Chapter6 plane geometry Module2 triangles lower | Books or Courses of MAT and Java

Jump to heading 14.Right Triangle

This can be determined by the Pythagorean theorem $a^{2} + b^{2} = c^{2}$ or by an angle of $90^{\circ}$ .

Jump to heading 15.Isosceles Right Triangle

A right triangle where the three sides are in the ratio $1 : 1 : \sqrt{2}$ or where $two sides are equal$ .

Jump to heading 16.Equilateral Triangle

The three sides are equal, or the three interior angles are equal, or the four centers coincide.

Jump to heading 17.Isosceles Triangle

A triangle with two equal sides or two equal angles.

Jump to heading 18.Focus 8

Triangle identification

The main approach is to use the conditions involving the relationships between the interior angles and the three sides of a triangle, combined with the properties of triangles to determine the shape of the triangle.
Focus should be placed on mastering the characteristics of Equilateral triangles, Isosceles triangles, Right triangles, Isosceles right triangles.
When the given conditions in a problem involve the side lengths of a triangle, the key is to use identity transformations to find the relationships among $a, b, c$ .

Jump to heading $20$ The three sides $a, b, c$ of $△ A B C$ satisfy $1 + \frac{b}{c} = \frac{b + c}{b + c - a}$ , then the triangle is $?$ .

$\begin{array}{lllll} (A) An isosceles triangle with side a as the leg & (B) An isosceles triangle with side a as the base \\ (C) Equilateral triangle & (D) Right triangle \\ (E) Obtuse triangle \end{array}$

Jump to heading Solution

$\begin{array}{ll} \frac{b + c}{c} = \frac{b + c}{b + c - a} \\ \frac{b + c}{c} = \frac{b + c}{b + c - a} = b + c - a = c & Since condition a + b > c satisfied,the numerator \neq 0 and equal \\ b = a \end{array}$

Jump to heading Conclusion

Derived Solution
$(A)$
According to the Solution, get $b = a$ , so choose $A$ .
Formula used
$\begin{array}{ll} a + b > c & Triangle Inequality theorem \end{array}$
Additionally, if the problem is $\frac{b}{c} - 1 = \frac{b - c}{b + c - a}$
$\begin{array}{ll} \frac{b - c}{c} = \frac{b - c}{b + c - a} \\ ① \frac{b - c}{c} = \frac{b - c}{b + c - a} = b - c = 0 & Assume that b - c = 0 \\ ② \frac{b - c}{c} = \frac{b - c}{b + c - a} = c = b + c - a & Assume that b - c \neq 0 \\ b = c \lor a = b \end{array}$

Jump to heading $21$ The three sides of $△ A B C$ are $a, b, c$ and they satisfy $4 a^{2} + 4 b^{2} + 13 c^{2} - 8 a c - 12 b c = 0$ , then $△ A B C = ?$ .

$\begin{array}{lllll} (A) Right triangle & (B) Isosceles triangle & (C) Equilateral triangle \\ (D) Isosceles right triangle & (E) Acute triangle \end{array}$

Jump to heading Solution

Factoring by grouping into a perfect square form, Non-negative $(x)^{2} \geq 0$
$\begin{array}{ll} 4 a^{2} - 8 a c + 4 c^{2} + 4 b^{2} - 12 b c + 9 c^{2} = 0 \\ 4 (a - c)^{2} + (2 b - 3 c)^{2} = 0 \\ a = c \land 2 b = 3 c & (x)^{2} = 0 \end{array}$

Jump to heading Conclusion

Derived Solution
$(B)$
According to the Solution, get $a = c \land 2 b = 3 c$ , so choose $B$ .
Formula used
$\begin{array}{ll} (a - b)^{2} = a^{2} - 2 a b + b^{2} & Perfect square formula \end{array}$

Jump to heading $22$ The three sides of $△ A B C$ are $a, b, c$ . If $a^{2} + 2 b c = b^{2} + 2 a c = c^{2} + 2 a b = 27$ , then $△ A B C = ?$ .

