What is the area (in sq units) of the largest triangle, that can be fitted into a rectangle of length $l$ units and width $w$ units?

\frac{l w}{2}

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\frac{l w}{3}

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\frac{l w}{6}

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\frac{l w}{4}

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Solution

The correct option is A $\frac{l w}{2}$

Let ABCD be the rectangle of length $l$ and width $w$ .
Now, we construct a triangle of maximum area inside it in all possible ways.
$∵$ We know that,
Area of triangle $= \frac{1}{2} \times B a s e \times H e i g h t$
So, for maximum area, base and height of maximum length is needed.
Here, maximum base length = $l$
and maximum height = $w$ .
$∴$ Area (maximum) of triangle $= \frac{1}{2} \times l \times w = \frac{l \times w}{2} s q u n i t s$ $.$

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Question 4

What is the area (in sq units) of the largest triangle, that can be fitted into a rectangle of length l units and width W units ?
(a) $\frac{l w}{2}$ (b) $\frac{l w}{3}$
(c) $\frac{l w}{6}$ (d) $\frac{l w}{4}$

Question 4

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What is the area (in sq units) of the largest triangle, that can be fitted into a rectangle of length l units and width w units?

What is the area (in sq units) of the largest triangle, that can be fitted into a rectangle of length $l$ units and width $w$ units?