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:
Option 1: $\frac{1}{4}\sqrt{4b^{2}-a^{2}}$ sq. units
Option 2: $\frac{a}{2}\sqrt{2a^{2}-b^{2}}$ sq. units
Option 3: $\frac{b}{4}\sqrt{4a^{2}-b^{2}}$ sq. units
Option 4: $\frac{b}{2}\sqrt{a^{2}-2b^{2}}$ sq. units
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Correct Answer: $\frac{b}{4}\sqrt{4a^{2}-b^{2}}$ sq. units
Solution : In an isosceles triangle with two sides of length $a$ units and a base of length $b$ units. The height can be found using the Pythagorean theorem, $ = \sqrt{a^2 - \left(\frac{b}{2}\right)^2}$ The area of the triangle, $= \frac{1}{2} \times$ base $\times$ height $= \frac{1}{2} \times b \times \sqrt{a^2 - \left(\frac{b}{2}\right)^2}$ $= \frac{1}{2} \times b \times \sqrt{a^2 - \left(\frac{b^2}{4}\right)}$ $= \frac{1}{2} \times b \times \sqrt{\left(\frac{4a^2-b^2}{4}\right)}$ $= \frac{1}{4} \times b \times \sqrt{(4a^2-b^2)}$ $= \frac{b}{4} \times \sqrt{(4a^2-b^2)}$ Hence, the correct answer is $ \frac{b}{4} \times \sqrt{(4a^2-b^2)}$ sq. units.
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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:
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 : In an isosceles triangle, the length of each equal side is twice the length of the third side. The ratios of areas of the isosceles triangle and an equilateral triangle with the same perimeter are:
Question : $\triangle PQR$ is an isosceles triangle and $PQ=PR=2a$ unit, $QR=a$ unit. Draw $PX \perp QR$, and find the length of $PX$.
Question : The perimeter of a rhombus is $2p$ units, and the sum of the lengths of the diagonals is $m$ units. The area of the rhombus is:
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