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 $\triangle ABC$ right angled at $B, AB=5\ cm$ and $\angle ACB=30^{o}$ then the length of the sides $BC$ is

  1. $5\sqrt {3}$
  2. $2\sqrt {3}$
  3. $10\ cm$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given:- $ABC$ is a right angled triangle in which $AB = 5 \; cm$ and 


$\angle{ACB} = 30°$

Using trigonometric ratio,

$\tan{C} = \cfrac{AB}{BC}$

$\Rightarrow \tan{30°} = \cfrac{5}{BC}$

$\Rightarrow BC = 5 \sqrt{3} \; cm$

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

ABC is a triangle, right-angled at B. M is a point on BC. Hence,
$AM^{2}\, +\, BC^{2}\, =\, AD^{2}\, +\, BM^{2}$
State true or false.

  1. True

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

Given, $\angle ABC = 90$, M is a point of BC
In $\triangle ABM$, 
$AB^2 + BM^2 = AM^2$ (Pythagoras theorem)
$AB^2 = AM^2 - BM^2$ (I)

In $\triangle ABC$,
$AB^2 + BC^2 = AC^2$ (Pythagoras Theorem)
$AB^2 = AC^2 - BC^2$ (II)

Equating I and II,
$AM^2 - BM^2 = AC^2 - BC^2$
thus, $AM^2 + BC^2 = AC^2 + BM^2$

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

A person wishes to fit three rods together in the shape of a right-angled triangle so that the hypotenuse is to be $4:cm$ longer than the base and $8:cm$ longer than the altitude. The lengths of the rods are:

  1. $3\:cm$, $4\:cm$, $5\:cm$
  2. $1.5\:cm$, $2\:cm$, $2.5\:cm$
  3. $6\:cm$, $8\:cm$, $10\:cm$
  4. $12\:cm$, $16\:cm$, $20\:cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the altitude$=x:cm$
$\therefore$ The Base$=(x+4):cm$
and the Hypotenuse$=(x+8):cm$

Using Pythtegores Theorem
$(x+8)^2=(x+4)^2+x^2$
$x^2-8x-48=0$
$(x-12)(x+4)=0$
$x=12$
$\therefore$ The sides are $12:cm$, $16:cm$, $20:cm$ 

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

What is the value of the hypotenuse of a right triangle whose sides are $12$ and $18$?

  1. $4.24$
  2. $3.46$
  3. $2.16$
  4. $21.63$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the sides be $a=12$ and $b=18$

Le the hypotenuse be $c$
Using Pythagoras theorem
${ c }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }\ { c }^{ 2 }={ (12) }^{ 2 }+{ (18) }^{ 2 }\ { c }^{ 2 }=144+324\ { c }^{ 2 }=468\ c=\sqrt { 468 } \ c=21.63$

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

A triangle whose lengths of sides are $5$ cm, $12$ cm and $13$ cm. The triangle is ____________.

  1. Obtuse-angled triangle

  2. Acute-angled triangle

  3. Right-angled triangle

  4. Triangle is not formed
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given sides of triangle are $5$ cm, $12$ cm, $13$ cm
Now applying pythagorus theorem:
$h^{2}= b^{2}+P^{2}$
$\Rightarrow 13^{2}= 12^{2}+5^{2}$
$ \Rightarrow 169 =144+25$
$ \Rightarrow 169 =169$
These three sides clearly satisfy ptyhagorous theorem.
Therefore, the triangle is 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

In any triangle $ABC$,  $AB^{2} + AC^{2} = 3 (AO^{2} + OC^{2})$.
where $O$ is mid-point of $BC$.

  1. True

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

By Apollonius theorem, AB^2 + AC^2 = 2(AO^2 + BO^2). Since O is the midpoint of BC, BO = OC. Thus, AB^2 + AC^2 = 2(AO^2 + OC^2). The given equation 3(AO^2 + OC^2) is incorrect.

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 perpendicular from A on side BC of a $\Delta ABC$ intersects BC at D such that $DB = 3CD$, then $2A{B^2} = A{C^2} + B{C^2}$.

