Jump to heading🔗 $\boxed{\text{1}}S=\frac{1}{2}ah$, where $h$ is the height of side $a$.

• Application
  When the base and height are known, this formula can be used to calculate the area.

Jump to heading🔗 $\boxed{\text{2}}S=\frac{1}{2}absinC$, where $C$ is the angle between sides $a,b$.

• Angle-sine Chart

  $C$	${30}^{\circ }\vee {150}^{\circ }$	${45}^{\circ }\vee {135}^{\circ }$	${60}^{\circ }\vee {120}^{\circ }$	${90}^{\circ }$
  $sinC$	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	1

• Application
  When two sides and the angle are known, this formula can be used to calculate the area.
  

Jump to heading🔗 $\boxed{\text{3}}S=\sqrt{p(p-a)(p-b)(p-c)}$, where $p=\frac{1}{2}(a+b+c)$.

• Application
  When the three sides of a triangle are known, this formula can be used to calculate the area.

$\begin{array}{l}S=\frac{1}{2}ah\{\begin{array}{ll}\frac{S_1}{S_2}=\frac{a_1}{a_2}& \text{h same}\\ \frac{S_1}{S_2}=\frac{h_1}{h_2}& \text{a same}\\ S_1=S_2& \text{a, h same}\end{array}\end{array}$

Jump to heading🔗 $\boxed{\text{12}}$Figure 6–13, if the area of $\mathrm{△}ABC$ is $1$, and the areas of $\mathrm{△}AEC,\mathrm{△}DEC,\mathrm{△}BED$ are equal, then the area of $\mathrm{△}AED$ is $?$.

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

• Show known conditions
  

  $\begin{array}{ll}{S}_{\mathrm{△}AEC}=S_{\mathrm{△}DEC}=S_{\mathrm{△}BED}=\frac{1}{3}& \end{array}$

• $\boxed{\text{1}}$Solve according to the position of point D

  $\begin{array}{ll}{S}_{\mathrm{△}BDE}=S_{\mathrm{△}CDE}\Rightarrow D\text{ is the midpoint of }BC& S_1=S_2\text{ The height and area of the pink and orange triangles are equal}\\ S_{\mathrm{△}ABD}=\frac{1}{2}{S}_{\mathrm{△}ABC}=\frac{1}{2}& AD\text{ is the median line }\mathrm{△}ABC\\ S_{\mathrm{△}AED}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\end{array}$

• $\boxed{\text{2}}$Solve according to the position of point E

  $\begin{array}{ll}{S}_{\mathrm{△}BCE}=2S_{\mathrm{△}ACE}& \text{The pink + orange triangle is 2x of the green triangle}\\ BE=2AE& \frac{S_1}{S_2}=\frac{a_1}{a_2}\text{ h same}\\ S_{\mathrm{△}AED}=\frac{1}{2}{S}_{\mathrm{△}BDE}=\frac{1}{6}& \frac{a_1}{a_2}=\frac{S_1}{S_2}\text{ h same}\end{array}$

Jump to heading🔗 $\boxed{\text{13}}$Figure 6–14, known $CD=5,DE=7,EF=15,FG=6$, the line segment $AB$ divides the figure into two parts, the area of the left part is 38, and the area of the right part is 65, then the area of triangle $ADG$ is $?$.

$\begin{array}{llllll}(\text{A})40& (\text{B})35& (\text{C})33& (\text{D})32& (\text{E})31& \end{array}$

• Show known conditions and assume that the unknown variable in the overall area
  

  $\begin{array}{l}\mathrm{△}ADE=x\mathrm{△}BCE=y\\ \mathrm{△}AEG=\frac{15+6}{7}x=3x\\ \mathrm{△}BEG=\frac{15}{7+5}y=\frac{5}{4}y\\ \{\begin{array}{l}x+y=38\\ 3x+\frac{5}{4}y=65\end{array}\\ y=38-x\\ 3x+\frac{5}{4}(38-x)=65\\ 3x+\frac{5}{4}\times 38-\frac{5}{4}\times x=65\\ 3x+\frac{190}{4}-\frac{5}{4}x=65\\ 4(3x)+4(\frac{190}{4})-4(\frac{5}{4}x)=4(65)\\ 12x-5x+190=260\\ 7x=260-190\\ 7x=70\\ x=\frac{70}{7}=10\\ y=38-10=28\\ S_{\mathrm{△}ADG}=x+3x=4\times 10=40\end{array}$

Jump to heading🔗 $\boxed{\text{14}}$If a triangle has two sides of length 4 and 6, and the area of the triangle is $6\sqrt{2}$, then the angle between the two sides is $?$.

