《2025年诚成教育学业评价八年级数学下册人教版》

23.(6分)如图,在四边形ABCD中,AB=AD=6,∠A=60°,∠ADC=150°,已知四边形ABCD 的周长为30,求四边形ABCD的面积.

答案：
解：如图，连接$BD$，过点$D$作$DE\perp AB$于点$E$. 　　　　　　 $\because AB = AD = 6,\angle A = 60^{\circ }$，$\therefore \triangle ABD$是等边三角形.$\therefore AE = BE = \frac{1}{2}AB = 3$，$BD = 6,\angle ADB = 60^{\circ }$.$\therefore DE=\sqrt{AD^{2}-AE^{2}} = 3\sqrt{3}$.$\therefore S_{\triangle ABD}=\frac{1}{2}AB\cdot DE=\frac{1}{2}\times 6\times 3\sqrt{3}= 9\sqrt{3}$.$\because \angle ADC = 150^{\circ }$，$\therefore \angle CDB = 150^{\circ } - 60^{\circ } = 90^{\circ }$，则$\triangle BCD$是直角三角形.又四边形$ABCD$的周长为$30$，$\therefore CD + BC = 30 - AD - AB = 30 - 6 - 6 = 18$.设$CD = x$，则$BC = 18 - x$.在$Rt\triangle BCD$中，$6^{2}+x^{2}=(18 - x)^{2}$，解得$x = 8$.$\therefore S_{\triangle BDC}=\frac{1}{2}\times 6\times 8 = 24$.$\therefore S_{四边形ABCD}=S_{\triangle ABD}+S_{\triangle BDC}=9\sqrt{3}+24$.

24.(7分)
已知 \start{cases}x = 2\\y = \sqrt{3}\end{cases}是关于$x,y$的二元一次方程$\sqrt{3}x = y + a$的解，求（a+1）(a−1)+7的值.

答案： $\because \begin{cases}x = 2\\y = \sqrt{3}\end{cases}$是关于$x,y$的二元一次方程$\sqrt{3}x = y + a$的解，$\therefore 2\sqrt{3}=\sqrt{3}+a$.$\therefore a = \sqrt{3}$.$\therefore (a + 1)(a - 1)+7=a^{2}-1 + 7= 3 - 1 + 7= 9$.