$
  2. $4 : 9$
  3. $81 : 16$
  4. $16 : 81$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If two triangles are similar to each other, then the ratio of the area of this triangle will be equal to the square of the ratio of the corresponding sides of this triangle.

$\therefore$ The ratio between area of these triangle$=\dfrac{(4)^2}{(9)^2}=\dfrac{16}{81}$

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 $\displaystyle 9\ { cm }^{ 2 }$ and $\displaystyle 16\ { cm }^{ 2 }$, respectively. The ratio of their corresponding heights is

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

${(\dfrac{{h1}}{{h2}})^2} = \dfrac{{S1}}{{S2}}$

Where h1 and h2 are the heights 

and S1, S2 are the areas of similar traingles

${{\rm{(}}\dfrac{{h1}}{{h2}})^2} = \dfrac{{9c{m^2}}}{{16c{m^2}}}$

$\dfrac{{h1}}{{h2}} = \sqrt {\dfrac{9}{{16}}} $

$\dfrac{{h1}}{{h2}} = \dfrac{3}{4}\ or\  {3:4}$