Question

Please Please Help! I really need help with this question!:Please explain step by step! Thank You. " In a scale drawing of a triangular banner, one side measure's 16 centimeters and the other two sides each measure 12 centimeters. On the actual banner, these two sides each measure 36 feet. What is the lenght of the remaining side of the actual banner?"

Answer 1

we know the sides on the isosceles triangle are 12, 12 and 16 on the drawing on paper.

we also know that the actual size for the 12-twin sides is 36, so we can set a proportion using those two and check at what ratio the drawing : actual are.

`\bf \cfrac{drawing}{actual}\qquad \cfrac{12}{36}\implies \cfrac{1}{3}\implies \stackrel{cm}{1}:\stackrel{ft}{3}`

since we know the drawing : actual ratio is really 1 : 3, then we can use that to find the actual size for the last side, we know on the drawing is 16 cm, so

`\bf \cfrac{drawing}{actual}\qquad 1:3\qquad \qquad \cfrac{\stackrel{drawing}{16}}{\stackrel{actual}{a}}=\cfrac{1}{3}\implies \cfrac{16\cdot 3}{1}=a\implies 48=a`

Answer 2

The three sides have the ratios

... (16 cm) : (12 cm) : (12 cm)

and

... (x ft) : (36 ft) : (36 ft)

Perhaps you can see that the numbers in the second ratio are 3 times the numbers in the first ratio, with cm changed to ft. If we multiply 16 cm by 3 and change cm to ft, we get

... x ft = (16 cm)*(3 ft/cm) = 48 ft

The length of the remaining side is 48 ft.