We are given a2 + b2 = c2 for △ABC and right △DEF constructed with legs a and b and hypotenuse n. Since △DEF is a right triangle, we know that a2 + b2 = n2 because of the . By substitution, c2 = n2 Using the square root property and the principle root, we can take the square root of both sides to get c = n. By , triangles ABC and DEF are congruent. Since it is given that ∠F is a right angle, then ∠ is also a right angle by CPCTC. Therefore, △ABC is a right triangle by .
We are given a2 + b2 = c2 for △ABC and right △DEF constructed with legs a and b and hypotenuse n. Since △DEF is a right triangle, we know that a2 + b2 = n2 because of the . By substitution, c2 = n2 Using the square root property and the principle root, we can take the square root of both sides to get c = n. By , triangles ABC and DEF are congruent. Since it is given that ∠F is a right angle, then ∠ is also a right angle by CPCTC. Therefore, △ABC is a right triangle by .
We are given a2 + b2 = c2 for △ABC and right △DEF constructed with legs a and b and hypotenuse n. Since △DEF is a right triangle, we know that a2 + b2 = n2 because of the . By substitution, c2 = n2 Using the square root property and the principle root, we can take the square root of both sides to get c = n. By , triangles ABC and DEF are congruent. Since it is given that ∠F is a right angle, then ∠ is also a right angle by CPCTC. Therefore, △ABC is a right triangle by .
Max has sticks of lengths 6, 7, 8, and 10 inches. He wants to make a right triangle for an art project. Which 3 sticks should he use?
Triangle JKL is triangle.
42 is 32 + 32.
Therefore, △JKL is .
Triangle XYZ is triangle.
Applying the same method, △ABC is .
52 is 32 + 42.
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