When the squares joined their vertices to form a night triangle, the combined area of the two smaller squares the same as
the wea of the target square
Which the values of the sea of the squares do NOT support this statement?

When three squares are joined at their vertices to form a right triangle, the combined area of the two smaller squares is the same as the area of the largest square.

Step-by-step explanation:

f the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle. for Example: ΔABC, if c2=a2+b2 then ∠C is a right triangle, ΔPQR being the right angle. hope this helps

When the squares joined their vertices to form a night triangle, the combined area of the two smaller squares the same as the wea of the target square Which the values of the sea of the squares do NOT support this statement?

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When the squares joined their vertices to form a night triangle, the combined area of the two smaller squares the same as
the wea of the target square
Which the values of the sea of the squares do NOT support this statement?