题目内容

As shown in the diagram（如图），the triangle PQR has PR=14cm and PQ=10cm．The side RQ produced meets the perpendicular PS at S，so that QS=5cm．The perimeter（周长） of triangle PQR is

cm．
如图，在△PQR中，PR=14cm，PQ=10cm．过点P作PS⊥RQ交RQ的延长线于点S，QS=5cm，则△PQR的周长为

cm．

分析：首先根据勾股定理计算出PS²=75，再利用勾股定理计算出SR的长，然后可得QR的长，进而得到△PQR的周长．

解答：解：∵∠S=90°，
∴PS²=PQ²-SQ²，
∵PQ=10cm，QS=5cm，
∴PS²=75，
∴SR=

142-75

=11cm，
∵SQ=5cm，
∴QR=11-5=6cm，
∴△PQR的周长为10+14+6=30cm，
故答案为：30．

点评：此题主要考查了勾股定理的应用，关键是掌握勾股定理：在任何一个直角三角形中，两条直角边长的平方之和一定等于斜边长的平方．

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（英汉词典：square正方形．intersect…at…与…相交于…）

A.
$数学公式$
B.
$数学公式$
C.
4
D.
5

As shown in the diagram(如图)，the triangle PQR has PR=14cm and PQ=10cm. The side RQ produced meets the perpendicular PS at S, so that QS=5cm. The perimeter(周长) of triangle PQR is _____cm．

As shown in figure point C is on the segment BG and quadrilateral ABCD is a square．AG intersects BD and CD at points E and F，respectively．If AE=5 and EF=3，then FG=（英汉词典：square正方形．intersect…at…与…相交于…）

试题分类

As shown in figure point C is on the segment BG and quadrilateral ABCD is a square．AG intersects BD and CD at points E and F，respectively．If AE=5 and EF=3，then FG=
（英汉词典：square正方形．intersect…at…与…相交于…）