# What is the maximum possible area of a rectangle that can fit inside the equilateral triangle of side length 10 so that one side of the rectangle lies on one side of the triangle?

Table of Contents

- 1 What is the maximum possible area of a rectangle that can fit inside the equilateral triangle of side length 10 so that one side of the rectangle lies on one side of the triangle?
- 2 How do you find the area of a rectangle inscribed in a triangle?
- 3 How many lines of symmetry does a isosceles rectangle have?
- 4 How do you find the maximum length of a rectangle?
- 5 How to prove that a triangle is an isosceles triangle?
- 6 What are the angles opposite to the equal sides of isosceles?
- 7 What is the greatest area of a rectangle inside a triangle?

## What is the maximum possible area of a rectangle that can fit inside the equilateral triangle of side length 10 so that one side of the rectangle lies on one side of the triangle?

21.65

Explanation: Maximum area of rectangle inscribed in an equilateral triangle of side 10 is 21.65.

## How do you find the area of a rectangle inscribed in a triangle?

If ABC is the triangle, let FG be that line, and FGHK be the rectangle. FG is half the length of AB, and FH, a side of the rectangle, is half the altitude CD of the triangle. So the area of the rectangle is 1/4 the base of the triangle times its height, so the rectangle is half the area of the triangle.

**What is the maximum area of a rectangle that can be inscribed in a right triangle?**

half

Therefore it is the case that if a rectangle is inscribed inside a right-angled triangle in this way, its greatest area will be exactly half that of the triangle. One of the first things we must do when taking an algebraic approach is to decide which length in the diagram to consider as our variable.

### How many lines of symmetry does a isosceles rectangle have?

1 line

Therefore, every isosceles triangle by definition has 1 line of symmetry.

### How do you find the maximum length of a rectangle?

Approach: For area to be maximum of any rectangle the difference of length and breadth must be minimal. So, in such case the length must be ceil (perimeter / 4) and breadth will be be floor(perimeter /4). Hence the maximum area of a rectangle with given perimeter is equal to ceil(perimeter/4) * floor(perimeter/4).

**Can isosceles triangle symmetrical?**

Every isosceles triangle has an axis of symmetry along the perpendicular bisector of its base. The two angles opposite the legs are equal and are always acute, so the classification of the triangle as acute, right, or obtuse depends only on the angle between its two legs.

## How to prove that a triangle is an isosceles triangle?

Theorem 2: Sides opposite to the equal angles of a triangle are equal. Proof: In a triangle ABC, base angles are equal and we need to prove that AC = BC or ∆ABC is an isosceles triangle. Construct a bisector CD which meets the side AB at right angles. Or ∆ABC is isosceles.

## What are the angles opposite to the equal sides of isosceles?

Theorem 1: Angles opposite to the equal sides of an isosceles triangle are also equal. Proof: Consider an isosceles triangle ABC where AC = BC. We need to prove that the angles opposite to the sides AC and BC are equal, that is, ∠CAB = ∠CBA. We first draw a bisector of ∠ACB and name it as CD.

**How do you find the area of the red rectangle?**

Therefore the area of the red rectangle is less than half of the area of the triangle A B C A B C. Suppose x = b 2 x = b 2. By showing that this splits the original triangle A B C A B C into four smaller, congruent triangles we can see that the area outside of the red rectangle is equal to the area inside of it.

### What is the greatest area of a rectangle inside a triangle?

The area of the original triangle \\(ABC\\) is given by \\(\\dfrac{ab}{2}\\). Therefore it is the case that if a rectangle is inscribed inside a right-angled triangle in this way, its greatest area will be exactly half that of the triangle.