$\begin{array}{lllll} (A) Isosceles triangle & (B) Isosceles right triangle & (C) Obtuse triangle \\ (D) Right triangle & (E) Equilateral triangle \end{array}$

Jump to heading Solution

$\begin{array}{ll} {\begin{cases} a^{2} + 2 b c = 27 ① \\ b^{2} + 2 a c = 27 ② \\ c^{2} + 2 a b = 27 ③ \end{cases} \\ a^{2} + b^{2} + c^{2} + 2 a b + 2 a c + 2 b c = 27 \times 3 & Add all terms together to form a perfect square \\ (a + b + c)^{2} = 81 \\ a + b + c = \sqrt{81} = 9 \\ (a^{2} + 2 b c) - (b^{2} + 2 a c) = 27 - 27 & ① - ② \\ a^{2} - b^{2} + 2 b c - 2 a c = 0 \\ (a + b) (a - b) + 2 c (b - a) = 0 \\ (a - b) (a + b - 2 c) = 0 \\ a = b \lor a + b = 2 c & (0) (a + b - 2 c) = 0 \lor (a - b) (0) = 0 \\ a = b {\begin{cases} a + b + c = 9 \\ \begin{array}{ll} c + 2 a = 9 \\ a^{2} + 2 a c = 27 & ② \\ c^{2} + 2 a^{2} = 27 & ③ \\ (c^{2} + 2 a^{2}) - (a^{2} + 2 a c) = 27 - 27 & ③ - ② \\ c^{2} - 2 a c + a^{2} = 0 \\ (c - a)^{2} = 0 \\ c = a & (0)^{2} = 0 \\ c + 2 c = 9 & a = c \\ 3 c = 9 \\ c = \frac{9}{3} = 3 \\ a = b = c = 3 \end{array} \end{cases} \\ a + b = 2 c {\begin{cases} a + b + c = 9 \\ \begin{array}{ll} 2 c + c = 9 \\ 3 c = 9 \\ c = \frac{9}{3} = 3 \\ a + b = 6 \\ b^{2} + 6 a = 27 & ② c = 3 \\ 9 + 2 a b = 27 & ③ c = 3 \\ a b = \frac{18}{2} = 9 & ③ \\ 6 \geq 2 \sqrt{9} & a + b \geq 2 \sqrt{a b} \\ 6 = 6 \Rightarrow a = b \land a + b = 6 \Rightarrow a = b = 3 \\ a = b = c = 3 \end{array} \end{cases} \end{array}$

Jump to heading Conclusion

Derived Solution
$(E)$
According to the Solution, get $a = b = c = 3$ , so choose $E$ .
Formula used
$\begin{array}{ll} {\begin{cases} (a + b)^{2} = a^{2} + 2 a b + b^{2} \\ (a - b)^{2} = a^{2} - 2 a b + b^{2} \end{cases} & Perfect square formula \\ a^{2} - b^{2} = (a - b) (a + b) & Difference of squares formula \\ a + b \geq 2 \sqrt{a b} & Arithmetic Mean-Geometric Mean Inequality \end{array}$

Jump to heading 19.Congruence of Triangle

Jump to heading $1$ Definition

If two triangles have the same shape and size, they're said to be congruent.

Jump to heading $2$ Identification

Congruence can be identified using side-side-side (SSS), side-angle-side (SAS), angle-side-angle (ASA), angle-angle-side (AAS).

Jump to heading $3$ Properties

If two triangles are congruent, their corresponding sides, corresponding angles, and areas are equal. In mathematics terms, the two triangles are equivalent, having the same side lengths, angles, and areas.

Jump to heading $4$ Applications

When folding, symmetry, or rotation occurs, congruence analysis can be used.

Jump to heading 20.Similarity of Triangle

Jump to heading $1$ Definition

If two triangles have the same shape and their sizes are proportional, they're said to be similar.

Proportionality of 1 implies congruence.

Jump to heading $2$ Identification

Similarity can be identified by two pairs of corresponding interior angles being equal.

Jump to heading $3$ Properties

Jump to heading In similar triangles, the ratios of corresponding sides are equal.
- Known as the similarity ratio: $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} = k$ .
In similar triangles, the ratio of the altitudes, medians, and angle bisectors are also equal to the similarity ratio.
In similar triangles, the ratio of the perimeters is equal to the similarity ratio.
- $\frac{C_{1}}{C_{2}} = k$ .
In similar triangles, the ratio of the area is equal to the square of the similarity ratio.
- $\frac{S_{1}}{S_{2}} = k^{2}$ .

Jump to heading $4$ Applications

When parallel lines appear, similarity should be used for analysis.

Remark: The properties of similar triangles can be fully extended to other similar figures, such as quadrilaterals.

Jump to heading 21.Focus 9

Triangle congruence

When folding, symmetry, or reflection is involved, congruent analysis should be used.

Jump to heading $23$ Figure 6–19, in triangle $△ A B C$ , $A D ⊥ B C$ at point $D$ , $C E ⊥ A B$ at point $E$ , $A D$ and $C E$ intersect at point $H$ , if $E H = E B = 3, A E = 4$ , then $C H = ?$ .

