1. **Problem statement:** Given triangle ABC with sides $a=7$ cm, $c=5$ cm, the distance from the incentre to side BC is 1.477 cm, and the area is 16.25 cm², find: - (a) side $b$ - (b) radius of the escribed circle tangent to side $a$ - (c) distance from the circumcenter to side $a$ 2. **Step 1: Find side $b$** - The distance from the incentre to side BC is the inradius $r$. - Area $A = r imes s$ where $s$ is the semiperimeter: $s = \frac{a+b+c}{2}$. - Given $r = 1.477$, $A = 16.25$, so: $$s = \frac{A}{r} = \frac{16.25}{1.477} \approx 11$$ - Using $s = \frac{a+b+c}{2}$: $$11 = \frac{7 + b + 5}{2} \Rightarrow 22 = 12 + b \Rightarrow b = 10$$ 3. **Step 2: Find radius of the escribed circle tangent to side $a$** - The radius $r_a$ of the escribed circle opposite side $a$ is: $$r_a = \frac{A}{s - a}$$ - Substitute values: $$r_a = \frac{16.25}{11 - 7} = \frac{16.25}{4} = 4.0625$$ 4. **Step 3: Find distance from circumcenter to side $a$** - The circumcenter is the intersection of perpendicular bisectors. - Distance from circumcenter to side $a$ equals the circumradius $R$ times $\sin A$. - Use Law of Cosines to find angle $A$ opposite side $a$: $$a^2 = b^2 + c^2 - 2bc \cos A$$ $$7^2 = 10^2 + 5^2 - 2 \times 10 \times 5 \cos A$$ $$49 = 100 + 25 - 100 \cos A$$ $$49 = 125 - 100 \cos A \Rightarrow 100 \cos A = 125 - 49 = 76$$ $$\cos A = 0.76$$ - Then $\sin A = \sqrt{1 - 0.76^2} = \sqrt{1 - 0.5776} = \sqrt{0.4224} \approx 0.65$ - Circumradius $R$ is given by: $$R = \frac{abc}{4A} = \frac{7 \times 10 \times 5}{4 \times 16.25} = \frac{350}{65} \approx 5.3846$$ - Distance from circumcenter to side $a$: $$d = R \sin A = 5.3846 \times 0.65 \approx 3.5$$ **Final answers:** - Side $b = 10$ cm - Escribed circle radius $r_a = 4.0625$ cm - Distance from circumcenter to side $a$ is approximately 3.5 cm