1. **Problem statement:** Given triangle ABC with sides $a=7$ cm, $c=5$ cm, the incenter's distance to side BC is $1.477$ cm, and the area is $16.25$ cm². Find: 1) side $b$ 2) radius of the excircle tangent to side $a$ 3) distance from the circumcenter to side $a$ 2. **Step 1: Find side $b$** - The incenter's distance to side BC is the inradius $r$ of the incircle. - Area $A = r imes s$ where $s$ is the semiperimeter: $s = \frac{a+b+c}{2}$. - Given $A=16.25$ and $r=1.477$, so: $$16.25 = 1.477 \times s \implies s = \frac{16.25}{1.477} \approx 11$$ - Since $s = \frac{7 + b + 5}{2} = \frac{12 + b}{2}$, then: $$11 = \frac{12 + b}{2} \implies 22 = 12 + b \implies b = 10$$ 3. **Step 2: Find the radius $r_a$ of the excircle tangent to side $a$** - The excircle radius opposite side $a$ is: $$r_a = \frac{A}{s - a}$$ - We have $s=11$, $a=7$, $A=16.25$, so: $$r_a = \frac{16.25}{11 - 7} = \frac{16.25}{4} = 4.0625$$ 4. **Step 3: Find the distance from the circumcenter to side $a$** - The circumcenter is the intersection of perpendicular bisectors. - The distance from the circumcenter to side $a$ equals the circumradius $R$ times $|\sin A|$ where $A$ is the angle opposite side $a$. - First, find $R$ using formula: $$R = \frac{abc}{4A}$$ - We know $a=7$, $b=10$, $c=5$, $A=16.25$, so: $$R = \frac{7 \times 10 \times 5}{4 \times 16.25} = \frac{350}{65} \approx 5.3846$$ - Next, find angle $A$ using Law of Cosines: $$a^2 = b^2 + c^2 - 2bc \cos A$$ $$7^2 = 10^2 + 5^2 - 2 \times 10 \times 5 \cos A$$ $$49 = 100 + 25 - 100 \cos A$$ $$49 = 125 - 100 \cos A \implies 100 \cos A = 125 - 49 = 76 \implies \cos A = 0.76$$ - Then: $$\sin A = \sqrt{1 - 0.76^2} = \sqrt{1 - 0.5776} = \sqrt{0.4224} \approx 0.6507$$ - Finally, distance from circumcenter to side $a$ is: $$d = R \sin A = 5.3846 \times 0.6507 \approx 3.503$$ **Final answers:** - Side $b = 10$ cm - Excircle radius tangent to side $a$: $4.0625$ cm - Distance from circumcenter to side $a$: $3.503$ cm