Question

Find the area of the triangle PQR whose vertices are P(-5, 7), Q(-4, -5), R(4, 5).​

Answer 1

To Find :-

• The area of the ΔPQR.

Solution :-

Given,

• ΔPQR whose vertices are P(-5, 7), Q(-4, -5), R(4, 5).

Here,

• x1 = -5
• y1 = 7
• x2 = -4
• y2 = -5
• x3 = 4
• y3 = 5

By using the formula,

ΔPQR = ¹/2 × [ x1 (y2 - y3) + x2 ( y3 - y1) + x3 ( y1 - y2) ]

[ Put the values ]

↪ ΔPQR = ¹/2 × [-5 (-5 - 5) + (-4)(5 - 7) + 4( 7-(-5)]

↪ ΔPQR = ¹/2 × [-5 (-10) + (-4) (-2) + 4 (12)]

↪ ΔPQR = ¹/2 × [50 + 8 + 48]

↪ ΔPQR = ¹/2 × [50 + 56]

↪ ΔPQR = ¹/2 × [106]

↪ ΔPQR = 53 square units

Therefore,

The area of the ΔPQR is 53 square units.

Answer 2

★Given:-

Vertices of Triangle PQR

•P(-5,7)

•Q(-4,-5)

•R(4,5)

★To Find out :-

•Area of the Triangle formed by these vertices

★Solution:-

Let • x1 =-5

• x2=-4

• x3= 4

• y1 = 7

• y2 = -5

• y3 = 5

》By using below formula

★Area of triangle PQR=1/2[ x1 (y2-y3)+ x2 (y3-y1)+ x3 (y1-y2)]

=1/2[ -5(-5-5) +(-4)(5-7)+4(7-(-5))]

=1/2 [-5×(-10)+(-4)(-2)+4×(12)]

=1/2[50+8+48]

=1/2×106

=53 sq. unit

Therefore ,Area of triangle PQR=53 sq. unit