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 congruence introduction to shapes similarity of triangles introduction to similar triangles

The ratio of the angles in $\triangle ABC$ is $2 : 3 : 4$. Which one of the following triangles is similar to $\triangle ABC ?$

  1. $ \triangle DEF $ has angles in the ratio $4 : 3 : 2.$
  2. $ \triangle PQR $ has angles in the ratio $1 : 2 : 3.$
  3. $ \triangle LMN $ has angles in the ratio $1 : 1 : 1.$
  4. $ \triangle STW $ has sides in the ratio $1 : 1 : 1.$
  5. $ \triangle XYZ $ has sides in the ratio $4 : 3 : 2.$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Similar triangles must have the same ratio of angles. The ratio 2:3:4 is equivalent to 4:6:8 or any scalar multiple, but the order of the ratio matters for similarity. Option A provides the same ratio 4:3:2, which represents the same set of interior angles as 2:3:4.

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

In triangle ABC, AB = AC = 8 cm, BC = 4 cm and P is a point in side AC such that AP = 6 cm. Prove that $\Delta\,BPC$ is similar to $\Delta\,ABC$. Also, find the length of BP.

  1. BP = 4 cm

  2. BP = 8 cm

  3. BP = 6 cm

  4. BP = 12 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: $\triangle ABC$, $AB = AC = 8$, $BC = 4$ and $AP = 6$

In $\Delta\,ABC$,
$\displaystyle\,\frac{AB}{BC}\,=\,\frac{8}{4}\,=\,2$,
In $\Delta\,BPC$,
$\displaystyle\,\frac{BC}{CP}\,=\,\frac{4}{2}\,=\,2$

Now, in $\triangle ABC$ and $\triangle BPC$
$\displaystyle\,\dfrac{AB}{BC}\,= \displaystyle\,\dfrac{BC}{CP}$
$\angle\,ABC\,=\,\angle\,C.$
Therefore, by SAS, $\Delta\,ABC \sim \Delta\,BPC$

Thus, $\dfrac{AB}{BP} = \dfrac{AC}{BC}$


$\dfrac{8}{BP} = \dfrac{8}{4}$
$BP = 4$ cm

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

In the given figure, $DE$ is parallel to $BC$ and the ratio of the areas of $\triangle ADE$ and trapezium $BDEC$ is $4:5.$ What is $DE : BC: ?$

  1. $1:2$
  2. $2:3$
  3. $4:5$

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

The ratio of area(ADE) to area(ABC) is 4/(4+5) = 4/9. Since the ratio of areas of similar triangles is the square of the ratio of their corresponding sides, DE/BC = sqrt(4/9) = 2/3.

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

If in $\triangle $s $ABC$ and $DEF,$ $\angle A=\angle E=37^{\circ}, AB:ED=AC:EF$ and $\angle F=69^{\circ},$ then what is the value of $\angle B: ?$

  1. $69^{\circ}$
  2. $74^{\circ}$
  3. $84^{\circ}$
  4. $94^{\circ}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In $\triangle ABC$ and $\triangle DEF$
$\angle A = \angle E =  37^{o}$
$\dfrac{AB}{ED} = \dfrac{AC}{EF}$
Thus, $\triangle ABC \sim \triangle EDF$ ....... (By SAS rule)
Thus, $\angle B = \angle D$

Now, $\triangle DEF$
$\angle D + \angle E + \angle F = 180$
$\angle D + 37 + 69 = 180$
$\angle D = 74^{\circ}$
Hence, $\angle B = \angle D = 74^{\circ}$

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

Two equilateral triangles with side $4 \ cm$ and $6 \ cm$ are _____ triangles.

  1. similar

  2. congruent

  3. both

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

Any two equilateral triangles are similar by SSS criteria..
$SSS$ similarity states that if the lengths of the corresponding sides of two triangles are proportional, then the triangles must be similar.
Two equilateral triangles with side $4 \ cm$ and $6 \ cm$ are similar triangles by $SSS$ similarlty. 

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

$\displaystyle \Delta ABC$ and $\displaystyle \Delta DEF$ are two similar triangles such that $\displaystyle \angle A={ 45 }^{ \circ  },\angle E={ 56 }^{ \circ  }$, then $\displaystyle \angle C$ =___.