  1. True

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

Let AD be the altitude. In right triangles ABD and ACD, AB^2 = AD^2 + BD^2 and AC^2 = AD^2 + CD^2. Given BD = 3CD, then AB^2 = AD^2 + 9CD^2 and AC^2 = AD^2 + CD^2. Subtracting gives AB^2 - AC^2 = 8CD^2. This does not lead to the provided equation.

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

State true or false
In a triangle ABC, P and Q are the id-points of AC and AB respectively and angle $BAC=90^o$.
then $BP^2 + CQ^2=5PQ^2$

  1. True

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

In $\triangle BAP$,
$BP^2 = AB^2+ AP^2$ (Pythagoras Theorem)

Since, Q is the mid-point of AB, then $AB = 2 AQ$
Hence, $BP^2 = (2 AQ)^2 + AP^2$
$BP^2 = 4 AQ^2 + AP^2$...(I)

In $triangle QAC$,
$QC^2 = AQ^2 + AC^2$ (Pythagoras theorem)
Since, P is the mid point of AC, $AC = 2 AP$
Hence, $QC^2 = AQ^2 + (2 AP)^2$ 
$QC^2 = AQ^2 + 4 AP^2$...(II)

By Adding I and II
$BP^2 + QC^2 = 4 AQ^2 + AP^2 + AQ^2 + 4 AP^2$
$BP^2 + QC^2 = 5 (AP^2 +AQ^2)$

In $\triangle QAP$,
$AQ^2 + AP^2 = PQ^2$
Hence, $BP^2 + QC^2 = 5 PQ^2$

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 quadrilateral ABCD, $\angle B\, =\, 90^{\circ}$ and $\angle D\, =\, 90^{o}$. Then:

  1. $2{AC}^{2}\, -\, {AB}^{2}\, =\, {BC}^{2}\, +\, {CD}^{2}\, +\, {DA}^{2}$
  2. $2{AC}^{2}\, =\, {BC}^{2}\, +\, {CD}^{2}\, +\, {DA}^{2}$
  3. $2{AC}^{2}\, -\, 2 {AB}^{2}\, =\, {BC}^{2}\, +\, {CD}^{2}\, +\, {DA}^{2}$
  4. $2{AC}^{2}\, =\, 2{BC}^{2}\, +\, {CD}^{2}\, +\, {DA}^{2} + {BC}^2 $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In quadrilateral $ABCD$, 

$\angle B = \angle D = 90^{\circ}$
Now, join $AC$
In $\triangle ABC$,
${AC}^2 = {AB}^2 + {BC}^2$
In $\triangle ADC$,
${AC}^2 = {AD}^2 + {CD}^2$
Add both the equations:
$2{AC}^2 = {AB}^2 + {BC}^2 + {AD}^2 + {CD}^2$
Hence, 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

A grassy land in the shape of a right angled triangle has its hypotenuse $1$ metre more than twice the shortest side. If the third side is $7$ metres more than the shortest side. The sides of the grassy land are:

  1. $8$m, $17$m, $15$m
  2. $2$m, $16$m, $13$m
  3. $10$m, $4$m, $5$m
  4. $7$m, $10$m, $14$m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the length of the shortest side be $x$ meters. Then,
Hypotenuse = $(2x +1)$ meters, Third side = $(x + 7)$ meters
(Hyppotenuse)$^2$ = sum of the square of the remaining two sides    ....[By pythagorous theroem]
$\Rightarrow (2x + 1)^2 = x^2 + (x + 7)^2$
$\Rightarrow 4x^2 + 4x + 1 = 2x^2 + 14x + 49$
$\Rightarrow 2x^2 - 10x - 48 = 0$
$\Rightarrow x^2 - 5x - 24 = 0$
$\Rightarrow x^2 - 8x + 3x - 24 = 0$
$\Rightarrow x(x - 8) + 3 (x - 8) = 0$
$\Rightarrow (x - 8) (x + 3) = 0$
$\Rightarrow x = 8, -3$
$\Rightarrow x = 8$             [$\because x = -3$ is not possible]
Hence, the lengths of the sides of the grassy land are $8$ m., $17$ m. and $15$ m.

