Geometry

Quadrilaterals

Hard

A parallelogram $ABCD$ has area $48 \mathrm{sqcm}$ . If the length of CD is $8 \mathrm{~cm}$ and that of $AD$ is s cm , then which one of the following is necessarily true?

$5 \le s \le 7$

$s \ne 6$

$s \le 6$

$s \ge 6$

✅ Correct Option: 4

We have a parallelogram $ABCD$ with:

Area = 48 sq cm

Length of $CD$ = 8 cm

Length of $AD$ = $s$ cm

We need to find what condition $s$ must satisfy.

The area of a parallelogram is given by:

Area = Base × Height

The height is always the perpendicular distance between two parallel sides, not the length of a slanted side.

Let's use $CD$ as our base (length = 8 cm).

To find the height, we drop a perpendicular from point $A$ to the line $CD$ . Let's call the foot of this perpendicular point $X$ .

The distance $AX$ represents the height of the parallelogram.

Using the area formula:

$48 = 8 \times AX$

$AX = \tfrac{48}{8} = 6$ cm

Now we have formed a right triangle $AXD$ where:

$AX = 6$ cm (height - one leg)

$XD$ is part of $CD$ (other leg)

$AD = s$ cm (hypotenuse)

In any right triangle, the hypotenuse is always longer than either leg.

Since $AD$ is the hypotenuse of right triangle $AXD$ :

$AD > AX$

$s > 6$

The side $AD$ is slanted, while $AX$ is the vertical height. The slanted distance must always be greater than the vertical distance.

Therefore, $s > 6$ cm is necessarily true.

In any parallelogram, each side must be greater than the perpendicular distance (height) between the parallel sides it connects.

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A parallelogram ABCDABCDABCD has area 48sqcm48 \mathrm{sqcm}48sqcm. If the length of CD is 8 cm8 \mathrm{~cm}8 cm and that of ADADAD is s cm , then which one of the following is necessarily true?

Solution

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CAT Past Year Questions (Topic-Wise)

A parallelogram $ABCD$ has area $48 \mathrm{sqcm}$ . If the length of CD is $8 \mathrm{~cm}$ and that of $AD$ is s cm , then which one of the following is necessarily true?