In Delta ABC,A(3, 0),B(0, 0) and C(0, -4). Re | vedclass

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(C) $1$. Given coordinates: $A(3, 0)$,$B(0, 0)$,$C(0, -4)$.$2$. Calculate lengths using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:$A

Vedclass · Accepted Answer

The area of $\Delta ABC$ is $12$.

Vedclass · Answer

(C) $1$. Given coordinates: $A(3, 0)$,$B(0, 0)$,$C(0, -4)$.$2$. Calculate lengths using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:$AB = \sqrt{(0-3)^2 + (0-0)^2} = \sqrt{9} = 3$.$BC = \sqrt{(0-0)^2 + (-4-0)^2} = \sqrt{16} = 4$.$AC = \sqrt{(0-3)^2 + (-4-0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.$3$. Check for right-angled triangle: $AB^2 + BC^2 = 3^2 + 4^2 = 9 + 16 = 25 = AC^2$. Since $AB^2 + BC^2 = AC^2$,$\Delta ABC$ is a right-angled triangle at $\angle B$. (Option $A$ is true).$4$. Midpoint of $\overline{AC} = (\frac{3+0}{2}, \frac{0-4}{2}) = (\frac{3}{2}, -2)$. (Option $B$ is true).$5$. Area of $\Delta ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes AB 	imes BC = \frac{1}{2} 	imes 3 	imes 4 = 6$. (Option $C$ states area is $12$,which is false).$6$. $AB=3, BC=4, AC=5$ is true. (Option $D$ is true).Therefore,the false statement is $C$.

In $\Delta ABC$,$A(3, 0)$,$B(0, 0)$ and $C(0, -4)$. Relative to this,the following information is given. Which of the information is false?

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