Question

in the diagram, ab = 10 and ac = 2√10. what is the perimeter of △abc? 10 units 10 + 2√10 units 20 units 20 + 2√10 units

Explanation:

Step1: Find the length of BC

Use the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
For points \(B(-2,-2)\) and \(C(5,-2)\), \(x_1=-2,y_1 = - 2,x_2=5,y_2=-2\).
\(BC=\sqrt{(5-(-2))^2+(-2-(-2))^2}=\sqrt{(5 + 2)^2+0^2}=\sqrt{49}=7\) (This is wrong. Wait, no. Wait, looking at the grid - assume \(B(-2,-2)\). Another way: Since \(B\) and \(C\) have the same \(y\) - coordinate (\(y=-2\)), the distance \(BC=\vert x_C-x_B\vert\). If \(B(-2,-2)\) and \(C(5,-2)\), then \(BC=\vert5-(-2)\vert = 7\) (wrong). Wait, no - wait, looking at the symmetry. Wait, actually, since \(A(3,4)\), \(B(-2,-2)\), \(C(5,-2)\).
The distance between \(B(-2,-2)\) and \(C(5,-2)\): \(d=\sqrt{(5 + 2)^2+(-2+2)^2}=7\) (no - actually, since \(y\) is same, \(d=\vert5-(-2)\vert=7\) (wrong). Wait, no - wait, using the distance formula for \(A(3,4)\) and \(B(-2,-2)\): \(AB=\sqrt{(3 + 2)^2+(4 + 2)^2}=\sqrt{25 + 36}=\sqrt{61}\) (no, but given \(AB = 10\)). Wait, no - the problem gives \(AB = 10\), \(AC=2\sqrt{10}\).
We need to find \(BC\). Since \(B\) and \(C\) are symmetric with respect to \(x = 3\) (the \(x\) - coordinate of \(A\)). The distance from \(B\) to \(x = 3\) is \(3-(-2)=5\), the distance from \(C\) to \(x = 3\) is \(5 - 3=2\) (no - wrong). Wait, using the distance formula for \(B\) and \(C\) (assuming \(B(-2,-2)\) from the grid - like points).
\(BC=\sqrt{(5+2)^2+(-2 + 2)^2}=7\) (no - wait, if \(A(3,4)\), \(B(-2,-2)\), \(C(5,-2)\)
\(AB=\sqrt{(3+2)^2+(4 + 2)^2}=\sqrt{25+36}=\sqrt{61}\) (but given \(AB = 10\)). Wait, no - the problem says \(AB = 10\), \(AC=2\sqrt{10}\).
We can also use the property of the triangle. Since \(B\) and \(C\) have \(y=-2\). The distance from \(A(3,4)\) to \(y=-2\) is \(4-(-2)=6\).
For \(AC\): using distance formula \(AC=\sqrt{(5 - 3)^2+(-2 - 4)^2}=\sqrt{4 + 36}=\sqrt{40}=2\sqrt{10}\) (matches). For \(AB\): \(\sqrt{(3+2)^2+(4 + 2)^2}=\sqrt{25 + 36}=\sqrt{61}\) (no - but problem says \(AB = 10\)). Wait, no - actually, if we assume that the triangle is such that the base \(BC\): since \(A\) is above \(x = 3\), and \(B\) and \(C\) are symmetric with respect to \(x = 3\) (because \(3=\frac{-2 + 5}{2}\)). The length of \(BC=\vert5-(-2)\vert=7\) (no - wait, the vertical distance from \(A\) to \(BC\) (where \(y=-2\)) is \(h = 4-(-2)=6\).
Using the Pythagorean theorem for \(AB\): Let the half - length of \(BC\) be \(x\). If \(AC = 2\sqrt{10}\), then for \(AC\): \(AC=\sqrt{x_1^2+6^2}\), where \(x_1\) is the horizontal distance from \(A\) to \(C\). \(x_1=5 - 3=2\), \(AC=\sqrt{4 + 36}=\sqrt{40}=2\sqrt{10}\). For \(AB\): the horizontal distance from \(A\) to \(B\) is \(3-(-2)=5\), \(AB=\sqrt{5^2+6^2}=\sqrt{25 + 36}=\sqrt{61}\) (no - but problem says \(AB = 10\)). Wait, no - the problem gives \(AB = 10\), \(AC=2\sqrt{10}\).
We know that the perimeter \(P=AB + AC+BC\).
Since \(B\) and \(C\) are symmetric about \(x = 3\) (the \(x\) - coordinate of \(A\)). The length of \(BC\) can be found by noting that if we consider the right - triangle formed by \(A\) and the projection on \(BC\).
The vertical distance from \(A\) to \(BC\) ( \(y=-2\)) is \(6\).
For \(AB = 10\), using the Pythagorean theorem (horizontal distance \(d_1\): \(AB=\sqrt{d_1^2+6^2}\), \(10=\sqrt{d_1^2+36}\), \(d_1^2=100 - 36 = 64\), \(d_1 = 8\).
For \(AC=2\sqrt{10}=\sqrt{40}\), horizontal distance \(d_2\): \(AC=\sqrt{d_2^2+6^2}\), \(40=d_2^2+36\), \(d_2 = 2\).
The length \(BC=d_1 + d_2=8 + 2=10\)

Step2: Calculate the perimeter

The perimeter \(P=AB + AC+BC\)
Given \(AB = 10\), \(AC=2\sqrt{10}\), \(BC = 10\)
\(P=10+2\sqrt{10}+10=20 + 2\sqrt{10}\)

Answer:

\(20 + 2\sqrt{10}\text{ units}\)