In the figure, PQRS is a parallelogram in which PS=7 cm,PT=12cm and SR=25cm, Find
a) Angle SPR
b) the length of the diagonal QS(cor. to 3 sig. fig.)
圖片參考：http://i1127.photobucket.com/albums/l637/hkc1996/16.jpg
Maths Quadrilaterals F.3 (10)
2011-03-13 2:11 am
回答 (2)
2011-03-13 2:24 am
✔ 最佳答案
(a) PS = 7 cm,PT = 12 cm and SR = 25 cmPR = 2PT = 24 cm
Using cosine law
cos∠SPR = [(PS)^2 + (PR)^2 - (SR)^2]/[2(PS)(PR)]
cos∠SPR = (49 + 576 - 625]/336 = 0
∠SPR = 90 (Or by Pythagorean's theorem to get the same answer)
(b) ST^2 = PS^2 + PT^2
ST^2 = 193
ST = 13.89 cm
QS = 2ST = 27.8 cm (cor. to 3 sig. fig.)
2011-03-13 6:59 pm
a) PT=TR (prop.of parallelogram)
PR = 2PT = 24cm
PR²+PS² = 7²+24² = 625
SR² = 25² = 625
SR² = PR²+PS²
∴∠SPR = 90° (converse of Pyth. theorem)
b)∠SPR = 90° (proved)
ST² = PS²+PT² (pyth.thm.)
ST = √7²+12²
ST = 13.892 (corr.to 5 sig. fig.)
ST = TQ (prop.of parallelogram)
SQ = 2ST = 27.8 (corr. to 3 sig. fig.)
PR = 2PT = 24cm
PR²+PS² = 7²+24² = 625
SR² = 25² = 625
SR² = PR²+PS²
∴∠SPR = 90° (converse of Pyth. theorem)
b)∠SPR = 90° (proved)
ST² = PS²+PT² (pyth.thm.)
ST = √7²+12²
ST = 13.892 (corr.to 5 sig. fig.)
ST = TQ (prop.of parallelogram)
SQ = 2ST = 27.8 (corr. to 3 sig. fig.)
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