Quantitative Aptitude

Geometry

1,950 Questions
All Circle Tangents and Chords 772 Coordinate Geometry 360 Circle and Arc Measures 216 Triangle Centers and Medians 175 Composite and Inscribed Shapes 174 Similarity, Congruence, and Proportionality 118 Right Triangles and Pythagorean Theorem 100 Standard 2D Mensuration 22 Polygon and Angle Sums 10 3D Solid Mensuration 3

Geometry Questions

Multiple choice maths constructions mid-point formula midpoints division of a line segment

The locus of mid points of chords to the circle $x^{2}+y^{2}-8x+6y+20=0$ which are parallel to the line $3x+4y+5=0$ 

  1. $3x+4y-25=0$
  2. $4x+3y+5=0$
  3. $4x-3y-25=0$
  4. $4x-3y+25=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The locus of the midpoints of chords of a circle parallel to a given line is a diameter of the circle perpendicular to the given line or parallel to the normal direction. Given the circle equation and the parallel line 3x + 4y + 5 = 0, calculating the center and finding the line passing through it with the perpendicular slope yields 4x - 3y - 25 = 0.

Multiple choice maths constructions mid-point formula midpoints division of a line segment

The locus of the middle points of chords of length $4$ on the circle $x^ {2}+y^ {2}=16$

  1. A straight line

  2. A circle of radius
  3. A circle of radius $2\sqrt {3}$

  4. An ellipse
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a circle of radius R = 4, the distance d of a chord of length L = 4 from the center is given by d = sqrt(R^2 - (L/2)^2). Here R = 4 and L/2 = 2, so d = sqrt(16 - 4) = sqrt(12) = 2(sqrt(3)). The locus of the midpoints of chords of constant length is a concentric circle whose radius is equal to this perpendicular distance from the center, which is 2(sqrt(3)).

Multiple choice maths properties of parallel lines and their transversal introduction to shapes similarity of triangles introduction to similar triangles

Ratio of two corresponding sides of two similar triangles is $4:9$. Then ratio of their area is ___.

  1. $\dfrac{16} {81}$
  2. $\dfrac{34} {81}$
  3. $\dfrac{81} {16}$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Ratio of areas of two similar triangles is equal to the squares of the ratio of their sides.

Ratio of sides $=\dfrac{4}{9}$
Ratio of areas $=\left( \dfrac { 4 }{ 9 }  \right) ^{ 2 }=\dfrac { 16 }{ 81 } $

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

The perimeter of two similar triangles are $24$ cm and $16$ cm, respectively. If one side of the first triangle is $10$ cm, then the corresponding side of the second triangle is

  1. $9$ cm
  2. $\dfrac{20}3$ cm
  3. $\dfrac{16}3$ cm
  4. $5$ cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
In similar triangles, ratio of the sides is equal to the ratio of the perimeters.
Thus, $\dfrac{p _1}{p _2} = \dfrac{s _1}{s _2}$
$\dfrac{24}{16} = \dfrac{10}{s _2}$
$s _2 = \dfrac{20}{3}$
Thus, side of the other triangle is $\dfrac{20}{3}$ cm.
Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

The area of two similar triangles $\displaystyle \Delta ABC$ and $\displaystyle \Delta DEF$ are 144 $\displaystyle cm^{2}$ and 81 $\displaystyle cm^{2}$ respectively If the longest side of larger $\displaystyle \Delta ABC$ be 36 cm then the longest side of the smaller triangle $\displaystyle \Delta DEF$ is

  1. 20 cm

  2. 26 cm

  3. 27 cm

  4. 30 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In similar triangle ABC & DBF
$\frac{AB}{De}=\frac{BC}{EF}=\frac{AC}{DF}=\frac{ratio ofArea of triangleABC}{ratio ofArea of triangleDEF}$ 
THEN $\frac{9}{12}=\frac{x}{36}$  (where x is longest side of the smaller triangle )
So x=27 cm

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

The perimeters of two similar triangles are $25\;cm$ and $15\;cm$ respectively. If one side of first triangle is $9\;cm$, then the corresponding side of the other triangle is

  1. $6.2\;cm$
  2. $3.4\;cm$
  3. $5.4\;cm$
  4. $8.4\;cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$The\quad perimeter\quad of\quad triangle\quad is\quad 25cm\quad and\quad 15cm.\ The\quad ratio\quad of\quad Perimeter\quad of\quad triangle\quad is\quad 25:15=5:3\ The\quad first\quad side\quad is\quad 9cm\quad ,let\quad the\quad other\quad side=x\ Hence,\quad \dfrac { 9 }{ x } =\dfrac { 5 }{ 3\  } \ \Rightarrow x=\dfrac { 3\times 9 }{ 5 } =\dfrac { 27 }{ 5 } =5.4\quad cm$

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

Two $\triangle sABC $ and DEF are similar. If $ar(DEF)= 243\ cm^2, ar(ABC)=108\ cm^2$ and $BC= 6\ cm$. Find $EF$.

