Math, asked by univaryash, 3 months ago

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class 8 ncert chater ii exercise 11.2
question 4​

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Answered by Anonymous
26

Answer:

Given :

Given :Diagonal of quadrilateral = 24 m and perpendicular dropped on it (diagonal) from the remaining opposite vertices are 8 m and 13 m.

Given :Diagonal of quadrilateral = 24 m and perpendicular dropped on it (diagonal) from the remaining opposite vertices are 8 m and 13 m.(As shown in figure)

Given :Diagonal of quadrilateral = 24 m and perpendicular dropped on it (diagonal) from the remaining opposite vertices are 8 m and 13 m.(As shown in figure)Find :

Given :Diagonal of quadrilateral = 24 m and perpendicular dropped on it (diagonal) from the remaining opposite vertices are 8 m and 13 m.(As shown in figure)Find :Area of the Quadrilateral.

Solution :

We know that

Area of quadrilateral = 1/2 × diagonal × (Sum of perpendicular on the quadrilateral from remaining opposite vertices)

Assume a quadrilateral ABCD such that

BD = diagonal of quadrilateral

AM and CN = perpendiculars on the diagonal

We have

Diagonal = BD = 24 m

Perpendicular on the quadrilateral (AM and AN) = 8 m and 13 m

Substitute the known values in above formula. To find the value of area of field (quadrilateral).

→ 1/2 × 24 × (8 + 13)

→ 12 × 21

→ 252 m²

∴ Area of quadrilateral is 252 m².

hope this helps you

Answered by danger7537
4

Step-by-step explanation:

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NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2

October 4, 2019 by Sastry CBSE

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2

Class 8 Maths Mensuration Exercise 11.1

Class 8 Maths Mensuration Exercise 11.2

Class 8 Maths Mensuration Exercise 11.3

Class 8 Maths Mensuration Exercise 11.4

Mensuration Class 8 Extra Questions

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.2

Ex 11.2 Class 8 Maths Question 1.

The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2 Q1

Solution:

Area of the trapezium = 12 × (a + b) × h

= 12 × (1.2 + 1) × 0.8

= 12 × 2.2 × 0.8

= 0.88 m2

Hence, the required area = 0.88 m2

Ex 11.2 Class 8 Maths Question 2.

The area of a trapezium is 34 cm2 and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel sides.

Solution:

Given: Area of trapezium = 34 cm2

Length of one of the parallel sides a = 10 cm

height h = 4 cm

Area of the trapezium = 12 × (a + b) × h

34 = 12 × (10 + b) × 4

⇒ 34 = (10 + b) × 2

⇒ 17 = 10 + b

⇒ b = 17 – 10 = 7 cm

Hence, the required length = 7 cm.

Ex 11.2 Class 8 Maths Question 3.

Length of the fence of a trapezium-shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2 Q3

Solution:

Given:

AB + BC + CD + DA = 120 m .

BC = 48 m, CD = 17 m, AD = 40 m

AB = 120 m – (48 m + 17 m + 40 m) = 120 – 105 m = 15 m

Area of the trapezium ABCD = 12 × (BC + AD) × AB

= 12 × (48 + 40) × 15

= 12 × 88 × 15

= 44 × 15 = 660 m2.

Hence, the required area = 660 m2

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