Mathematics · Quantitative Aptitude

Triangle Properties

243 Questions

Triangle properties encompass the rules governing the sides, angles, and area of different types of triangles. Key areas include the Pythagorean theorem, similar triangles, and centroid calculations. These concepts form a foundational part of geometry in various competitive examinations.

Similar trianglesArea and perimeterPythagorean theoremTriangle inequalityEquilateral properties

Triangle Properties Questions

Multiple choice maths triangle inequality construction of parallel lines and triangles triangle inequality related to lines and triangles sum of the lengths of two sides of a triangle

The number of triangles with any three of the length $1, 4, 6$ and $8 $ cm as sides is:

  1. $4$
  2. $2$
  3. $1$
  4. $0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Only $1.$ Since, the sum of any two sides of a triangle must be greater than the third side.

$1,4,6$ no, because $1+4<6$
$1,4,8$ no, because $1+4<8$
$1,6,8$ no, because $1+6<8$
$4,6,8$ yes, because $4+6>8 , 4+8>6 , 8+6>4$
Option $C$ is correct.

Multiple choice maths triangle inequality construction of parallel lines and triangles triangle inequality related to lines and triangles sum of the lengths of two sides of a triangle

Which of the following sets of side lengths will not form a triangle?

  1. $11$ cm, $10$ cm, $11$ cm
  2. $3$ m, $3$ m , $3$ m
  3. $9$ mm, $9$ mm, $12$ mm
  4. $3$ cm, $4$ cm, $7$ cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The sum of any two sides of a triangle is greater than the third side. 

Here, if we consider $3$ cm, $4$ cm, $7$ cm as side lengths then the sum of two sides $(3 + 4)$ cm is equal to the third side and not greater than the third side i.e., $7$ cm.
Thus, the side lengths $3$ cm, $4$ cm , $7$ cm will not form a triangle.
Hence, option D is correct. 

Multiple choice maths triangle inequality construction of parallel lines and triangles triangle inequality related to lines and triangles sum of the lengths of two sides of a triangle

The length of two sides of a triangle are $20 $ mm and $29 $ mm. Which of the following can be the value of third side to form the triangle?

  1. $6 $ mm
  2. $7 $ mm
  3. $23 $ mm
  4. $8 $ mm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that $(29-20) $ mm should smaller than the third side. 

Thus, the third side is greater than $9 $ mm.
Also, third side should be less than sum of $20$ and $29 $ mm  i.e. $49
$ mm.
Thus, $23 $ mm can be the length of third side to form a triangle.

Multiple choice maths triangle inequality construction of parallel lines and triangles triangle inequality related to lines and triangles sum of the lengths of two sides of a triangle

The lengths of two sides of a triangle are $7 $ cm and $10 $ cm. What is the possible value range of the third side?

  1. $3 $ cm $<$ third side $< 10 $ cm
  2. $7 $ cm $<$ third side $< 10 $ cm
  3. $3 $ cm $<$ third side $< 17 $ cm
  4. $7 $ cm $<$ third side
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that:
(i) The sum of lengths of any two sides of a triangle is greater than the third side. Thus, we know that $(7 + 10) $ cm is greater than the third side.
Therefore, third side is less than $17 $ cm.
(ii) The difference of lengths of any two sides of triangle is smaller than the third side. Thus $(10 - 7) \ cm$ is smaller than the third side.
Therefore, third side is greater than $3 $ cm
Thus, $3 $ cm $<$ third side $< 17 $ cm.

Multiple choice maths triangle inequality construction of parallel lines and triangles triangle inequality related to lines and triangles sum of the lengths of two sides of a triangle

The lengths of two sides of a triangle are $3 $ cm and $4 $ cm. Which of the following, can be the length of third side to form a triangle?

  1. $0.5 $ cm
  2. $5 $ cm
  3. $8 $ cm
  4. $10$ cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

(I) We know that $(3 + 4) $ cm is greater than third side.
Thus, the third side is smaller than $7$ cm
(ii) we know that $(4 - 3) $ cm is smaller than third side .
Thus, the third side is greater than $1 $cm.
Therefore, $1 $ cm < third side $< 7 $ cm.

Thus, $5  $ cm can be the length of third side for a triangle. 

Multiple choice maths triangle inequality construction of parallel lines and triangles triangle inequality related to lines and triangles sum of the lengths of two sides of a triangle

Find all possible lengths of the third side, if sides of a triangle have $3$ and $9$.

  1. $6 < x < 12$
  2. $5 < x < 12$
  3. $6 < x < 10$
  4. $6 < x < 11$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The Triangle Inequality theorem states that the sum of any $2$ sides of a triangle must be greater than the measure of the third side.
So, difference of two sides $< x <$ sum of two sides, will give you the possible length of a triangle.
Therefore, $9 - 3 < x < 9 + 3$
$6 < x < 12$ is the possible length of the third side of a triangle.
For checking the possible length: Take $3, 9, 7$
$3 + 9 > 7 (a + b > c)$
$9 + 7 > 3 (b + c > a)$
$3 + 7 > 9 (a + c > b)$
Which satisfy the triangle inequality theorem.

Multiple choice maths triangle inequality construction of parallel lines and triangles triangle inequality related to lines and triangles sum of the lengths of two sides of a triangle

Find all possible lengths of the third side, if sides of a triangle have $2$ and $5$.

