Mathematics Part ii Solutions Solutions for Class 8 Math Chapter 3 Area Of Quadrilaterals are provided here with simple step-by-step explanations. These solutions for Area Of Quadrilaterals are extremely popular among class 8 students for Math Area Of Quadrilaterals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part ii Solutions Book of class 8 Math Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part ii Solutions Solutions. All Mathematics Part ii Solutions Solutions for class 8 Math are prepared by experts and are 100% accurate.

Page No 112:

Question 1:

One side of a parallelogram is of length 24 centimetres and the height from this side is 18 centimetres. What is its area?

 

Answer:

Let the given parallelogram be ABCD with DC = 24 cm and height AE = 18 cm.

We know that the area of a parallelogram is the product of the length of one side and its distance to the opposite side.

∴ Area of the parallelogram ABCD = DC × AE 

= 24 cm × 18 cm

= 432 cm2

 

Page No 112:

Question 2:

Two adjacent sides of a parallelogram are 16 and 24 centimetres long. The height from the longer side is 15 centimetres. What is the height from the shorter side?

 

Answer:

Let ABCD be the given parallelogram with AB = DC = 24 cm, BC = AD = 16 cm and AE = 15 cm.

Construction: Draw the height AF from the shorter side AD.

We know that the area of a parallelogram is the product of the length of one side and its distance to the opposite side.

∴ Area of the parallelogram ABCD = DC × AE 

= 24 cm × 15 cm

= 360 cm2

Area of parallelogram ABCD can also be written in terms of the product of the length of other side and its distance to the opposite side.

∴ Area of the parallelogram ABCD = BC × AF

⇒ 360 cm2 = 16 cm × AF

Thus, the height from the shorter side is 22.5 cm.

 

Page No 112:

Question 3:

The sides of a parallelogram are 25 centimetres and 15 centimetres. The height to the 25 centimetres side is 15 centimetres. What is the speciality of this parallelogram?

 

Answer:

The sides of a parallelogram are 25 cm and 15 cm.

The height to the 25 cm side is 15 cm which is equal to the other side of the parallelogram.

Therefore, the other side of the parallelogram is itself the height of the parallelogram.

Thus, the given parallelogram is a rectangle. 

 



Page No 113:

Question 1:

12 centimetres, 25 centimetres

 

Answer:

First diagonal, d1 = 12 cm

Second diagonal, d2 = 25 cm

Area of the rhombus =

 

Page No 113:

Question 2:

14.5 centimetres, 18 centimetres

 

Answer:

First diagonal, d1 = 14.5 cm

Second diagonal, d2 = 18 cm

Area of the rhombus =

 

Page No 113:

Question 3:

15.5 centimetres, 10.5 centimetres

 

Answer:

First diagonal, d1 = 15.5 cm

Second diagonal, d2 = 10.5 cm

Area of the rhombus =

 



Page No 115:

Question 1:

Find the area of a trapezium with the parallel sides 24 centimetres and 18 centimetres long and the distance between them, 16 centimetres.

 

Answer:

We know that the area of trapezium is half the product of the sum of its parallel sides and the distance between them.

Here, the parallel sides are 24 cm and 18 cm and the distance between them is 16 cm.

∴ Area of trapezium =

 

Page No 115:

Question 2:

Find the area of the trapezium shown below:

 

Answer:

In ΔADC, right-angled at D:

Now, in trapezium ABCD, the parallel sides are AB = 4 cm, DC = 12 cm and the height AD = 5 cm.

We know that the area of trapezium is half the product of the sum of its parallel sides and the distance between them.

∴ Area of trapezium ABCD =

 

Page No 115:

Question 3:

In the trapezium ABCD shown below, AB and CD are parallel. Using the given measurements (in centimetres), find its area:

 

 

Answer:

In ΔAPD, right-angled at P:

Also, PQ = DC = 6 cm

∴ AB = AP + PQ + PB = (3 + 6 + 3) cm = 12 cm

Now, in trapezium ABCD, the parallel sides are AB = 12 cm, DC = 6 cm and the height DP = 4 cm.

We know that the area of trapezium is half the product of the sum of its parallel sides and the distance between them.

∴ Area of trapezium ABCD =

 



Page No 116:

Question 1:

In the trapezium PQRS shown below, PQ is parallel to RS. Find its area.

