Find the area of triangle PQR whose vertices areP(-5,7) ,Q(-4,-5) and R(4,5)
Answers
Given : A triangle PQR whose vertices are P(-5,7) ,Q(-4,-5) and R(4,5)
To find : Area of Triangle PQR
Solution:
P = (-5 , 7)
Q = ( -4 , - 5)
R = ( 4 , 5)
Area of Triangle PQR
= (1/2) | -5(-5 - 5) -4(5 - 7) + 4(7 -(-5))|
= (1/2) | 50 + 8 + 48 |
= (1/2) | 106 |
= 106/2
= 53
Area of Triangle PQR = 53 sq units
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Answer and Step-by-step Explanation:
Given:
✰ Point P( -5, 7 )
✰ Point Q( -4, -5 )
✰ Point R ( 4, 5 )
To find:
✠ The area of triangle PQR
Solution:
Consider,
In point P( -5, 7 )
✫ -5 = x₁
✫ 7 = y₁
In point Q( -4, -5 )
✫ -4 = x₂
✫ -5 = y₂
In point R ( 4, 5 )
✫ 4 = x₃
✫ 5 = y₃
Now,
Area of ∆PQR = 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]
Area of ∆PQR = 1/2 [ -5( -5 - 5 ) -4( 5 - 7 ) +4( 7 + 5 )]
Area of ∆PQR = 1/2 [ ( 25 + 25 ) + ( - 20 + 28 ) + ( 28 + 20 ) ]
Area of ∆PQR = 1/2 [ 50 + 8 + 48 ]
Area of ∆PQR = 1/2 [ 58 + 48 ]
Area of ∆PQR = 1/2 [ 58 + 48 ]
Area of ∆PQR = 1/2 [ 106 ]
Area of ∆PQR = 1/2 × 106
Area of ∆PQR = 53 sq.unit
∴ The ∆ of triangle PQR = 53 sq.unit
Signs:
+ + = +
- - = +
- + = -
+ - = -
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