Mathematics · Quantitative Aptitude

Geometry of Triangles and Angles

846 Questions

Triangle and angle geometry covers angle sums, properties of equilateral shapes, and right triangles. These principles form the basis of advanced quantitative aptitude sections. Practicing spatial problems helps secure points in exams.

Triangle angle sumsPythagorean theoremProperties of equilateral trianglesComplementary and supplementary anglesRight triangle formulas

Geometry of Triangles and Angles Questions

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

$\Delta ABC \sim  \Delta PQR$ and $\displaystyle\frac{A( \Delta ABC)}{A( \Delta PQR)}=\dfrac{16}{9}$. If $PQ=18$ cm and $BC=12$ cm, then $AB$ and $QR$ are respectively:

  1. $9$ cm, $24$ cm
  2. $24$ cm, $9$ cm
  3. $32$ cm, $6.75$ cm
  4. $13.5$ cm, $16$ cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle\frac{16}{9}=\left[\frac{AB}{PQ}\right]^2=\left[\frac{BC}{QR}\right]^2$

$\displaystyle\Rightarrow \frac{16}{9}=\left[\frac{AB}{18}\right]^2$ and $\displaystyle\frac{16}{9}=\left[\frac{12}{QR}\right]^2$

$\displaystyle \Rightarrow \frac{4}{3}=\frac{AB}{18}$ and $\displaystyle \frac{4}{3}=\frac{12}{QR}$

$\Rightarrow AB=24$ cm, $QR=9$ cm.

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

If $\triangle ABC\sim \triangle  PQR,$  $ \cfrac{ar(ABC)}{ar(PQR)}=\cfrac{9}{4}$,  $AB=18$ $cm$ and $BC=15$ $cm$, then $QR$ is equal to:

  1. $10$ $cm$
  2. $12$ $cm$
  3. $\cfrac{20}{3}$ $cm$
  4. $8$ $cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $\triangle ABC \sim \triangle PQR$,

Then, $\dfrac{ar(ABC)}{ar(PQR)} = \dfrac{AB^2}{PQ^2} = \dfrac{BC^2}{QR^2} = \dfrac{AC^2}{PR^2}$
$\dfrac{9}{4} = \dfrac{BC^2}{QR^2}$

$\dfrac{9}{4} = \dfrac{15^2}{QR^2}$
$QR^2 = \dfrac{4 \times 225}{9}$
$QR^2 = 100$
$QR = 10 \ cm$

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

In $\Delta ABC$, a line is drawn parallel to $BC$ to meet sides $AB$ and $AC$ in $D$ and $E$ respectively. If the area of the $\Delta ADE$ is $\dfrac 19$ times area of the $\Delta ABC$, then the value of $\dfrac {AD}{AB}$ is equal to:

  1. $\dfrac 13$
  2. $\dfrac 14$
  3. $\dfrac 15$
  4. $\dfrac 16$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

By theorem on ratio of areas of similar triangles, we get

$\dfrac {A(\triangle ADE)}{A(\triangle ABC)} = \left(\dfrac {AD}{DB}\right)^2$

$\therefore \dfrac 19 = \dfrac {AD^2}{DB^2}$

$\therefore \dfrac {AD}{DB}= \dfrac 13$.

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

If $\triangle ABC$ and $\triangle PQR$ are similar and $\dfrac {BC}{QR} = \dfrac {1}{3}$ find $\dfrac {area (PQR)}{area (BCA)}$

  1. $9$
  2. $3$
  3. $\dfrac {1}{3}$
  4. $\dfrac {1}{9}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\triangle ABC$ & $\triangle PQR$ are similar.

$\therefore \cfrac{AB}{PQ}=\cfrac{BC}{QR}=\cfrac{1}{3}$
$\therefore \cfrac{\text{Area}(PQR)}{\text{Area}(BCA)}={(\cfrac{QR}{BC}})^{2}$
$\cfrac { { Area }(PQR) }{ { Area }(BCA) } ={ (\cfrac { 3 }{ 1 } ) }^{ 2 }=9$

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

In $\Delta ABC$, $D$ is a point on $BC$ such that $3BD = BC$. If each side of the triangle is $12 cm$, then $AD$ equals:

  1. $4\sqrt { 5 } cm$
  2. $4\sqrt { 6 } cm$
  3. $4\sqrt { 7 } cm$
  4. $4\sqrt { 11 } cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given $\triangle ABC$ with $D$ a point on $BC$ such that $3BD=BC$

$\therefore$ $BD=\dfrac{BC}{3}=4cm$
$AD\  \bot\ BC$
Let's take a point $E$ on $BC$ which makes a right angle triangle $ADE$ and $AEB$ at $E$ such that $BE=\dfrac{1}{2}BC=6cm$
$\therefore\ DE=BE-BD=2cm$.
$\therefore\ AE^2=AB^2-BE^2=144-36=108$
$\because\ AED=90^{o}$
$\therefore\ AD^2=AE^2+DE^2=108+4=112$
$\therefore\ AD=4\sqrt{7}cm$.