  1. $\displaystyle { 56 }^{ \circ }$
  2. $\displaystyle { 45 }^{ \circ }$
  3. $\displaystyle { 101 }^{ \circ }$
  4. $\displaystyle { 79 }^{ \circ }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\Delta ABC \sim \Delta DEF$        ...Given

$\Rightarrow \angle A = \angle D$                 ...C.A.S.T.
$\Rightarrow \angle B = \angle E$                 ...C.A.S.T.
$\Rightarrow \angle C = \angle F$                 ...C.A.S.T.
$\therefore \angle B = \angle E = 56^o$
In $\Delta ABC$,
$\angle A + \angle B + \angle C = 180^o$        ....Angle sum property of triangles
$\Rightarrow 45^o+56^o+\angle C = 180^o$
$\Rightarrow \angle C = 79^o$

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

If a triangle with side lengths as $5, 12$, and $15$ cm is similar to a triangle which has longer side length as $24$ cm, then the perimeter of the other triangle is:

  1. $38.4$
  2. $44$
  3. $51.2$
  4. $58$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The longer side of the bigger triangle is $24$ cm.

The longer side of the smaller triangle is $15$ cm.
They are in ratio $24:15 = \cfrac{24}{15} = 1.6$
Thus, their perimeters also would be in the ratio $1.6$
The perimeter of the smaller triangle is $5 + 12 + 15 = 32$ cm
Implies the perimeter of the bigger triangle would be $32 \times 1.6 = 51.2$ cm

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

In $\Delta ABC$, DE is || to BC, meeting AB and AC at D and E. If AD = 3 cm, DB = 2 cm and AE = 2.7 cm, then AC is equal to:

  1. $6.5$ cm
  2. $4.5$ cm
  3. $3.5$ cm
  4. $5.5$ cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
In $\triangle$$ ADE$ and $\triangle$$ ABC$,

$\angle$$ADE=$$\angle$$ABC  $  (corresponding angles)

$\angle$$AED=$$\angle$$ACB$    (corresponding angles)

so,$\triangle$$ ADE$ $\sim$$\triangle$$ ABC$

so, $\dfrac{AD}{AB}=\dfrac{AE}{AC}$ 

$AC=2.7$$\times$$\dfrac{5}{3}$$=$$ 4.5$ cm
Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

Point L, M and N lie on the sides AB, BC and CA of the triangle ABC such that $\ell (AL) : \ell (LB) = \ell (BM) : \ell (MC) = \ell (CN) : \ell (NA) = m : n$, then the areas of the triangles LMN and ABC are in the ratio

  1. $\dfrac{m^2}{n^2}$
  2. $\dfrac{m^2 - mn + n^2}{(m + n)^2}$
  3. $\dfrac{m^2 - n^2}{m^2 + n^2}$
  4. $\dfrac{m^2 + n^2}{(m + n)^2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Dividing the sides of a triangle in the ratio m:n creates smaller corner triangles. The area of each corner triangle relative to the main triangle can be found using the fraction of the sides bounding the vertex, resulting in the area ratio formula (m^2 - mn + n^2) / (m + n)^2 for the inner triangle LMN.

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

If in a $\triangle ABC,{a}^{2}+{b}^{2}+{c}^{2}=8{R}^{2},$ where $R=$ circumradius,then the triangle is

