Mensuration Practice Test 5

By Pythagoras theorem,
${\left(\mathit{A}\mathit{B}\right)}^2$ = ${\left(\mathit{A}\mathit{P}\right)}^2+{\left(\mathit{B}\mathit{P}\right)}^2$
${\left(\mathit{A}\mathit{B}\right)}^2$ = ${\left(8\right)}^2+{\left(6\right)}^2$
= 64 + 36 = 100
AB = 10 cm
Now In ΔANP and ΔBNP
${\left(\mathit{A}\mathit{P}\right)}^2$ = ${\left(\mathit{P}\mathit{N}\right)}^2+\mathit{A}{\mathit{N}}^2$ ... (i)
${\left(\mathit{P}\mathit{B}\right)}^2$ = ${\left(\mathit{P}\mathit{N}\right)}^2+{\left(\mathit{B}\mathit{N}\right)}^2$ ... (ii)
Subtracting (ii) from (i),
$\mathit{A}{\mathit{P}}^2-\mathit{P}{\mathit{B}}^2$ = $\mathit{A}{\mathit{N}}^2-\mathit{B}{\mathit{N}}^2$
${\left(8\right)}^2-{\left(6\right)}^2$ = ${\left(\mathit{A}\mathit{B}-\mathit{B}\mathit{N}\right)}^2-\mathit{B}{\mathit{N}}^2$

64 − 36 = $\mathit{A}{\mathit{B}}^2+\mathit{B}{\mathit{N}}^2$ − 2AB.BN − $\mathit{B}{\mathit{N}}^2$
28 = AB2 − 2AB.BN
28 = 100 − 2 ×10.BN
72 = 20BN
BN = 3.6 cm.