1.) triangular corner lot has perpendicular sides of lengths 120 feet and 160 feet. Find the dimensions of the largest rectangular building that can be constructed on the lot with sides parallel to the streets.

let the length of the rectangle parallel to the 160 ft side be x

let its height be y
I see 3 similar triangles, the original and the 2 smaller triangles
by ratios:
y/(160-x) = 120/160 = 3/4
4y = 480 - 3x
y = 120 - 3x/4

area of rectangle = A
= xy
= x(120 - 3x/4) = 120x - 3x^2/4
dA/dx = 120 - 3x/2
= 0 for a max of A
120 - 3x/2 = 0
3x = 240
x = 80 , as expected
then y = 120 - 3(80)/4 = 60 , also as expected

the rectangle should be 80 ft by 60 ft , (the midpoints of the sides)

Well, if the lot is triangular, I hope the building won't be triangular too, otherwise that would be quite the unique architectural choice! Now, let me do some math clowning.

To find the largest rectangular building that can be constructed on the lot with sides parallel to the streets, we need to figure out the dimensions.

The streets form a right angle, and we have two sides perpendicular to each other, measuring 120 feet and 160 feet.

To maximize the area of the rectangle, we want the length and width to be as big as possible.

Since we don't have a third side of the triangle, let's assume we extend the 120-foot side to complete the rectangle.

So, the length of the building would be 160 feet, which is the longer side of our triangular lot, and the width would be 120 feet, the shorter side of the triangular lot.

Therefore, the largest rectangular building that can be constructed on the lot with sides parallel to the streets would have dimensions of 160 feet by 120 feet. Happy building!

To find the dimensions of the largest rectangular building that can be constructed on the triangular corner lot with sides parallel to the streets, we need to determine the maximum possible area.

Since the sides of the triangular lot are perpendicular, we can consider the triangle as a right triangle. The two perpendicular sides have lengths 120 feet and 160 feet.

To find the maximum area, we need to determine the longest side of the rectangular building. This side will be the hypotenuse of the right triangle.

Using the Pythagorean theorem, we can find the length of the hypotenuse:
c^2 = a^2 + b^2
c^2 = 120^2 + 160^2
c^2 = 14400 + 25600
c^2 = 40000
c = √40000
c ≈ 200 feet

Therefore, the longest side of the rectangular building that can be constructed on the lot is approximately 200 feet.

Since the sides of the rectangular building must be parallel to the streets, we can consider one side of the rectangle as the longer side (200 feet) and the other side as the shorter side.

To find the dimensions of the rectangle, we need to determine the other side.

Since one side of the rectangle is 200 feet, and we have a right triangle with sides 120 feet and 160 feet, the other side of the rectangle can be found using the Pythagorean theorem:

a^2 = c^2 - b^2
a^2 = 200^2 - 120^2
a^2 = 40000 - 14400
a^2 = 25600
a = √25600
a ≈ 160 feet

Therefore, the dimensions of the largest rectangular building that can be constructed on the triangular corner lot with sides parallel to the streets are approximately 200 feet by 160 feet.

To find the dimensions of the largest rectangular building that can be constructed on a triangular corner lot, we need to maximize the area of the rectangle. In this case, the two perpendicular sides of the triangular lot, measuring 120 feet and 160 feet, will act as the potential sides of the rectangle.

Let's assume the dimensions of the rectangle are x and y, with x representing the length parallel to the 120-foot side of the lot and y representing the length parallel to the 160-foot side of the lot.

We can visualize the problem by drawing a diagram. Start by drawing the triangular lot with perpendicular sides of 120 feet and 160 feet. Then draw a rectangle inside the triangular lot with sides x and y, parallel to the streets. The rectangle should fit snugly within the triangular lot.

To maximize the area, we need to consider the property of a triangle that tells us that the height of the triangle will bisect the base when the area is maximized. In this case, the base of the triangle is the 160-foot side of the lot, and the height is the 120-foot side of the lot.

Since the rectangle has to fit within the triangular lot, the length x can be at most 120 feet, and the length y can be at most 160 feet.

To maximize the area of the rectangle, we need to choose the larger dimension for the rectangle. So, we will set x equal to 120 feet since it is smaller than y.

Therefore, the dimensions of the largest rectangular building that can be constructed on the given triangular corner lot with sides measuring 120 feet and 160 feet are 120 feet (length parallel to the 120-foot side) and y feet (length parallel to the 160-foot side), where y should be less than or equal to 160 feet.

Note: Without the actual value of y given, we cannot determine the exact dimensions of the rectangle.