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The answers to questions 3, 6 and 9 of mathematics p48 in the first volume of ninth grade published by People's Education Press.
Question 3: The sum of two right angles of a right triangle is 14cm, and the area is 24cm? Find the length of two right-angled sides. Solution: Let the shorter right-angle side length be xcm. Then 1/2x( 14-x)=24, (x-6)(x-8)=0, x 1=6, x2=8,14-x = 8; When x=8, 14-x=6 (not in line with the meaning of the question, omitted) A: The lengths of the two right-angled sides are 6 cm and 8 cm respectively. Question 6: Solution, assuming that * * * has X teams. Then the solution of (x-1)+(x-2)+(x-3)+3+2+1= 90/2 is X 1 = 10, and X2 =-9. Question 9: Solution: If the width of the vertical bar is 2xcm and the width of the horizontal bar is 3xcm, then

2? 30? 3x+2? 20? 2x-4? 2x? 3x= 1/4×20×30x≈0.6 or x≈ 10.2 (truncated) 3× 0.6 ≈1.8,2× 0.6 ≈1.2.

Horizontal bar width 1.8cm and vertical bar width 1.2cm, it's up to you to play for so long! ! ! I have to go to bed.