3. Types of Quadrilaterals

The following diagram contains the different types of Quadrilaterals and their properties. Note that the properties of a particular figure flow downward and apply for figures connected through the arrows.

We will discuss more about the types of quadrilaterals, with their respective figures in the upcoming pages.

3.1 Quadrilateral

1) A Quadrilateral is a shape enclosed by 4 intersecting lines forming 4 vertices, 4 edges and 4 angles. 2) Sum of the interior angles of a quadrilateral equals $360^{\mathrm{o}}$. i.e., $ \angle A + \angle B + \angle C + \angle D = \bm {360^{\mathrm{o}}} $ 3) When the mid-points of adjacent sides of a quadrilateral are joined,we get a parallelogram.
4) Area $= \dfrac {1}{2} \times d \times (h_1 + h_2)$ where, $d$ is the length of the diagonal; and $h_1$ and $h_2$ are the perpendiculars drawn to the diagonal from the other 2 vertices.

Example 10

ABCD as shown below is a quadrilateral with area of 72 sq. units. If BD = 12 and AE = 3, then CF = ?

Solution

AE and CF are altitudes drawn to the diagonal BD.

Area of Quadrilateral $ = \dfrac{1}{2} \times d \times (h_1 + h_2) $⇒ $72 = \dfrac{1}{2} \times 12 \times $ ( 3 + CF )

⇒ 3 + CF = 12 ⇒ CF = 9

Answer: $9$

3.2 Parallelogram

A Parallelogram is a quadrilateral where the opposite sides are parallel. Other properties are as follows:

1) Opposite sides are equal, i.e., AB $=$ CD and BC $=$ AD 2) Opposite angles are equal, i.e., $ \angle $ A $ = $ $ \angle $ C and $ \angle $ B $ = $ $ \angle $ D
3) Adjacent angles are supplementary, i.e. $ \angle $ A + $ \angle $ B $ = 180^{\mathrm{o}}$ 4) The diagonals bisect each other, i.e., AO $ = $ OC and BO $ = $ OD 5) Sum of squares of diagonals $=$ Sum of squares of sides, i.e., AC$^2$ + BD$^2$ $=$ AB$^2$ + BC$^2$ + CD$^2$ + DA$^2$
6) Perimeter $= 2 \times$ sum of a pair of adjacent sides $= 2 \times $ (AB + BC)
7) Each Diagonal divides the parallelogram into 2 congruent triangles, which have the same base and altitude. (Area formulae are derived by adding the areas of the 2 congruent triangles formed by a diagonal) 8) Where $b$ and $h$ are the base and altitude/height drawn to the base, Area of a parallelogram $= 2 \times \dfrac{1}{2} bh = bh$
9) Where $a$ and $b$ are two adjacent sides and θ is an angle of the parallelogram, Area of a parallelogram $= 2 \times \dfrac{1}{2} ab sinθ = ab sinθ$
10) If lines are drawn to the 4 vertices from any point O inside a parallelogram, then 4 triangles are formed such that sum of areas of opposite triangles are equal, i.e., Area ($\triangle $AOB) + Area($\triangle$COD) = Area ($\triangle$ BOC) + Area ($\triangle$ DOA)
11) When the mid-points of a parallelogram are joined, we get another parallelogram. PQRS is also a parallelogram. 12) All the properties of a parallelogram also apply to a rectangle, a rhombus and a square.

Example 11

In a parallelogram ABCD, E is a point on CD such that AE $=$ CE. If $\angle$ACB $= 100^{\mathrm{o}}$ and $\angle$DAE $= 50^{\mathrm{o}}$ , then $\angle$AED = ?

Solution

As $\triangle$AEC is isosceles, $\angle$EAC $=$ $\angle$ECA $= y$

As AB | | CD and AC is a transversal,

$\angle$DAC $=$ $\angle$BCA (Alternate Interior angles)

⇒ $50^{\mathrm{o}} + y = 100^{\mathrm{o}}$ ⇒ $ y = 50^{\mathrm{o}}$

$\angle$AED $= 2y$ $= 100^{\mathrm{o}}$ (Exterior angle of $\triangle$ AEC)

Answer: $100^{\mathrm{o}}$

Example 12

In parallelogram ABCD, the points P, Q, R and S lie on AB, BC, CD and DA respectively such that AP : PB = CQ : BQ = CR : DR = AS : DS = 1 : 3. If the area of ABCD is 24 cm$^2$, then what is the area of PQRS?

Solution

We can subtract the areas of the 4 triangles ASP, BPQ, CQR and DRS from the area of ABCD, i.e. 24 cm$^2$. The diagonal BD cuts ABCD into 2 congruent triangles with equal area of 12 cm$^2$ each. As two sides are in the same proportion and $\angle$A is common, using SAS, $\triangle$ ASP $\sim $ $\triangle $ADB
∴ $ \dfrac {\mathrm{Area( \triangle ASP)}}{\mathrm{Area (\triangle ADB)}} = \dfrac {\mathrm {AS^2}}{\mathrm {AD^2}}$ =$(\dfrac{\mathrm {AS}}{\mathrm {AS} + \mathrm {DS}})^2 = ( \dfrac {1}{4})^2 = {\dfrac{1}{16}}$ ⇒ $\dfrac{\mathrm {Area ( \triangle ASP)}}{12} = \dfrac{1}{16}$ ⇒ ${\mathrm {Area ( \triangle ASP)}} = \dfrac{3}{4}$
$ \dfrac{\mathrm {Area (\triangle DRS)}}{\mathrm {Area(\triangle DCA)}}=\dfrac{ \mathrm{DS^2}}{\mathrm{AD^2}}$ = $(\dfrac{\mathrm {DS}}{\mathrm {AS} + \mathrm{DS}})^2 = (\dfrac {3}{4})^2 = \dfrac{9}{16}$ ⇒ $\dfrac{\mathrm {Area (\triangle DRS)}}{12} = \dfrac {9}{16}⇒ {\mathrm {Area (\triangle DRS)}}$ = $\dfrac{27}{4}$

$ \triangle $ ASP and $ \triangle $ CQR are congruent, and $ \triangle $ DRS and $ \triangle $ BPQ are congruent. Area of PQRS = $24-( \dfrac{3}{4} + \dfrac{3}{4} + \dfrac{27}{4} + \dfrac{27}{4}) = 24 - 15 = 9 $cm$^2$

Answer: $9$ cm$^2$