Question

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luas PQR adalah....cm2
a.11/25
b.1
c.10
d.17 1/2



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Answer 1

Cari terlebih dahulu nilai x dengan Menggunakan dalil pythagoras:[tex] \sf {RQ}^{2} = {PQ}^{2} + {PR}^{2} [/tex][tex] \sf {(\sqrt{5})}^{2} = {4x - 3}^{2} + {2x}^{2} [/tex][tex] \sf 5 = 16x^2 - 24x+ 9 + 4x^2 [/tex][tex] \sf16x^2 - 24x + 9 + 4x^2- 5 = 0 [/tex][tex] \sf20x^2 - 24x + 4 = 0 [/tex][tex] \sf5x^2 - 6x + 1 = 0 [/tex][tex] \sf(5x - 1)(x + 1) = 0 [/tex][tex] \sf x_1 = \frac{1}{5}~~~~~~~~~~~ x_2 = 1[/tex]untuk [tex]\rm x = \frac{1}{5}[/tex] tidak memenuhi karena panjang PQ bernilai negatif....PQ = (4x - 3) cm = (4(1) - 3) cm = (4 - 3) cm = 1 cmPR = 2x cm = 2(1) cm = 2 cm...L= ½ × PR × PQL= ½ × 2 cm × 1 cmL= 1 cm × 1 cmL = 1 cm²OPSI B