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 maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

In a $\Delta ABC,\,AB=AC=2.5\;cm,\,BC=4\;cm$. Find its height from $A$ to the opposite base.

  1. $1.5\;cm$
  2. $1\;cm$
  3. $2\;cm$
  4. $3\;cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$In\triangle ABC $


In order to find height we need to consider that $  AD\bot BC$

$ Hence\quad in\quad right\quad angled \ \triangle ADC$

$ { AC }^{ 2 }={ AD }^{ 2 }+{ DC }^{ 2 }(Phythagoreas\quad Theorm)$

$ \Rightarrow { AD }^{ 2 }={ AC }^{ 2 }-{ DC }^{ 2 }$

$ \Rightarrow { AD }^{ 2 }={ (2.5) }^{ 2 }-({ 2) }^{ 2 }$

$ \Rightarrow { AD }^{ 2 }=6.25-4$

$ \Rightarrow { AD }^{ 2 }=2.25$

$ \Rightarrow AD=1.5$

$ Hence\quad option\quad (A)\quad is\quad right\quad answer$

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

If the sides of a right angled triangle are $x, 3x + 3$  and $3x + 4$, then $x$ is equal to:

  1. $-1$
  2. $7$
  3. $6$

  4. Both A and B
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

As the sides are of a right angled triangle, we have
$ {Hypotenuse}^{2} = {Side1}^{2} + {Side2}^{2} $ where hypotenuse is the largest side.
$ {(3x+4)}^{2}  = {(3x+3)}^{2} + {x}^{2} $
$ => 9{x}^{2} + 16 + 24x = 9{x}^{2} + 9 + 18x + {x}^{2}  $
$ => {x}^{2} - 6x - 7 = 0 $
$ => {x}^{2} - 7x + x - 7 = 0 $
$ => x(x-7) + (x-7) = 0 $
$ => (x-7)(x+1) = 0 $
$ => x = 7, -1 $
As the side cannot be negative, $ x = 7 $.

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

In a field of shape of a right angled triangle, the farmer wants to measure the $3$ sides but being a huge field, he was only able to measure $2$ sides, $1$ side of which was $6$ km and other was $8$ km. Can you find the length of $3^{rd}$ side for him?

  1. $10$ km
  2. $8$ km
  3. $14$ km
  4. $13$ km
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The field is in the shape of a right angled triangle.

Using Pythagoras theorem,
$6^2 + 8^2 = \mbox{(3rd side)}^2$
$\mbox{(3rd side)}^2 = 36 + 64$
$\mbox{(3rd side)}^2 = 100$
$\therefore \mbox{3rd side} = 10$ km
So, option A is correct.

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

Which of the following can't be  the lengths of the sides of a right-angled triangle?

  1. $5$ inches, $12$ inches, $13$ inches
  2. $\displaystyle\frac{1}{3}$ of a foot, $\displaystyle\frac{1}{4}$ of a foot, $\displaystyle\frac{1}{5}$ of a foot
  3. $9$cm, $40$cm, $41$cm
  4. $\displaystyle\frac{3}{4}$ of a foot, $1$ foot, $15$ inches
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a set of numbers to be sides of a right angled triangle, the square of the longest side must equal the sum of the squares of the other two sides.

In option A, $13^2 = 169, 5^2 + 12^2 = 169$ so possible.
In option B, $\left (\dfrac{1}{3}\right)^2 = \dfrac{1}{9}, \left (\dfrac{1}{4}\right)^2 + \left (\dfrac{1}{5}\right)^2 = \dfrac{41}{400}$
Since, these two values are not equal, this set cannot form a right angled triangle.

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

The ratio of the two legs of a right-angled triangle is $3:1$. If the lengths of the legs are whole numbers, what can be the possible value of the hypotenuse?

  1. $\sqrt{40}$
  2. $\sqrt{47}$
  3. $\sqrt{55}$
  4. $\sqrt{63}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The legs of a right angled triangle are in the ratio $3:1$ and they are whole numbers.

The possibilities are as listed below:
$3,1$ - The hypotenuse becomes $\sqrt{(3)^2 + (1)^2} = \sqrt{10}$
$6,2$ - The hypotenuse becomes $\sqrt{(6)^2 + (2)^2} = \sqrt{40}$
$9,3$ - The hypotenuse becomes $\sqrt{(9)^2 + (3)^2} = \sqrt{90}$
Thus, the hypotenuse will always be square root of a multiple of $10$, which is in option A.

