Standardized Test Practice

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Presentation transcript:

Standardized Test Practice EXAMPLE 2 Standardized Test Practice SOLUTION By counting, PQ = 4 and QR = 3. Use the Distance Formula to find PR. d = y 2 – 1 ( ) x +

Standardized Test Practice EXAMPLE 2 Standardized Test Practice = + 1 – 4 ( ) 2 – 1 (– 5 ) ) PR = 4 2 + (– 3 ) = 25 5 = By the SSS Congruence Postulate, any triangle with side lengths 3, 4, and 5 will be congruent to PQR. The distance from (–1, 1) to (–1, 5) is 4. The distance from (–1, 5) to (–4, 5) is 3. The distance from (– 1, 1) to (–4, 5) is = 4 2 + (– 3 5 (–4) – (–1) ( ) 5 – 1) 25 The correct answer is A. ANSWER

GUIDED PRACTICE for Example 2 4. has vertices J(–3, –2), K(0, –2), and L(–3, –8). RST has vertices R(10, 0), S(10, – 3), and T(4, 0). Graph the triangles in the same coordinate plane and show that they are congruent. JKL ANSWER KJ = SR = 3. JL = RT = 6. LK = TS = 3 5.