Question : If the length of each of two equal sides of an isosceles triangle is 10 cm and the adjacent angle is $45^{\circ}$, then the area of the triangle is:
Option 1: $20\sqrt{2}$ sq. cm
Option 2: $12\sqrt{2}$ sq. cm
Option 3: $25\sqrt{2}$ sq. cm
Option 4: $15\sqrt{2}$ sq. cm
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Correct Answer: $25\sqrt{2}$ sq. cm
Solution : $AD = AB \sin 45^\circ$ $= 10\times\frac{1}{\sqrt2}$ $= 5\sqrt2$ cm $\therefore$ Area of $\triangle ABC = \frac{1}{2}\times BC \times AD$ $=\frac{1}{2}\times 10 \times 5\sqrt2$ $= 25\sqrt2$ sq. cm Hence, the correct answer is $25\sqrt2$ sq. cm.
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Question : In a right-angled triangle $\Delta PQR, PR$ is the hypotenuse of length 20 cm, $\angle PRQ = 30^{\circ}$, the area of the triangle is:
Question : The ratio of the length of each equal side and the third side of an isosceles triangle is 3 : 5. If the area of the triangle is $30 \sqrt{11}$ cm2, then the length of the third side (in cm) is:
Question : The area of the parallelogram whose length is 30 cm, width is 20 cm and one diagonal is 40 cm is:
Question : The lengths of the two parallel sides of a trapezium are 28 cm and 40 cm. If the length of each of its other two sides is 12 cm, then the area (in cm2) of the trapezium is:
Question : If for an isosceles triangle, the length of each equal side is $a$ units and that of the third side is $b$ units, then its area will be:
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