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 business maths basic concepts in geometry conditional statements and converse euclid's postulates introduction to euclid's geometry

The converse of "If in a triangle $ABC, AB=AC$, then $\angle B=\angle C$", is

  1. lf in a triangle $ABC, \angle B=\angle C$, then $AB=AC$.
  2. lf in a triangle$ABC, AB\neq AC$, then $\angle B\neq\angle C$.
  3. lf in a triangle $ABC, \angle B\neq\angle C$, then $AB\neq AC$.
  4. lf in a triangle $ABC, \angle B\neq\angle C$, then $AB=AC$ .
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Take $p:AB=AC$

and $q:\angle B=\angle C$
So the given statement is symbolically represented as $p\rightarrow q$
Now by definition, Converse of a conditional statement $p\rightarrow q$ is $q\rightarrow p$
So $q\rightarrow p$ is given by 
"If in a triangle $ABC, \angle B=\angle C,$ then $AB=AC.$"

Multiple choice business maths basic concepts in geometry conditional statements and converse euclid's postulates introduction to euclid's geometry

The converse of "if in a triangle $ABC, AB>AC$, then $\angle C=\angle B$", is

  1. lf in a triangle $ABC, \angle C=\angle B$, then $AB>AC$.
  2. lf in a triangle$ABC, AB\not\simeq AC$, then $\angle C\not\simeq \angle B$.
  3. lf in a triangle $ABC, \angle C\not\simeq \angle B$, then $ AB\not\simeq AC$.
  4. lf in a triangle $ABC, \angle C\not\simeq \angle B$, then $AB>AC$.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Take $p:AB>AC$

and $q: \angle C=\angle B$
So the given statement is symbolically represented as $p\rightarrow q$
Now by definition, Converse of a conditional statement $p\rightarrow q$ is $q\rightarrow p$
Thus $q\rightarrow p$ is given by
"If in a $\triangle ABC, \angle C=\angle B$ then $AB>AC$."

Multiple choice physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

A book is lying on a table, what is the angle between the book on the table and the weight of the book?

  1. $0^o$
  2. $45^o$
  3. $90^o$
  4. $180^o$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Weight always point towards the center of the earth that is perpendicular to the surface of the earth that is towards the table and $perpendicular$ to surface of the table.


Now as the book is lying on the table i.e. book is $parallel$ to the table and the weight is perpendicular to the table so the required angle is $90^0$
Option C is correct.

Multiple choice maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

Use a transformation matrix to find the image of $D(-7,6)$ after a rotation of $180^0$ counterclockwise around the origin.

  1. $(7,6)$
  2. $(-7,-6)$
  3. $(7,-6)$
  4. $(-7,6)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The transformation matrix for rotation  is $\begin{bmatrix} cos\theta  & -sin\theta  \ sin\theta  & cos\theta  \end{bmatrix}$

For $\theta=180^{0}$ , the transformation matrix will be $\quad \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}$
So the image of point $(-7,6)$ is $\quad \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}\begin{bmatrix} -7 \ 6 \end{bmatrix}=\begin{bmatrix} 7 \ -6 \end{bmatrix}$
Therefore the correct option is $C$

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

Let  $ABC$ be triangle. Let $A$ be the point $(1,2),y=x$be the perpendicular bisector of $AB$ and $x-2y+1=0$ be the angle bisector of $\angle C$. If equation of $BC$ is given by $ax+by-5=0$, then the value of $a+b$ is 

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

$\begin{array}{l} { m _{ AB } }=-1 \ y-2=-1\left( { x-1 } \right)  \ y-2=-x+1 \ x+y=3 \ \underline { x-y=0 }  \ 2x=3 \ \therefore x=\frac { 3 }{ 2 } \, \, \, \, \, ,y=\frac { 3 }{ 2 }  \ \frac { { 1+h } }{ 2 } =\frac { 3 }{ 2 } \, \, \,  \ h=2 \ \frac { { 2+k } }{ 2 } =\frac { 3 }{ 2 }  \ k=1\, \, \, \, \, \, \, \, \, \, \, B\left( { 2,1 } \right)  \ As\, \, image\, \, through\, \, x-2y+1=0 \ { m _{ AD } }=-2 \ y-2=-2\left( { x-1 } \right)  \ y-2=-2x+2 \ 2x+y=4\times 2\, \, \, \, \, \, \, x-2y+1=0 \ 4x+2y=8 \ \underline { x-2y=-1 }  \ 5x=7 \ x=\frac { 7 }{ 5 } \, \, \, \, \, \, y=\frac { { x+1 } }{ 2 } =\frac { { 12 } }{ { 5\times 2 } } =\frac { 6 }{ 5 }  \ \frac { { m+1 } }{ 2 } =\frac { 7 }{ 5 } \, \, \, \, \, \, \, \, \frac { { n+1 } }{ 2 } =\frac { 6 }{ 5 }  \ m+1=\frac { { 14 } }{ 5 } \, \, \, \, \, \, \, n+2=\frac { { 12 } }{ 5 }  \ m=\frac { 9 }{ 5 } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, n=\frac { { 12 } }{ 5 } -2=\frac { 2 }{ 5 }  \ E\left( { \frac { 9 }{ 5 } ,\frac { 2 }{ 5 }  } \right)  \ BC=BE\, \, \, \, \, \, B\left( { 2,1 } \right) \, \, \, E\left( { \frac { 9 }{ 5 } ,\frac { 2 }{ 5 }  } \right)  \ y-1=\left( { \frac { { 1-\frac { 2 }{ 5 }  } }{ { 2-\frac { 9 }{ 5 }  } }  } \right) \left( { x-2 } \right)  \ 3x-y=5 \ a=3\, \, \, \, b=-1 \ a+b=2 \end{array}$