$\begin{array}{lllll} (A) 1 & (B) \frac{4}{3} & (C) \frac{5}{3} & (D) \sqrt{3} & (E) 2 \end{array}$

Jump to heading Solution

Find the equal side and the equal angle (Acute angle) in two congruent right triangles
$\begin{array}{ll} Equal angle (Acute angle) {\begin{cases} \begin{array}{ll} ∠ 1 + ∠ 2 = 90^{\circ} \\ ∠ 3 + ∠ 4 = 90^{\circ} \\ ∠ 2 = ∠ 3 & Vertical angles are equal \end{array} \end{cases} ⟹ ∠ 1 = ∠ 4 \\ Equal side {\begin{cases} \begin{array}{ll} E H = E B \end{array} \end{cases} ⟹ △ A E H ≅ △ C E B \\ A E = C E = 4 \\ C H = C E - E H = 4 - 3 = 1 \end{array}$

Jump to heading Conclusion

Derived Solution
$(A)$
According to the Solution, get $C H = C E - E H = 1$ , so choose $A$ .
Formula used
$\begin{array}{ll} Right triangle congruence {\begin{cases} Find a side and an angle (Acute angle) \\ Find their two sides(No needto find the third side because the Pythagorean theorem is satisfied) \end{cases} \end{array}$

Jump to heading $24$ Figure 6–20, in a right triangle $A B C$ , the hypotenuse $A B = 13$ and the leg $A C = 5$ , the leg $A C$ is folded onto the hypotenuse $A B$ so that it coincides with the hypotenuse, and point $C$ coincides with point $E$ , the fold is $A D$ , what is the area of the shaded region in the figure $?$ .

$\begin{array}{lllll} (A) 20 & (B) \frac{40}{3} & (C) \frac{38}{3} & (D) 14 & (E) 12 \end{array}$

Jump to heading Solution

Show known conditions
$\begin{array}{ll} B C = 12 & 5,12,13 \\ △ A C D ≅ △ A E D \\ A E = A C = 5 \\ B E = 13 - 5 = 8 \end{array}$

$1$ Solve for the $E D$ length of the shadow
$\begin{array}{ll} △ B D E \sim △ B A C \\ \frac{D E}{A C} = \frac{B E}{B C} \Rightarrow \frac{D E}{5} = \frac{8}{12} \\ D E \times 12 = 8 \times 5 \\ D E \times 12 = 40 \\ D E = \frac{40}{12} = \frac{10}{3} \\ S_{△ B D E} = \frac{1}{2} \times \frac{10}{3} \times 8 = \frac{80}{6} = \frac{40}{3} \end{array}$

$2$ Solve using the area ratio equal to the similarity ratio squared
$\begin{array}{ll} \frac{S_{△ B D E}}{S_{△ B A C}} = (\frac{8}{12})^{2} = (\frac{2}{3})^{2} = \frac{4}{9} \\ S_{△ B A C} = \frac{1}{2} \times 5 \times 12 = 30 \\ \frac{S_{△ B D E}}{30} = \frac{4}{9} \\ S_{△ B D E} \times 9 = 30 \times 4 \\ S_{△ B D E} \times 9 = 120 \\ S_{△ B D E} = \frac{120}{9} = \frac{40}{3} \end{array}$

Jump to heading Conclusion

Derived Solution
$(B)$
According to the Solution, get $S_{△ B D E} = \frac{40}{3}$ , so choose $B$ .
Formula used
$\begin{array}{ll} (5, 12, 13) & Commonly used pythagorean numbers \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} = k & Similar triangle properties \\ \frac{S_{1}}{S_{2}} = k^{2} & Similar triangle properties \\ S = \frac{1}{2} a h & Triangle area formula \end{array}$

Jump to heading 22.Focus 10

Triangle similarity

When parallelism occurs, similarity analysis is used. For the similar figure, the area ratio is equal to the square of a similarity ratio.
Congruent shapes look at the sides.
Similar shapes look at angles.

Jump to heading $25$ There are $?$ correct options below.