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

In $\Delta ABC \sim  \Delta PQR$, $M$ is the midpoint of $BC$ and $N$ is the midpoint of $QR$. If the area of $\Delta ABC =$ $100$ sq. cm and the area of $\Delta PQR =$ $144$ sq. cm. If $AM = 4$ cm, then $PN$ is:

  1. $4.8$ cm
  2. $12$ cm
  3. $4$ cm
  4. $5.6$ cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\dfrac { ar(\triangle ABC) }{ ar(\triangle PQR) } =\dfrac { 100 }{ 144 } $

If triangles are similar then ratio of their areas is equal to ratio of square of their corresponding sides
$\dfrac { AB^{ 2 } }{ PQ^{ 2 } } =\dfrac { 100 }{ 144 } \ \dfrac { AB }{ PQ } =\dfrac { 10 }{ 12 } $
$AM$ and $PN$ are medians
Therefore, $ \dfrac { AM }{ PN } =\dfrac { AB }{ PQ } $
$\Rightarrow  \dfrac { 4 }{ PN } =\dfrac { 10 }{ 12 } \ \Rightarrow PM=4.8$

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

Which among the following is/are correct?
(I) If the altitudes of two similar triangles are in the ratio $2:1$, then the ratio of their areas is $4 : 1$.
(II) $PQ \parallel BC$ and $AP : PB=1:2$. Then, $\dfrac{A(\triangle APQ)}{A(\triangle ABC)}=\dfrac{1}{4}$

  1. $(I)$
  2. $(II)$
  3. Both $(I)$ and $(II)$

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

Option A: This statement is correct. The ratio of the altitudes of the similar triangles  is  $2:1$

Ratio of the areas of the similar triangles $=$ Square of the ratio of  the  altitudes.

$\therefore$ Ratio  of  the  areas $ =  { \left( \dfrac { 2 }{ 1 }  \right)  }^{ 2 }=  4:1$


Option B: If  $PQ\parallel BC$,  then $\triangle APQ \sim \triangle ABC$ by AA test of similarity.

Hence, $\dfrac {A( \triangle APQ)}{A(\triangle ABC)}=\dfrac { AP^2 }{ AB^2 }$

If $AP = x$ and $BP = 2x$, then $AB = 3x$.

$\therefore \dfrac {A( \triangle APQ)}{A(\triangle ABC)}=\dfrac 19$

So, the given statement is false

Multiple choice maths axioms, postulates and theorems euclid's fifth postulate conditional statements and converse euclid's postulates

Select the correct statement for the following:

$A$: The angles of an equilateral triangle are equal.
$B$: Angles opposite to two congruent sides of a triangle are congruent.

  1. $A$ is a theorem and $B$ is its corollary
  2. $B$ is a theorem and $A$ is its corollary
  3. $A$ and $B$ are both theorems
  4. $A$ and $B$ are both corollaries
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$B$ is called the  Triangle theorem.  $A$ can be proved using this theorem, as in an equilateral triangle, all sides are equal and the angles opposite to these congruent sides are Equal. So, $A$ is the corollary of $B$

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The points $(10,7,0)$, $(6,6-1)$ and $(6,9,-4)$ form a 

  1. Right -angled triangle

  2. Isosceles triangle
  3. Both $(1)$ & $(2)$

  4. Equilateral triangle
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$P(10, 7, 0),\ Q(6, 6, -1),\ R(6, 9, -4)$
$PQ=\sqrt {4^2+(-1)^2 +(-1)^2}$
$=\sqrt {16+1+1}=3\sqrt 3$
$QR=\sqrt {0^2+3^2+-3^2}$
$PR=\sqrt {4^2+2^2+(-4)^2}$
$=\sqrt {16+4+16}=6$
$PQ^2+QR^2=(3\sqrt 2)^2+(3\sqrt 2)^2=6^2=PR^2$
$\therefore \ \triangle PQR$ is a right angle $D$ of $Q,\ PQ=QR$
$\triangle PQR$ is a isoscels traingle
$(C)$ Both $(1)$ & $(2)$


Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

If $(1, 1, a)$ is the centroid of the triangle formed by the points $(1, 2, -3)$ , $(\mathrm{b}, 0, 1)$ and $(-1, 1, -4)$ then $a-b$ $=$

  1. $-5$
  2. $-7$
  3. $5$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The coordinates of the vertices of the triangle are given by $(1,2,-3) , (b,0,1), (-1,1,-4)$

Accordingly the coordinates of the centroid of this triangle will be given by 

($ \dfrac{b}{3}, 1 , -2 $)