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

If  in $\triangle ABC$  and $\triangle EDA,$ $\displaystyle BC\bot AB,AE\bot AB$ and $\displaystyle DE\bot AC$ then $\displaystyle DE.BC=AD.AB$ 

  1. True

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

In $\displaystyle \Delta ABC$ and $\displaystyle \Delta EDA$,
We have
$\displaystyle \angle ABC=\angle ADE$ [Each equal to $\displaystyle { 90 }^{ o }$]
$\displaystyle \angle ACB=\angle EAD$ [Alternate angles]
$\displaystyle \therefore $ By AA Similarity
$\displaystyle \Delta ABC\sim \Delta EDA$
$\displaystyle \Rightarrow \frac { BC }{ AB } =\frac { AD }{ DE } $
$\displaystyle \Rightarrow DE.BC=AD.AB$.
Hence proved.

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

If in $\displaystyle \Delta ABC$ and $\displaystyle \Delta DEF,\frac { AB }{ DE } =\frac { BC }{ FD } $, then they will be similar if :

  1. $\displaystyle \angle B=\angle E$
  2. $\displaystyle \angle A=\angle D$
  3. $\displaystyle \angle B=\angle D$
  4. $\displaystyle \angle A=\angle F$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If two sides of a triangle are proportional to the corresponding two sides in another triangle, and their included angles are equal, then the two triangles are similar by SAS rule.

If $\quad \dfrac { AB }{ DE } = \dfrac { BC }{ FD } $, then for two triangles ABC and DEF to be similar, the included angle must be equal. In this case, the included angles are $\quad \angle B\quad and\quad \angle D$.

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

Given $\Delta ABC-\Delta PQR$. If $\dfrac{AB}{PQ}=\dfrac{1}{3}$, then find $\dfrac{ar\Delta ABC}{ar\Delta PQR'}$.

  1. $\dfrac{1}{9}$
  2. $\dfrac{1}{8}$
  3. $\dfrac{8}{9}$
  4. $\dfrac{9}{1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\dfrac{AB}{PQ}=\dfrac{1}{3}$
$\dfrac{ar\Delta ABC}{ar\Delta PQR}=\left(\dfrac{AB}{PQ}\right)^2=\left(\dfrac{1}{3}\right)^2=\dfrac{1}{9}$.
Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

$\Delta DEF -\Delta ABC$; If DE $:$ AB $=2:3$ and ar($\Delta$DEF) is equal to $44$ square units, then find ar($\Delta$ABC) in square units.

  1. $99$
  2. $33$
  3. $11$
  4. $66$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The ratio of areas of similar triangles is the square of the ratio of their corresponding sides. (DE/AB)^2 = (2/3)^2 = 4/9. Area(DEF)/Area(ABC) = 4/9. 44/Area(ABC) = 4/9, so Area(ABC) = 44 * 9 / 4 = 99.

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

Given, $\Delta$ABC$-\Delta$PQR. If $\dfrac{ar(\Delta ABC)}{ar(\Delta PQR)}=\dfrac{9}{4}$ and $AB=18$cm, then find the length of PQ.

  1. $19$
  2. $12$
  3. $32$
  4. $44$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The ratio of areas is the square of the ratio of corresponding sides. Area(ABC)/Area(PQR) = (AB/PQ)^2. 9/4 = (18/PQ)^2. Taking the square root, 3/2 = 18/PQ. PQ = 18 * 2 / 3 = 12.

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

ABC is an isosceles triangle right angled at B. Similar triangles ACD and ABE are constructed in sides AC and AB. Find the ratio between the areas of $\triangle ABE$ and $\triangle ACD$.

  1. $2:1$
  2. $1:1$
  3. $1:2$

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

In $\triangle ABC$

$\implies { AB }^{ 2 }+{ BC }^{ 2 }={ AC }^{ 2 }$
$\implies\quad { AB }^{ 2 }+{ AB }^{ 2 }={ AC }^{ 2 }$
$\implies\quad { AC }^{ 2 }={ 2AB }^{ 2 }\quad -(1)$
Ratio of areas of similar triangle is equal to ratio of squares of their corresponding sides.
$\implies\quad \cfrac { Area\quad (\triangle ABE) }{ Area\quad (\triangle ACD) } =\cfrac { { AB }^{ 2 } }{ { AC }^{ 2 } } $
 using(1)
$\implies\quad \cfrac { Area\quad (\triangle ABE) }{ Area\quad (\triangle ACD) } =\cfrac { { AB }^{ 2 } }{ { 2AB }^{ 2 } } =\cfrac { 1 }{ 2 } $

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

If $\Delta ABC \sim \Delta PQR$ and $\displaystyle {{PQ} \over {AB}} = {5 \over 2}$ then area $(\Delta ABC):$ area $(\Delta PQR) = ?$

  1. $\displaystyle {{25} \over 4}$
  2. $\displaystyle {4 \over {25}}$
  3. $\displaystyle {5 \over 2}$
  4. $\displaystyle {{25} \over 2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\Delta ABC\sim \Delta PQR$

Also $\dfrac{PQ}{AB}=\dfrac{5}{2}$

If triangles are similar then the ratio of their is equal to ratio of square of the sides

$\dfrac{ar(ABC)}{ar(PQR)}=\dfrac{AB^2}{PQ^2}=\dfrac{4}{25}$.