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

$\angle B$ in $\triangle ABC$ and $\angle S$ in $\triangle RST$ are right angles. The lengths of sides $AC$ and $RT$ are equal. Determine the relation between the following.

A: The length of side $AB$.
B: The length of side $RS$.

  1. The quantity in statement A is greater than B.

  2. The quantity in statement B is greater than A.

  3. The two quantities are equal.

  4. The relationship cannot be determined from the given information.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In $\triangle ABC$ and $\triangle RST$,

hypt. $AC=$ hypt. $RT$

$\therefore AC^2 = RT^2$

$\therefore { AB }^{ 2 }+{ BC }^{ 2 }={ RS }^{ 2 }+{ ST }^{ 2 }$.

which does not imply that $AB=RS$.
So, the relationship cannot be determined.

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

In $\triangle ABC,\angle ABC={ 90 }^{ o }$. If $AC=(x+y)$ and $BC=(x-y)$, then the length of $AB$ is:

  1. ${ x }^{ 2 }-{ y }^{ 2 }$
  2. $2xy$
  3. $2\sqrt { xy } $
  4. ${ x }^{ 2 }+{ y }^{ 2 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, $\angle ABC=90^{\circ},$

Also, $AC=(x+y)$ and $BC=(x-y)$
We need to find the length of $AB$.
Now as $\angle ABC=90^{\circ},$ at $B$ we can use Pythagoras theorem
$\therefore$  $AB^2+BC^2=AC^2$
$\implies$  $AB^2=AC^2-BC^2$
$\implies$  $AB^2=(x+y)^2-(x-y)^2$
Using the formula for $(a+b)^2$ and $(a-b)^2$
$\implies$ $AB^2=x^2+2xy+y^2-(x^2-2xy+y^2)$
By cancelling the like terms we get,
$AB^2=4xy$
Taking square root on both the sides we get,
$AB=2\sqrt{xy}$.
Hence, the answer is C,

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

In a triangle $ABC$ with $\angle A = 90^o$, $P$ is a point on $BC$ such that $PA : PB = 3:4$. If $AB=\sqrt{7}$ and $AC=\sqrt{5}$, then $BP:PC$ is 

  1. $2:1$
  2. $4:3$
  3. $4:5$
  4. $8:7$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In right triangle ABC, AB^2 + AC^2 = BC^2. 7 + 5 = 12, so BC = sqrt(12) = 2*sqrt(3). P is on BC such that PA/PB = 3/4. This is a geometric problem that requires calculating the position of P. Based on the ratio, the answer is 2:1.

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

Triangle $ABC$ is right angled at $A$. The points $P$ and $Q$ are on the hypotenuse $BC$ such that $BP = PQ = QC$.
If $AP = 3$ and $AQ = 4$, then the length $BC$ is equal to

  1. $\sqrt { 27 } $
  2. $\sqrt { 36 } $
  3. $\sqrt { 45 } $
  4. $\sqrt { 54 } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$BP=PQ=QC=x(let)$

$ AP=3;AQ=4$
$ In\triangle AQB$
$ AP\quad is\quad median\quad by\quad Apollonius\quad Thm$
$ { AB }^{ 2 }+{ AQ }^{ 2 }=2({ AP }^{ 2 }+{ PQ }^{ 2 })$
$\implies\quad { AB }^{ 2 }+16=2(9+{ x }^{ 2 })$
$\implies\quad { AB }^{ 2 }=2{ x }^{ 2 }+2\quad -(1)$
$ Similarly,in\triangle APC$
$ AQ\quad is\quad median$
$ So,$
$ { AC }^{ 2 }+{ AP }^{ 2 }=2({ AQ }^{ 2 }+{ QC }^{ 2 })$
$ \therefore { AC }^{ 2 }+9=2(16+{ x }^{ 2 })$
$\implies\quad { AC }^{ 2 }=2{ x }^{ 2 }+23\quad -(2)$
$ (1)+(2)$
$ { AB }^{ 2 }+{ AC }^{ 2 }=4{ x }^{ 2 }+25$
$\implies\quad { BC }^{ 2 }=4{ x }^{ 2 }+25\quad [In\triangle ABC,using\quad pythagoras\quad thm]$
$\implies\quad { (3x) }^{ 2 }=4{ x }^{ 2 }+25$
$\implies\quad 9{ x }^{ 2 }=4{ x }^{ 2 }+25$
$\implies\quad x=\sqrt { 5 } $
BC=3x=$\sqrt { 45 } $

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

Given that in a right angled triangle the length of two sides are 11 and 60. Find the perimeter of the triangle.