$\begin{array}{lllll} (1) All right triangles are similar & (2) All isosceles triangles are similar \\ (3) All parallelograms are similar & (4) All equilateral triangles are similar \\ (5) All trapezoids are similar & (6) All squares are similar \\ (7) All rectangles are similar & (8) All circles are similar \\ (A) 1 & (B) 2 \\ (C) 3 & (D) 4 \\ (E) 5 \end{array}$

Jump to heading Solution

$\begin{array}{ll} (1) All right triangles are similar & △ 30^{\circ} 60^{\circ} 90^{\circ} \neq △ 45^{\circ} 45^{\circ} 90^{\circ} Two pairs of corresponding interior angles are not equal ❌ \\ (2) All isosceles triangles are similar & Two pairs of corresponding interior angles are not equal ❌ \\ (3) All parallelograms are similar & Four pairs of corresponding interior angles are not equal ❌ \\ (4) All equilateral triangles are similar & △ 60^{\circ} 60^{\circ} 60^{\circ} = △ 60^{\circ} 60^{\circ} 60^{\circ} Two pairs of corresponding interior angles are equal ✅ \\ (5) All trapezoids are similar & The corresponding interior angles are not equal and the corresponding sides are not proportional ❌ \\ (6) All squares are similar & Four pairs of corresponding interior angles 90^{\circ} are equal ✅ \\ (7) All rectangles are similar & The corresponding sides are not proportional ❌ \\ (8) All circles are similar & The circles are similar because they have the same shape, just different sizes ✅ \end{array}$

Jump to heading Conclusion

Derived Solution
$(C)$
According to the Solution, get $(4), (6), (8)$ , so choose $C$ .
Formula used
$\begin{array}{ll} Similar triangle identification \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} = k & Similar triangle properties \end{array}$

Jump to heading $26$ Figure 6–21, in $△ A B C, D E, F G, B C$ are parallel to each other, $A D = D F = F B$ , then $S_{△ A D E} : S_{quadrilateral D E G F} : S_{quadrilateral F G C B} = ?$ .

$\begin{array}{lllll} (A) 1 : 3 : 5 & (B) 1 : 2 : 5 & (C) 1 : 3 : 4 & (D) 1 : 3 : 6 & (E) 2 : 3 : 5 \end{array}$

Jump to heading Solution

Show known conditions
$\begin{array}{ll} △ A D E \sim △ A F G \sim △ A B C \\ S_{△ A D E} : S_{△ A F G} : S_{△ A B C} = A D^{2} : A F^{2} : A B^{2} \\ S_{△ A D E} : S_{△ A F G} : S_{△ A B C} = 1^{2} : 2^{2} : 3^{2} = 1 : 4 : 9 \\ S_{△ A D E} : S_{quadrilateral D E G F} : S_{quadrilateral F G C B} = 1 : (4 - 1) : (9 - 4) \\ S_{△ A D E} : S_{quadrilateral D E G F} : S_{quadrilateral F G C B} = 1 : 3 : 5 \end{array}$

Jump to heading Conclusion

Derived Solution
$(A)$
According to the Solution, get $S_{△ A D E} : S_{quadrilateral D E G F} : S_{quadrilateral F G C B} = 1 : 3 : 5$ , so choose $A$ .
Formula used
$\begin{array}{ll} Similar triangle definition \\ \frac{S_{1}}{S_{2}} = k^{2} & Similar triangle properties \end{array}$

Jump to heading $27$ Figure 6–22, in $△ A B C, D, E, F$ are points on $A B, A C, B C$ respectively, and $D E ∥ B C, E F ∥ A B, A D : D B = 2 : 3, B C = 20$ , then $C F = ?$ .

$\begin{array}{lllll} (A) 15 & (B) \frac{40}{3} & (C) \frac{38}{3} & (D) 14 & (E) 12 \end{array}$

Jump to heading Solution

Show known conditions
$\begin{array}{ll} D E F B is a parallelogram \\ △ A D E \sim △ A B C \\ \frac{D E}{B C} = \frac{A D}{A B} \\ \frac{D E}{20} = \frac{2}{5} \\ D E \times 5 = 2 \times 20 \\ D E \times 5 = 40 \\ D E = \frac{40}{5} = 8 \\ B F = D E = 8 \\ C F = B C - B F = 20 - 8 = 12 \end{array}$

Jump to heading Conclusion

Derived Solution
$(E)$
According to the Solution, get $C F = 12$ , so choose $E$ .
Formula used
$\begin{array}{ll} \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} = k & Similar triangle properties \\ Similar triangle applications \end{array}$

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Jump to heading🔗 14.Right Triangle

Jump to heading🔗 15.Isosceles Right Triangle

Jump to heading🔗 16.Equilateral Triangle

Jump to heading🔗 17.Isosceles Triangle

Jump to heading🔗 18.Focus 8

Jump to heading🔗 20The three sides a,b,c of △ABC satisfy 1+bc=b+cb+c−a, then the triangle is ?.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading🔗 21The three sides of △ABC are a,b,c and they satisfy 4a2+4b2+13c2−8ac−12bc=0, then △ABC=?.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading🔗 22The three sides of △ABC are a,b,c. If a2+2bc=b2+2ac=c2+2ab=27, then △ABC=?.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading🔗 19.Congruence of Triangle

