Question

12. а.
Find the area of the quadrilateral those verties
(-2,-3), (о, ч) , (-ч, 2), (0,-1) ​

Answer 1

Answer:

ANSWER

ANSWERThe coordinates of A,B,C & D are A(1,1),B(7,3),C(12,2),D(7,21).

ANSWERThe coordinates of A,B,C & D are A(1,1),B(7,3),C(12,2),D(7,21).Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD

ANSWERThe coordinates of A,B,C & D are A(1,1),B(7,3),C(12,2),D(7,21).Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACDArea of triangle = 21∣[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]∣

ANSWERThe coordinates of A,B,C & D are A(1,1),B(7,3),C(12,2),D(7,21).Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACDArea of triangle = 21∣[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]∣Area of triangle ABC = 21∣[1(3−2)+7(2−1)+12(1−3)]∣

ANSWERThe coordinates of A,B,C & D are A(1,1),B(7,3),C(12,2),D(7,21).Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACDArea of triangle = 21∣[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]∣Area of triangle ABC = 21∣[1(3−2)+7(2−1)+12(1−3)]∣=21∣[1+7−24]∣

ANSWERThe coordinates of A,B,C & D are A(1,1),B(7,3),C(12,2),D(7,21).Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACDArea of triangle = 21∣[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]∣Area of triangle ABC = 21∣[1(3−2)+7(2−1)+12(1−3)]∣=21∣[1+7−24]∣=21(16)=8 sq. units

ANSWERThe coordinates of A,B,C & D are A(1,1),B(7,3),C(12,2),D(7,21).Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACDArea of triangle = 21∣[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]∣Area of triangle ABC = 21∣[1(3−2)+7(2−1)+12(1−3)]∣=21∣[1+7−24]∣=21(16)=8 sq. unitsArea of triangle ACD = 21∣[1(2−21)+12(21−1)+7(1−2)]∣

ANSWERThe coordinates of A,B,C & D are A(1,1),B(7,3),C(12,2),D(7,21).Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACDArea of triangle = 21∣[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]∣Area of triangle ABC = 21∣[1(3−2)+7(2−1)+12(1−3)]∣=21∣[1+7−24]∣=21(16)=8 sq. unitsArea of triangle ACD = 21∣[1(2−21)+12(21−1)+7(1−2)]∣=21∣[−19+240−7]∣

ANSWERThe coordinates of A,B,C & D are A(1,1),B(7,3),C(12,2),D(7,21).Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACDArea of triangle = 21∣[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]∣Area of triangle ABC = 21∣[1(3−2)+7(2−1)+12(1−3)]∣=21∣[1+7−24]∣=21(16)=8 sq. unitsArea of triangle ACD = 21∣[1(2−21)+12(21−1)+7(1−2)]∣=21∣[−19+240−7]∣=21(214)=107 sq. units

ANSWERThe coordinates of A,B,C & D are A(1,1),B(7,3),C(12,2),D(7,21).Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACDArea of triangle = 21∣[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]∣Area of triangle ABC = 21∣[1(3−2)+7(2−1)+12(1−3)]∣=21∣[1+7−24]∣=21(16)=8 sq. unitsArea of triangle ACD = 21∣[1(2−21)+12(21−1)+7(1−2)]∣=21∣[−19+240−7]∣=21(214)=107 sq. unitsHence, area of quadrilateral ABCD = 107+8=115 sq. units

Step-by-step explanation:

mark me as brainlist please