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 diagonalWe 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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