Mathematics · Quantitative Aptitude

Geometry of Triangles and Angles

758 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

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

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