Question

5
The
area of the tiangle ABC with verlices A(3,0),
B (7,0) and (8,4) is​

Answer 1

Area of △ABC whose Vertices A ≡(x1,y1),B≡(x2,y2)andC≡(x3,y3) are given by

Area of △ABC whose Vertices A ≡(x1,y1),B≡(x2,y2)andC≡(x3,y3) are given by△ =∣∣∣12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]∣∣∣

Area of △ABC whose Vertices A ≡(x1,y1),B≡(x2,y2)andC≡(x3,y3) are given by△ =∣∣∣12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]∣∣∣Here , x1=3,y1=0,x2=7,y2=0x3=8andy3=4

Area of △ABC whose Vertices A ≡(x1,y1),B≡(x2,y2)andC≡(x3,y3) are given by△ =∣∣∣12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]∣∣∣Here , x1=3,y1=0,x2=7,y2=0x3=8andy3=4∴ triangle=|(1)/(2)[3(0-4)+7(4-0)+8(0-0)]|=| (1)/(2)(-12+28+0)|(1)/(2)(16)|=8Hence,therequirredareaoftriangleABC` is 8.

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Answer 2

Step-by-step explanation:

we use the formula 1/2{x1 (y2 - y3) + x2 ( y3- y1 ) + x3 ( y1 - y2 )

= 1/2 { 3 (0 - 4) + 7 (4 - 0) + 8 (0 - 0)

= 1/2 { -12 + 28 + 0}

= 1/2 × 16

= 8 unit square