Euler's Eyes
2026-10-11
Exercises from chapter 4 — Triangles
Exercises and resolutions
Exercise 80 — Classify as true (V) or false (F)
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All isosceles triangle is equilateral.
Falso. An isosceles can have a different base than the other sides.
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All equilateral triangle is isosceles.
Verdadeiro. It has (at least) two congruent sides.
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A scalene triangle can be isosceles.
Falso. Scalene does not have two congruent sides.
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Every isosceles triangle is an acute triangle.
Falso. There are isosceles rectangles and obtuse angles.
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Every right triangle is scalene triangle.
Falso. The isosceles rectangle (45°, 45°, 90°) is a counterexample.
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There are right-angled and isosceles triangles.
Verdadeiro. Congruent legs.
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There is isosceles triangle obtuse angle.
Verdadeiro. Obtuse vertex angle, such as 120°, 30°, 30°.
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Every acute triangle is either isosceles or equilateral.
Falso. There are scalene acutangles, such as 50°, 60°, 70°.
Exercise 81 — Classify as true (V) or false (F)
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All isosceles triangles are congruent.
Falso. They can have different sizes and shapes.
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All equilateral triangles are congruent.
Falso. They have the same shape, but not the same size.
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All right triangles are congruent.
Falso. The cathets may vary.
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All isosceles right triangles are congruent.
Falso. The size of the peccary varies.
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All acute triangles are congruent.
Falso. Evidently not.
If $\triangle ABC$ is isosceles with base $\overline{BC}$, determine $x$, being $AB = 2x - 7$ and $AC = x + 5$.
Ver resposta
$x = 12$
The triangle $ABC$ is equilateral, with $AB = 15 - y$, $BC = 2x - 7$ and $AC = 9$. Determine $x$ and $y$.
Ver resposta
$x = 8$ e $y = 6$
If $\triangle ABC$ is isosceles with base $\overline{BC}$, determine $BC$, being $AB = 3x - 10$, $BC = 2x + 4$ and $AC = x + 4$.
Ver resposta
$BC = 18$ ($x = 7$)
If $\triangle ABC$ is isosceles with base $\overline{BC}$, determine $x$, being $\hat{B} = 2x - 10^{\circ}$ and $\hat{C} = 30^{\circ}$.
Ver resposta
$x = 20^{\circ}$
If $\triangle ABC$ is isosceles with base $\overline{AC}$, determine $x$, being $\hat{A} = x + 30^{\circ}$ and $\hat{C} = 2x - 20^{\circ}$.
Ver resposta
$x = 50^{\circ}$
If $\triangle ABC$ is isosceles with base $\overline{BC}$, determine $x$ and $y$.
Ver resposta
$x = 85^{\circ}$ e $y = 50^{\circ}$
Determine $x$ and $y$, knowing that the triangle $ABC$ is equilateral.
Ver resposta
a) $x = 4$ e $y = 9$ · b) $x = 4$ e $y = 3$
If the perimeter of a equilateral triangle is 75 cm, how long is each side?
Ver resposta
25 cm
If the perimeter of a isosceles triangle is 100 m and the base measures 40 m, how long are each of the other sides?
Ver resposta
30 m e 30 m
Determine the perimeter of the triangle $ABC$ in the following cases:
a) equilateral triangle with $AB = x + 2y$, $AC = 2x - y$ and $BC = x + y + 3$; b) isosceles triangle base $\overline{BC}$ with $AB = 2x + 3$, $AC = 3x - 3$ and $BC = x + 3$.
Ver resposta
a) 45 · b) 39
In a isosceles triangle, the semiperimeter is worth 7,5 m. Calculate the sides of this triangle, knowing that the sum of the congruent sides is four times the base.
Ver resposta
3 m, 6 m e 6 m
The pairs of triangles for each item are congruent. Indicate the case of congruence. (Equal dashes indicate congruent sides; equal arcs indicate congruent angles.)
Ver resposta
a) LAL · b) LLL · c) LAAₒ · d) LAAₒ · e) LAAₒ · f) LAL, ALA or LAAₒ · g) special case
Consider the triangles $T_1, T_2, \ldots, T_{12}$. Find the pairs of congruent triangles and indicate the case of congruence. Play: click on two triangles to form a pair.
