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)
sinBAC 27 = sin 30 o 14 sinBAC= sin 30 o 14 ×27 sinBAC=0.9643  BAC= 74.64 o BAC ( obtuse )= 180 o 74.64 o  = 105.36 o

(b)
AB parallel with DC BAC=ACD BCD= 105.36 o + 30 o    = 135.36 o DC AB = 3 7 DC= 3 7 ×14 cm   =6 cm B D 2 = 27 2 + 6 2 2( 27 )( 6 )cosBCD B D 2 =765324cos 135.36 o B D 2 =995.54 BD=31.55 cm

(c)
ABC= 180 o 30 o 105.36 o    = 44.64 o A C 2 = 27 2 + 14 2 2( 27 )( 14 )cosABC A C 2 =925756cos 44.64 o A C 2 =387.08 AC=19.67 cm Area ABC = 1 2 ( 14 )( 27 )sin 44.64 o =132.80  cm 2 Area ACD = 1 2 ( 19.67 )( 6 )sin 105.36 o =56.90  cm 2 Area of quadrilateral ABCD =132.80+56.90 =189.7  cm 2


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)
P R 2 = 7 2 + 8 2 2( 7 )( 8 )cos 80 o P R 2 =11319.4486 PR= 93.5514 PR=9.6722 cm

(a)(ii)
In cyclic quadrilateral PQR+PSR=180 PQR+80=180 PQR= 100 o sinQPR 3 = sin100 9.6722 sinQPR=0.3055 QPR= 17 o 47 PRQ= 180 o 100 o 17 o 47           = 62 o 13

(b)(i)
Area of PRS = 1 2 ×7×8sin 80 o =27.5746  cm 2

(b)(ii)

Area of PRS=27.5746 1 2 ×9.6722×h=27.5746                    h= 27.5746×2 9.6722                      =5.7018 cm Shortest distance=5.7018 cm

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