Question 166
Um triângulo tem lados medindo 7 cm, 24 cm e 25 cm. A área desse triângulo, em cm², é
(A) 42 (B) 84 (C) 87,5 (D) 168 (E) 175

Um triângulo tem lados medindo 5 cm, 12 cm e 13 cm. A área desse triângulo é
(A) 20 cm$^2$ (B) 25 cm$^2$ (C) 30 cm$^2$ (D) 35 cm$^2$ (E) 40 cm$^2$

A triangle has sides measuring 5 cm, 12 cm, and 13 cm. What is the area, in square centimeters, of this triangle?
(A) 20
(B) 25
(C) 30
(D) 35
(E) 40

In triangle ABC, $\overline { \mathrm { AB } } = 5$, $\overline { \mathrm { AC } } = 6$, and $\cos ( \angle \mathrm { BAC } ) = - \frac { 3 } { 5 }$. Find the area of triangle ABC. [3 points]

15. In $\triangle A B C$, the sides opposite to angles $A , B , C$ are $a , b , c$ respectively. If $b = 6 , a = 2 c , B = \frac { \pi } { 3 }$, then the area of $\triangle A B C$ is $\_\_\_\_$ .

18. (12 points) In $\triangle A B C$ , the sides opposite to angles $A , B , C$ are $a , b , c$ respectively. Given that When $t = 0$, $S = 3$; when $t = \pm 1$, $S = 4\sqrt{2}$.
Therefore, the area of quadrilateral $ADBE$ is $3$ or $4\sqrt{2}$.

In $\triangle A B C$, the sides opposite to angles $A , B , C$ are $a , b , c$ respectively. Given $B = 150 ^ { \circ }$ ,
(1) If $a = \sqrt { 3 } c , b = 2 \sqrt { 7 }$ , find the area of $\triangle A B C$ ;
(2) If $\sin A + \sqrt { 3 } \sin C = \frac { \sqrt { 2 } } { 2 }$ , find $C$ .

In the quadrangular pyramid $P - ABCD$ , the base $ABCD$ is a square with $AB = 4$ , $PC = PD = 3$ , $\angle PCA = 45^{\circ}$ , then the area of $\triangle PBC$ is
A. $2\sqrt{2}$
B. $3\sqrt{2}$
C. $4\sqrt{2}$
D. $5\sqrt{2}$

In $\triangle ABC$, $a = 7$, $A$ is an obtuse angle, $\sin 2B = \frac { \sqrt { 3 } } { 7 } b \cos B$.
(1) Find $\angle A$;
(2) Choose one condition from conditions (1), (2), and (3) below as a given condition and find the area of $\triangle ABC$.
(1) $b = 7$; (2) $\cos B = \frac { 13 } { 14 }$; (3) $c \sin A = \frac { 5 } { 2 } \sqrt { 3 }$. Note: If conditions (1), (2), and (3) are solved separately, only the first solution will be graded.

Three triangles are formed by drawing lines from the vertices of a triangle $ABC$ to the opposite sides, each making equal angles with the sides. Let $\Delta_1, \Delta_2, \Delta_3$ be the areas of the three smaller triangles formed at the vertices with the circumradius equal to 1.
a) Express the total area $\Delta = \Delta_1 + \Delta_2 + \Delta_3$ in terms of $A, B, C$.
b) Find the angles $A, B, C$ that maximize $\Delta$.
c) Verify that the maximum occurs for an isosceles triangle and prove $C = A$ by calculus.

A regular pentagon is inscribed in a circle of radius $r$ and another regular pentagon is circumscribed about the same circle. Find the ratio of the area of the inscribed pentagon to the area of the circumscribed pentagon.
(A) $\sin^2 36^\circ$ (B) $\cos^2 36^\circ$ (C) $\tan^2 36^\circ$ (D) $\cos^2 54^\circ$

Consider a paper in the shape of an equilateral triangle $A B C$ with circumcenter $O$ and perimeter 9 units. If we fold the paper in such a way that each of the vertices $A , B , C$ gets identified with $O$, then the area of the resulting shape in square units is:
(A) $\frac { 3 \sqrt { 3 } } { 4 }$
(B) $\frac { 4 } { \sqrt { 3 } }$
(C) $\frac { 3 \sqrt { 3 } } { 2 }$
(D) $3 \sqrt { 3 }$.

The sides of a right angled triangle are in arithmetic progression. If the triangle has area 24, then what is the length of its smallest side?