Jump to heading🔗 1Definition

Jump to heading🔗 2Identification

Jump to heading🔗 3Properties

Jump to heading🔗 4Applications

Jump to heading🔗 20.Similarity of Triangle

Jump to heading🔗 1Definition

Jump to heading🔗 2Identification

Jump to heading🔗 3Properties

Jump to heading🔗 In similar triangles, the ratios of corresponding sides are equal.

Jump to heading🔗 4Applications

Jump to heading🔗 21.Focus 9

Jump to heading🔗 23Figure 6–19, in triangle △ABC, AD⊥BC at point D, CE⊥AB at point E, AD and CE intersect at point H, if EH=EB=3,AE=4, then CH=?.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading🔗 24Figure 6–20, in a right triangle ABC, the hypotenuse AB=13 and the leg AC=5, the leg AC is folded onto the hypotenuse AB so that it coincides with the hypotenuse, and point C coincides with point E, the fold is AD, what is the area of the shaded region in the figure ?.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading🔗 22.Focus 10

Jump to heading🔗 25There are ? correct options below.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading🔗 26Figure 6–21, in △ABC,DE,FG,BC are parallel to each other, AD=DF=FB, then S△ADE:SquadrilateralDEGF:SquadrilateralFGCB=?.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading🔗 27Figure 6–22, in △ABC,D,E,F are points on AB,AC,BC respectively, and DE∥BC,EF∥AB,AD:DB=2:3,BC=20, then CF=?.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

Jump to heading 14.Right Triangle

Jump to heading 15.Isosceles Right Triangle

Jump to heading 16.Equilateral Triangle

Jump to heading 17.Isosceles Triangle

Jump to heading 18.Focus 8

Jump to heading $20$ The three sides $a, b, c$ of $△ A B C$ satisfy $1 + \frac{b}{c} = \frac{b + c}{b + c - a}$ , then the triangle is $?$ .

Jump to heading Solution

Jump to heading Conclusion

Jump to heading $21$ The three sides of $△ A B C$ are $a, b, c$ and they satisfy $4 a^{2} + 4 b^{2} + 13 c^{2} - 8 a c - 12 b c = 0$ , then $△ A B C = ?$ .

Jump to heading Solution

Jump to heading Conclusion

Jump to heading $22$ The three sides of $△ A B C$ are $a, b, c$ . If $a^{2} + 2 b c = b^{2} + 2 a c = c^{2} + 2 a b = 27$ , then $△ A B C = ?$ .

Jump to heading Solution

Jump to heading Conclusion

Jump to heading 19.Congruence of Triangle

Jump to heading $1$ Definition

Jump to heading $2$ Identification

Jump to heading $3$ Properties

Jump to heading $4$ Applications

Jump to heading 20.Similarity of Triangle

Jump to heading $1$ Definition

Jump to heading $2$ Identification

Jump to heading $3$ Properties

Jump to heading In similar triangles, the ratios of corresponding sides are equal.

Jump to heading $4$ Applications

Jump to heading 21.Focus 9

Jump to heading $23$ Figure 6–19, in triangle $△ A B C$ , $A D ⊥ B C$ at point $D$ , $C E ⊥ A B$ at point $E$ , $A D$ and $C E$ intersect at point $H$ , if $E H = E B = 3, A E = 4$ , then $C H = ?$ .

Jump to heading Solution

Jump to heading Conclusion

Jump to heading Solution

Jump to heading Conclusion

Jump to heading 22.Focus 10

Jump to heading $25$ There are $?$ correct options below.

Jump to heading Solution

Jump to heading Conclusion

Jump to heading $26$ Figure 6–21, in $△ A B C, D E, F G, B C$ are parallel to each other, $A D = D F = F B$ , then $S_{△ A D E} : S_{quadrilateral D E G F} : S_{quadrilateral F G C B} = ?$ .

Jump to heading Solution

Jump to heading Conclusion

Jump to heading $27$ Figure 6–22, in $△ A B C, D, E, F$ are points on $A B, A C, B C$ respectively, and $D E ∥ B C, E F ∥ A B, A D : D B = 2 : 3, B C = 20$ , then $C F = ?$ .

Jump to heading Solution

Jump to heading Conclusion