A triangle has a base of 16 inches and an altitude of 8 inches. Find the dimensions of the largest rectangle that can be inscribed in the triangle if the base of the rectangle coincides with the base of the triangle.
I suspect it will be 4x8, but let's see what the calculations show.
If triangle ABC has vertices at (0,0),(16,0) and (h,8) then the sloping sides are lines with the equations
y = 8/h x
y = 8(x-16)/(h-16)
Now, suppose the inscribed rectangle has height k. The the line y=k intersects the sides where x is
hk/8 and k(h-16)/8 + 16
The length of the rectangle is thus 16-2k
The area of the rectangle is
a = k(16-2k) = 16k-2k^2
The vertex of this parabola is where k=4.
Thus the inscribed rectangle of maximal area is half the height of the triangle, making its base half the base of the triangle. The rectangle is thus 4x8.
Why did the rectangle join the circus? Because it wanted to be in-SCALED!
In this case, since the base of the rectangle coincides with the base of the triangle, the base of the rectangle will also be 16 inches. Now let's find the height of the largest rectangle that can be inscribed in the triangle.
Since the altitude of the triangle is 8 inches, it splits the triangle into two smaller triangles. Both of these smaller triangles are similar to the larger triangle.
Using the property of similar triangles, we can say that the shorter side of the triangle is to the shorter side of the smaller triangle as the longer side of the triangle is to the longer side of the smaller triangle.
Therefore, the height of the smaller triangle is half the height of the larger triangle. In this case, the height of the smaller triangle is 4 inches.
Since the base of the rectangle coincides with the base of the triangle, the rectangle will also have a base of 16 inches.
So, the dimensions of the largest rectangle that can be inscribed in the triangle are 16 inches for the base and 4 inches for the height.
To find the largest rectangle that can be inscribed in the given triangle, we need to determine the dimensions of the rectangle.
Let's begin by drawing the given triangle:
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The base of the triangle, which coincides with the base of the rectangle, has a length of 16 inches. The altitude of the triangle, which is perpendicular to the base, has a length of 8 inches.
Since the rectangle is inscribed in the triangle, its width will extend from one side of the triangle to the other, while its length will be perpendicular to the base of the triangle. Let's assume the width of the rectangle is 'w', and the length is 'l'.
We now need to find the relationship between the width and length of the rectangle.
Considering the given triangle, we know that its base has a length of 16 inches and its altitude measures 8 inches. The concept of similar triangles states that the ratios of corresponding sides of two similar triangles are equal.
In this case, we can consider the small triangle formed by the base and altitude of the larger triangle, and the smaller triangle formed by the corresponding sides of the rectangle within the larger triangle.
Let's label the sides of the large triangle as follows:
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A B
And the sides of the rectangle as follows:
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A B
Using the concept of similar triangles, we can set up the following ratios:
Length ratio: l / 8 = B / 16
Width ratio: w / 16 = A / 8
Now, we can solve these ratios to find the relationship between the width and length of the rectangle.
Cross-multiplying the length ratio:
l = (B/16) * 8
l = B/2
Cross-multiplying the width ratio:
w = (A/8) * 16
w = 2A
Now, we can substitute the values of 'l' and 'w' back into the length ratio:
l / 8 = B / 16
(B/2) / 8 = B / 16
Multiplying both sides by 8:
B/2 = B/2
This equation implies that the base 'B' of the rectangle can be any value. Therefore, the base of the rectangle can have any length as long as it does not exceed the base of the given triangle (16 inches).
To maximize the area of the rectangle, we can make the width 'w' of the rectangle as large as possible. Since w = 2A, and A cannot exceed the altitude of the triangle (8 inches), the largest value for A (and thus, w) is 8 inches.
Therefore, the dimensions of the largest rectangle that can be inscribed in the triangle are:
Width (w) = 2A = 2 * 8 = 16 inches
Length (l) = B/2 = 16/2 = 8 inches
To find the dimensions of the largest rectangle that can be inscribed in a triangle, we need to determine the rectangle's height and width within the given triangle.
To understand this, let's first visualize the problem. In this case, the base of the triangle measures 16 inches, and the altitude (or height) measures 8 inches. We need to find the dimensions (height and width) of the largest rectangle that can fit inside this triangle, with the base of the rectangle coinciding with the base of the triangle.
To solve this, we can observe that the largest rectangle is a square, as it will maximize the area within the given triangle. Therefore, the width and height of the rectangle will be equal.
To determine the dimensions of this square (rectangle), we need to find the length of its sides.
Here's how we can do it:
Step 1: Split the triangle into two smaller right-angled triangles by drawing a perpendicular line from the top vertex (where the altitude intersects the base) to the base. This line bisects the base into two equal parts, each measuring 8 inches.
Step 2: Now, we have a right-angled triangle with a base of 8 inches and a height of 8 inches. Since the base of the rectangle coincides with the base of the triangle, the width of the rectangle will be 8 inches.
Step 3: To find the height of the rectangle, we can use the Pythagorean theorem. The hypotenuse of the right-angled triangle is the diagonal of the square (rectangle) we are trying to find.
Applying the Pythagorean theorem:
hypotenuse^2 = base^2 + height^2
Let the height of the rectangle be h inches.
8^2 + h^2 = (16/2)^2 (since the base of the right-angled triangle and the rectangle are equal)
64 + h^2 = 64
h^2 = 64 - 64
h^2 = 0
h = √0 or 0.
Step 4: The height of the rectangle is found to be 0. However, for a rectangle to exist, it should have both non-zero height and width. Since the height is 0, it means that no rectangle can be inscribed within the given triangle when the base of the rectangle coincides with the base of the triangle.
Therefore, there are no dimensions for the largest rectangle that fits these conditions.