VERTICES OF A RIGHT TRIANGLE WORKSHEET

Vertices of a Right Triangle Worksheet : 

Worksheet given in this section will be much useful for the students who would like to practice problems on determining if the given points are the vertices of a rectangle. 

Vertices of a Right Triangle Worksheet - Problems

Problem 1 :

Examine whether the given points  A (-3,-4) and B (2,6) and C(-6,10) forms a right triangle.

Problem 2 :

Examine whether the given points  P (7,1) and Q (-4,-1) and R (4,5) forms a right triangle.

Problem 3 :

Examine whether the given points  P (4,4) and Q (3,5) and R (-1,-1) forms a right triangle.

Problem 4 :

Examine whether the given points  A (2,0) and B (-2,3) and C (-2,-5) forms a right triangle.

Problem 5 :

Examine whether the given points  A (0,0) and B (5,0) and C (0,6) forms a right triangle.

Problem 6 :

Examine whether the given points  P (4,4) and Q (3,5) and R (-1,-1) forms a right triangle.

Vertices of righttriangle worksheet - Solution

Problem 1 :

Examine whether the given points  A (-3,-4) and B (2,6) and C(-6,10) forms a right triangle.

Solution :

Distance Between Two Points (x ₁, y₁) and (x₂ , y₂)

√(x₂ - x₁) ² + (y₂ - y₁) ²

The three points are  A (-3,-4) and B (2,6) and C(-6,10)

Distance between the points A and B

= √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = -3, y₁ = -4, x₂ = 2  and  y₂ = 6.

=  √(2-(-3))² + (6-(-4))²

=   √(2+3)² + (6+4)²

=  √5² + 10²

=   √25 + 100 

=   √125 units

Distance between the points B and C

= √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 2, y₁ = 6, x₂ = -6  and  y₂ = 10

=    √(-6-2)² + (10-6)²

=    √(-8)² + (4)²

=    √64 + 16

=    √80 units

Distance between the points C and A

= √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = -6, y₁ = 10, x₂ = -3  and  y₂ = -4.

=    √(-3-(-6))² + (-4-10)²

=    √(-3+6)² + (-14)²

=    √3² + (-14)²

=    √9 + 196 

=    √205 units

AB = √125 units

BC = √80 units

CA = √205 units

(CA)² = (AB)² + (BC)²

(√205)²  = (√125)² + (√80)²

205 = 125 + 80

205 = 205

Therefore A,B and C forms a right triangle. 

Problem 2 :

Examine whether the given points  P (7,1) and Q (-4,-1) and R (4,5) forms a right triangle.

Solution :

Distance Between Two Points (x ₁, y₁) and (x₂ , y₂)

√(x₂ - x₁)² + (y₂ - y₁)²

The three points are  P (7,1) and Q (-4,-1) and R (4,5)

Distance between the points P and Q

= √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 7, y₁ = 1, x₂ = -4  and  y₂ = -1.

=  √(-4-7)² + (-1-1)²

=   √(-11)² + (-2)²

=   √121 + 4

=   √125 units

Distance between the points Q and R

= √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = -4, y₁ = -1, x₂ = 4  and  y₂ = 5.

=   √(4-(-4))² + (5-(-1))²

=    √(4+4)² + (5+1)²

=    √8² + 6²

=    √64 + 36

=    √100 units

Distance between the points R and P

= √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 4, y₁ = 5, x₂ = 7  and  y₂ = 1.

=  √(7-4)² + (1-5)²

=    √(3)² + (-4)²

=    √9 + 16 

=    √25 units

PQ = √125 units

QR = √100 units

RP = √25 units

(PQ)² = (QR)² + (RP)²

(√125)²  = (√100)² + (√25)²

125 = 100 + 25

125 = 125

Hence, the given points  P,Q and R forms a right triangle.

Problem 3 :

Examine whether the given points  P (4,4) and Q (3,5) and R (-1,-1) forms a right triangle.

Solution :

Distance Between Two Points (x ₁, y₁) and (x₂ , y₂)

√(x₂ - x₁)² + (y₂ - y₁)²

The three points are  P (4,4) and Q (3,5) and R (-1,-1)

Distance between the points P and Q

= √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 4, y₁ = 4, x₂ = 3 and y₂ = 5.

=  √(3-4)² + (5-4)²

=  √(-1)² + (1)²

=  √1 + 1

=  √2 units

Distance between the points Q and R

= √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 3, y₁ = 5, x₂ = -1  and  y₂ = -1.

=  √(-1-3)² + (-1-5)²

=   √(-4)² + (-6)²

=  √16 + 36

=  √52 units

Distance between the points R and P

= √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = -1, y₁ = -1, x₂ = 4  and  y₂ = 4.

