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[Eighth grade mathematics] Factorizing and judging the shape of triangles.
Method Guidance: Factorizing and judging the shape of triangle is essentially to examine students' mastery of factorization. This kind of topic is often finally decomposed into the form of the sum of two squares and the form of two products, namely the following form:

Combining the relationship of three sides of a triangle to express the relationship, we can judge the relationship of the sides and then determine the shape of the triangle.

The main reason is that students can't skillfully solve factorization problems in groups. Strengthening training and summarizing the same type of problems and solving methods can help us to answer skillfully faster and more accurately.

Sample demonstration

1.? It is known that A, B and C are the lengths ABC of three sides of a triangle, and

Can you determine the shape of triangle ABC?

2.? It is known that the sides of triangle ABC are A, B, C, and satisfy

Try to judge the shape of triangle ABC?

3.? It is known that A, B and C are the lengths of three sides of △ABC. When b2+2ab=c2+2ac,?

(1) Try to determine which triangle △ABC belongs to; ?

(2) If a=4 and b=3, find the perimeter of △ABC.

4.? It is known that the sides of triangle a, b and c are a, b and C.

Try to judge the relationship between the value of algebraic expression and 0? 、

Reference answer

1.

2.

3.

△ABC is an isosceles triangle for the following reasons: ∵a, B and C are the lengths of three sides △∴b=c, b2+2ab=c2+2ac, ∴ B2+2ab-2ac = 0, and the factorization is: (b-c) (b+c+2a).

4.

try one's hand (at)

practise

1.? It is known that a, b and c are three sides of △ABC, which satisfies? Try to judge the shape of △ABC. ?

2.? It is known that A, B and C are three sides of △ABC. Please explain that the value of (A2+B2-C2) 2-4a2b2 must be negative.

3.? If the three-side lengths A, B and C of △ABC satisfy A2+B2+C2-AB-BC-AC = 0, please judge the shape of △ABC.

Practice answering

1.

Answer: A2C2-B2C2-A4+B4 = 0C2 (A2-B2)-(A2+B2) (A2-B2) = 0 (A2-B2) (C2-A2-B2) = 0A2-B2 = 0 or c2=a2+b2∴a=b or C2 =

2.

Answer: (A2+B2-C2) 2-4a2b2 = (A2+B2-C2+2ab) (A2+B2-C2-2ab) = [(A+B) 2-C2].

3.

Answer: ∵ A2+B2+C2-AB-BC-AC = 0? ∴2a2+2b2+2c2﹣2ab﹣2bc﹣2ac=0,a2+b2﹣2ab+b2+c2﹣2bc+a2+c2﹣2ac=0,∴(a﹣b)2+(b﹣c)2+(c﹣a)2=0,∴a﹣b=0,b﹣c=0,c﹣a=0,

∴△abc ∴a=b=c is an equilateral triangle.