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Quadrilaterals
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CAT 2025 Lesson : Quadrilaterals - Rhombus & Rectangle

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3.3 Rhombus

1) A parallelogram where all sides are equal.

2) Opposite sides are parallel.

3) Opposite angles are equal.
4) Diagonals are the angle bisectors.

5) Diagonals bisect each other at right angles.

6) A circle can be inscribed in a rhombus.
7) Where d1d_1 and d2 d_2 are the lengths of the two diagonals, Area of Rhombus =12d1d2 = \dfrac{1}{2} d_1 d_2 .

8) If the mid-points of adjacent sides of a rhombus are joined, we get a rectangle. In this figure, P, Q, R and S are the midpoints of the sides and PQRS is a rectangle.

Example 13

In the parallelogram ABCD, AC ⊥\bot BD. If ∠\angleADC =60o= 60^{\mathrm{o}} and BD=6 = 6 cm, then CD == ?

Solution

In a parallelogram, diagonals bisect each other.

∴ DO == BO == 3 cm

ABCD is a rhombus as the diagonals are perpendicular.

In a rhombus, the diagonals are angle bisectors.

∴ ∠\angle ADO == ∠\angleCDO =30o= 30^{\mathrm{o}}

△\triangleCOD is a 30o−60o−90o30^{\mathrm{o}} - 60^{\mathrm{o}} - 90^{\mathrm{o}} triangle.

∴ OC : OD : CD == 1 : 3\sqrt{3} : 2

ODCD=32⇒CD=3×23=23 \dfrac{\text OD}{\text CD} = \dfrac{\sqrt{3}}{2}⇒ CD = \dfrac{ 3 \times 2}{\sqrt{3}} = 2\sqrt{3}

Answer: 232\sqrt{3}


3.4 Rectangle

1) A parallelogram where all angles are equal, i.e. 90o\bm{90^{\mathrm{o}}}.

2) Opposite sides are equal and parallel.

3) Diagonals are equal and bisect each other.
4) A circle can be circumscribed over a rectangle. O is the circumcentre and OA is the circumradius.

5) The two sides are referred to as length (ll) and breadth (bb). i.e., AB == CD =l= l and BC == DA =b= b

6) Perimeter =2(l+b)= 2(l + b)
7) Area of a rectangle =lb= lb

8) Applying Pythagoras theorem, length of diagonal == l2+b2 \sqrt{ l^2 + b^2}

9) If the mid-points of adjacent sides of a rectangle are joined, we get a rhombus. PQRS is a rhombus.

Example 14

A rectangular sheet of paper, when halved by folding it at the mid-point of its longer side, results in a rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter sides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the smaller rectangle?
[CAT 2004]

(1)42(2)22(3)2(4)Noneoftheabove (1) 4 \sqrt{2 } (2) 2 \sqrt{2} (3) \sqrt{2} (4) \mathrm {None of the above}

Solution

Let AB =l= l be the longer side of the original rectangle ABCD.

E and F are mid-points of AB and CD. The rectangle is folded along EF. The smaller rectangle we get after folding is AEFD.

The proportion of length to breadth can remain the same only if the 2 becomes the length and l2\dfrac{l}{2} becomes the breadth of the new rectangle.

As the proportion of length to breadth are the same,

l2=2l/2⇒l=2\dfrac{l}{2} = \dfrac{2}{l/2} ⇒ l =2

Area of the smaller rectangle =2×l2=l=22= 2\times \dfrac{l}{2} = l = 2\sqrt{2}

Answer:(22) 222 \sqrt{2}


Example 15

ABCD is a rectangle with AD = 10. P is a point on BC such that ∠APD = 90°. If DP = 8 then length of BP is
[XAT 2008]

(1)6.4(2)5.2(3)4.8(4)3.6(5)Noneoftheabove(1) 6.4 (2) 5.2 (3) 4.8 (4)3.6 (5)\mathrm {None of the above}

Solution

△\triangle APD is right-angled at P.

∴ AP =102−82=6= \sqrt{10^2 - 8^2} = 6 We drop a perpendicular from P to AD, which meets AD at Q.

Product of base and altitude of a triangle are always equal.

⇒ PQ ×\times AD == AP ×\times PD

⇒ PQ ×10=6×8\times 10 = 6 \times 8

⇒ PQ =4.8= 4.8

In right-angled △\triangle PQA , QA =62−4.82=3.6= \sqrt{6^2 - 4.8^2} = 3.6


As ABPQ is a rectangle, AQ == BP == 3.6

Answer:(44) 3.63.6


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