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9.4.4 Solution of Triangles, Long Questions (Question 7 & 8)


Question 7:
Diagram below shows a quadrilateral ABCD where the sides AB and CD are parallel. ∠BAC is an obtuse angle.

Given that AB = 14 cm, BC = 27 cm, ∠ACB = 30o and AB : DC = 7 : 3.
Calculate
(a) ∠BAC.
(b) the length, in cm, of diagonal BD.
(c) the area, in cm2, of quadrilateral ABCD.


Solution:
(a)
sinBAC27=sin30o14sinBAC=sin30o14×27sinBAC=0.9643 BAC=74.64oBAC (obtuse)=180o74.64o =105.36o

(b)
AB parallel with DCBAC=ACDBCD=105.36o+30o   =135.36oDCAB=37DC=37×14 cm  =6 cmBD2=272+622(27)(6)cosBCDBD2=765324cos135.36oBD2=995.54BD=31.55 cm

(c)
ABC=180o30o105.36o   =44.64oAC2=272+1422(27)(14)cosABCAC2=925756cos44.64oAC2=387.08AC=19.67 cmArea ABC=12(14)(27)sin44.64o=132.80 cm2Area ACD=12(19.67)(6)sin105.36o=56.90 cm2Area of quadrilateral ABCD=132.80+56.90=189.7 cm2


Question 8:
Diagram below shows a cyclic quadrilateral PQRS.

(a) Calculate
(i) the length, in cm, of PR,
(ii) ∠PRQ.
(b) Find
(i) the area, in cm2, of ∆ PRS,
(ii) the short distance, in cm, from point S to PR.


Solution:
(a)(i)
PR2=72+822(7)(8)cos80oPR2=11319.4486PR=93.5514PR=9.6722 cm

(a)(ii)
In cyclic quadrilateralPQR+PSR=180PQR+80=180PQR=100osinQPR3=sin1009.6722sinQPR=0.3055QPR=17o47PRQ=180o100o17o47  =62o13

(b)(i)
Area of PRS=12×7×8sin80o=27.5746 cm2

(b)(ii)

Area of PRS=27.574612×9.6722×h=27.5746   h=27.5746×29.6722 =5.7018 cmShortest distance=5.7018 cm

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