  1. $9$
  2. $81$
  3. $3$
  4. $72$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given:-
$\triangle{ABC} \simeq \triangle{DEF}$
$ar \left( DEF \right) = 243 {cm}^{2}$
$ar \left( ABC \right) = 108 {cm}^{2}$
$BC = 6 cm$

To Find:- $EF = ?$

As we know that,
$\because \; \triangle{ABC} \simeq \triangle{DEF}$

$\cfrac{ar \left( \triangle{ABC} \right)}{ar \left( \triangle{DEF} \right)} = {\left( \cfrac{AB}{DE} \right)}^{2} = {\left( \cfrac{BC}{EF} \right)}^{2} = {\left( \cfrac{AC}{DF} \right)}^{2}$

$\therefore \; \cfrac{ar \left( \triangle{ABC} \right)}{ar \left( \triangle{DEF} \right)} = {\left( \cfrac{BC}{EF} \right)}^{2}$

$\Rightarrow \; \cfrac{108}{243} = \cfrac{{6}^{2}}{{EF}^{2}}$

$\Rightarrow \; {EF}^{2} = \cfrac{243}{108} \times 36$

$\Rightarrow \; EF = \sqrt{81}$

$\Rightarrow \; EF = 9$

Hence, the correct answer is $9$.
Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

The areas of two similar triangles $\triangle{ABC}$ and $\triangle{DEF}$ are $144\ cm^{2}$ and $81\ cm^{2}$ respectively. If the longest side of larger $\triangle{ABC}$ be $36\ cm$, then, the largest side of the similar triangle $\triangle{DEF}$ is

  1. $20\ cm$
  2. $26\ cm$
  3. $27\ cm$
  4. $30\ cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Therefore, the ratio of their sides is the square root of the ratio of their areas, which is sqrt(144/81) = 12/9 = 4/3. Setting up the proportion 36/x = 4/3 yields x = 27 cm for the smaller triangle's corresponding side.

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

The perimeter of two similar triangle are $30\ cm$ and $20\ cm$. If one side of first triangle is $12\ cm$ determine the corresponding side of second triangle.

  1. $8\ cm$
  2. $4\ cm$
  3. $3\ cm$
  4. $16\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the two similar triangles be $\triangle ABC$ and $\triangle DEF$

$\therefore \dfrac {AB}{DE} = \dfrac {BC}{EF} = \dfrac {AC}{DF} = \dfrac {P _{1}}{P _{2}}$

$\Rightarrow \dfrac {AB}{DE} = \dfrac {P _{1}}{P _{2}}$

$\Rightarrow \dfrac {12}{DE} = \dfrac {30}{20}$

$\Rightarrow DE = 8\ cm$

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

The perimeter of two similar triangles $\triangle ABC$ and $\triangle DEF$ are $36$ cm and $24$ cm respectively. If $DE=10 $ cm, then $AB$ is :

  1. $12$ cm
  2. $20$ cm
  3. $15$ cm
  4. $18$ cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given that triangles $ABC$ and $DEF$ are similar.

Also given, $DE=10$ cm and perimeters of triangles $ABC$ and $DEF$ are $36$ cm and $24$ cm.
So, the corresponding sides of the two triangles is equal to the ratio of their perimeters.

Hence, $\dfrac {\text{perimeter of} \ ABC}{ \text{perimeter of } \ DEF}$ $=\dfrac {AB}{DE}$
Therefore, $\dfrac {36}{24}=\dfrac {AB}{10}$ 
$\Rightarrow AB=\dfrac {36\times 10}{24}$
$\Rightarrow AB=15$ cm

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The number of normals to the ellipse $\dfrac { { x }^{ 2 } }{ 25 } +\dfrac { { y }^{ 2 } }{ 16 } =1$ which are tangents to the circle ${ x }^{ 2 }+{ y }^{ 2 }=9$ is

  1. 1

  2. 2

  3. 3

  4. 0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A normal to an ellipse is tangent to a circle if the distance from the center to the normal equals the radius. For the given ellipse and circle, there is only one such normal.

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The number of tangents to the circle ${x}^{2}+{y}^{2}=3$ that are normals to the ellipse $\cfrac{{x}^{2}}{9}+\cfrac{{y}^{2}}{4}$ is

  1. one

  2. two

  3. three

  4. zero
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let $y=mx+c$ is tangent to $x^2+y^2=3$ 
Then by condition of tangency
$\left|\dfrac{c}{\sqrt{m^2+1}}\right|=\sqrt{3}$
$\Rightarrow c^2=3(m^2+1)$        ...(i)
$y=mx+c$ is Normal to $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$
in $c=\dfrac{(b^2-a^2)m}{\sqrt{a^2+b^2m^2}}$

$\Rightarrow c=\dfrac{(4-9)m}{\sqrt{9+4m^2}}$

$\Rightarrow c^2=\dfrac{25m^2}{4m^2+9}$

$\Rightarrow 3(m^2+1)(4m^2+9)=25m^2$
let $m^2=t$
$\Rightarrow 3(t+1)(4t+9)=25t$
$\Rightarrow$ since, $D<0$.
Hence, There is no rout:
Hence, there is no tangent to circle which is Normal to ellipse.