Question

GIVE ME PROPER SOLUTION .​

Answer 1

SOLUTION :-

=> (I) Given,

=> • Taking A as origin, we will take AD as X-axis and AB as Y-axis.

=> •It can be observed that the coordinates of point P, Q and R are (4,6) , (3,2) and (6,5) respectively.

=> •Let p(x1,y1) = (4,6)

=>• Let Q(x2,y2) = (3,2)

=>•Let R(x3,y3) = (6,5)

=> Area of triangle = ½[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]--------(1)

=>By substituting the values of vertices P,Q,R in equ. (1)

=>Area of triangle PQR = ½[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]

=>Area of triangle PQR = ½[4(2-5)+3(5-6)+6(6-2)]

=> Area of triangle PQR = ½[-12-3+24]

=>.°. Area of triangle PQR = 9/2 square units.

=>(II) Given,

=> • Taking C as origin, ,CB as X-axis and CD as Y-axis.

=> • The coordinates of point P, Q and R are (12,2) (13,6) and (10,3) respectively.

=> •Let p(x1,y1) = (12,2)

=>• Let Q(x2,y2) = (13,6)

=>•Let R(x3,y3) = (10,3)

=> Area of triangle = ½[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]--------(1)

=>By substituting the values of vertices P,Q,R in equ. (1)

=>Area of triangle PQR = ½[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]

=>Area of triangle PQR = ½[12(6-3)+13(3-2)+10(2-6)]

=> Area of triangle PQR = ½[36-13+40]

=>.°. Area of triangle PQR = 9/2 square units.

It can observed that the area of the triangle is same in both the cases.

Answer 2

Step-by-step explanation:

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