Ver resposta
$T_1 \equiv T_8$ (LAL) · $T_2 \equiv T_7$ (LAL) · $T_3 \equiv T_5$ (LAL) · $T_4 \equiv T_{11}$ (ALA) · $T_6 \equiv T_{10}$ (LLL) · $T_9 \equiv T_{12}$ (LAAₒ)
In cases a), b) and c), select the congruent triangles and indicate the case of congruence.
Ver resposta
a) I ≡ II (LAL) · b) I ≡ III (ALA) · c) I ≡ III (special case)
Indicate the congruent triangles in the figures below, citing the case of congruence. In c), $\overline{AC} \equiv \overline{DF}$ and $\overline{AB} \equiv \overline{DE}$.
Ver resposta
a) $\triangle ABC \equiv \triangle ADC$ (LAL) · b) $\triangle ACB \equiv \triangle ECD$ (LAAₒ) · c) $\triangle CAB \equiv \triangle FDE$ (LAL) · d) $\triangle EBA \equiv \triangle ECD$ (LLL), or $\triangle ECA \equiv \triangle EBD$ (LLL)
Why is ALL or LLA not a case of congruence between triangles?
Ver resposta
Because there are triangles that have ALL (or LLA) congruent and are not congruent. See the “LLA?” in the figure of congruence cases: triangles $A^{\prime}B^{\prime}C^{\prime}$ and $A^{\prime}B^{\prime}C^{\prime\prime}$ have common $\hat{B}$, common $\overline{AB}$ and $\overline{AC} \equiv \overline{AC^{\prime\prime}}$ but $\overline{BC} \neq \overline{BC^{\prime\prime}}$.
In the figure, the triangle $ABC$ is congruent to the triangle $DEC$. Determine $\alpha$ and $\beta$, being $\hat{A} = 3\alpha$, $\hat{B} = \beta + 48^{\circ}$, $\hat{E} = 5\beta$ and $\hat{D} = 2\alpha + 10^{\circ}$.
Ver resposta
$\alpha = 10^{\circ}$ e $\beta = 12^{\circ}$
In the figure, the triangle $ABD$ is congruent to the triangle $CBD$. Calculate $x$ and $y$ and the sides of the triangle $ACD$, being $AB = x$, $BC = 2y$, $CD = 3y + 8$ and $DA = 2x$.
Ver resposta
$x = 16$, $y = 8$ e $AD = CD = AC = 32$
In the figure, the triangle $CBA$ is congruent to the triangle $CDE$. Determine the value of $x$ and $y$ and the ratio between the perimeters of these triangles, being $AB = 35$, $AC = 2x - 6$, $CE = 22$ and $DE = 3y + 5$.
Ver resposta
$x = 14$, $y = 10$ and the ratio between the perimeters is 1.
In the figure, the triangle $PCD$ is congruent to the triangle $PBA$. Determine $x$ and $y$ and the ratio between the perimeters of the triangles $PCA$ and $PBD$, being $AB = 15$, $CD = x + 5$, $AP = 2y + 17$ and $PD = 3y - 2$.
Ver resposta
$x = 10$, $y = 19$ and the ratio between the perimeters is 1.
In the figure, the triangles $ABC$ and $CDA$ are congruent. Calculate $x$ and $y$.
Ver resposta
$x = 60^{\circ}$ e $y = 9^{\circ}$
In the figure, knowing that $C$ is the midpoint of $\overline{BE}$, prove that the triangles $ABC$ and $DEC$ are congruent.
Ver resposta
Use ALA: $\hat{B} \equiv \hat{E}$ (straight), $\overline{BC} \equiv \overline{CE}$ and $B\hat{C}A \equiv E\hat{C}D$ (o.p.v.).
In the figure, knowing that $\alpha \equiv \beta$ and $\gamma \equiv \delta$, prove that the triangles $ABC$ and $CDA$ are congruent.
Ver resposta
Use ALA: $\alpha \equiv \beta$, $\overline{AC}$ common and $\gamma \equiv \delta$.
If $\alpha \equiv \beta$ and $\varphi \equiv \theta$, demonstrate that the triangle $ABC$ is congruent to the triangle $ABD$.