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 hypotenuse of a right angled triangle is $25$cm. The other two sides are such that one is $5$cm longer than the other. Their lengths (in cm) are:

  1. $10, 15$
  2. $20, 25$
  3. $15, 20$
  4. $25, 30$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given: Hypotenuse of right triangle $= 25$ cm
Let the other sides be $x$ and $x + 5$
Thus, By Pythagoras Theorem:
$25^2 = x^2 + (x + 5)^2$
$625 = x^2 + x^2 + 25 + 10x$
$2x^2 + 10x - 600 = 0 $
$x^2 + 5x - 300 = 0 $
$x^2 + 20x - 15x - 300 = 0$
$(x + 20) (x - 15) = 0 $
$x = 15, -20$
Thus, the other two sides are $15$ cm and $20$ cm.

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

Hypotenuse of a right triangle is $25cm$ and out of the remaining two sides, one is longer than the other by $5cm$. Find the lengths of the other two sides.

  1. $10$cm and $20$ cm
  2. $15$cm and $20$ cm
  3. $25$cm and $20$ cm
  4. $5$cm and $20$ cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let one side be $xcm$. Then the other side will be $(x+5)cm$. Therefore, from Pythagoras theorem
${x}^{2}+{(x+5)}^{2}={25}^{2}$
$\Rightarrow { x }^{ 2 }+{ x }^{ 2 }+10x+25=625$
$\Rightarrow { x }^{ 2 }+5x-300=0\quad \quad \Rightarrow { x }^{ 2 }+20x-15x-300=0$
$\Rightarrow x(x+20)-15(x+20)=0$
$\Rightarrow (x-15)(x+20)=0\quad \Rightarrow \quad x=15\quad or\quad x=-20$
Rejecting $x=-20$, we have length of one side $=15cm$ and that of the other side $=(15+5)cm=20cm$

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 lengths of the sides of a right-angled triangle are all given in natural numbers. If two of these numbers are odd and they differ by $50$, then the least possible value for the third side is:

  1. $61$
  2. $60$
  3. $51$
  4. $50$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We can observe,considering general numbers for right angled triangle,
$5^2 = 3^2 + 4^2$ (5, 3 are odd and 4 lies between 5, 3)
and  $13^2 = 5^2 + 12^2$ (5, 13 are odd and 12 lies between 5, 13)
By trial and error
$61^2 = 60^2 + 11^2$
Since (61, 11 are odd and 61 hypotenuse)
$61 -11 = 50$
Hence, option 'B' 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

One side other than the hypotenuse of a right-angled isosceles triangle is $4$ cm. The length of the perpendicular on the hypotenuse from the opposite vertex is:

  1. $8$cm
  2. $4\sqrt { 2 } $cm
  3. $4$ cm
  4. $2\sqrt { 2 } $cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let assume A, B, C be the vertices of the given triangle and right-angled at A i.e, $\angle A = 90^{\circ}$, $AD \perp BC$.......(D is the point of intersecton of perpendicular from A on BC)
By Pythagoras theorem,
$BC^2 = AB^2 + AC^2$
$BC^2 = 4^2 + 4^2$
$BC = 4\sqrt{2}$ cm
Area of triangle = $\frac{1}{2} base \times height$
Thus, $\frac{1}{2} AB \times AC = \frac{1}{2} AD \times BC$
$4 \times 4 = 4\sqrt{2} \times AD$
$AD = 2 \sqrt{2}$ cm
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 length of the hypotenuse of a right angled $\Delta$ whose two legs measure $12 \ cm$ and $0.35 \ m$ is:

  1. $37 \ cm$
  2. $3.72 \ cm$
  3. $0.372 \ cm$
  4. $37 \ m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$0.35 \ m = 0.35 \times 100 \ cm = 35 \ cm.$
We have,
$(hypotenuse)^2\, =\, (side)^2\, +\, (side)^2$
$=\, (12)^2\, +\, (35)^2$
$= 144 + 1225$
$= 1369$
Hypotenuse $=\sqrt{1369} = 37 \ cm.$