question_answer

Find the unknowns length x in the following figures:

(a)

(b)

(c)

(d)

Answer:

(a) \[\Delta ABC\]is right-angled at B.

∴ By Pythagoras theorem, we have

\[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\]

\[{{x}^{2}}={{8}^{2}}+{{9}^{2}}\]

\[{{x}^{2}}={{6}^{4}}+{{3}^{6}}\]

\[x=\sqrt{100}=10\]

(b) \[\Delta ABC\]is right-angled at B.

∴ By Pythagoras theorem, we have

\[A{{C}^{2}}=A{{B}^{2}}+BC\]

\[\Rightarrow {{x}^{2}}={{12}^{2}}+{{5}^{2}}\]

\[\Rightarrow {{x}^{2}}=144+25\]

\[\Rightarrow {{x}^{2}}=169\]

\[\Rightarrow {{x}^{2}}=\sqrt{169}=13\]

(c) DABC is right-angled at B.

∴ By Pythagoras theorem, we have

\[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\]

\[\Rightarrow {{x}^{2}}={{8}^{2}}+{{15}^{2}}\]

\[\Rightarrow {{x}^{2}}=64+225\]

\[\Rightarrow {{x}^{2}}=289\]

\[x=\sqrt{289}=17\]

(d) DABC is right-angled at B.

∴ By Pythagoras theorem, we have

\[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\]

\[\Rightarrow {{x}^{2}}={{24}^{2}}+{{7}^{2}}\]

\[\Rightarrow {{x}^{2}}=576+49\]

\[\Rightarrow {{x}^{2}}=625\]

\[x=\sqrt{625}=25\]