=  √(4-(-1))² + (4-(-1))²

=   √(4+1)² + (4+1)²

=   √5² + (5)²

=  √25 + 25 

=  √50 units

PQ = √2 units

QR = √52 units

RP = √50 units

(QR)² = (PQ)² + (RP)²

(√52)²  = (√2)² + (√50)²

52 = 2 + 50

52 = 52

Hence, the given points P,Q and R forms a right triangle.

Problem 4 :

Examine whether the given points  A (2,0) and B (-2,3) and C (-2,-5) forms a right triangle.

Solution :

Distance Between Two Points (x ₁, y₁) and (x₂ , y₂)

√(x₂ - x₁)² + (y₂ - y₁)²

The three points are  A (2,0) and B (-2,3) and C (-2,-5)

Distance between the points A and B

= √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 2, y₁ = 0, x₂ = -2  and  y₂ = 3.

=  √(-2-2))² + (3-0)²

=    √(-4)² + (3)²

=   √16 + 9

=   √25  units

Distance between the points B and C

= √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = -2, y₁ = 3, x₂ = -2  and  y₂ = -5.

=  √(-2-(-2))² + (-5-3)²

=    √(-2+2)² + (-8)²

=    √0² + 64

=    √64 units

Distance between the points C and A

= √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = -2, y₁ = -5, x₂ = 2  and  y₂ = 0.

=  √(2-(-2))² + (0-(-5))²

=   √(2+2)² + (0+5)²

=    √4² + (5)²

=    √16 + 25 

=    √41 units

AB = √25 units

BC = √64 units

CA = √41 units

(BC)² = (AB)² + (CA)²

(√64)²  = (√25)² + (√41)²

64 = 25 + 41

64 ≠ 66

Hence, the given points A,B and C will not form a right triangle.

Problem 5 :

Examine whether the given points  A (0,0) and B (5,0) and C (0,6) forms a right triangle.

Solution :

To show that the given points forms a right triangle we need to find the distance between three points. The sum of squares of two sides is equal to the square of remaining side.

Distance Between Two Points (x ₁, y₁) and (x₂ , y₂)

√(x₂ - x₁)² + (y₂ - y₁)²

The three points are  A (0,0) and B (5,0) and C (0,6)

Distance between the points A and B

= √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 0, y₁ = 0, x₂ = 5  and  y₂ = 0.

=    √(5-0)² + (0-0)²

=    √(5)² + (0)²

=    √5² + 0²

=    √25 + 0 

 =    √25 units

Distance between the points B and C

= √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 5, y₁ = 0, x₂ = 0  and  y₂ = 6.

=    √(0-5)² + (6-0)²

=    √(-5)² + (6)²

=    √25 + 36

=    √61 units

Distance between the points C and A

= √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 0, y₁ = 6, x₂ = 0  and  y₂ = 0.

=    √(0-0)² + (0-6)²

=    √(0)² + (-6)²

=    √0 + 36

=    √36 units

AB = √25 units

BC = √61 units

CA = √36 units

(BC)² = (AB)² + (CA)²

(√61)²  = (√25)² + (√36)²

61 = 25 + 36

61 = 61

Hence, the given points A,B and C forms a right triangle.

Problem 6 :

Examine whether the given points  P (4,4) and Q (3,5) and R (-1,-1) forms a right triangle.

Solution :

To show that the given points forms a right triangle we need to find the distance between three points. The sum of squares of two sides is equal to the square of remaining side.

Distance Between Two Points (x ₁, y₁) and (x₂ , y₂)

√(x₂ - x₁)² + (y₂ - y₁)²

The three points are  P (4,4) and Q (3,5) and R (-1,-1)

Distance between the points P and Q

= √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 4, y₁ = 4, x₂ = 3  and  y₂ = 5

=    √(3-4)² + (5-4)²

=    √(-1)² + (1)²

=    √1 + 1

=    √2 units

Distance between the points Q and R

= √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 3, y₁ = 5, x₂ = -1  and  y₂ = -1.

=    √(-1-3)² + (-1-5)²

=    √(-4)² + (-6)²

=    √16 + 36

=    √52 units

Distance between the points R and P

= √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = -1, y₁ = -1, x₂ = 4  and  y₂ = 4.

=    √(4-(-1))² + (4-(-1))²

=    √(4+1)² + (4+1)²

=    √5² + (5)²

=    √25 + 25 

=    √50 units

PQ = √2 units

QR = √52 units

RP = √50 units

(QR)² = (PQ)² + (RP)²

(√52)²  = (√2)² + (√50)²

52 = 2 + 50

52 = 52

Hence, the given points P,Q and R forms a right triangle.

After having gone through the stuff given above, we hope that the students would have understood how to examine whether the points given are the vertices of a rectangle. 

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