Ver resposta
Use ALA: $\alpha \equiv \beta$, $\overline{AB}$ common and $\varphi \equiv \theta$.
In the figure, being $\overline{BF} \equiv \overline{CD}$, $m(A\hat{B}C) = m(F\hat{D}E)$ and $m(B\hat{A}C) = m(D\hat{E}F)$, prove that $\overline{AC} \equiv \overline{EF}$.
Ver resposta
$\triangle CBA \equiv \triangle FDE$ (LAAₒ): $\overline{BC} \equiv \overline{DF}$ (as $BF = CD$ and $FC$ is common), $\hat{B} \equiv \hat{D}$ and $\hat{A} \equiv \hat{E}$.
Being $\overline{AB} \equiv \overline{AE}$, $m(B\hat{A}D) = m(C\hat{A}E)$, $m(A\hat{B}C) = 90^{\circ}$ and $m(A\hat{E}D) = 90^{\circ}$, prove that $\overline{BC} \equiv \overline{DE}$. (In the book picture, $C$ and $D$ are on the same side, with $\overline{AC}$ cutting into $\overline{ED}$.)
Ver resposta
$\triangle BAC \equiv \triangle EAD$ (ALA): $B\hat{A}C \equiv E\hat{A}D$ (adding $C\hat{A}D$ to the congruent angles $B\hat{A}D$ and $C\hat{A}E$… or subtracting), $\overline{AB} \equiv \overline{AE}$ and $\hat{B} \equiv \hat{E}$ (straight angles).
Show that the median relative to the base of a isosceles triangle is also angle bisector.
Ver resposta
Use LLL: with $M$ base midpoint, $\triangle ABM \equiv \triangle ACM$.
Prove that angle bisector relative to the base of isosceles triangle is also median.
Ver resposta
Use LAL: with $S$ the foot of angle bisector, $\triangle ABS \equiv \triangle ACS$, then $BS = SC$.
Prove that the medians relative to the congruent sides of isosceles triangle are congruent.
See the resolution of the book
Hypothesis: $\overline{AB} \equiv \overline{AC}$; $\overline{BM}$ and $\overline{CN}$ are medians. Thesis: $\overline{BM} \equiv \overline{CN}$.
In triangles $BAM$ and $CAN$: $\overline{BA} \equiv \overline{CA}$, $\hat{A}$ is common and $\overline{AM} \equiv \overline{AN}$ (halves of congruent sides). By LAL, $\triangle BAM \equiv \triangle CAN$, then $\overline{BM} \equiv \overline{CN}$.
Prove that the angle bisectors relative to the congruent sides of a isosceles triangle are congruent.
Ver resposta
Use ALA on the triangles formed by angle bisectors and the base.
Prove that, if angle bisector relative to a side of a triangle is also median relative to that side, then that triangle is isosceles.
Ver resposta
Extend the median: take $P$ on the ray $\overrightarrow{AM}$ with $\overline{MP} \equiv \overline{AM}$. Then $\triangle AMB \equiv \triangle PMC$ (LAL), then $B\hat{A}M \equiv C\hat{P}M$ and $\overline{AB} \equiv \overline{PC}$. Since $\overline{AM}$ is angle bisector, $C\hat{P}M \equiv C\hat{A}M$, and $\triangle ACP$ is isosceles: $\overline{AC} \equiv \overline{PC} \equiv \overline{AB}$.
With segments of 8 cm, 5 cm and 18 cm can a triangle be constructed? Why? (Test on the triangular inequality figure.)
Ver resposta
No, because the condition $|8 - 5| < 18 < 8 + 5$ is false.
Two sides, $\overline{AB}$ and $\overline{BC}$, of a triangle $ABC$ measure respectively 8 cm and 21 cm. How much can the third side measure, knowing that it is a multiple of 6?
Ver resposta
18 cm or 24 cm
Determine the variation range of $x$, knowing that the sides of a triangle are expressed by $x + 10$, $2x + 4$ and $20 - 2x$.
Ver resposta
$\dfrac{6}{5} < x < \dfrac{26}{3}$
If two sides of isosceles triangle measure 38 cm and 14 cm, what could be the measurement of the third side?
Ver resposta
38 cm (the triangle 14, 14, 38 does not exist).