As shown in the figure, the Wang family owns a triangular piece of land $\triangle A B C$ , where $\overline { B C } = 16$ meters. The government plans to requisition the trapezoid $D B C E$ portion to develop a road with straight lines $D E , B C$ as edges, with road width $h$ meters, leaving the Wang family with only $\frac { 9 } { 16 }$ of the original land area. After negotiation, the plan is changed to develop a road with parallel lines $B E , F C$ as edges, with the same road width, where $\angle E B C = 30 ^ { \circ }$ . Only the $\triangle B C E$ region needs to be requisitioned. According to this agreement, the Wang family's remaining land $\triangle A B E$ has (15-1)(15-2)(15-3) square meters.

As shown in the figure, consider a rectangular stone block with a vertex $A$ and a face containing point $A$. Let the midpoints of the edges of this face be $B , E , F , D$ respectively. Another face of the rectangular block containing point $B$ has its edge midpoints as $B , C , H , G$ respectively. Given that $\overline { B C } = 8$ and $\overline { B D } = \overline { D C } = 9$. The stone block is now cut to remove eight corners, such that the cutting plane for each corner passes through the midpoints of the three adjacent edges of that corner.
Find the area of $\triangle B C D$. (Non-multiple choice question, 4 points)

$ABC$ is a right triangle\ $\mathrm { AB } \perp \mathrm { AC }$\ $\mathrm { DE } \perp \mathrm { BC }$\ $| \mathrm { AD } | = | \mathrm { DB } | = 3$ units\ In triangle $ABC$, $D$ and $E$ lie on sides $AB$ and $BC$ respectively.\ If the area of triangle $ABC$ is 6 times the area of triangle $BDE$, what is $| AC |$ in units?\ A) $2 \sqrt { 3 }$\ B) $3 \sqrt { 2 }$\ C) $2 \sqrt { 6 }$\ D) 3\ E) 6

ABC and BDE are equilateral triangles\ $[ \mathrm { BD } ] \perp [ \mathrm { AC } ]$\ $[ \mathrm { BF } ] \perp [ \mathrm { DE } ]$\ $[ \mathrm { FH } ] \perp [ \mathrm { BE } ]$\ $| \mathrm { AB } | = 16$ units
Accordingly, what is the area of triangle BFH in square units?\ A) $12 \sqrt { 3 }$\ B) $15 \sqrt { 3 }$\ C) $18 \sqrt { 3 }$\ D) $20 \sqrt { 3 }$\ E) $24 \sqrt { 3 }$

$ABC$ right triangle\ $[ \mathrm { AC } ] \perp [ \mathrm { BC } ]$\ $[AB]$ // $[DE]$\ $[BC]$ // $[FH]$\ $| \mathrm { AD } | = | \mathrm { DH } | = | \mathrm { HC } |$\ $| \mathrm { GE } | = 4$ units\ $| \mathrm { GF } | = 2$ units
Accordingly, what is the area of triangle ABC in square units?\ A) $9 \sqrt { 3 }$\ B) $12 \sqrt { 3 }$\ C) $15 \sqrt { 3 }$\ D) $18 \sqrt { 3 }$\ E) $20 \sqrt { 3 }$

ABCD right trapezoid, ABD equilateral triangle\ $[AB]$ // $[DC]$\ $| \mathrm { BF } | = 4 | \mathrm { DF } |$\ $| \mathrm { AB } | = 8$ units
Accordingly, what is the area of right trapezoid ABCE in square units?\ A) $10 \sqrt { 3 }$\ B) $12 \sqrt { 3 }$\ C) $16 \sqrt { 3 }$\ D) $18 \sqrt { 3 }$\ E) $20 \sqrt { 3 }$

ABCD kite\ $[ \mathrm { AC } ] \perp [ \mathrm { BD } ]$\ $| \mathrm { AB } | = | \mathrm { BC } |$\ $| \mathrm { AD } | = | \mathrm { DC } |$\ $| \mathrm { BE } | = 4 | \mathrm { ED } |$\ $| \mathrm { AC } | = 16$ units
The area of kite ABCD in the figure is 160 square units.\ Accordingly, what is the perimeter of kite ABCD in units?\ A) $20 \sqrt { 5 }$\ B) $24 \sqrt { 5 }$\ C) $28 \sqrt { 5 }$

In a triangle $ABC$, the length of side $AB$ is equal to half the length of side $BC$.
If two of the altitudes of this triangle have lengths 4 units and 10 units, which of the following could be the length of the other altitude?
I. 2 units II. 5 units III. 8 units
A) Only I B) Only II C) Only III D) I and II E) II and III