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darleen cashillo lesson 11-4 practice 9. construct viable arguments. on…

Question

darleen cashillo
lesson 11-4 practice

  1. construct viable arguments. on a coordinate plane, plot triangles

abc and def with vertices a(-3, -1), b(-1, 2), c(1, 1), d(3, -4),
e(1, -1), and f(-1, -2). then prove △abc ≅ △def.
for each pair of triangles, write the congruence criterion, if any, that can be
used to show the triangles are congruent.
10.
image of a rectangle with a diagonal and right angles, and two sides marked equal
11.
image of two triangles with a common side segment, angles marked equal, and two sides marked equal
12.
image of a parallelogram with a diagonal and two right angles

  1. (overline{pq}) bisects ( angle spt ).

image of a triangle with vertex p, base sqt, and q on the base, with pq as an angle bisector

Explanation:

Step1: Calculate side lengths of △ABC

Use the distance formula \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).
For \(AB\): \(A(-3, -1)\), \(B(-1, 2)\)
\(AB = \sqrt{(-1 - (-3))^2 + (2 - (-1))^2} = \sqrt{(2)^2 + (3)^2} = \sqrt{4 + 9} = \sqrt{13}\)
For \(BC\): \(B(-1, 2)\), \(C(1, 1)\)
\(BC = \sqrt{(1 - (-1))^2 + (1 - 2)^2} = \sqrt{(2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}\)
For \(AC\): \(A(-3, -1)\), \(C(1, 1)\)
\(AC = \sqrt{(1 - (-3))^2 + (1 - (-1))^2} = \sqrt{(4)^2 + (2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}\)

Step2: Calculate side lengths of △DEF

For \(DE\): \(D(3, -4)\), \(E(1, -1)\)
\(DE = \sqrt{(1 - 3)^2 + (-1 - (-4))^2} = \sqrt{(-2)^2 + (3)^2} = \sqrt{4 + 9} = \sqrt{13}\)
For \(EF\): \(E(1, -1)\), \(F(-1, -2)\)
\(EF = \sqrt{(-1 - 1)^2 + (-2 - (-1))^2} = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}\)
For \(DF\): \(D(3, -4)\), \(F(-1, -2)\)
\(DF = \sqrt{(-1 - 3)^2 + (-2 - (-4))^2} = \sqrt{(-4)^2 + (2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}\)

Step3: Compare side lengths

\(AB = DE = \sqrt{13}\), \(BC = EF = \sqrt{5}\), \(AC = DF = 2\sqrt{5}\).
By SSS (Side - Side - Side) congruence criterion, if three sides of one triangle are equal to three sides of another triangle, the triangles are congruent. So \(\triangle ABC \cong \triangle DEF\) by SSS.

Answer:

\(\triangle ABC \cong \triangle DEF\) by the SSS (Side - Side - Side) congruence criterion.