$\begin{array}{llllll}(\text{A}){30}^{\circ }& (\text{B}){45}^{\circ }\vee {135}^{\circ }& (\text{C}){60}^{\circ }\vee {120}^{\circ }& (\text{D}){75}^{\circ }& (\text{E}){90}^{\circ }& \end{array}$

$\begin{array}{l}S=\frac{1}{2}\times 4\times 6\times sinC=6\sqrt{2}\\ 12sinC=6\sqrt{2}\\ sinC=\frac{6\sqrt{2}}{12}=\frac{\surd 2}{2}\\ sinC=\frac{\surd 2}{2}={45}^{\circ }\vee {135}^{\circ }\end{array}$

Jump to heading🔗 $\boxed{\text{15}}$If a triangle has two sides of length 4 and 6, and the length of the third side is changing, then the maximum area of the triangle is $?$.

$\begin{array}{llllll}(\text{A})18& (\text{B})16& (\text{C})14& (\text{D})12& (\text{E})10& \end{array}$

$\begin{array}{l}{S}_{max}=\frac{1}{2}\times 4\times 6\times sin{90}^{\circ }\\ sin{90}^{\circ }=12& \text{When the area of the triangle reaches its maximum }sinC=1\end{array}$

Jump to heading🔗 $\boxed{\text{16}}$If the three sides of a triangle are 7,8,9 then its area is $?$.

$\begin{array}{llllll}(\text{A})16\sqrt{2}& (\text{B})12\sqrt{3}& (\text{C})18\sqrt{3}& (\text{D})12\sqrt{5}& (\text{E})18\sqrt{5}& \end{array}$

$\begin{array}{l}p=\frac{7+8+9}{2}=12\\ S=\sqrt{12(12-7)(12-8)(12-9)}\\ S=\sqrt{12\times 5\times 4\times 3}=12\sqrt{5}\end{array}$

• Two triangles are called congruent-angle triangles if they share an equal or supplementary angle.
• Their area ratio is the same as the ratio of the products of the sides around those angles.
• Figure 6–11, in $\mathrm{\angle }ABC$ and $\mathrm{\angle }ADE$, the sine of $\mathrm{\angle }\mathbf{A}$ is the same (same angles), so $\boxed{{\displaystyle S_{\mathrm{△}ABC}:S_{\mathrm{△}ADE}=(AB\cdot AC):(AD\cdot AE)}}$.
  
  $\frac{S_{\mathrm{△}ADE}}{S_{\mathrm{△}ABC}}=\frac{\cancel{\frac{1}{2}}AD\times AE\times \cancel{sinA}}{\cancel{\frac{1}{2}}AB\times AC\times \cancel{sinA}}=\frac{AD\times AE}{AB\times AC}$

Jump to heading🔗 $\boxed{\text{17}}$Figure 6–15, in $\mathrm{△}ABC$, points $D,E$ lie onside $AB,AC$ respectively, and $AD:AB=2:5,AE:AC=4:7,S_{\mathrm{△}ADE}=16$, then the area of $\mathrm{△}ABC$ is $?$.

$\begin{array}{llllll}(\text{A})56& (\text{B})65& (\text{C})66& (\text{D})70& (\text{E})72& \end{array}$

• Show known conditions
  

  $\begin{array}{l}\frac{S_{\mathrm{△}ADE}}{S_{\mathrm{△}ABC}}=\frac{AD\times AE}{AB\times AC}=\frac{2}{5}\times \frac{4}{7}=\frac{8}{35}\\ S_{\mathrm{△}ADE}=16\\ \frac{16}{S_{\mathrm{△}ABC}}=\frac{8}{35}\\ 16\times 35=S_{\mathrm{△}ABC}\times 8\\ 560=S_{\mathrm{△}ABC}\times 8\\ S_{\mathrm{△}ABC}=\frac{560}{8}=70\end{array}$

Jump to heading🔗 $\boxed{\text{18}}$Figure 6–15, in $\mathrm{△}ABC$, point $D$ lies on the extension of side $BA$, and point $E$ lies onside $AC$, give that $AB:AD=5:2,AE:EC=3:2,S_{\mathrm{△}ADE}=12$, then the area of $\mathrm{△}ABC$ is $?$.

