I. Area\[=\frac{x\sqrt{4{{y}^{2}}-{{x}^{2}}}}{4}\] |
II. Perimeter \[=(2y+x)\] |
III. Height \[=\frac{1}{2}\sqrt{4{{y}^{2}}-{{x}^{2}}}\] |
A) I only
B) I and II only
C) II and III only
D) I, II and III
Correct Answer: D
Solution :
For an isosceles triangles having base x and each of equal side as y, we must haveI. Area\[=\frac{x\sqrt{4{{y}^{2}}-{{x}^{2}}}}{4}\] |
II. Perimeter\[=(2y+x)\] |
III. Height\[=\frac{1}{2}\sqrt{4{{y}^{2}}-{{x}^{2}}}\] |
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