S.chand books class 8 maths solution chapter 25 Areas of Rectilinear Figures exercise 25 A

 EXERCISE 25 A

---

Q1 | Ex-25A | Class 8 |S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 |myhelper

OPEN IN YOUTUBE

Question 1

(i) Find the perimeter and area of a square whose one side is 6.5cm long.
(ii) Find the area of a square whose perimeter is 124 cm.
(iii) Find the perimeter and diagonal of a square whose area is 64 cm²

Sol :

(i) 

Perimeter of square=4×side

=4×6.5=26 cm

Area of square=(side)²

=(6.5)²

=6.5×6.5=42.25 cm²

(ii)

Perimeter of square=4×side

4×side=124 cm

side$=\frac{124}{4}$

side=31 cm

Area of square=(side)²

=(31)²

=961 cm²

(iii)

Area of square=64 cm²

side²=64

side=√64=8 cm

Perimeter of square=4×side

=4×8 =32 cm

Diagonal $=a \sqrt{2}=8 \sqrt{2}$ cm

---

Q2 | Ex-25A | Class 8 |S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 |myhelper

OPEN IN YOUTUBE

Question 2

Find in hectares the area of a square-shaped field 14km long.

Sol :

Area of Square=(Side)²

=(14)²

=196 km²

100 ha=1 km²

196×100 ha=196×1 km²

19600 ha=196km²

---

Q3 | Ex-25A | Class 8 |S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 |myhelper

OPEN IN YOUTUBE

Question 3

Copy and complete the following table for rectangles:

	Length	Breadth	Area	Perimeter
(i)	4.4cm		77cm2
(ii)	20mm			104mm
(iii)	1.5km	90m	....ha
(iv)	40m		1 are

Sol :

(i)

Area of rectangle=77 cm²

l×b=77

4.4×b=77

$b=\frac{77}{4.4}$

$b=\frac{70}{4}$

b=17.50

Perimeter of rectangle=2(l+b)

=2×(4.4+17.5)

=2×(21.9)

=43.8 cm

(ii)

Perimeter of rectangle=104 mm

2(l+b)=104

$b=\frac{104}{2}-20$

b=52-20=32 mm

Area of rectangle=l×b

=20×32=640 mm²

(iii)

90 m

1 km=1000 m

$\frac{1}{1000} km=1m$

$\frac{90}{1000} km= 90 m$

0.09 km=90 m

10 ha=1 km

0.09×10 ha=0.09 km

b=0.9 ha=0.09 km

1.5 km

10 ha=1 km

10×1.5 ha=1.5 km

l=15 ha=1.5 km

Area of rectangle=l×b=15×0.9

=13.5 ha

Perimeter of rectangle=2(l+b)

=2(15+0.9)

=2×15.9

=31.8 ha or 3180 m

(iv)

1 are=100m²

Area of rectangle=1 are

l×b=100m²

40×b=100

$b=\frac{100}{40}$

b=2.5 m

Perimeter of rectangle=2(l+b)

=2(40+2.5)

=2×42.5

=85 m

---

Q4 | Ex-25A | Class 8 |S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 |myhelper

OPEN IN YOUTUBE

Question 4

Find the area of a rectangle plot, one side of which is 48 m and its diagonal is 50 m

Sol :

Area of rectangle plot=l×b

=48 m×b

Diagonal$=\sqrt{l^{2}+b^{2}}$

$50=\sqrt{(48)^{2}+b^{2}}$

Squaring both sides

$2500=48^{2}+b^{2}$

2500-2304=b²

196=b²

b=14 m

Area =l×b=14×48

=672 m²

---

Q5 | Ex-25A | Class 8 |S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 |myhelper

OPEN IN YOUTUBE

Question 5

The perimeter of a rectangle is 98 cm and its breadth is 9cm. Find its
(i) Length
(ii) Length of diagonal
(iii) Area

Sol :

