# NCERT Solutions Class 8 Maths Chapter 11 Mensuration Exercise 11.2

### Introduction :

In this exercise we will come to know about the area of Trapezium, quadrilateral and polygon. To find the area of trapezium we should know the length of the parallel sides and the perpendicular distance between these two parallel sides. Half the product of the sum of the length of parallel sides and the perpendicular distance between them gives the area of trapezium.

To find the area of a general quadrilateral, we can split the quadrilateral into two triangles by drawing of its diagonals.

Similarly, to find the area of a polygon, we can split a quadrilateral into triangles and find its area.

NCERT Class 9 Maths Chapter 5 Introduction to Euclid's Geometry :

- Class 8 Maths Chapter 11 Mensuration Exercise 11.1
- Class 8 Maths Chapter 11 Mensuration Exercise 11.2
- Class 8 Maths Chapter 11 Mensuration Exercise 11.3
- Class 8 Maths Chapter 11 Mensuration Exercise 11.4

#### NCERT Solutions Class 8 Maths Exercise 11.2

**Q1.** 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.

**Solution: **

Given: side a =1m

side b = 1.2m

height=0.8m

So, the area of trapezium table is 0.88m^{2}.

**Q2.**The area of a trapezium is 34 cm^{2} 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 :**

Area= 34 cm^{2}

Side a = 10cm

Height = 4cm

Area of Trapezium =\(\displaystyle \frac{1}{2}\times (a+b)\times h\)

\(\displaystyle \begin{array}{l}34=\frac{1}{2}\times (10+b)\times 4\\34=(10+b)\times 2\\\frac{{34}}{2}=(10+b)\\\,17=10+b\\\,\,b=\,17-10\\b\,=\,7cm\end{array}\)

so, the length of other parallel side is 7cm.

**Q3. **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.

**Solution :**

Perimeter of Trapezium = AB+BC+CD+DA

120 = AB+48+17+40

120 = AB +105

AB = 120 - 105

AB = 15cm

Now, we can find the area of Trapezium

Area of Trapezium = \(\displaystyle =\frac{1}{2}\times (a+b)\times h\)

\(\displaystyle \begin{array}{l}=\frac{1}{2}\times (48+40)\times 15\\=\frac{1}{2}\times (88)\times 15\\=44\times 15\\=660{{m}^{2}}\end{array}\)

So, the area of field is 660m^{2.}

**Q4. **The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.

**Solution :**

Given: base=24m

\(\displaystyle \begin{array}{l}{{h}_{1}}=13m\\{{h}_{2}}=8m\end{array}\)

Area of Quadrilateral = \(\displaystyle (area\,of\,\Delta ABD)+(area\,of\,\Delta BCD)\)

\(\displaystyle \begin{array}{l}=\frac{1}{2}\times (area\,of\,\Delta ABD)+(area\,of\,\Delta BCD)\\=\frac{1}{2}\times (BD\times {{h}_{1}})+\frac{1}{2}(BD\times {{h}_{2}})\\=\frac{1}{2}\times (24\times 13)+\frac{1}{2}(24\times 8)\\=(12\times 13)+(12\times 8)\\=12(13+8)\\=\,12\times \,21\\=\,252{{m}^{2}}\end{array}\)

So, the area of Field is 252m^{2.}

**Q5. **The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

Solution:

Given : \(\displaystyle \begin{array}{l}{{D}_{1}}=7.5cm\\{{D}_{{2\,}}}=\,12cm\end{array}\)

Area of Rhombus = \(\displaystyle \frac{1}{2}\times {{D}_{1}}\times {{D}_{{2\,}}}\)

\(\displaystyle \begin{array}{l}=\frac{1}{2}\times 7.5\times 12\\=7.5\times 6\\=45c{{m}^{2}}\end{array}\)

so, the area of rhombus is 45cm^{2.}

**Q6. **Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.

**Solution :**

Given: side=5cm

altitude=4.8cm

\(\displaystyle {{D}_{1}}=8cm\)

To Find: length of other diagonal(\(\displaystyle {{D}_{{2\,}}}\))

As rhombus is special parallelogram so we use area of parallelogram to find the area of rhombus.

