Bantu jawab soal berikut pada gambar di bawah !

misalkan di tengah-tengah HG = D


Luas penampang = Luas segitiga = 12
1/2 * (2x + 2) t = 12
(x + 1) t = 12
t = 12/(x + 1)    ..... (1)

perhatikan sgitiga AEH dan CDH sebangun (sudut,sudt,sudt), maka berlaku :

AE/ HD = HE/CD
x / 1 = 3 / (t - 3)
x = 3/(t - 3)
substitusi nilai t pada pers (1) ke x, get
x = 3/(12/(x+1))
x = 3(x+1)/12
x = (x+1)/4
4x = x + 1
x = 1/3

substitusi x = 1/3 ke pers (1), get
t = 12/(x +1) = 12/(1/3 + 1) = 12/(4/3) = 12 * 3/4 = 9

jadi alas, (2x + 2) = 2(1/3) + 2 = 2/3 + 2 = 8/3
dan tinggi t = 9