Let $l$ be the length of each equal side of an isosceles triangle. If the length of each equal side is doubled, keeping its height unchanged, then the difference of the squares of bases of the new triangle and the given triangle is

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4 l^{2}

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9 l^{2}

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12 l^{2}

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Solution

The correct option is D $12 l^{2}$
Base length of old triangle $= 2 \sqrt{l^{2} - h^{2}}$
Base length of new triangle $= 2 \sqrt{4 l^{2} - h^{2}}$
Difference of their squares
$= 4 (4 l^{2} - h^{2} - l^{2} + h^{2}) = 12 l^{2}$ .

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Let l be the length of each equal side of an isosceles triangle. If the length of each equal side is doubled, keeping its height unchanged, then the difference of the squares of bases of the new triangle and the given triangle is

The correct option is D 12l2Base length of old triangle =2√l2−h2Base length of new triangle =2√4l2−h2Difference of their squares=4(4l2−h2−l2+h2)=12l2.

Let $l$ be the length of each equal side of an isosceles triangle. If the length of each equal side is doubled, keeping its height unchanged, then the difference of the squares of bases of the new triangle and the given triangle is

The correct option is D $12 l^{2}$
Base length of old triangle $= 2 \sqrt{l^{2} - h^{2}}$
Base length of new triangle $= 2 \sqrt{4 l^{2} - h^{2}}$
Difference of their squares
$= 4 (4 l^{2} - h^{2} - l^{2} + h^{2}) = 12 l^{2}$ .