Area of rhombus = side \(\displaystyle \times \) Altitude

=5\(\displaystyle \times \)4.8

=24cm^{2}

Now we have the area of rhombus, so we can find the other diagonal

Area of Rhombus =\(\displaystyle \frac{1}{2}\times {{D}_{1}}\times {{D}_{{2\,}}}\)

\(\displaystyle \begin{array}{l}24=\frac{1}{2}\times {{D}_{1}}\times {{D}_{{2\,}}}\\24=\frac{1}{2}\times 8\times {{D}_{{2\,}}}\\24=4\times {{D}_{{2\,}}}\\{{D}_{{2\,}}}=\frac{{24}}{4}\\{{D}_{{2\,}}}\,=\,6cm\end{array}\)

So, the length of other diagonal is 6cm.

**Q7.** The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m2 is ₹ 4.

**Solution :**

Given: length of first diagonal = 45cm

length of second diagonal = 30cm

Number of tiles = 3000

To find : cost of polishing the floor

\(\displaystyle \begin{array}{l}Area\,of\,1\,tile=\frac{1}{2}\times {{d}_{1}}\times {{d}_{2}}\\=\frac{1}{2}\times 45\times 30\\=45\times 15\\=675c{{m}^{2}}\end{array}\)

If the area of one tile is = 675cm^{2}

Then, area of 3000 tiles are = 3000 \(\displaystyle \times \) area of one tile

=3000 \(\displaystyle \times \) 675

= 2025000cm^{2}

Now, if the cost of polishing per m^{2 }is =₹ 4

Then cost of polishing the floor is = 4 \(\displaystyle \times \) 202.5 (covert cm^{2} into m^{2)}

= ₹ 810.

so, the cost of polishing the floor is ₹ 810.

**Q8. **Mohan wants to buy a trapezium-shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

**Solution :**

Given: area=1050m^{2}

height = 100m

To find : length of the side along the river

Let the side along the road be x

Then the side along the river is = 2x

so,

\(\displaystyle \begin{array}{l}Area\,of\,trapezium=\frac{1}{2}\times (a+b)\times h\\10500\,\,=\frac{1}{2}\times (2x+x)\times 100\\10500\,\,=\frac{1}{2}\times (3x)\times 100\\10500\,\,=3x\,\times \,50\\10500\,\,=\,150x\\x=\,\frac{{10500\,}}{{150}}\\x\,=\,70m\end{array}\)

now, the side along the road (x) is = 70m

then the side along the road (2x) is = 2 \(\displaystyle \times \) 70

= 140m

Hence, the length of the side along the river is = 140m.

**Q9. **The top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.

**Solution :**

Given: height =4m

length= 11m

breadth = 5m

Area of octagonal surface = area of trapezium ABCH + area of rectangle + area of trapezium GDEF

= \(\displaystyle \frac{1}{2}\times (a+b)\times h\) + l \(\displaystyle \times \) b + \(\displaystyle \frac{1}{2}\times (a+b)\times h\)

As area of trapezium ABCH is = area of trapezium GDEF

\(\displaystyle \begin{array}{l}=\frac{1}{2}\times (a+b)\times h\\=\frac{1}{2}\times (11+5)\times 4\\=\frac{1}{2}\times (16)\times 4\\\,=8\times 4\\\,\,=\,32{{m}^{2}}\end{array}\)

Area of Rectangle = length \(\displaystyle \times \) breadth

= 11\(\displaystyle \times \) 5

=55m^{2}

Now, the area of octagonal surface is = 32 + 55 + 32 (as both trapezium are equal)

=64+55

=119m^{2}

So, the area of octagonal surface is 119m^{2}.

**Q10. **There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways.

Find the area of this park using both ways. Can you suggest some other way of finding its area?

**Solution :**

(i) by Jyoti's diagram,

Area of pentagonal shape = Area of trapezium ABCD + Area of trapezium ADEF

= 2(\(\frac { 1 }{ 2 }\) (a + b) × h)

= 2 × \(\frac { 1 }{ 2 }\) (a + b) × h

= (15 + 30) × 7.5

= 45 × 7.5

= 337.5m^{2}

(ii) by Kavita's diagram

Area of the pentagonal shape = Area of ∆ABE + Area of square BCDE

= \(\frac { 1 }{ 2 }\) × b × h + 15 × 15

= \(\frac { 1 }{ 2 }\) × 15 × 15 + 225

= 112.5 + 225

= 337.5 m^{2}

^{}

Yes, we can also find the other way to calculate the area of the given pentagonal shape.

Join CE to divide the figure into three parts, i.e., trapezium ABCE .

Area of ABCDE = Area of ∆EDC + ∆BCE + ∆ABE.

NCERT Class 9 Maths Chapter 5 Introduction to Euclid's Geometry :