Maharashtra State Board course SSC (English Medium) Class 10th Board Exam
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For Finding Ab and Bc with the Help of Information Given in the Figure, Complete Following Activity. Ab = Bc .......... - Geometry

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Question

For finding AB and BC with the help of information given in the figure, complete following activity.

AB = BC ..........

\[\therefore \angle BAC = \]

\[ \therefore AB = BC =\] \[\times AC\]

\[ =\] \[\times \sqrt{8}\]

\[ =\] \[\times 2\sqrt{2}\]

 =

Solution

In ∆ABC,
∠B = 90, AC =\[\sqrt{8}\] AB = BC, ∴ ∠A = ∠C = 45

By 45∘ − 45 − 90 theorem,

\[AB = BC = \frac{1}{\sqrt{2}} \times AC\]
\[ = \frac{1}{\sqrt{2}} \times \sqrt{8}\]
\[ = \frac{1}{\sqrt{2}} \times 2\sqrt{2}\]
\[ = 2\]

Hence, AB = 2 and BC = 2.
Hence, the completed activity is

AB = BC .......... Given

\[\therefore \angle BAC = {45}^o \]
\[ \therefore  AB = BC = \frac{1}{\sqrt{2}} \times AC\]
\[ = \frac{1}{\sqrt{2}} \times \sqrt{8}\]
\[ = \frac{1}{\sqrt{2}} \times 2\sqrt{2}\]
\[ = 2\]

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Solution For Finding Ab and Bc with the Help of Information Given in the Figure, Complete Following Activity. Ab = Bc .......... Concept: Pythagoras Theorem.
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