Question

calculate the sides of a triangle with vertices (8,11),(23,3),(2,3)​

Answer 1

Answer:(i) Let A(2,3), B(−1,0) and C(2,−4) are the vertices of △ABC

Area of triangle=  

2

1

​

[x  

1

​

(y  

2

​

−y  

3

​

)+x  

2

​

(y  

3

​

−y  

1

​

)+x  

3

​

(y  

1

​

−y  

2

​

)]

Here (x  

1

​

,y  

1

​

)=(2,3)

(x  

2

​

,y  

2

​

)=(−1,0)

(x  

3

​

,y  

3

​

)=(2,−4)

=  

2

1

​

[2(0+4)−1(−4−3)+2(3−0)]

=  

2

1

​

[8+7+6]

=  

2

21

​

 sq.unit

(i) Let A(−5,1), B(3,−5) and C(5,2) are the vertices of △ABC

Area of triangle=  

2

1

​

[x  

1

​

(y  

2

​

−y  

3

​

)+x  

2

​

(y  

3

​

−y  

1

​

)+x  

3

​

(y  

1

​

−y  

2

​

)]

Here (x  

1

​

,y  

1

​

)=(−5,−1)

=(x  

2

​

,y  

2

​

)=(3,−5)

=(x  

3

​

,y  

3

​

)=(5,2)

=  

2

1

​

[−5(−5−2)+3(2+1)+5(−1+5)]

=  

2

1

​

[35+9+20]

=  

2

64

​

=32 sq.unit

Step-by-step explanation:

Answer 2

Answer:

The sides of the triangle are 10, 17, and 21.

Step-by-step explanation:

The length of the line joining two points $(a,b)$ and $(x,y)$ on the plane is given by the formula,

$length = \sqrt{(x-a)^2+(y-b)^2}$

Given that the three vertices of the triangle are (8,11), (23,3), and (2,3). So the length of its sides can find using the above formula.

The length of the side joining the points (8,11) and (23,3) is,

$S_1= \sqrt{(23-8)^2+(3-11)^2}=\sqrt{15^2+(-8)^2}=\sqrt{225+64}=\sqrt{289}= 17$

The length of the side joining the points (23,3) and (2,3) is,

$S_2=\sqrt{(23-2)^2+(3-3)^2}=\sqrt{21^2}=21$

The length of the side joining the points (8,11) and (2,3) is,

$S_3=\sqrt{(8-2)^2+(11-3)^2}=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$

So the sides of the triangle are 10, 17, and 21.