[Pythagorean Theorem (direct and inverse).] [ Theorem on the sum of the squares of the diagonals ]

The height of a rhombus, drawn from the vertex of the obtuse angle, divides its side into segments of length $a$ and $b$. Find the diagonals of the rhombus.
Let the height $B H$, drawn from the vertex $B$ of the obtuse angle of the rhombus $A B C D$, divide the side $A D$ into segments $A H=a$ and $D H=b$. By the Pythagorean theorem,

$B H^{2}=(a+b)^{2}-a^{2}=2 a b+b^{2}, B D^{2}=2 a b+b^{2}+b^{2}=2 b(a+b), A C^{2}=4 A B^{2}-B D^{2}=4(a+b)^{2}-2 b(a+b)=2(2 a+b)$ $(a+b)$.

Similarly for the case $A H=b, D H=a$.

## Answer

$\sqrt{2 b(a+b)}, \sqrt{2(a+b)(2 a+b)}$ or $\sqrt{2 a(a+b)}, \sqrt{2(a+b)(2 b+a)}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c1015ea0b2d2805d8876g-01.jpg?height=174&width=2026&top_left_y=1148&top_left_x=16)

The side of the triangle is 2, and the adjacent angles are $30^{\circ}$ and $45^{\circ}$. Find the other sides of the triangle.

## Hint

Draw the height from the vertex of the largest angle of the triangle.

## Solution

Let $A D=h$ be the height of the given triangle $A B C, \angle B=45^{\circ}, \angle C=30^{\circ}$. Then $B D=A D=h, C D=h \sqrt{3}, h$ $+h \sqrt{3}=B C+C D=2$, from which $h=\frac{2}{\sqrt{3}+1}=\sqrt{3}-1$. Therefore, $A B=B D \sqrt{2}=\sqrt{2}(\sqrt{3}-1)$, $A C=$ $2 h=2(\sqrt{3}-1)$.

## Answer

$2(\sqrt{3}-1), \sqrt{2}(\sqrt{3}-1)$

![](https://cdn.mathpix.com/cropped/2024_05_06_c1015ea0b2d2805d8876g-01.jpg?height=171&width=2030&top_left_y=2347&top_left_x=11)

The cosine of the angle at the base of an isosceles triangle is $3 / 5$, and the height dropped to the base is $h$. Find the height dropped to the lateral side.

## Solution

If $\cos \alpha=3 / 5$, then $\sin \alpha=4 / 5$, so the lateral side of the triangle is $5 / 4$, and the base is $6 / 4$. According to the formula for the area of a triangle, the heights are inversely proportional to the corresponding sides, so the required

## Answer

$6 h / 5$