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9th Class Mathematics Areas of Parallelograms and Triangles Question Bank Area of Parallelogram & Triangle

  • question_answer
    In the given figure an isosceles triangle, the measure of each of equal sides is 10 cm and the angle between them is 45°, the area of the triangle is

    A)  25 \[c{{m}^{2}}\]                  

    B)  \[\frac{25}{2}\sqrt{2}\,c{{m}^{2}}\]

    C)  \[25\sqrt{2}\]\[c{{m}^{2}}\]    

    D)  \[25\sqrt{3}\]\[c{{m}^{2}}\]

    Correct Answer: C

    Solution :

    (c): \[AB=AC=10\]cm \[Area=\frac{1}{2}bc\,sinA=\frac{1}{2}\times 10\times 10\,sin{{45}^{{}^\circ }}\] \[=\frac{50}{\sqrt{2}}=\frac{50\times \sqrt{2}}{\sqrt{2}\times \sqrt{2}}=25\sqrt{2}\]\[c{{m}^{2}}\]       


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