Question

directions: if each quadrilateral below is a rectangle, find the missing measures.
1.
vw =
wx =
yw =
zx =
vx =
2.
gf =
ge =
df =
hf =
dg =
*gh = 14
3.
m∠1 =
m∠2 =
m∠3 =
m∠4 =
m∠5 =
m∠6 =
m∠7 =
m∠8 =
m∠9 =
m∠10 =
m∠11 =
4.
m∠bcd =
m∠abd =
m∠cbe =
m∠ade =
m∠aeb =
m∠dea =
5.
m∠jmk =
m∠jkh =
m∠hlk =
m∠hjl =
m∠lhk =
m∠jlk =

1. find wz.
2. if sq = 11x - 26 and pr = 5x + 28, find pr.

Explanation:

Problem 1

Step1: 矩形对边相等，所以\(VW = YX = 31\)，\(WX = VY = 19\)

Step2: 矩形对角线相等且互相平分，\(YW\)是对角线，根据勾股定理\(YW=\sqrt{19^{2}+31^{2}}=\sqrt{361 + 961}=\sqrt{1322}\approx36.36\)，\(ZX=\frac{1}{2}YW\approx18.18\)，\(VX = YW\approx36.36\)（这里可能题目希望用矩形性质，对角线相等且平分，所以\(YW = VX\)，且\(ZX=\frac{1}{2}VX\)，不过更准确的是，矩形对角线相等，所以\(YW = VX\)，且\(ZX=\frac{1}{2}VX\)，不过可能题目中\(VY = 19\)，\(YX = 31\)，所以\(VW = 31\)，\(WX = 19\)，\(YW\)是对角线，长度为\(\sqrt{19^{2}+31^{2}}\)，\(ZX=\frac{1}{2}YW\)，\(VX = YW\)）

Step1: 矩形对边相等，所以\(GF = DE = 11\)

Step2: 矩形对角线相等且平分，\(GE\)是对角线，\(GH = 14\)，所以\(GF = 11\)，\(GE = DF\)，\(HF = GH = 14\)？不，矩形\(DGFE\)，\(DE = GF = 11\)，\(DG = EF\)，\(GH = 14\)，所以\(GF = 11\)，\(GE\)是对角线，\(DF = GE\)，\(HF = \frac{1}{2}DF\)？不对，矩形对角线相等且平分，所以\(GE = DF\)，且\(H\)是中点，所以\(GH = HF = 14\)？不，\(GH = 14\)，所以\(HF = 14\)，\(DF = GE = \sqrt{11^{2}+(14 + 14)^{2}}=\sqrt{121+784}=\sqrt{905}\approx30.08\)，\(DG = EF = 14 + 14 = 28\)？可能我错了，重新看，矩形\(DGEF\)？不，图中\(D\)、\(E\)、\(F\)、\(G\)，\(DE = 11\)，\(GH = 14\)，所以\(GF = DE = 11\)，\(GE\)是对角线，\(DF = GE\)，\(HF = GH = 14\)？不，矩形对边相等，所以\(GF = DE = 11\)，\(DG = EF\)，对角线相等且平分，所以\(GE = DF\)，\(H\)是中点，所以\(GH = HF = 14\)，\(DG = EF = 14 + 14 = 28\)，\(GE = \sqrt{11^{2}+28^{2}}=\sqrt{121 + 784}=\sqrt{905}\approx30.08\)，所以\(GF = 11\)，\(GE\approx30.08\)，\(DF\approx30.08\)，\(HF = 14\)，\(DG = 28\)

Step1: 矩形对边相等，所以\(WZ = XY\)，\(WX = ZY\)，所以\(7x - 6 = 3x + 14\)

Step2: 解方程\(7x - 3x = 14 + 6\)，\(4x = 20\)，\(x = 5\)

Step3: 代入\(WZ = 7x - 6 = 7\times5 - 6 = 29\)

Answer:

\(VW = 31\)，\(WX = 19\)，\(YW=\sqrt{1322}\approx36.36\)，\(ZX=\frac{\sqrt{1322}}{2}\approx18.18\)，\(VX=\sqrt{1322}\approx36.36\)（如果题目希望用整数，可能我理解错了，重新看，矩形对边相等，所以\(VW = YX = 31\)，\(WX = VY = 19\)；对角线相等且平分，所以\(YW = VX\)，且\(ZX=\frac{1}{2}VX\)，\(YW\)的长度用勾股定理\(19^{2}+31^{2}=361 + 961 = 1322\)，所以\(YW=\sqrt{1322}\)，不过可能题目是整数，可能我错了，再想，矩形的对角线相等，所以\(YW = VX\)，且\(ZX=\frac{1}{2}VX\)，\(VY = 19\)，\(YX = 31\)，所以\(VW = 31\)，\(WX = 19\)，\(YW\)是对角线，长度为\(\sqrt{19^{2}+31^{2}}\)，\(ZX=\frac{1}{2}YW\)，\(VX = YW\)）

Problem 2