Twelfth question in junior high school mathematics and 12 questions in junior high school mathematic

Assuming ad=x, then cd=ad=x, ab=ac=2x, the title says divide the circumference into two parts: 9cm and 12cm

1) When the upper part is 9cm and the lower part is 12cm.

ad+ab=x+2x=9 so x=3 cd=3 ab=2x=6cd+bc=12 bc=12-3=9

So each side ab=ac=6 bc=9

2) When the upper part is 12cm and the lower part is 9cm.

ad+ab=x+2x=12 so x=4 cd=4 ab=2x=8cd+bc=9 bc=9-4=5

So each side ab=ac=8 bc=5

Half term exam! But I don't remember!

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<> remember to give it to me.

Solution: (1) Proof: The quadrilateral ABCD is a square, and the properties of the combined square are:

ad=cd, adp= dcg=90°, cdg+ adh=90°, dh ap, dah+ adh=90°, equal substitution: cdg= dah, adp dcg, dp=cg

2) PQR is an isosceles triangle

qpr= dpa, pqr= cqe, cq=dp, cq=cg, qce= gce, ce=ce, ceq ceg, which are equal by the corresponding angles of the congruent triangle: cqe= cge, pqr= cge, qpr= dpa, adp dcg, pqr= qpr, i.e.: pqr is an isosceles triangle

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The midpoint of the trapezoidal diagonal, parallel to the two bases, is equal to half of the difference between the two bases. This is equivalent to a theorem at all.

Look at the detailed process picture, the picture is not very standard, make do with it.

Just like the auxiliary line you added, extend the EF to BC and FE to AD. Then use the median line theorem to find it.

The original formula can be formed as: a 2 + 2a + 1 + b 2-6b + 9 = 0

then (a+1) 2+(b-3) = 0

The solution is a=

The deformation yields (a+1) 2+(b-3) 2=0

Get a=-1, b=3 so a2-b 2=-8