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 similarity areas of similar figures areas of similar triangles relations between the areas of triangles

The areas of two similar triangles are $121$ cm$^{2}$ and $64$ cm$^{2}$, respectively. If the median of the first triangle is $12.1$ cm, then the corresponding median of the other is:

  1. $6.4$ cm
  2. $10$ cm
  3. $8.8$ cm
  4. $3.2$ 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 ratio of the squares of the corresponding medians. Therefore,
$\displaystyle \frac{121}{64}=\frac{\left ( 12.1 \right )^{2}}{x^{2}},$ where $x$ is the median of the other $\triangle .$
$\Rightarrow $ $\displaystyle x^{2}=\frac{\left ( 12.1 \right )^{2}\times 64}{121}\Rightarrow x=\sqrt{\frac{121}{100}\times 64}$
   $\displaystyle =\frac{11}{10}\times 8=8.8$ cm.

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

If the sides of two similar triangles are in the ratio $1:7$, find the ratio of their areas.

  1. $7:1$
  2. $1:7$
  3. $1:49$
  4. $1:14$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that the relation between area of two similar triangle:
If two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides. 

Given, sides of two similar triangles are in the ratio $1:7$.
So, the ratio of their areas $= 1:49$.

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

The corresponding sides of two similar triangles are in the ratio $a : b$. What is the ratio of their areas?

  1. $a : b$
  2. $2a : 2b$
  3. $a^{2} : b^{2}$
  4. $\dfrac {1}{a} : \dfrac {1}{b}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given two triangles are similar, then the ratio of the areas $=a^2:b^2$

Eg: The ratio of the sides of a similar triangle is $4:9$
Scale factor for the sides of these triangles $k=\cfrac 49$
$\therefore$ Ratio of area will be:
$k^2=\cfrac {area of \triangle A}{area of \triangle B}=(\cfrac 49)^2=\cfrac {16}{81}$

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

The ratio of areas of two similar triangles is $81 : 49$. If the median of the smaller triangle is $4.9\ cm$, what is the median of the other?

  1. $4.9\ cm$
  2. $6.3\ cm$
  3. $7\ cm$
  4. $9\ cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Area of $\triangle ABC= \cfrac 12 \times base \times height$

In similar triangles, $\cfrac {base 1}{base 2}=\cfrac {height 1}{height 2}=\cfrac {side 1}{side}$
$\therefore \cfrac {Area 1}{Area 2}= (\cfrac {Median 1}{Median 2})^2$
Ratio of Medians $=\sqrt{\cfrac {81}{49}}=\cfrac 97 >1$ 
$\therefore$ Altitude of smaller triangle $=4.9 \times \cfrac 97=6.3$

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

In similar triangles $\triangle ABC$ and $\triangle FDE, DE = 4 cm, BC = 8 cm$ and area of $\triangle FDE = 25 cm^2$. What is the area of $\Delta ABC$?

  1. 144 cm$^2$
  2. 121 cm$^2$
  3. 100 cm$^2$
  4. 81 cm$^2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

  $DE=4\,cm,\,BC=8\,cm$ and $ar(\triangle FDE)=25\,cm^2$                  [ Given ]


$\Rightarrow$  $\triangle ABC\sim\triangle FDE$             [ Given ]

$\Rightarrow$  $\dfrac{ar(\triangle ABC)}{ar(\triangle FDE)}=\dfrac{(BC)^2}{(DE)^2}$                       [ By area of similar triangle theorem ]

$\Rightarrow$  $\dfrac{ar(\triangle ABC)}{25}=\dfrac{(8)^2}{(4)^2}$

$\Rightarrow$  $ar(\triangle ABC)=\dfrac{64}{16}\times 25$

$\Rightarrow$  $ar(\triangle ABC)=4\times 25$
$\therefore$  $ar(\triangle ABC)=100\,cm^2$

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

The areas of two similar triangles are $81\ cm^{2}$ and $49\ cm^{2}$. If the altitude of the bigger triangle is $4.5\ cm$, find the corresponding altitude of the smaller triangle.