  1. equilateral

  2. isosceles

  3. right angled

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

Since, ${a}^{2}+{b}^{2}+{c}^{2}=8{R}^{2}$
Using sine rule, we have
${\left(2R\sin{A}\right)}^{2}+{\left(2R\sin{B}\right)}^{2}+{\left(2R\sin{C}\right)}^{2}=8{R}^{2}$
$\Rightarrow 4{R}^{2}\left[{\left(\sin{A}\right)}^{2}+{\left(\sin{B}\right)}^{2}+{\left(\sin{C}\right)}^{2}\right]=8{R}^{2}$
$\Rightarrow {\sin}^{2}A+{\sin}^{2}B+{\sin}^{2}C=2$
$\Rightarrow 1-{\cos}^{2}A+{\sin}^{2}B+1-{\cos}^{2}C=2$
$\Rightarrow -{\cos}^{2}A+{\sin}^{2}B-{\cos}^{2}C=2-2=0$
$\Rightarrow {\cos}^{2}A-{\sin}^{2}B+{\cos}^{2}C=0$
$\Rightarrow \cos{\left(A+B\right)}\cos{\left(A-B\right)}+{\cos}^{2}C=0$
Since, $A+B+C=\pi \Rightarrow A+B=\pi-C$
$\Rightarrow \cos{\left(\pi-C\right)}\cos{\left(A-B\right)}+{\cos}^{2}C=0$
$\Rightarrow  -\cos{C}\cos{\left(A-B\right)}+{\cos}^{2}C=0$
$\Rightarrow  \cos{C}\left[-cos{\left(A-B\right)}+\cos{C}\right]=0$
$\Rightarrow \cos{C}=0, -cos{\left(A-B\right)}+\cos{C}=0$
$\Rightarrow C=\dfrac{\pi}{2}$ or $-cos{\left(A-B\right)}+\cos{\left(\pi-\left(A+B\right)\right)}=0$
$\Rightarrow -cos{\left(A-B\right)}-cos{\left(A+B\right)}=0$
Using transformation angle formula, we have
$\Rightarrow 2\cos{A}\cos{B}=0$
$\therefore \cos{A}=0,\cos{B}=0$
Hence $\angle{A}=\dfrac{\pi}{2} $or $\angle{B}=\dfrac{\pi}{2}, $ or  $\angle{C}=\dfrac{\pi}{2}$

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

In $\triangle ABC,$ which of the following statements are true:

  1. maximum value of $\sin{2A}+\sin{2B}+\sin{2C}$ is same as the maximum value of $\sin{A}+\sin{B}+\sin{C}$
  2. $R\ge 2r,$ where $R$ is circumradius and $r$ is the inradius.
  3. ${R}^{2}\ge \dfrac{abc}{\left(a+b+c\right)}$
  4. $\triangle ABC$ is right angled if $r+2R=s,$ where $s$ is semi perimeter.
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

Option$\left(a\right)$
$\because$ Maximum value of $\sin{2A}+\sin{2B}+\sin{2C}$ and  $\sin{A}+\sin{B}+\sin{C}$ is same that is $3$.
Option$\left(b\right)$
$\because \sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}\le \dfrac{1}{8}$
$\Rightarrow \dfrac{r}{4R}\le \dfrac{1}{8}$
$\Rightarrow R\ge 2r$
Option$\left(c\right)$
$\because \dfrac{abc}{a+b+c}=\dfrac{4R\triangle}{2s}$
                 $=2R.r=R\left(2r\right)\le {R}^{2}$
$\therefore {R}^{2}\ge \dfrac{abc}{a+b+c}$   ($\because R\ge 2r$)
Option$\left(d\right)$
$\angle{B}={90}^{0}$
$\therefore r=\left(s-b\right)\tan{\dfrac{B}{2}}=s-b$
$R=\dfrac{b}{2\sin{B}}=\dfrac{b}{2}$
$\Rightarrow 2R=b$
$\therefore r+2R=s-b+b=s$

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

There exist a triangle $ABC$ satisfying

  1. $\tan{A}+\tan{B}+\tan{C}=0$
  2. $\dfrac{\sin{A}}{2}=\dfrac{\sin{B}}{3}=\dfrac{\sin{C}}{7}$
  3. ${\left(a+b\right)}^{2}={c}^{2}+ab$ and $\sqrt{2}\left(\sin{A}+\cos{A}\right)=\sqrt{3}$
  4. $\sin{A}+\sin{B}=\left(\dfrac{\sqrt{3}+1}{2}\right), \cos{A}\cos{B}=\dfrac{\sqrt{3}}{4}=\sin{A}\sin{B}$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