The side $\overline{AB}$ of a triangle $ABC$ is expressed by an integer. Determine its maximum value, knowing that the sides $\overline{AC}$ and $\overline{BC}$ measure respectively 27 cm and 16 cm and that $\hat{C} < \hat{A} < \hat{B}$.
Ver resposta
15 cm
Show that a right-angled triangle has two acute angles.
See the resolution of the book
Consider the external angle $\gamma^{\prime}$ adjacent to the right angle; as is supplementary of a rectum, $\gamma^{\prime} = 90^{\circ}$. With $\alpha$ and $\beta$ being the non-right internal angles, according to the external angle theorem, $\gamma^{\prime} > \alpha$ and $\gamma^{\prime} > \beta$. Therefore $\alpha < 90^{\circ}$ and $\beta < 90^{\circ}$: the triangle has two acute angles.
Show that the hypotenuse of a right triangle is greater than each of the legs.
Ver resposta
Use the previous exercise: the right angle is the largest angle, and the largest side is opposite the largest angle.
Show that the obtuse triangle has two acute angles.
Ver resposta
In the same way as exercise 118, using the external angle adjacent to the obtuse (which is acute).
Show that the side opposite the obtuse angle of an obtuse triangle is greater than each of the other sides.
Ver resposta
Use the above: the obtuse angle is the largest angle.
Show that the hypotenuse of a right triangle is greater than the semisum of the legs.
Ver resposta
Hypotenuse > legs: $a > b$ and $a > c$ $\Rightarrow$ $2a > b + c$ $\Rightarrow$ $a > \dfrac{b + c}{2}$.
Prove that any side of a triangle is smaller than the semiperimeter.
Ver resposta
By the triangular inequality, $a < b + c$; adding $a$: $2a < a + b + c$, then $a < \dfrac{a + b + c}{2}$.
If $P$ is an internal point of a triangle $ABC$, show that $B\hat{P}C$ is greater than $B\hat{A}C$.
Ver resposta
Extend $\overline{AP}$ until you cut $\overline{BC}$. Use the exterior angle theorem twice: $B\hat{P}C$ is exterior to a triangle that contains an exterior angle greater than $\hat{A}$.
If $P$ is an internal point of a triangle $ABC$, show that $PB + PC < AB + AC$.
See the resolution of the book
Extend $\overline{BP}$ until you cut $\overline{AC}$ at point $Q$. Let them be $x = PB$, $y = PC$, $t = PQ$, and $b^{\prime} = AQ$, $b^{\prime\prime} = QC$ (with $b^{\prime} + b^{\prime\prime} = b$). By the triangular inequality in triangles $ABQ$ and $PQC$:
$$c + b^{\prime} > x + t \quad\text{e}\quad t + b^{\prime\prime} > y.$$Adding: $c + b^{\prime} + b^{\prime\prime} + t > x + y + t$, that is, $c + b > x + y$: $PB + PC < AB + AC$.
If $P$ is an internal point of a triangle $ABC$ and $x = PA$, $y = PB$ and $z = PC$, show that $x + y + z$ is between the semiperimeter and the perimeter of the triangle.
Ver resposta
Use the previous exercise three times (adding, $2(x + y + z) < 2p$) and the triangular inequality in the triangles $PAB$, $PBC$, $PCA$ (adding, $2(x + y + z) > a + b + c$).
Show that the perimeter of the triangle $MNP$ is smaller than the perimeter of the triangle $ABC$ in the figure.
Ver resposta
Consider the triangles $APN$, $BPM$ and $MNC$ and apply the triangular inequality to each one.
If $m_a$ is the median relative to side $a$ of a triangle with sides $a$, $b$ and $c$, then $m_a < \dfrac{b + c}{2}$.
Ver resposta
Consider $\triangle ACA^{\prime}$, where $A^{\prime}$ is the symmetric of $A$ with respect to the midpoint of $\overline{BC}$: $AA^{\prime} = 2m_a$, $CA^{\prime} = c$. Then $2m_a < b + c$ and $2m_a > |b - c|$.
Prove that the sum of the medians of a triangle is less than the perimeter and greater than the semiperimeter.
Ver resposta
Use the previous result for the three medians and add them up. For the other inequality, use the triangle inequality on the triangles formed by centroid.