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INEX Using cooranates to Prove Geometric Theorems: Mastery Test 3 Select the correct answer from each drop-down menu. \( \triangle A B C \) has the points \( A(1,7), B(-2,2) \), and \( C(4,2) \) as its vertices. The measure of the longest side of \( \triangle A B C \) is \( \square \) units. \( \triangle A B C \) is \( \square \) triangle. If \( \triangle A B D \) is formed with the point \( D(1,2) \) as its \( \square \) \( x \), then \( \triangle A B D \) is \( \square \) triangle. The length of side \( A D \) is units. 5 \( 29^{\wedge}(1 / 2) \) \( 34^{\wedge}(1 / 2) \) 6 Reset Next

Ask by Graham Coles. in the United States
Mar 18,2025

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- The longest side of \(\triangle ABC\) is \(6\) units. - \(\triangle ABC\) is an **isosceles** triangle. - Point \(D(1,2)\) is the **midpoint** of side \(BC\). - \(\triangle ABD\) is a **right** triangle. - The length of side \(AD\) is \(5\) units.

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Mind Expander

To determine the longest side of triangle \( ABC \), we first calculate the lengths of each side using the distance formula: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). For sides: - \( AB = \sqrt{((-2)-1)^2 + (2-7)^2} = \sqrt{(-3)^2 + (-5)^2} = \sqrt{34} \) - \( BC = \sqrt{(4 - (-2))^2 + (2 - 2)^2} = \sqrt{(6)^2 + (0)^2} = 6 \) - \( AC = \sqrt{(4-1)^2 + (2-7)^2} = \sqrt{(3)^2 + (-5)^2} = \sqrt{34} \) From this, sides \( AB \) and \( AC \) are both \( \sqrt{34} \) units long, and side \( BC \) is the longest at \( 6 \) units. Now switching gears to triangle \( ABD \), where point \( D(1,2) \) is chosen. We find that: - \( AD = \sqrt{(1-1)^2 + (2-7)^2} = \sqrt{(0)^2 + (-5)^2} = 5 \) - While keeping \( AB \) and \( BD \) lengths similar (previously calculated). You could conclude if \( AB \), \( AD \), and \( BD \) show a right angle since \( AB = 5 \) and height \( AD \) aligns vertically with horizontal changes to \( BD \). Enjoy piecing together the geometric puzzle using your calculations and observe the relationships glowing with logic! It’s like detective work, piecing together how the properties of shapes unfold!

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