1. an architect wants to design a rectangular room sothat its length is 8 meters

Question

1. an architect wants to design a rectangular room sothat its length is 8 meters more than its width and its perimeteris greater than 56 meteres. If each of the dimensions of theroom mujst be a whole number of meters, waht are the smallestpossible measures of the length and width? 2. Each leg of an isosceles triangle measures 5 cm morethan the base of the triangle. If the perimeter of the triangle isgreater than 70 cm, what are the smallest possible integralmeasures of the sides of the triangle? 1. an architect wants to design a rectangular room sothat its length is 8 meters more than its width and its perimeteris greater than 56 meteres. If each of the dimensions of theroom mujst be a whole number of meters, waht are the smallestpossible measures of the length and width? 2. Each leg of an isosceles triangle measures 5 cm morethan the base of the triangle. If the perimeter of the triangle isgreater than 70 cm, what are the smallest possible integralmeasures of the sides of the triangle?

Explanation / Answer

Let a is the length of the rectangularroom.       b is the width of therectangular room. We have:             a = b +8                                (1) and the perimeter is greater than 56 =>           2a + 2b > 56 =>          2(a+b ) > 56 =>             a + b >28                       (2) Sustitute (1) into (2)                   b+ 8 + b > 28              =>2b + 8    > 28            => 2b           >20             =>   b           >10 => the smallest possible whole number for b( or thewidth) is 11meters =>   a = b+ 8 = 11+8 = 19meters => the smallest possible whole number for thelength is 19 meters. Question 2: Let a is the base of the isosceles triangle.       b is the leg (orthe side) of the isosceles triangle. We have:                b = 5 + a and the perimeter is greater than 70cm =>          a + b+ b       > 70 =>          a + 2b          >70 Sustitute b = 5+a into the above:    =>        a + 2 (5 + a) > 70 =>            a+ 10 + 2a    > 70 =>          3a   + 10        > 70 =>                 3a           > 60 =>                        a      > 20 => The smallest possible intergral measures of thesides of the triangle is 21 cm =>   b = 5+ a = 5+21 = 26cm => The smallest possible intergral measures of thesides of the triangle is 26cm Let a is the base of the isosceles triangle.       b is the leg (orthe side) of the isosceles triangle.       b is the leg (orthe side) of the isosceles triangle. We have:                b = 5 + a and the perimeter is greater than 70cm =>          a + b+ b       > 70 =>          a + 2b          >70 Sustitute b = 5+a into the above:    =>        a + 2 (5 + a) > 70 =>            a+ 10 + 2a    > 70 =>          3a   + 10        > 70 =>                 3a           > 60 =>                        a      > 20 => The smallest possible intergral measures of thesides of the triangle is 21 cm =>   b = 5+ a = 5+21 = 26cm => The smallest possible intergral measures of thesides of the triangle is 26cm

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.