A polygon that has got 3 sides is a triangle. To draw a triangle, we need 3 independent measures.

**Construction
of Quadrilaterals**

We have already
learnt how to draw triangles in the earlier classes. A polygon that has got 3 sides
is a triangle. To draw a triangle, we need 3 independent measures. Also, there is
only one way to construct a triangle, given its 3 sides. For example, to construct
a triangle with sides 3*cm*, 5*cm* and 7*cm*, there is only one way to do it.

Now, let
us move on to quadrilaterals. A polygon that is formed by 4 sides is called a quadrilateral.
Isn’t it? But, a quadrilateral can be of different shapes. They need not look like
the same for the given 4 measures. For example, some of the quadrilaterals having
their sides as 4 *cm*, 5 *cm*, 7 *cm* and 9 *cm* are given below

So, to construct
a particular quadrilateral, we need a 5^{th} measure. That can be its diagonal
or an angle measure. Moreover, even if 2 or 3 sides are given, using the measures
of the diagonals and angles, we can construct quadrilaterals.

**Note**

We can split any quadrilateral into two triangles by drawing a diagonal.

In the above figures, a quadrilateral is split in two ways by its
diagonals. So, if a diagonal is given, first draw the lower triangle with two sides
and one diagonal. Then, draw the upper triangle with other two measures.

(i) • A polygon in which atleast one interior angle is more than
180°, is called a concave polygon. In the given polygon, interior angle at C is
more than 180°.

• A polygon in which each interior angle is less than 180°, is called
a convex polygon. In the given polygon, all interior angles are less than 180°.

(ii) Look at the following quadrilaterals.

Though, we can construct a quadrilateral in two ways as shown above,
we do not take into account the concave quadrilaterals in this chapter. Hence, the
construction of only convex quadrilaterals are treated here.

**Note**

Consider the given Quadrilateral ABCD. In which AC is a diagonal
(*d*), BE (*h*_{1}) and DF (*h*_{2})
are the perpendiculars drawn from the vertices B and D on diagonal AC.

Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD

We shall
now see a few types on constructing these quadrilaterals when its:

(i) 4 sides
and a diagonal are given.

(ii) 3
sides and 2 diagonals are given.

(iii) 4 sides
and an angle are given.

(iv) 3
sides and 2 angles are given.

(v) 2 sides
and 3 angles are given.

__1. Constructing
a quadrilateral when its 4 sides and a diagonal are given__

**Example 5.21**

Construct
a quadrilateral *DEAR* with *DE*=6 *cm*,
*EA* = 5 *cm*, *AR* = 5.5*cm*, *RD*
= 5.2 *cm *and *DA* = 10 *cm*. Also find its
area.

*Solution:*

**Given:**

*DE *= 6* cm*,* EA *= 5* cm*,*
AR *= 5.5* cm*,

*RD *= 5.2* cm *and a diagonal

*DA *= 10* cm*

**Steps:**

4. Draw a
line segment *DE* = 6 *cm*.

5. With *D* and *E* as centres, draw arcs of radii 10 *cm* and 5 *cm* respectively and
let them cut at *A*.

6. Join *DA* and *EA*.

7. With *D* and *A* as centres, draw arcs of radii 5.2 *cm* and 5.5 *cm* respectively
and let them cut at *R*.

8. Join *DR* and *AR* .

*9. DEAR *is the required quadrilateral.

** **

__2. Construct
a quadrilateral when its 3 sides and 2 diagonals are given__

**Example 5.22**

Construct
a quadrilateral *NICE* with *NI*=4.5 *cm*, *IC*= 4.3 *cm*, *NE*=
3.5 *cm*, *NC*= 5.5 *cm* and *IE* = 5 *cm*. Also find its area.

*Solution:*

**Given:**

*NI *= 4.5* cm*,* IC*= 4.3* cm*,

*NE *= 3.5* cm *and two diagonals,

*NC *= 5.5* cm *and* IE *= 5* cm*

**Steps:**

1. Draw a
line segment *NI* = 4.5 *cm*.

2. With *N* and *I* as centres, draw arcs of radii 5.5 *cm* and 4.3 *cm* respectively
and let them cut at *C*.

3. Join *NC* and *IC*.

4. With *N* and *I* as centres, draw arcs of radii 3.5 *cm* and 5 *cm* respectively and
let them cut at *E*.

5. Join *NE*, *IE*
and *CE* .

*6. NICE *is the required quadrilateral.