  1. $2 < x < 7$
  2. $3 > x < 7$
  3. $3 < x > 7$
  4. $3 < x < 7$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The Triangle Inequality theorem states that the sum of any $2$ sides of a triangle must be greater than the measure of the third side.
So, difference of two sides $< x <$ sum of two sides, will give you the possible length of a triangle.
Therefore, $5 - 2 < x < 5 + 2$
$3 < x < 7$ is the possible length of the third side of a triangle.
For checking the possible length: Take $2, 5, 4$
$2 + 5 > 4 (a + b > c)$
$5 + 4 > 2 (b + c > a)$
$2 + 4 > 5 (a + c > b)$
Hence, the above condition satisfied the triangle inequality theorem.

Multiple choice maths triangle inequality construction of parallel lines and triangles triangle inequality related to lines and triangles sum of the lengths of two sides of a triangle

A triangle has side lengths of $6$ inches and $9$ inches. If the third side is an integer, calculate the minimum possible perimeter of the triangle (in inches).

  1. $4$
  2. $15$
  3. $8$
  4. $19$
  5. $29$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the third side be $x$.
Sum of any two sides of a triangle is greater than the third side. 

Hence, $6+x>9$ or $x>3$ and $6+9>x$ or $x<15$.
Therefore, $x\epsilon (3,15)$
Hence, the minimum possible integral value of $x$ is $4$. 
Thus the minimum possible length of the third side is $4$. 
Hence, the minimum possible perimeter is $4+6+9=19$ units.

Multiple choice maths construction of triangles constructions of triangles construction of triangles - ii constructions

Construct a triangle $PQR$, whose perimeter is $22 cm$ and whose sides are in the ratio $2 : 4 : 5$. Measure the sides of the triangle.

  1. $5, 7, 10$
  2. $4, 8, 10$
  3. $5, 8, 9$

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

Let the sides be $2x,4x$ and $5x$

Perimeter $=22cm$
$2x+4x+5x=22$
$11x=22$
$x=2$
So the sides are 
$2x=2\times2=4cm$
$4x=4\times 2=8cm$
$5x=5\times 2=10cm$
Option $B$ is correct.

Multiple choice maths construction of triangles constructions of triangles construction of triangles - ii constructions

For constructing a triangle whose perimeter and both base angles are given, the first step is to:

  1. Draw a base of any length
  2. Draw the base of length $=$ perimeter

  3. Draw the base angles from a random line.
  4. Draw a base of length $=\dfrac13 \times$ perimeter.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Below are the steps for constructing a triangle whose perimeter and base angles are given.

Step 1 : Draw a line segment/base equal to perimeter 

Step 2 : From any point X draw ray at one of the given base angles. From any point Y draw ray at second of the given base angles

Step 3 : draw angle bisector of X and Y, two angle bisectors intersect each other at point A

Step 4 : Draw line bisector of XA and AY respectively these two line bisectors intersect XY at point B and C

Step 5 : Join A to B and A to C 

Step 6 : Triangle ABC is the required triangle.


Hence, the answer is 'Draw the base of length $=$ perimeter'.

Multiple choice combining transformations transformations vectors and transformations maths

The area of triangle formed by the lines $x+y-3=0$, $x-3y+9=0$ and $3x-2y+1=0$ is:

  1. $\cfrac { 16 }{ 7 } $ sq. units
  2. $\cfrac { 10 }{ 7 } $ sq. units
  3. $4$ sq. units
  4. $9$ sq. units
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$L _1:x+y-3=0$

$L _2:x-3y+9=0$
$L _3:3x-2y+1=0$

$A$ is intersection point of $L _1$ and $L _2$ which is $(0,3)$ on solving $L _1$ and $L _2$.

$B$ is intersection point of $L _2$ and $L _3$ which is $(1,2)$ on solving $L _2$ and $L _3$.

$C$ is intersection point of $L _3$ and $L _1$ which is $(\dfrac{15}{7},\dfrac{26}{7})$ on solving $L _3$ and $L _1$.

Now, area of $\Delta ABC=\dfrac{1}{2}|(x _1y _2-x _2y _1)+(x _2y _3-x _3y _2)+(x _3y _1-x _1y _3)|$

$=\dfrac{1}{2}|(0-3)+(\dfrac{26}{7}-\dfrac{30}{7})+(\dfrac{45}{7}-0)|$

$=\dfrac{1}{2}|-3-\dfrac{4}{7}+\dfrac{45}{7}|=\dfrac{1}{2}\times \dfrac{20}{7}$

$=\dfrac{10}{7} sq. units$

Multiple choice mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

If the area of the triangle with vertices $(2, 5), (7, k)$ and $(3, 1)$ is $10$, then find the value of $k$.

  1. $-5$ or $35$
  2. $5$ or $-35$
  3. $15$ or $-5$
  4. $-5$ or $-25$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
If $(x _1,y _1), (x _2, y _2)$ ans $(x _3, y _3)$ are the vertices of a triangle, then its area is given by $\pm \dfrac {1}{2}[x _1(y _2-y _3)+x _2(y _3-y _1)+x _3(y _1-y _2)]$ 
Given vertices are $(2,5), (7,k), (3,1)$ and area is $10$.
Therefore, $\pm 10 = \dfrac {1}{2}[2(k-1)+7(1-5)+3(5-k)]$
$\Rightarrow \pm 20=2k-2-28+15-3k$
$\Rightarrow \pm 20=-k-15$
$\Rightarrow k = 5$ or $-35$