Perimeter of rectangle=98 cm

2(l+b)=98

2(l+9)=98

l+9=49

l=49-9=40 cm

(i) Length = 40 cm

(ii) Length of diagonal$=\sqrt{l^{2}+b^{2}}$

$=\sqrt{(40)^{2}+9^{2}}$

$=\sqrt{1600+81}$

$=\sqrt{1681}$

=41 cm

(iii)

Area of rectangle=l×b

=40×9

=360 cm²

---

Q6 | Ex-25A | Class 8 |S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 |myhelper

OPEN IN YOUTUBE

Question 6

The area of a square is 1 hectare. Calculate its:
(i) perimeter
(ii) length of diagonal

Sol :

Area of square=1 ha

side²=1

side=1

(i) Perimeter of square=4×side

=4×1= 4 ha or 400 m

(ii) Length of diagonal$=side \sqrt{2}$

$=1\sqrt{2}$ ha

or $=100\sqrt{2}$ m

---

Q7 | Ex-25A | Class 8 |S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 |myhelper

OPEN IN YOUTUBE

Question 7

How many tiles of area 15cm² will be needed to cover a floor 5.85m by 6.90m ?

Sol :

Area of rectangle floor=(5.85×6.90) m²

=40.365 m²

1 m²=10000 cm²

=403650 cm²

Area of single tile=15 cm²

Number of tiles$=\frac{403650}{15}$

=26910 tiles

---

Q8 | Ex-25A | Class 8 |S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 |myhelper

OPEN IN YOUTUBE

Question 8

A room of dimensions 5.4m by 4.8m . A carpet is laid centrally, leaving a margin of 40cm all around between the carpet and the walls of the room. Find the area of the carpet in square meters.

Sol :

Area of room=5.4 m×4.8 m

=25.92 m²

40 cm=0.40 m

Area of carpet=l×b

=(5.4-0.4-0.4)×(4.8-0.4-0.4)

=(5.4-0.8)×(4.8-0.8)

=(4.6)×(4)

=18.4 m²

---

Q9 | Ex-25A | Class 8 |S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 |myhelper

OPEN IN YOUTUBE

Question 9

Find the areas of the shaded parts in figs.(i) to (iii) . All corners are right-angled.
(i) 

Sol :
Area of inner rectangle=4.2×6.7

=28.14 m²

Area of outer rectangle=6.2×8 .7
=53.94 m²

Shaded region=53.94-28.14=25.8 m²


(ii) 

Sol :

Area of bigger rectangle=19.6×15

=294 m²

Area of white region=l×b

=(19.6-5.4)(15-2-2)

=14.2×11

=156.2 m²

Shaded region=294-156.2

=137.8 m²



(iii)

Sol :

Area of a single small rectangular region

=l×b

$=4.2 \times\left(\frac{6.7-1.5}{2}\right)$

$=4.2 \times \frac{52}{2}$

=4.2×2.6

=10.92 m²

Area of shaded region=4×10.92

=43.68  m²

---

Q10 | Ex-25A | Class 8 |S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 |myhelper

OPEN IN YOUTUBE

Question 10

An open tank is 3m long, 2m wide and 1m50cm deep. Find the amount of paint required to paint its inside and outside, if 1 kg of paint cover 20 square meter (Neglect the thickness of the material)

Sol :

Dimension of  open tank=3m×2m×1.5 m

Total area=l×b+2(l×h)+2(b×h)

=3×2+2(3×1.5)+2(2×1.5)

=6+9+6=21 m²

Total area=2×21=42 m²

Paint 1 kg=20 m²

Total paint required $=\frac{42}{20}$

=2.1 kg

---

Q11 | Ex-25A | Class 8 |S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 |myhelper

OPEN IN YOUTUBE

Question 11

A photograph 25cm by 20cm is mounted on a card so that there is margin of 2.5cm all the way round. What fraction of the card is covered ?

Sol :

Dimension of photograph=25 cm×20 cm

Length=25+2.5+2.5=30 cm

Breadth=20+2.5+2.5=25 cm

Area of card=30×25=750 cm²

Area of photograph=25×20=500 cm²

Fraction of area covered$=\frac{500}{750}=\frac{2}{3}$