6.3 Converse of the Pythagorean Theorem Đảo của định lí Pythagoras

Since $3\sqrt{7} = \sqrt{63}$ and $\sqrt{49} < \sqrt{63} < \sqrt{121},$ we have $7 < 3\sqrt{7} < 11.$ So the longest side is $c = 3\sqrt{7},$ the other sides are $a = 4$ and $b = 7,$ and a triangle can be formed because $a + b = 11 > c.$ Compare $c^2$ with $a^2 + b^2$:

Since $c^2 < a^2 + b^2,$ the triangle is acute (Theorem 6-5).

Answer: acute.

Since $\sqrt{5} < \sqrt{11} < \sqrt{16} = 4,$ the longest side is $c = 4$; the other sides are $a = \sqrt{5}$ and $b = \sqrt{11}.$ A triangle can be formed because $a + b > 2 + 3 = 5 > 4.$ Compare $c^2$ with $a^2 + b^2$:

Since $c^2 = a^2 + b^2,$ the triangle is right (Theorem 6-3).

Answer: right.

The diagonals cut the kite into four small triangles; each diagonal also cuts it into two halves, which gives four larger triangles (sides $13,$ $20,$ $21$ twice, $13,$ $13,$ $24$ and $20,$ $20,$ $24$). Test each kind:

so the two upper and the two lower small triangles are right (Theorem 6-3). For the larger triangles:

so none of them is right.

Answer: $4$ right triangles.

The diagram has three triangles: the left one, the right one and the whole triangle (its base is $17 + 17 = 34$).

Left triangle, sides $16,$ $17,$ $17$: $16^2 + 17^2 = 256 + 289 = 545 \ne 289 = 17^2,$ so it is not right.

Right triangle, sides $17,$ $17,$ $30$: $17^2 + 17^2 = 578 \ne 900 = 30^2,$ so it is not right.

Whole triangle, sides $16,$ $30,$ $34$:

so it is right.

Answer: $1$ right triangle.

Since $x > 3,$ $x^2 - 9 > 0,$ so all three sides are positive. We show that $c^2 = a^2 + b^2.$

So $c^2 = a^2 + b^2,$ and by Theorem 6-3 the triangle is a right triangle (the right angle is opposite $c$).

Since $x > 5,$ $x^2 - 25 > 0,$ so all three sides are positive. We show that $c^2 = a^2 + b^2.$

So $c^2 = a^2 + b^2,$ and by Theorem 6-3 the triangle is a right triangle (the right angle is opposite $c$).

1. Draw right $\triangle$ with legs $a$ and $b$ and the right angle at $K$; call its hypotenuse $c.$ By Theorem 6-2, $c^2 = a^2 + b^2.$

2. $t^2 > a^2 + b^2$ (given), so $t^2 > c^2$ and, since lengths are positive, $t > c.$

3. The two triangles have two pairs of equal sides ($a$ and $b$), and the third side $t$ is longer than $c.$ By the SSS Inequality Theorem, $m\angle J > m\angle K = 90^\circ,$ so $\angle J$ is obtuse.

Theorem 6-5: if the square of the longest side $c$ of a triangle is less than the sum of the squares of the other two sides $a$ and $b,$ the triangle is acute. Let $\angle C$ be the angle opposite $c.$

1. Draw right $\triangle$ with legs $a$ and $b$ and the right angle at $K$; call its hypotenuse $x.$ By Theorem 6-2, $x^2 = a^2 + b^2.$

2. $c^2 < a^2 + b^2$ (given), so $c^2 < x^2$ and $c < x.$

3. The two triangles have the sides $a$ and $b$ in common, and $c < x.$ By the SSS Inequality Theorem, $m\angle C < m\angle K = 90^\circ,$ so $\angle C$ is acute.

4. $c$ is the longest side, so $\angle C$ is the largest angle of the triangle (the larger side is opposite the larger angle). The other two angles are smaller than $\angle C,$ so they are acute too: the triangle is acute.