Question

In $\Delta$ ABC, D is a point on side BC. Find AB, if AC=3 cm, AD=3cm, BD=8cm and CD=1cm.

• A. 7 cm
• B. 9 cm
• C. 11 cm
• D. 6 cm

Answer 1

The correct option is B 9 cm


Drop a perpendicular from A on DC at E. Now,
$\Delta AED\cong \Delta AEC$ by RHS congruency,
so that, $DE=EC=\left(\frac{1}{2}\right)DC=0.5cm$
also
ar$\left(\Delta ADC\right)=\left(\frac{1}{2}\right)\times DC\times AE={(s\times (s-a)\times (s-b)\times (s-c))}^{-2}$ where, s is the semi-perimeter of the triangle
or, $A{E}^2=\left(\frac{4\times \left(3.5\right)\times \left(0.5\right)\times \left(0.5\right)\times \left(2.5\right)}{1}\right)c{m}^2=8.75c{m}^2$
Now in right $\Delta AEB,A{B}^2=E{B}^2+A{E}^2={(ED+DB)}^2+A{E}^2={\left(8.5\right)}^2+8.75=81c{m}^2$,
therefore, AB = 9 cm