A rectangular picture frame is 40cm long and 30cm wide, leaving an uniform margin of width x cm. The area of the picture is 704cm2. Write expression for the length and width of the picture. Hence find the dimensions of the picture.

Huu hapa ni ufumbuzi wa tatizo lako: Step 1: Andika urefu na upana wa picha kwa kutumia x. Kiwango cha urefu wa fremu ni 40 cm na upana ni 30 cm. Kwa kuwa picha imewekwa katikati na kuna ukingo wa x cm pande zote, urefu na upana wa picha utapungua kwa 2x. Urefu wa picha (L_p) utakuwa: L_p = (40 - 2x) cm Upana wa picha (W_p) utakuwa: W_p = (30 - 2x) cm Step 2: Unda mlinganyo kwa kutumia eneo la picha. Eneo la picha limetolewa kama 704 cm^2. Eneo la mstatili ni urefu × upana. (40 - 2x)(30 - 2x) = 704 Step 3: Tatua mlinganyo wa quadratic ili kupata thamani ya x. Panua mlinganyo: 1200 - 80x - 60x + 4x^2 = 704 4x^2 - 140x + 1200 = 704 Toa 704 kutoka pande zote mbili: 4x^2 - 140x + 1200 - 704 = 0 4x^2 - 140x + 496 = 0 Gawanya mlinganyo mzima kwa 4: x^2 - 35x + 124 = 0 Tumia fomula ya quadratic x = -b ± sqrt(b^2 - 4ac)2a: x = -(-35) ± sqrt((-35)^2 - 4(1)(124))2(1) x = 35 ± sqrt(1225 - 496)2 x = 35 ± sqrt(729)2 x = (35 ± 27)/(2) Kuna thamani mbili zinazowezekana kwa x: x_1 = (35 + 27)/(2) = (62)/(2) = 31 x_2 = (35 - 27)/(2) = (8)/(2) = 4 Step 4: Chagua thamani sahihi ya x. Ikiwa x = 31 cm, basi urefu wa picha ungekuwa 40 - 2(31) = 40 - 62 = -22 cm, ambayo haiwezekani. Kwa hiyo, thamani sahihi ya x ni 4 cm. Step 5: Tafuta vipimo vya picha. Tumia x = 4 cm katika fomula za urefu na upana wa picha: Urefu wa picha: L_p = 40 - 2(4) = 40 - 8 = 32 cm Upana wa picha: W_p = 30 - 2(4) = 30 - 8 = 22 cm Vipimo vya picha ni: Urefu: 32 cm Upana: 22 cm Tuma swali linalofuata 📸