** **

__3. Construct
a quadrilateral when its 4 sides and one angle are given__

**Example 5.23**

Construct
a quadrilateral *MATH* with *MA*=4 *cm*,
*AT*= 3.6 *cm*, *TH* = 4.5 *cm*, *MH*=
5* cm *and* *∠*A *=* *85º* *. Also find its area.

*Solution:*

**Given:**

*MA*=4* cm*,* AT*= 3.6* cm*,

*TH *= 4.5* cm*,* MH*= 5* cm *and* *∠*A *=* *85º

**Steps:**

1. Draw a
line segment *MA* = 4 *cm*.

2. Make ∠*A* = 85º .

3. With *A* as centre, draw an arc of radius 3.6 *cm*. Let it cut the ray *AX* at *T*.

4. With *M* and *T* as centres, draw arcs of radii 5 *cm* and 4.5 *cm* respectively
and let them cut at *H*.

5. Join *MH* and *TH*.

*6. MATH *is the required quadrilateral.

**Calculation of Area:**

Area of the
quadrilateral *MATH* = 1/2 ×
*d* × ( *h*_{1} + *h*_{2}
) *sq. units*

= 1/2 ×
5.1×
(3.9 +
2.8)

= 2.55 ×
6.7 = 17.09 *cm*^{2}

^{}

** **

__4. Construct
a quadrilateral when its 3 sides and 2 angles are given__

**Example 5.24**

Construct
a quadrilateral *ABCD* with *AB*=7 *cm*,
*AD*= 5 *cm*, *CD* = 5 *cm*, ∠*BAC* = 50º and ∠*ABC* = 60º. Also find its area.

*Solution:*

**Given:**

*AB*=7* cm*,* AD*= 5* cm*,* CD *= 5* cm *and two angles ∠*BAC* = 50º and* *∠*ABC* = 60º

**Steps:**

1. Draw a
line segment *AB* = 7 *cm.*

2. At *A* on *AB*
, make ∠*BAY* = 50º and at *B* on *AB*,
make ∠*ABX* = 60º . Let them intersect
at *C*.

3. With *A* and *C* as centres, draw arcs of radius 5 *cm* each. Let them intersect at *D*.

4. Join AD
and CD.

*5. ABCD *is the required quadrilateral.

**Calculation of Area:**

Area of the quadrilateral
ABCD = 1/2 × *d* × ( *h*_{1} + *h*_{2} ) *sq. units*

= 1/2 × 6.4 × (3.8 +
5.3)

= 3.2 × 9.1 = 29.12
*cm*^{2}

^{}

** **

__5. Construct
a quadrilateral when its 2 sides and 3 angles are given__

**Example 5.25**

Construct
a quadrilateral *PQRS* with *PQ=QR*=5 *cm*, ∠*QPR* = 50°, ∠*PRS* = 40° and ∠*RPS *=* *80°. Also find its area.

*Solution:*

**Given:**

*PQ= *5*
cm*,* QR*=5* cm*,* *∠*QPR
*=* *50°,

∠*PRS *=* *40°* *and* *∠*RPS *=* *80°

**Steps:**

1. Draw a
line segment *PQ* = 5 *cm*.

2. At *P* on *PQ*,
make ∠*QPX* = 50°.

3. With *Q* as centre, draw an arc of radius 5 *cm*. Let it cut *PX* at *R*.

4. At *R* on *PR*,
make ∠*PRS* = 40° and at *P* on *PR*, make ∠*RPS* = 80°. Let them
intersect at *S*.

*5. PQRS *is the required quadrilateral.

**Calculation of Area: **

Area of
the quadrilateral PQRS = 1/2 × *d* × ( *h*_{1} + *h*_{2} ) *sq. units*

= 1/2
× 6.4 × (4.7 + 3.8)

= 3.2
× 8.5 = 27.2 *cm*^{2}

^{}

**Think**

Is it
possible to construct a quadrilateral *PQRS*
with *PQ* = 5 *cm*, *QR* = 3 *cm*, *RS*
= 6 *cm*, *PS* = 7 *cm* and *PR* = 10 *cm*. If not, why?

**Solution:**

The lower triangle cannot be constructed as the sum of two sides

5 + 3 = 8 < 10 cm. So this quadrilateral cannot be constructed.

Tags : Questions with Answers, Solution | Geometry | Chapter 5 | 8th Maths , 8th Maths : Chapter 5 : Geometry

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8th Maths : Chapter 5 : Geometry : Construction of Quadrilaterals | Questions with Answers, Solution | Geometry | Chapter 5 | 8th Maths

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