Given an isosceles triangle with a base of 12 and a side length of 18. What segments need to be marked off from the vertex of the triangle on its sides so that connecting their ends forms a trapezoid with a perimeter of 40?

#
The smaller base of the trapezoid cuts off a similar triangle from the given triangle.

## Solution

Let $ABC$ be the given triangle, $AB=BC=18, AC=12$. Denote the sought segments $BM$ and $BN$ by $x$. Then $AM=CN=18-x$.

From the similarity of triangles $MBN$ and $ABC$, we find that $MN=\frac{BM \cdot AC}{AB}=2x / 3$.

The perimeter of trapezoid $AMNC$ is $2AM + AC + MN = 2(18 - x) + 12 + 2x / 3 = 40$. From this, $x=6$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96d74c2697cd905bd948g-16.jpg?height=323&width=552&top_left_y=2480&top_left_x=753)

Answer

![](https://cdn.mathpix.com/cropped/2024_05_06_96d74c2697cd905bd948g-17.jpg?height=103&width=1092&top_left_y=157&top_left_x=11)

Inside a right angle, a point $M$ is given, the distances from which to the sides of the angle are 4 and 8. A line passing through point $M$ cuts off a triangle from the right angle with an area of 100. Find the legs of the triangle.

## Solution

Denote the sought legs by $x$ and $y$. Then $xy=200$. Moreover, from the similarity of right triangles, it follows that $8 / y - 4 = x / y$. Hence, $x = 8y / (y - 4)$.

Substituting $x$ into the first equation, we get $8y^2 / (y - 4) = 200, y^2 = 25y - 100, y^2 - 25y + 100 = 0$.

Therefore, $y_1 = 20, x_1 = 10$ or $y_2 = 5, x_2 = 40$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96d74c2697cd905bd948g-17.jpg?height=234&width=389&top_left_y=749&top_left_x=846)

Answer

20, 10 or $5, 40$.