NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.1 (Ex. 11.1) Mensuration updated for academic session 2022-2023 free to use or download in PDF file format. Get here all the solutions in text and videos format free. All the questions of mensuration ex. 11.1 of grade 8 is done step by step using simple formulae. Solutions are made easy to understand and are free to use online or offline.

## Class 8 Maths Chapter 11 Exercise 11.1 Solution

 Class: 8 Mathematics Chapter: 11 Mensuration Exercise: 11.1 PDF Explanation and Video Solution

### Class 8 Maths Chapter 11 Exercise 11.1 Solution in Videos

Class 8 Maths Chapter 11 Exercise 11.1 Solution
Class 8 Maths Chapter 11 Exercise 11.1 Explanation

#### Trapezium:

A trapezium is a quadrilateral having one pair of parallel opposite sides. suppose, ABCD is a trapezium in which AB || DC Each of the two parallel sides of a trapezium is called its base. Thus, AB and DC are the bases of trap. ABCD. Draw CL Ʇ AB and DM Ʇ AB. Let CL = DM = h. Then, h is called the height or altitude of trapezium ABCD. A trapezium ABCD is said to be an isosceles trapezium of its nonparallel sides AD and BC are equal.

##### Area of a Trapezium:

Let ABCD be a trapezium in which AB || DC and let h be the height. Then, area of trap. ABCD = {1/2 x (AB + DC) x h} sq. units.
PROOF:
Join AC. Draw CL Ʇ AB and AM Ʇ CD (produced ). Let CL = AM = h Area of trap. ABCD = ar (DABC) + ar (DACD)
= {1/2 x AB x CL} + {1/2 x DC x AM}
= {1/2 x AB x h} + {1/2 x DC x h}
= {1/2 x (AB + DC) x h} sq. units.
So, area of a trapezium = ½ x (sum of parallel sides) x (distance between them).

### Two parallel sides of a trapezium are of lengths 27 cm and 19 cm respectively. and the distance between them is 14 cm. Find the area of the trapezium.

Area of the trapezium = 1/2 x (sum of parallel sides) x (distance between them)
= {1/2 x (27+19) x 14} cm² = {1/2 x 46 x 14} cm² = (23 x 14) cm² = 322 cm².

### The parallel sides of a trapezium are 25 cm and 13 cm; its nonparallel sides are equal, each being 10 cm. Find the area of the trapezium.

Let ABCD be the given trapezium in which AB || DC, AB = 25 cm, DC = 13 cm AD = BC = 10 cm.
Draw CL Ʇ AB and CM || DA, meeting AB at L and M respectively.
Clearly, AMCD is a parallelogram.
So, AM = DC = 13 cm. MB
= (AB – AM) = (25 – 13) cm = 12 cm.
Now, CM = DA = 10 cm and CB = 10 cm.
So, triangle CMB is an isosceles triangle and CL Ʇ MB
This means, L is the midpoint of MB
So, ML = LB = x MB = ½ x 12 cm = 6 cm.
Form right triangle CLM, we have:
CL2 = (CM2 – ML2) = {(10)2 – 62} cm = (100 – 36) cm = 64 cm
So, CL = √64 cm = 8 cm
Hence, height of the trapezium = 8 cm.
So, area of trap. ABCD = ½ x (25 + 13) x 8 cm² = ½ x 38 x 8 cm²
= (38 x 4) cm = 152 cm²

### How find the area of a trapezium?

The area of a trapezium is computed with the following formula: Area = 1 2 × Sum of parallel sides × Distance between them.

### What is a trapezium in math?

A trapezoid, also known as a trapezium, is a flat closed shape having 4 straight sides, with one pair of parallel sides.

### What is the area and perimeter of trapezium? Find the area of the trapezium, in which the sum of the bases (parallel sides) is 60 cm and its height is 20 cm.

We know that, Area of a Trapezium, A = h(a+b)/2 square units.
A = 20 (60)/2 = 600 cm2
Perimeter = Sum of all the sides = AB + BC + CD + DA.
Perimeter = 2 x 60 = 120 cm

### Is a trapezium a rhombus?

No, because a trapezoid has only one pair of parallel sides. … If their two pairs of sides are equal, it becomes a rhombus, and if their angles are equal, it becomes a square.      