A) \( \Large 10\ \sqrt{2} m\) |
B) \( \Large 10\ \sqrt{3} m\) |
C) 20 m |
D) \( \Large 20\ \sqrt{2} m\) |
D) \( \Large 20\ \sqrt{2} m\) |
Let x be the base and side of isosceles \( \Large \triangle ABC \).
By hypothesis,
\( \Large \frac{1}{2}x \times x = 800 \)
=> \( \Large x^{2} = 1600 \)
=> x = 40 m
Therefore, Hypotenuse, BC = 2 x = 40 x/2 m
If s be the side of the maximum square, then
\( \Large s^{2}+\frac{1}{2} \left(x-s\right)s+\frac{1}{2} \left(x-s\right)2=800 \)
=> \( \Large s^{2}+xs-s^{2} = 800 \)
=> \( \Large xs = 800 \)
=> \( \Large s = \frac{800}{40} = 20 m \)
Therefore, Length of diagonal of the square = \( \Large 20 \sqrt{2} m \)
1). If perimeter of a triangle is 100 m and its sides are in the ratio 1 : 2 : 2, then area of the triangle (in \( \Large m^{2} \)) is
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2). In the given figure, AD is the internal bisector and AE is the external bisector of \( \Large \angle \)BAC of any \( \Large \triangle \) ABC. Then which one of the following statements is not correct?
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3). In the given figure, \( \Large \angle \)ABC = \( \Large \angle \)ADB = \( 90^{\circ} \), which one of the following statements does not hold good?
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4). Area of an equilateral triangle of side x is
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5). In the given figure, \( \Large \triangle ABC \) is an equilateral triangle. O is the point of intersection of the medians. If AB = 6 cm, then OB is equal to
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6). If an isosceles right triangle has an area 200 sq. cm, then area of a square drawn on hypotenuse is
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7). \( \Large \triangle ABC\ and\ PQR \) are congruent if
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8). \( \Large \triangle \)PQR and \( \Large \triangle \)LMN are similar. If 3 PQ = LM and MN = 9 cm, then QR is equal to
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9). If D, E, F are mid-points of the sides BC, CA and AB respectively of a triangle ABC, then which one of the following is not correctly matched?
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10). If lengths of two sides of a triangle are given, then its area is greater when
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