  1. $3 cm$
  2. $2.5 cm$
  3. $4 cm$
  4. $3.5 cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given: Area of $two$ similar triangle $81{cm}^{2}$ and $49{cm}^{2}$

Altitude of bigger triangle $=4.5cm$
For similar triangle,
${\text{Ratio on sides}}^{2}=\text {Ratio of their Area}$
$\therefore$ $\cfrac { { 4.5 }^{ 2 } }{ { x }^{ 2 } } =\cfrac { 81 }{ 49 } $
$\cfrac { 45\times 45 }{ { x }^{ 2 }\times 100 } =\cfrac { 81 }{ 49 } $
$100{x}^{2}=25\times 49$
${x}^{2}=\cfrac{25\times 49}{100}$
${x}^{2}=\cfrac{49}{4}$
$x=\cfrac{7}{2}$
$x=3.5cm$

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

The areas of two similar triangles are 100 $cm^2$ and 64 $cm^2$. If the median of greater side of first triangle is 13 cm, find the corresponding median of the other triangle.

  1. 20 cm

  2. 15 cm

  3. 10 cm

  4. 16 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given area of two similar triangles are $100$ sq cm and $64$ sq cm
The areas of two Similar-Triangles are in the ratio of the squares of the corresponding medians
The ratio of area of triangle $=\dfrac{100}{64}=\dfrac{25}{16}$
Median of greater triangle is $13$ cm and let other median is $x$ cm
$\therefore \dfrac{(13)^{2}}{(x)^{2}}=\dfrac{25}{16}$

$\Rightarrow \dfrac{169}{x^{2}}=\dfrac{25}{16}$

$\Rightarrow 25x^{2}=169\times 16$

$\Rightarrow x^{2}=\dfrac{2704}{25}=108.16$

$\Rightarrow x=10cm$

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

Triangles ABC and DEF are similar. If their areas are 64 $cm^2$ and 49 $cm^2$ and if AB is 7 cm, then find the value of DE.

  1. 8 cm
  2. $\dfrac{49}{8}$ cm
  3. $\dfrac{8}{49}$ cm
  4. $\dfrac{64}{7}$cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\Delta ABC \Delta DEF$
$\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF}$.
We know that,
$\dfrac{Area of \Delta  ABC}{ Area of \Delta  DEF} = $ $\Rightarrow \dfrac{64}{49} = \left ( \dfrac{7}{DE} \right )^2$
$\Rightarrow \left ( \dfrac{8}{7} \right )^2 = \left ( \frac{7}{DE} \right )^2 \Rightarrow \left ( \dfrac{8}{7} \right )^2 = \left ( \dfrac{7}{DE} \right )^2 = \dfrac{49}{8}$cm 

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

The area of two similar triangles are $200$ and $128$, then the ratio of their corresponding altitude is __________

  1. $25:16$
  2. $5:4$
  3. $4:5$
  4. $16:25$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since we know that ratio of areas of two similar triangles is equal to the square of the ratio of their altitude
therefore
Ratio of their altitude=$\sqrt {\dfrac{{200}}{{128}}} $
$ = \sqrt {\dfrac{{100}}{{64}}} $
$ = \dfrac{{10}}{8}$
$ = \dfrac{5}{4}$
$ = 5:4$

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

Intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre.

  1. True

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

For any two parallel tangents to a circle, the segment of a third tangent intercepted between them subtends a 90-degree angle at the center because the radii to the points of tangency are perpendicular to the tangents.

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

The chord of contact of the pair of tangents to the circle $x^2+y^2=1$ drawn from any point on the line $2x+y=4$ passes through a fixed point. 

  1. True

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

If chords are drawn to the circle from a fixed point $(x _1,y _1)$ and then tangents are drawn at point of contact, the point of intersection of all tangents lie on a fixed point.


