Question

the area of triangle adc is twice that of triangle bdc. given that the length of ad is 8, bd =

Explanation:

Step1: Recall area - formula

The area of a triangle is $A=\frac{1}{2}bh$. Triangles $ADC$ and $BDC$ have the same height (the perpendicular distance from $C$ to $AB$). Let the height of both $\triangle ADC$ and $\triangle BDC$ with respect to the bases $AD$ and $BD$ be $h$.

Step2: Set up area - ratio equation

The area of $\triangle ADC=\frac{1}{2}AD\times h$ and the area of $\triangle BDC=\frac{1}{2}BD\times h$. Given that $A_{ADC} = 2A_{BDC}$, we have $\frac{1}{2}AD\times h=2\times\frac{1}{2}BD\times h$.

Step3: Simplify the equation

Cancel out $\frac{1}{2}h$ on both sides of the equation $\frac{1}{2}AD\times h = 2\times\frac{1}{2}BD\times h$. We get $AD = 2BD$.

Step4: Solve for $BD$

Since $AD = 8$, substituting $AD$ into $AD = 2BD$, we can solve for $BD$ by dividing both sides of the equation by 2. So $BD=\frac{AD}{2}$.

Answer:

4