第61页 - 同步练习八年级数学苏科版江苏凤凰科学技术出版社

 连接 $AC，$在 $Rt\triangle ACD$ 中，$\because AC^{2}=CD^{2}+AD^{2}=3^{2}+4^{2}=25，$$\therefore AC=5。$

 $\because AC^{2}+BC^{2}=5^{2}+12^{2}=169，$$AB^{2}=13^{2}=169，$$\therefore AC^{2}+BC^{2}=AB^{2}，$$\therefore \angle ACB=90^{\circ}。$

 该区域面积$=S_{\triangle ACB}-S_{\triangle ACD}=\frac{1}{2}\times AC\times BC - \frac{1}{2}\times AD\times CD=\frac{1}{2}\times5\times12 - \frac{1}{2}\times4\times3=30 - 6=24(m^{2})。$

 铺满这块空地共需花费$=24\times30=720$（元）。

 答：用该草坪铺满这块空地共需花费720元。

 $\because \triangle ABC$是等边三角形，$\therefore \angle CAB=60^{\circ}。$

 $\because \triangle PAC$绕点$A$按逆时针方向旋转后得到$\triangle P'AB，$

 $\therefore \angle PAP'=\angle CAB=60^{\circ}，$$AP=AP'，$$P'B=PC=10。$

 $\therefore \triangle APP'$是等边三角形，$\therefore PP'=AP=6，$$\angle APP'=60^{\circ}。$

 $\because PB^{2}+PP'^{2}=8^{2}+6^{2}=64 + 36=100=10^{2}=P'B^{2}，$

 $\therefore \triangle BPP'$是直角三角形，$\angle P'PB=90^{\circ}。$

 $\therefore \angle APB=\angle APP'+\angle P'PB=60^{\circ}+90^{\circ}=150^{\circ}。$

 综上，$PP'$的长为$6，$$\angle APB$的大小为$150^{\circ}。$

3,4,5

$解：(2)当 k 大于 2 时,k^{2}+\left \lbrack \left( \frac{1}{2}k\right)^{2}-1\right\rbrack^{2}=\left \lbrack \left( \frac{1}{2}k\right)^{2}+1\right\rbrack^{2}.\ $

$证明:\because左边=k^{2}+\left \lbrack \left( \frac{1}{2}k\right)^{2}-1\right\rbrack^{2}=k^{2}+\left \lbrack \frac{1}{4}k^{2}-1\right\rbrack^{2}=k^{2}+\frac{1}{16}k^{4}+1-\frac{1}{2}k^{2}=\frac{1}{16}k^{4}+\frac{1}{2}k^{2}+1;$

$右边=\left \lbrack \left( \frac{1}{2}k\right)^{2}+1\right\rbrack^{2}=\left \lbrack \frac{1}{4}k^{2}+1\right\rbrack^{2}=\frac{1}{16}k^{4}+\frac{1}{2}k^{2}+1.$

$\therefore左边=右边,$

$\therefore等式成立$

【答案】：
（1）3,4,5 （2）当 k 大于 2 时,$k^{2}+\left \lbrack \left( \frac{1}{2}k\right)^{2}-1\right\rbrack^{2}=\left \lbrack \left( \frac{1}{2}k\right)^{2}+1\right\rbrack^{2}$. 证明:$\because$左边$=k^{2}+\left \lbrack \left( \frac{1}{2}k\right)^{2}-1\right\rbrack^{2}=k^{2}+\left \lbrack \frac{1}{4}k^{2}-1\right\rbrack^{2}=k^{2}+\frac{1}{16}k^{4}+1-\frac{1}{2}k^{2}=\frac{1}{16}k^{4}+\frac{1}{2}k^{2}+1$;右边$=\left \lbrack \left( \frac{1}{2}k\right)^{2}+1\right\rbrack^{2}=\left \lbrack \frac{1}{4}k^{2}+1\right\rbrack^{2}=\frac{1}{16}k^{4}+\frac{1}{2}k^{2}+1$.$\therefore$左边=右边,$\therefore$等式成立

【解析】：
（1）6,8,10
（2）当$k$是大于2的偶数时，$k^{2}+\left[\left(\frac{1}{2}k\right)^{2}-1\right]^{2}=\left[\left(\frac{1}{2}k\right)^{2}+1\right]^{2}$
证明：左边$=k^{2}+\left[\left(\frac{1}{2}k\right)^{2}-1\right]^{2}=k^{2}+\left(\frac{1}{4}k^{2}-1\right)^{2}=k^{2}+\frac{1}{16}k^{4}-\frac{1}{2}k^{2}+1=\frac{1}{16}k^{4}+\frac{1}{2}k^{2}+1$
右边$=\left[\left(\frac{1}{2}k\right)^{2}+1\right]^{2}=\left(\frac{1}{4}k^{2}+1\right)^{2}=\frac{1}{16}k^{4}+\frac{1}{2}k^{2}+1$
$\because$左边=右边
$\therefore$等式成立