 

 

Answer:

In ΔSAP, right-angled at A:

Also, AB = SR = 10 cm

∴ PQ = PA + AB + BQ = (9 + 10 + 16) cm = 35 cm

Now, in trapezium PQRS, the parallel sides are SR = 10 cm, PQ = 35 cm and the height SA = 12 cm.

We know that the area of trapezium is half the product of the sum of its parallel sides and the distance between them.

∴ Area of trapezium ABCD =

 



Page No 119:

Question 1:

One diagonal of a quadrilateral is 35 centimetres long and the perpendiculars to it from the opposite vertices are 15 centimetres and 19 centimetres long. What is the area of this quadrilateral?

 

Answer:

Length of the diagonal, d = 35 cm

Length of the first perpendicular, p1 = 15 cm

Length of the second perpendicular, p2 = 19 cm

Area of the quadrilateral =

=

 

Page No 119:

Question 2:

The area of a quadrilateral is 1470 square centimeters and the length of one of its diagonals is 42 centimetres. Calculate the sum of the lengths of the perpendiculars to this diagonal from the opposite vertices.

 

Answer:

Area of the quadrilateral = 1470

Length of the diagonal, d = 42 cm

Let the length of the first and the second perpendiculars be p1 and p2 respectively.

Area of the quadrilateral =

Therefore, the sum of the lengths of the perpendiculars to the given diagonal from the opposite vertices is 70 cm.

 

Page No 119:

Question 3:

The diagonals of a quadrilateral are perpendicular to each other and their lengths are 16 centimetres and 10 centimetres. What is its area?

 

Answer:

Let the given quadrilateral be ABCD with AC = 10 cm, BD = 16 cm.

Given: AC and BD are perpendicular to each other.

Area of ΔABD =

Area of ΔBCD =

Now, area of the quadrilateral ABCD = Area of ΔABD + Area of ΔBCD

= (8 cm × AE) + (8 cm × CE)

= 8 cm × (AE + CE)

= 8 cm × AC 

= 8 cm × 10 cm

= 80 cm2

 

Page No 119:

Question 4:

In the figure below, the diagonals of the quadrilateral ABCD are perpendicular to each other.

 

Answer:

It is given that diagonals of the quadrilateral ABCD are perpendicular to each other.

Also, AC = 26 cm and BD = 20 cm


 

Area of ΔABD =

Area of ΔBCD =

Now, area of the quadrilateral ABCD = Area of ΔABD + Area of ΔBCD

= (10 cm × AE) + (10 cm × CE)

= 10 cm × (AE + CE)

= 10 cm × AC

= 10 cm × 26 cm

= 260 cm2

 



Page No 120:

Question 1:

Prove that for any quadrilateral in which the diagonals are perpendicular to each other, the area is half the product of the lengths of the diagonals.

 

Answer:

Let the given quadrilateral be ABCD with AC = d1 units, BD = d2 units.

Also, AC and BD are perpendicular to each other.

Area of ΔABD =

Area of ΔBCD =

Now, area of the quadrilateral ABCD = Area of ΔABD + Area of ΔBCD

Hence, we can say that for any quadrilateral in which the diagonals are perpendiculars to each other, the area is half the product of the lengths of the diagonals.

 

Page No 120:

Question 2:

Prove that for any quadrilateral with a pair of opposite sides parallel, the area is half the product of the sum of the lengths of the parallel sides and the distance between them.

 

Answer:

Let the given quadrilateral be ABCD in which AD is parallel to BC.

Also, AD = a units, BC = b units 

Construction: Draw a line DE parallel to AB and height DF on side BC such that DF = h units.

As AD || BE and DE || AB, ADEB is a parallelogram.

The following diagram depicts the given quadrilateral ABCD with all the measurements in units.

Here, BE = AD = a units (As ADEB is a parallelogram)

EC = BC − BE = (b − a) units

The perpendicular distance between two parallel lines is same, therefore the height of the parallelogram ADEB and triangle DEC is same.

Area of the parallelogram ADEB = BE × DF 

= a units × h units 

= ah sq. units

Area of the ΔDEC =

Area of the quadrilateral ABCD = Area of the parallelogram ADEB + Area of ΔDEC

= ah sq. units +

Hence, we can say that if a pair of opposite sides of a quadrilateral is parallel then its area is half the product of the sum of the lengths of the parallel sides and the distance between them.

 



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