Question

Please solve 16th question

Answer 1

Using Pythagoras in ABC

${ab}^{2} + {bc}^{2} = {ac}^{2} \\ {ab}^{2} + {bc}^{2} = {(ad + cd)}^{2}$

${ad}^{2} + 2 {bd}^{2} + {cd}^{2} = {(ad + cd)}^{2}$

..... (1)

Similarly in ADB and BDC

We get,

${ad}^{2} + {bd}^{2} = {ab}^{2}$

.... (2)

${cd}^{2} + {bd}^{2} = {bc}^{2}$

..... (3)

Adding 2 and 3

${ad}^{2} + 2{bd}^{2} + {cd}^{2} = {ab}^{2} + {bc}^{2}$

Now from(1)

${ad}^{2} + 2{bd}^{2} + {cd}^{2} = {(ad + cd)}^{2}$

${8}^{2} + 2 {bd}^{2} + {2}^{2} = {(8 + 2)}^{2} \\ 100 = 64 + 2 {bd}^{2} + 4 \\ 2 {bd}^{2} = 100 - 68 \\ 2 {bd}^{2} = 32 \\ {bd}^{2} = 16 \\ bd = 4cm$