The perimeter of the rectangle below is 120 units. Find the length of side AB .?
I'm not sure you have this question right... If you have a rec DCAB then "side" DA would not actually be a side. It's the diagonal of the rectangle making it the hypotenuse of the right triangle DAB. To solve that you would need to find the length of side DB. By a^2 + b^2 = c^2 So it would be. DB^2 + BA(4z)^2 = DA(5z-3)^2 Solve for DB and the use the formula 2DB + 2BA = 120 That will tell you what z is and then you just plug that back in to find the sides.
Find the length and width of a rectangle that has a perimeter P and a maximum area?
that's a situation in Differential Calculus and being so, the prognosis/answer ought to contain a Differential Calculus idea (differentiation) and not in any respect trial-and-mistakes. the first formula to think about is section = A = LW the position L = length of the rectangle W = width of the rectangle considering the fact that Perimeter = P = 36 = 2(L + W), then W = 18 - L and substituting this contained in the formula for the section, A = L(18 - L) = 18L - L^2 Differentiating the above, dA/dL = 18 - 2L and for dA/dL to be optimal, it will be set equivalent to 0. therefore, dA/dL = 0 = 18 - 2L fixing for L, L = 9 considering the fact that L = 9 ft, then W = 18 - L = 18 - 9 W = 9 ft to envision even if the calculated dimensions will certainly yield the optimal section, take the 2d spinoff of the unique equation for the section, d^2A/dL^2 = -2L and considering the fact that d^2A/dL^2 < 0 (detrimental value) then the calculated values are the size which will yield the optimal section. answer: Dimensions are: length = Width = 9 ft
Let the length of the rectangle be x and that of the breadth be y.Since the length is twice of its breadth,therefore x/y= 2/1.=> x=2y. Perimeter of the rectangle is 60m. => 2(x+y)= 60. => x+y= 30. Area of the rectangle = xy . Substituting the value of x as 2y in the equation x+y = 30 we get 3y = 30 . =>y= 10. Since x=2y,the value of x=20. Area of the rectangle = 20 x 10 =200 m sq units.
Perimeter of Rectangle, Find side length help?
Okay... the rectangle ABCD is made up of sides AB, BC, CD, and AD. If DA is 2y+3, then the side opposite (BC) it is also 2y+3 because a rectangle is made up of two parallel lines intersecting at right angles. The same applies to AB and CD. The total is 106. 106 = 2y+3, 2y+3, 3y, and 3y. 106 = 10y+6 (I added everything together) 100 = 10y (I subtracted 6 from each side) 10 = y (I divided each side by 10) So, BC is the side opposite DA, making it equal to 2y+3. 2(10)+3 simplifies to 20 + 3, simplifies to 23. Your answer is 23.
The perimeter of the rectangle below is 130 units. Find the length of the side QR.?
PS = 5x QR = 4x +2 Perimeter: 2(L + W) = 130 2(5x + 4x +2) = 130 10x +8x +4 = 130 18x = 130 -4 18x = 126 x = 126/18 x = 7 QR = 4x +2 = 4(7) + 2 = 30 units
Pythagorean Theoem(Side 1)^2 + (side 2) ^2 = (Hyptenuse)^2x^2 + 126^2 = 130^2x^2 + 126^2 - 126^2 = 130^2 - 126^2x^2 = 16900 - 15876x^2 = 1024 SqRt(x^2) = SqRt(1024)x = 32��C 8�0Source: Joe
Perimeter of a rectangle = 2 x (Length + breadth) = pLet initial length be 'l'and initial breadth be 'b'Therefore , 2 x ( l + b) = 25 cmNow dimensions are tripled. So the new length is 3 times the original length i.e. 3l and similarly breadth is 3b So, new perimeter = 2 x (New Length + New Breadth) = 2 x ( 3l + 3b) = 3 x ( 2 x (l+b)) = 3 x original perimeter = 3 x 25 cm = 75 cm
The perimeter P, i.e., the distance around a rectangle, is given by the formula: P = 2l + 2w, where l and w are the length and the width of the rectangle, respectively.Since the perimeter P is 10 times the width w, then we can write this as:P = 10w, andsubstituting into the perimeter formula, we get:10w = 2l + 2wNow, solving for l, we first subtract 2w from both sides of the equation as follows:10w- 2w = 2l + 2w - 2wNow, collecting like-terms on both the right and left sides, we get:8w = 2l + 02l = 8wNow, dividing both sides by 2 to isolate l and thus solve for l, we have:(2l)/2 = (8w)/2(2/2)l = (8/2)w(1)l = 4wl = 4wCHECK: P = 2l + 2w10w = 2(4w) + 2w10w = 8w + 2w10w = 10wTherefore, the length of the given rectangle is indeed 4 times the width, i.e., l = 4w.
Find the maximum area for a rectangle with a perimeter of 36 feet. state the length and width?
This is a problem in Differential Calculus and being so, the analysis/solution should involve a Differential Calculus concept (differentiation) and not trial-and-error. The first formula to consider is Area = A = LW where L = length of the rectangle W = width of the rectangle Since Perimeter = P = 36 = 2(L + W), then W = 18 - L and substituting this in the formula for the area, A = L(18 - L) = 18L - L^2 Differentiating the above, dA/dL = 18 - 2L and for dA/dL to be maximum, it should be set equal to 0. Hence, dA/dL = 0 = 18 - 2L Solving for L, L = 9 Since L = 9 feet, then W = 18 - L = 18 - 9 W = 9 feet To check whether the calculated dimensions will indeed yield the maximum area, take the second derivative of the original equation for the area, d^2A/dL^2 = -2L and since d^2A/dL^2 < 0 (negative value) then the calculated values are the dimensions that will yield the maximum area. ANSWER: Dimensions are: Length = Width = 9 feet
The area A of a rectangle is given by the formula: A = lw, where w = the width of the rectangle, and l = the length of the rectangle.The perimeter P of a rectangle is the distance around a rectangle and is given by the following formula:P = 2l + 2wSince the length l is 2 larger than the width, then we can write the following equation which relates l and w: l = w + 2Now, substituting P = 32 and l = w + 2 into the perimeter formula, we have:P = 2l + 2w32 = 2(w + 2) + 2w32 = 2(w) + 2(2) + 2w32 = 2w + 4 + 2w32 = 2w + 2w + 432 = 4w + 432 - 4 = 4w + 4 - 432 - 4 = 4w + 028 = 4w4w = 28 (Since equality is symmetric, i.e., if a = b, then b = a)4w/4 = 28/4(4/4)w = 28/4(1)w = 7w = 7Therefore, ...l = w + 2l = 7 + 2l = 9CHECK:P = 2l + 2w32 = 2(9) + 2(7)32 = 18 + 1432 = 32Therefore, the area of the given rectangle is:A = lw = (9)(7)A = 63 square units