The fixed point is called pole and fixed line is called polar.


Equation of polar is $T=0$.

$C:x^2+y^2-1=0$

Equation of polar is $T=0$.

$xx _1+yy _1-1=0$

The line is identical to given line $2x+y-4=0$.

By comparing coefficients, we get,
$\dfrac{x _1}{2}=\dfrac{y _1}{1}=\dfrac{-1}{-4}$

$x _1=\dfrac{1}{2},y _1=\dfrac{1}{4}$

Hence, the fixed point is $(\dfrac{1}{2}, \dfrac{1}{4})$.

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

From a point $P$ which is at a distance of $13$ cm from the centre $O$ of a circle of radius $5$ cm, the pair of tangents $PQ$ and $PR$ to the circle are drawn. Then the area of the quadrilateral $PQOR$ is:

  1. $60$ cm$^{2}$
  2. $65$ cm$^{2}$
  3. $30$ cm$^{2}$
  4. $32.5$ cm$^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The radius perpendicular tangent at the pt. of contact, therefore, $OQ\perp PQ$ and $OR\perp PR$
In rt. $\triangle OPQ$, we have
$PQ=\sqrt{OP^{2}-OQ^{2}}$
   $=\sqrt{169-25}=\sqrt{144}=12$ cm
$\Rightarrow $ $PR=12$ cm (Two tangents from the same external pt. to a circle are equal)
Now area of quad. $PQOR=2\times $Area of $\triangle POQ$
   $\displaystyle =\left ( 2\times \frac{1}{2}\times 12\times 5 \right )$ cm$^{2}=60$ cm$^{2}$

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

Circles ${ C } _{ 1 },{ C } _{ 2 },{ C } _{ 3 }$ have their centres at $\left( 0,0 \right) ,\left( 12,0 \right) ,\left( 24,0 \right) $ and have radii $1,2$ and $4$ respectively. Line ${t} _{1}$ is a common internal tangent to ${C} _{1}$ and ${C} _{2}$ and has a positive slope and line ${t} _{2}$ is a common internal tangent to ${C} _{2}$ and ${C} _{3}$ and has a negative slope. Given that lines ${t} _{1}$ and ${t} _{2}$ intersect at $(x,y)$ and that $x=p-q\surd r$, where $p,q$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p+q+r$.

  1. $p+q+r=26$
  2. $p+q+r=24$
  3. $p+q+r=28$
  4. $p+q+r=27$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

From a point outside a circle, one tangent and one secant are drawn. The length of exterior part of secant is $7$ cm and that of interior part is $9$ cm. Find the length of tangent segment.

  1. $10.6$ cm
  2. $10.9$ cm
  3. $11.2$ cm
  4. $11.6$ cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let length of tangent be $l$

Length of exterior part of secant $=m=7 $ cm
Length of interior part of secant $=n=9 $ cm
Now using the secant intersection theorem, we have
${ l }^{ 2 }=m(m+n)\ \Rightarrow { l }^{ 2 }=7(7+9)\ \Rightarrow { l }^{ 2 }=112 $
$\Rightarrow l=\sqrt { 112 } =10.6$ cm
Option A is correct.

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

Draw a circle of radius 4 cm. Construct a pair of tangents to it, the angle between which is $60^0$. Also justify the construction. Measure the distance between the centre of the circle and the point of intersection of tangents.

  1. 4 cm

  2. 6 cm

  3. 8 cm

  4. 10 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Join O and P.
Now, in triangles OPQ and OPR,
OP = OP (Common)
OQ = OR (radius of circle)
PQ = PR (tangents from single point)

Hence OPQ and OPR are congruent triangles.
$\angle OPQ = \angle OPR = 30^{\circ}$


Thus in triangle OPQ, $\dfrac{OQ}{OP} = Sin 30$

OP = $\dfrac{4}{Sin30}$

OP = 8 cm