Option$\left(a\right)$ SInce
$\tan{A}+\tan{B}+\tan{C}=\tan{A}\tan{B}\tan{C}$
But here, $\tan{A}+\tan{B}+\tan{C}=0,$ is impossible.
Option$\left(b\right)$
$\dfrac{\sin{A}}{2}=\dfrac{\sin{B}}{3}=\dfrac{\sin{C}}{7}$
or $\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{7}$
$\Rightarrow \dfrac{a+b}{5}=\dfrac{c}{7}$
$\Rightarrow \dfrac{a+b}{c}=\dfrac{5}{7}<1$
$\therefore a+b<c$ is impossible.
Option$\left(c\right)$
${\left(a+b\right)}^{2}={c}^{2}+ab$
$\Rightarrow {a}^{2}+{b}^{2}+2ab={c}^{2}+ab$
$\Rightarrow {a}^{2}+{b}^{2}-{c}^{2}=-ab$
$\Rightarrow \dfrac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}=\dfrac{-1}{2}$
$\Rightarrow \cos{C}=\dfrac{-1}{2}$
$\therefore \angle{C}={120}^{0}$
and $\sqrt{2}\left(\sin{A}+\cos{A}\right)=\sqrt{3}$
$\Rightarrow \sqrt{2}\left[\sqrt{2}\left(\dfrac{1}{\sqrt{2}}\sin{A}+\dfrac{1}{\sqrt{2}}\cos{A}\right)\right]=\sqrt{3}$
We know that $\sin{\dfrac{\pi}{4}}=\cos{\dfrac{\pi}{4}}=\dfrac{1}{\sqrt{2}}$
$\Rightarrow 2\left[\sin{\left(A+\dfrac{\pi}{4}\right)}\right]=\sqrt{3}$
$\Rightarrow \left[\sin{\left(A+\dfrac{\pi}{4}\right)}\right]=\dfrac{\sqrt{3}}{2}$
$\Rightarrow \left[\sin{\left(A+\dfrac{\pi}{4}\right)}\right]=\sin{\dfrac{\pi}{3}}$
$\Rightarrow A+\dfrac{\pi}{4}=\dfrac{\pi}{3}$
$\therefore A=\dfrac{\pi}{3}-\dfrac{\pi}{4}=\dfrac{\pi}{12}$ is possible.
Option$\left(d\right)$
$\because \sin{A}+\sin{B}=\dfrac{\sqrt{3}+1}{2}$                 ................$\left(1\right)$
and $\cos{A}\cos{B}=\dfrac{\sqrt{3}}{4}=\sin{A}\sin{B}$
$\therefore \cos{A}\cos{B}-\sin{A}\sin{B}=\dfrac{\sqrt{3}}{4}-\dfrac{\sqrt{3}}{4}=0$
$\Rightarrow \cos{\left(A+B\right)}=0$
$\Rightarrow A+B=\dfrac{\pi}{2}$
$\therefore B=\dfrac{\pi}{2}-A$
From eqn$\left(1\right)$ 
$\sin{A}+\cos{A}=\dfrac{\sqrt{3}+1}{2}$
$\Rightarrow \sqrt{2}\left[\sin{A}\dfrac{1}{\sqrt{2}}+\cos{A}\dfrac{1}{\sqrt{2}}\right]=\dfrac{\sqrt{3}+1}{2}$
We know that $\sin{\dfrac{\pi}{4}}=\cos{\dfrac{\pi}{4}}=\dfrac{1}{\sqrt{2}}$
$\Rightarrow \sqrt{2}\left[\sin{A}\cos{\dfrac{\pi}{4}}+\cos{A}\sin{\dfrac{\pi}{4}}\right]=\dfrac{\sqrt{3}+1}{2}$
$\Rightarrow \sin{\left(A+\dfrac{\pi}{4}\right)}=\dfrac{\sqrt{3}+1}{2\sqrt{2}}$
$\Rightarrow \sin{\left(A+\dfrac{\pi}{4}\right)}=\sin{\dfrac{5\pi}{12}}$
$\Rightarrow A+\dfrac{\pi}{4}=\dfrac{5\pi}{12}$
$\Rightarrow A=\dfrac{5\pi}{12}-\dfrac{\pi}{4}=\dfrac{\pi}{6}$
$\therefore B=\dfrac{\pi}{2}-A=\dfrac{\pi}{2}-\dfrac{\pi}{6}=\dfrac{2\pi}{6}=\dfrac{\pi}{3}$
Thus, $\angle{C}=\dfrac{\pi}{2}$ is possible.