  1. $132$
  2. $145$
  3. $89$
  4. $200$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\begin{array}{l} { 11^{ 2 } }+{ 60^{ 2 } }={ 61^{ 2 } } \\ \Rightarrow Perimeter\, \, of\, \, \Delta =132\, \, cm \\ \left\{ { \because sides\, \, 11,60\, \, \& \, \, 61 } \right\}  \end{array}$
Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

In a right angled triangle the hypotenuse is  $2\sqrt{2}$ times the length of the perpendicular drawn from the opposite vertex on the hypotenuse. The the other two angles are

  1. $\left( \dfrac { \pi }{ 3 } ,\dfrac { \pi }{ 6 } \right)$
  2. $\left( \dfrac { \pi }{ 4 } ,\dfrac { \pi }{ 4 } \right)$
  3. $\left( \dfrac { \pi }{ 8} ,\dfrac { 3\pi }{ 8 } \right)$
  4. $\left( \dfrac { \pi }{ 12 } ,\dfrac { 5\pi }{ 12 } \right)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the altitude to the hypotenuse be h. Hypotenuse c = 2*sqrt(2)*h. In a right triangle, h = (a*b)/c. Also, h = c*sin(A)*cos(A). So 2*sqrt(2)*h = c, meaning sin(A)cos(A) = 1/(2*sqrt(2)). This leads to the angles pi/8 and 3pi/8.

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

In a right angled triangle, the square of the hypotenuse is equal to twice the product of the other two sides. One of the acute angles of the triangle is:

  1. $40^0$
  2. $42^0$
  3. $44^0$
  4. $45^0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $\triangle ABC$ be right angled at B.

Then, by Pythagoras theorem,

$AC^2 = AB^2+BC^2$

But $AC^2 = 2 \times AB . BC$     .....Given

Hence, $AB^2 + BC^2 = 2 \times AB. BC$

$AB^2 + BC^2 - 2 \times AB. BC = 0$

$(AB - BC)^2 = 0$

So, $AB = BC$

Therefore, $\triangle ABC$ is right isosceles triangle.

So, $\angle A = \angle C = 45^o$.

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

Diagonals $\overline{AC}$ and $\overline{BD}$ of quadrilateral $ABCD$ are perpendicular. $AD=DC=8, AC=BC=6, m\angle ADC = 60^o$. The area of $ABCD$ is

  1. $4\sqrt{5}+8\sqrt{3}$
  2. $16\sqrt{3}$
  3. $32\sqrt{3}$
  4. $8\sqrt{5}+16\sqrt{3}$
  5. $48$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The altitude to the base of an isosceles triangle also bisects the vertex angle, so $m\angle ADE=30$.

With the hypotenuse of the triangle having a length of $8$, $AE=4$ and $DE=4\sqrt { 3 }$.
$\triangle AEC$ is a right angle with leg $4$ and hypotenuse $6$.
Use the pythagorean theorem to determine that 
$BE=\sqrt { 6^{ 2 }-4^{ 2 } } =\sqrt { 36-16 } =\sqrt { 20 } =2\sqrt { 5 }$
The area of the quadrilateral with perpendicular diagonals is equal to half the product of the diagonals, so the area of $ABCD$ is:
$A=\dfrac { 1 }{ 2 } \times 8\times (2\sqrt { 5 } +4\sqrt { 3 } )=4(2\sqrt { 5 } +4\sqrt { 3 } )=8\sqrt { 5 } +16\sqrt { 3 }$  

Multiple choice physics trigonometrical ratios angles and sides naming the sides in a right angled triangle angle and their measurement

The area of the semicircle drawn on the hypotenuse of a right angled triangle is equal to the difference of the areas of the semicircles drawn on the other two sides of the triangles.

  1. True

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

The Pythagorean theorem states that a^2 + b^2 = c^2. The area of a semicircle is (pi * r^2) / 2. For a triangle with sides a, b, and hypotenuse c, the areas are (pi * (a/2)^2) / 2, (pi * (b/2)^2) / 2, and (pi * (c/2)^2) / 2. Since a^2 + b^2 = c^2, the sum of the areas of the semicircles on the legs equals the area of the semicircle on the hypotenuse, not the difference.