the length of a rectangle is 8cm more than the width and its area is 172cm^2 .Find
a) the width od the rectangle;
b) the length of the diagonal of the rectangle, giving your answer correct to 2 decimal places
Q1. A wire of length 200 cm is cut into two parts and each part is bent to form a square. If the area of the larger square is 9 times that of the smaller square, find the perimeter of the larger square.
Q2. A wire of length 100cm is cut into two parts and each part is bent to form a square. If the sum of the areas of the two squares is 425cm^2, find the lengths of the sides of the two squares.
Q3. A group of students is on a sightseeing tour. the total fare is $120 and this is to be shared equally among the students. If two more students join the tour, each will pay $2 less. Find the original number of students in the group
I agree with steve
To solve this problem, we can use the formula for the area of a rectangle: Area = length × width.
a) Let's assume the width of the rectangle is "x". According to the problem, the length of the rectangle is 8cm more than the width. So, the length can be represented as "x + 8".
The formula for the area of a rectangle is:
Area = length × width
Given that the area is 172cm^2, we can set up the equation:
172 = (x + 8) × x
To solve this equation, we need to expand the brackets and then rearrange it:
172 = x^2 + 8x
0 = x^2 + 8x - 172
Now we can solve this quadratic equation. We can either factorize it or use the quadratic formula.
Using factoring, we can find two numbers whose sum is 8 and whose product is -172. These numbers are 14 and -6.
So, the quadratic equation can be factorized as:
(x + 14)(x - 6) = 0
Setting each factor equal to zero, we find two possible solutions:
x + 14 = 0 or x - 6 = 0
Solving these equations, we find:
x = -14 or x = 6
Since we are dealing with lengths, the width cannot be negative. Therefore, the width of the rectangle is 6cm.
b) To find the length of the diagonal of the rectangle, we can use the Pythagorean theorem. According to the theorem, in a right triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.
In this case, the width and length of the rectangle form the two sides of the right triangle, and the diagonal is the hypotenuse.
The formula for the Pythagorean theorem is:
diagonal^2 = width^2 + length^2
Substituting the values we know, we get:
diagonal^2 = 6^2 + (6 + 8)^2
diagonal^2 = 36 + 196
diagonal^2 = 232
To find the length of the diagonal, we take the square root of both sides:
diagonal = √232
Calculating the square root of 232 gives us approximately 15.26 cm.
Therefore, the length of the diagonal of the rectangle is approximately 15.26 cm.
w(w+8) = 172
w = 2(√47 - 2)
Now you can get the other values.
Q1 since the larger is 9 times the area, it has 3 times the perimeter. So, the two squares have perimeters 50 and 150.
Q2
x^2 + y^2 =425
4x+4y = 100
The squares have sides 5 and 20
Q3
If there are x students paying $y each, then
xy = 120
(x+2)(y-2) = 120
By inspection, x=10,y=12