# NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.2

NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.2 (Ex. 11.2) Mensuration in Hindi Medium and English Medium free download in PDF for session 2020-2021 CBSE exams. Here we will learn to find the area of quadrilaterals in various ways.

Exercise 11.2 of grade 8 mathematics deals the questions based on application of areas of quadrilateral. Sometime it can be calculated using triangles otherwise it depends on the situation.## Class 8 Maths Chapter 11 Exercise 11.2 Solution

Class: 8 | Mathematics |

Chapter: 11 | Mensuration |

Exercise: 11.2 | English and Hindi Medium Solutions |

### CBSE NCERT Class 8 Maths Chapter 11 Exercise 11.2 Solution in Hindi and English Medium

### Class 8 Maths Chapter 11 Exercise 11.2 Solution in Videos

#### Area of a Polygon

We know that the plots and fields are generally in the form of regular or irregular polygons. We find their areas by dividing them into triangles, rectangles, parallelograms and trapeziums.

##### Area of a Quadrilateral

Let ABCD be a given quadrilateral in which BD is one of its diagonals.

Let AL Ʇ BD and CM Ʇ BD.

Let AL = h and CM = h. Then,

area of quad. ABCD

= ar(DABD) + ar(DBCD)

= ½ x BD x AL + ½ x BD x CM

= ½ x BD x h + ½ x BD x h

= ½ x BD x (h + h) sq. units.

##### In the given figure, ABCD is a quadrilateral in which BD = 14 cm, AL Ʇ BD, CM Ʇ BD such that AL = 6 cm and CM = 8 cm. Find the area of quad. ABCD.

Area of quad. ABCD = ar (triangle ABD) + ar (triangle BCD)

= {½ x BD x AL + ½ x BD x CM} cm²

= {½ x 14 x 6 + ½ x 14 x 8} cm²

= (42 + 56) cm2 = 98 cm²

Area of Irregular Polygon

##### Find the area of the given pentagon ABCDE in which each one of BF, CH and EG is perpendicular to AD such that AF = 9 cm, AG = 13 cm, AH = 19 cm, AD = 24 cm, BF = 6 cm, CH = 8 cm and EG = 9 cm.

Area of the given pentagon ABCDE

= {1/2 x AF x BF + ½ x (BF + CH) x FH} + {½ x HD x CH + ½ x AD x EG}

= {1/2 x AF x BF} + {½ x (BF + CH) x (AH – AF)} + {1/2 x (AD – AH) x CH} + {1/2 x AD x EG}

= {1/2 x 9 x 6} + {½ x (6 + 8) x (19 – 9)} + {1/2 x (24 – 19) x 8} + {1/2 x 24 x 9} cm²

= (27 + 70 + 20 + 108) cm² = 225 cm².

Area of Regular Polygons

Q Find the area of the given regular hexagon ABCDF in which each side measures 5 cm, height BE = 11 cm and width FD = 8 cm.

Join BE. Then, = ar(hexagon. ABCDF) = 2 x ar(trap. ABEF) + ar(trap. BCDE)

= 2 x ½ x (AF + BE) x FL, where FL Ʇ BE

= 2 x ½ x (5 + 11) x 4 cm² (Because FL = 1/2FD = 4 cm)

= 64 cm²

##### What are the five properties of trapezium?

Properties of a Trapezium

A trapezium has two parallel sides and two non-parallel sides. The diagonals of regular trapezium bisect each other. The length of the mid-segment is equal to half the sum of the parallel bases, in a trapezium.

##### Are diagonals equal in a trapezium?

Answer. Diagonals are equal only in the special case of trapezium that is known as a trapezoid (or isosceles trapezium). Furthermore, a characteristic property of trapezoid is that the diagonals happen to be equal.

##### Can a trapezium have a right angle?

A trapezium cannot have four right angles. A trapezium is a quadrilateral, which means that it has four sides and four angles.

##### What angles does a trapezium have?

A trapezium has four angles. Regardless of whether you use the British or U.S. definition, a trapezium is a quadrilateral. This means that it is a polygon with four sides. Since all polygons have the same number of sides as they have angles, a trapezoid has four sides and four angles.