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How to write on pages 62 and 63 of the first volume of the third grade of Mathematics Beijing Normal University Edition?
Solutions; (1) Let the width of gold paper edge be xcm.

Judging from the meaning in the title: 40×90=(90+2x)(40+2x)×72%.

Ordered: x squared +65x-350=0.

Solution: x 1=-70 (irrelevant, omitted); x2=5

∴: The width of the edge of the golden paper should be 5cm.

(2) If the side parallel to the wall is XM, the side perpendicular to the wall is (20-x/2).

x(20-x/2)= 180

Solution: x 1=20+2√ 10.

x2=20-2√ 10

The chicken farm area can reach180m2.

x(20-x/2)=200

Solution: x 1=x2=20.

The chicken farm covers an area of 200 square meters.

x(20-x/2)=250

There is no real solution to this equation.

The area of a chicken farm cannot reach 250 square meters.

(3) Let the radius of the bottom surface of the cylinder be r.

2∏R× 15+2∏R squared =200∏.

Ordered: r square+15R- 100=0.

Solution: R 1=-20 (irrelevant, omitted), R2=5.

The radius of the bottom of the cylinder is 5 cm.

(4) To draw a figure in this question, first take a straight line parallel to the X axis through point B, and then make a vertical line with the straight line through points A and P respectively.

This can form a right-angled trapezoid, and then subtract the area of two right-angled triangles from the area of the right-angled trapezoid to be equal to 18. List.

The equation will do.

Here's a hint for you, and you can do the rest yourself. I believe you can do it.