In Delta DEF, m angle D = 90^{circ}. If EF = | vedclass

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(A) In $\Delta DEF$,since $m \angle D = 90^{\circ}$,the triangle is a right-angled triangle.According to the Pythagorean theorem,the square of the hyp

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$2 \sqrt{5}$

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(A) In $\Delta DEF$,since $m \angle D = 90^{\circ}$,the triangle is a right-angled triangle.According to the Pythagorean theorem,the square of the hypotenuse is equal to the sum of the squares of the other two sides.Here,$EF$ is the hypotenuse because it is opposite to the right angle at $D$.Therefore,$EF^2 = DE^2 + DF^2$.Given $EF = 6$ and $DF = 4$,we substitute these values into the equation:$6^2 = DE^2 + 4^2$$36 = DE^2 + 16$$DE^2 = 36 - 16$$DE^2 = 20$$DE = \sqrt{20} = \sqrt{4 	imes 5} = 2 \sqrt{5}$.

In $\Delta DEF, m \angle D = 90^{\circ}$. If $EF = 6$ and $DF = 4$,then $DE = \dots$

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