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Question

In a given square $$ABCD$$, if the area of triangle $$ABD$$ is $$36\ cm^{2}$$, find
(i) the area of triangle $$BCD$$; (ii) the area of the square $$ABCD$$


Solution

$$ABCD$$ is a square
$$BD$$ is the diagonal
We know that the diagonal divides the square into two congruent triangles.
$$\therefore$$ $$\triangle ABD $$ $$\cong \triangle BCD$$
$$\therefore$$ Area of $$\triangle ABD$$ =area of $$\triangle BCD$$ 
Area of $$\triangle ABD$$ = $$36cm^2$$ (given)
$$\therefore$$ Area of $$\triangle BCD$$ = $$36cm^2$$
Area of the square $$ABCD$$ = Area of $$\triangle ABD$$ + Area of $$\triangle BCD$$ = $$36cm^2$$ +$$36cm^2$$
Area of the square $$ABCD$$ = $$72cm^2$$

Mathematics

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