Hence, $ \dfrac{b}{3} $ $= 1$ Or, $b= 3$

and $a = -2$

So $a- b = -5$

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The circum centre of the triangle formed by the points $(2, 5, 1), (1, 4, -3)$ and $(-2, 7, -3)$ is

  1. $(6,0,1)$
  2. $(0,6,-1)$
  3. $(-1,6,2)$
  4. $(6,1,-2)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the points $A(2,5,1) B(1,4,-3)$ and $C(-2,7,-3)$
Now using distance formula in $3D$, we have
$AB {=}$$\sqrt{18}$, $BC {=}$$\sqrt{18}$, and $AC {=}$$\sqrt{36}$
Since, ${AB}^{2}+{BC}^{2}$${=}$${AC}^{2}$
Hence, it is right angle triangle and as we know that the circumcentre of right angled triangle is at the midpoint of hypotenuse i.e $AC$.
Therefore by using section formula (1:1), circumcentre ${=}(0,6,-1)$

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

Assertion (A): The points $A(2,9,12) ,B(1,8,8) ,C(2,11,8) D(1,12,12)$ are the vertices of a rhombus
Reason (R): $AB = BC = CD = DA$ and $AC = BD$

  1. Both A and R are individually true and R is the correct explanation of A

  2. Both A and R individually true but R is not the correct explanation of A

  3. A is true but R is false

  4. Both A and R false
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given: The points $A(2,9,12) ,B(1,8,8) ,C(2,11,8) D(1,12,12)$ are the vertices of a rhombus. 
So using distance formula Reason is not true. 
Thus both A and R false. 
Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

In a $\triangle {ABC}$, side $AB$ has the equation $2x+3y=29$ and the side $AC$ has the equation $x+2y=16$. If the mid point of $BC$ is $(5,6)$, then the equation of $BC$ is

  1. $2x+y=7$
  2. $x+y=1$
  3. $2x-y=17$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let co-ordinates of $B$ be $(x _1, y _1)$ & $C$ be $(x _2, y _2)$

$\therefore$ $(5,6)$ is the mid point,

so, $\dfrac{x _1 + x _2}{2} = 5, \dfrac{y _1 + y _2}{2} = 6$

$\Rightarrow x _1 + x _2 = 10, y _1 + y _2 = 12$

$B(x _1, y _1)$ lies on the line $2x + 3y = 29$

$\therefore 2x _1 + 3y _1 = 29$  ----(1)

$C(x _2, y _2)$ lies on the line $x + 2y = 16,$

$\therefore x _2 + 2y _2 = 16$  ----(2)

$\therefore$ putting $x _1, y _1$ in the form of $x _2, y _2$ in (1)

$2(10 - x _2) + 3(12 - y _2) = 29$  {$x _1 = 10 - x _2, y _1 = 12 - y _2$}

$\Rightarrow 20 - 2x _2 + 36 - 3y _2 = 29$

$\Rightarrow 2x _2 + 3y _2 = 27$  ----(3)

on subtracting $(3)$ and $(2)$ $\times$ $2$

$-y _2 = -5$

$y _2 = 5$

Putting $y _2 \,  in (2)$

$x _2 + 2(5) = 16$

$x _2 = 6$

$x _1 = 10 - x _2$

      $= 4$

$y _1 = 12 - 5 = 7$

Equation :

$\dfrac{x - x _1}{x _2 - x _1} = \dfrac{y - y _1}{y _2 - y _1}$

$\Rightarrow \dfrac{x - 4}{2} = \dfrac{y - 7}{-2}$

$\Rightarrow -x + 4 = y - 7$

$\Rightarrow x + y = 11$
Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

If $A= \left ( 5,-1,1 \right ),B= \left ( 7,-4,7 \right ),C= \left ( 1,-6,10 \right ),D= \left ( -1,-3,4 \right )$. Then $ABCD$ is a

  1. square

  2. rectangle

  3. rhombus

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

AB${=}$ $\sqrt{{(7-5)}^{2}+{(-4+1)}^{2}+{(7-1)}^{2}}$
AB${=}$ $\sqrt{{(2)}^{2}+{(-3)}^{2}+{(6)}^{2}}$
AB${=}$ $\sqrt{49}$
AB${=}$ 7
Similarly you find that BC${=}$ $\sqrt{49}$  CD${=}$ 7  and DA${=}$7
Hence all sides of quadrilateral are equal, Now we check the diagonals
AC${=}$ $\sqrt{{(1-5)}^{2}+{(-6+1)}^{2}+{(10-1)}^{2}}$
AC${=}$ $\sqrt{122}$
similarly BD${=}$ $\sqrt{74}$ 
Diagonals are not equal
direction ratio of line passing through AC is (-4,-5,9)
direction ratio of line passing through  BD is (-8,1,-3), As the dot product dr of AC and BD are equal to 0 which means AC is perpendicular to BD,
All sides are equal and diagonal are not equal but bisect each other at right angle
hence it is rhombus