$\begin{array}{llllll}(\text{A})30& (\text{B})35& (\text{C})43& (\text{D})48& (\text{E})50& \end{array}$

• Show known conditions
  

  $\begin{array}{l}\frac{S_{\mathrm{△}ADE}}{S_{\mathrm{△}ABC}}=\frac{AD\times AE}{AB\times AC}=\frac{2\times 3}{5\times (3+2)}=\frac{6}{25}\\ S_{\mathrm{△}ADE}=12\\ \frac{12}{S_{\mathrm{△}ABC}}=\frac{6}{25}\\ 12\times 25=S_{\mathrm{△}ABC}\times 6\\ 300=S_{\mathrm{△}ABC}\times 6\\ S_{\mathrm{△}ABC}=\frac{300}{6}=50\end{array}$

• Figure 6–12, within triangle $ABC$, the lines $AD,AE,CF$ are concurrent at point $O$, then $S_{\mathrm{△}ABO}:S_{\mathrm{△}ACO}=BD:DC$.
  
• This theorem provides a new method for converting an area ratio into a segment ratio, because the shapes of $\mathrm{△}ABO$ and $\mathrm{△}ACO$ resemble a Swallow's tail, the theorem is known as the Swallowtail Theorem. it is widely applied in various geometric problems; its particular significance lies in the fact that it can exist within any triangle, offering a way to relate the areas of sub-triangles to the corresponding segments on the base.
• Derivation process: Since $\mathrm{△}ABD$ and $\mathrm{△}ACD$ have equal altitude.
  $\begin{array}{ll}{S}_{\mathrm{△}ABD}:S_{\mathrm{△}ACD}=BD:CD& \mathrm{△}ABD\text{ and }\mathrm{△}ACD\text{ have equal altitude}\\ S_{\mathrm{△}BOD}:S_{\mathrm{△}COD}=BD:CD& \mathrm{△}BOD\text{ and }\mathrm{△}COD\text{ have equal altitude}\\ S_{\mathrm{△}ABO}:S_{\mathrm{△}ACO}=BD:CD& (S_{\mathrm{△}ABD}-S_{\mathrm{△}BOD}):(S_{\mathrm{△}ACD}-S_{\mathrm{△}COD})=BD:CD\end{array}$

Jump to heading🔗 $\boxed{\text{19}}$Figure 6–17, in triangle $ABC$, $BD:DC=4:9,CE:EA=4:3$, then $AF:FB=?$.

$\begin{array}{llllll}(\text{A})27:17& (\text{B})27:14& (\text{C})25:16& (\text{D})28:15& (\text{E})27:16& \end{array}$

• Show known conditions
  

  $\begin{array}{l}\mathrm{①}\frac{BD}{CD}=\frac{S_{\mathrm{△}AOB}}{S_{\mathrm{△}AOC}}\\ \mathrm{②}\frac{CE}{AE}=\frac{S_{\mathrm{△}BOC}}{S_{\mathrm{△}AOB}}\\ \mathrm{③}\frac{AF}{BF}=\frac{S_{\mathrm{△}AOC}}{S_{\mathrm{△}BOC}}\\ \frac{BD}{CD}\times \frac{CE}{AE}\times \frac{AF}{BF}=\frac{\cancel{S_{\mathrm{△}AOB}}}{\cancel{S_{\mathrm{△}AOC}}}\times \frac{\cancel{S_{\mathrm{△}BOC}}}{\cancel{S_{\mathrm{△}AOB}}}\times \frac{\cancel{S_{\mathrm{△}AOC}}}{\cancel{S_{\mathrm{△}BOC}}}=1\\ \frac{4}{9}\times \frac{4}{3}\times \frac{AF}{BF}=1\\ \frac{27}{16}(\frac{16}{27})\times \frac{AF}{BF}=\frac{27}{16}(1)\\ \frac{AF}{BF}=\frac{27}{16}\end{array}$