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(D) From statement $I$,the hypotenuse $b = 5\, cm$.From statement $III$,one of the angles is $60^{\circ}$.In a right-angled triangle $ABC$ (right-angl

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Both $I$ and $III$

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(D) From statement $I$,the hypotenuse $b = 5\, cm$.From statement $III$,one of the angles is $60^{\circ}$.In a right-angled triangle $ABC$ (right-angled at $B$),if $\angle C = 60^{\circ}$,then $\angle A = 30^{\circ}$.Using trigonometry,$\cos 60^{\circ} = \frac{BC}{AC} = \frac{a}{b}$.Thus,$a = b \cos 60^{\circ} = 5 	imes \frac{1}{2} = 2.5\, cm$.Also,$\sin 60^{\circ} = \frac{AB}{AC} = \frac{c}{b}$.Thus,$c = b \sin 60^{\circ} = 5 	imes \frac{\sqrt{3}}{2} = 2.5\sqrt{3}\, cm$.The area of the triangle is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes a 	imes c$.Area $= \frac{1}{2} 	imes 2.5 	imes 2.5\sqrt{3} = \frac{6.25\sqrt{3}}{2} = 3.125\sqrt{3}\, cm^{2} = \frac{25\sqrt{3}}{8}\, cm^{2}$.Therefore,statements $I$ and $III$ together are sufficient to answer the question.

What is the area of a given right-angled triangle?
$I.$ The length of the hypotenuse is $5\, cm$.
$II.$ The perimeter of the triangle is four times that of its base.
$III.$ One of the angles of the triangle is $60^{\circ}$.

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What is the area of a given right-angled triangle?$I.$ The length of the hypotenuse is $5\, cm$.$II.$ The perimeter of the triangle is four times that of its base.$III.$ One of the angles of the triangle is $60^{\circ}$.