# Please ans Q.27. If the perimeter of an isosceles right triangle is 2p, then its area is (A) $\left(2+\sqrt{2}\right){p}^{2}$ (B) $\left(2-\sqrt{2}\right){p}^{2}$ (C) $\frac{\left(3-2\sqrt{2}\right){p}^{2}}{4}$ (D) $\left(3-2\sqrt{2}\right){p}^{2}$

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