Mathematics Solutions Solutions for Class 8 Math Chapter 1 Rational And Irrational Numbers are provided here with simple step-by-step explanations. These solutions for Rational And Irrational Numbers are extremely popular among Class 8 students for Math Rational And Irrational Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 8 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

Page No 95:

Question 1:

If base of a parallelogram is 18 cm and its height is 11 cm, find its area.

Answer:

Area of a parallelogram = Base×Height=18×11=198 cm2

Page No 95:

Question 2:

If area of a parallelogram is 29.6 sq cm and its base is 8 cm, find its height.

Answer:

Area of a parallelogram = 29.6 sq cm
Area of a parallelogram = Base×Height
29.6=8×hh=29.68=3.7 cm
Height = 3.7 cm

Page No 95:

Question 3:

Area of a parallelogram is 83.2 sq cm. If its height is 6.4 cm, find the length of its base.

Answer:

Area of a parallelogram = Base×Height
83.2=b×6.4b=13
Length of base = 13 cm



Page No 97:

Question 1:

Lengths of the diagonals of a rhombus are 15 cm and 24 cm, find its area.

Answer:

Area of rhombus = 12×Product of diagonals
Area of rhombus=12×15×24=180 cm2
 

Page No 97:

Question 2:

Lengths of the diagonals of a rhombus are 16.5 cm and 14.2 cm, find its area.

Answer:

Area of rhombus = 12×Product of diagonals

Area of rhombus=12×16.5×14.2=117.15 cm2

Page No 97:

Question 3:

If perimeter of a rhombus is 100 cm and length of one diagonal is 48 cm, what is the area of the quadrilateral ?

Answer:


Perimeter of the rhombus = 100 cm
4×side=100side=1004=25 cm
Thus, each side of the rhombus = 25 cm.
Diagonals of a rhombus bisect each other at 90°.
So, AO = OC = 482=24 cm
In AOB
We apply Pythagoras theorem,
AO2+OB2=AB2242+OB2=252OB2=625-576=49OB=7 cm
So, DB = 2×OB=2×7=14 cm
Area of rhombus = 12×Product of diagonals=12×14×48=336 cm2
 

Page No 97:

Question 4:

If length of a diagonal of a rhombus is 30 cm and its area is 240 sq cm, find its perimeter.

Answer:


Let the other diagonal be d cm
Area of a rhombus = 12Product of diagonals
240=12×30×dd=240×230=16
AC = 30 cm
DB = 16 cm
Diagonals of a rhombus bisect at right angles. 
In AOB,
AO2+OB2=AB2152+82=AB2AB2=225+64=289AB=289=17 cm
Thus, the side of the rhombus = 17 cm
Perimeter = 4×17=68 cm



Page No 99:

Question 1:

In ☐ ABCD, l(AB) = 13 cm, l(DC) = 9 cm, l(AD) = 8 cm, find the area of ☐ ABCD.

Answer:



Draw perpendicular from c to line AB. Name the point E. 
CE = AD = 8 cm
EB = AB-AE=AB-CD=13-9=4 cm
Area of rectangle AECD = l×b=9×8=72 cm2
Area of Triangle BEC = 12×b×h=12×8×4=16 cm2
Area of ☐ ABCD = Area of AECD + Area of triangle BEC = 72 + 16 = 88 cm2

Page No 99:

Question 2:

Length of the two parallel sides of a trapezium are 8.5 cm and 11.5 cm respectively and its height is 4.2 cm, find its area.

Answer:


Area of trapezium = 12sum of parallel sides×height
=12×8.5+11.5×4.2=12×20×4.2=42 cm2

Page No 99:

Question 3:

☐ PQRS is an isosceles trapezium l(PQ) = 7 cm. seg PM ⊥ seg SR, l(SM) = 3 cm, Distance between two parallel sides is 4 cm, find the area of ☐ PQRS

Answer:


Draw a perpendicular from Q to line MR. Where it meets the line MR name it point N. 
MN = PQ = 7 cm
In PMS, 
PM2+SM2=PS242+32=PS2PS2=16+9=25PS=5 cm
PQRS is an isosceles trapezium so, PS = QR = 5 cm
PM = QN = 4 cm
So, NR = SM = 3 cm
SR = SM + MN + NR = 3 + 7 + 3 = 13 cm
Area of trapezium PQRS = 12×sum of parallel sides×height
=12×7+13×4=40 cm2



Page No 101:

Question 1:

Sides of a triangle are cm 45 cm, 39 cm and 42 cm, find its area.

Answer:

a = 45 cm
b = 39 cm
c = 42 cm
s=a+b+c2=45+39+422=63
Area=ss-as-bs-c=6363-4563-3963-42=63×18×24×21=756 cm2
 

Page No 101:

Question 2:

Look at the measures shown in the adjacent figure and find the area of ☐ PQRS.

Answer:


Join PR.
In triangle PSR,
Applying Pthygoras theorem,
PS2+SR2=PR2362+152=PR2PR2=1296+225=1521PR=1521=39 m
In triangle PQR,
s=56+25+392=60Area=6060-5660-2560-39=60×4×35×21=176400=420 m2
Area of triangle PSR = 12×15×36=270 m2
Area of PQRS = Area of triangle PSR + Area of triangle PQR
= 270 + 420
= 690 m2

Page No 101:

Question 3:

Some measures are given in the adjacent figure, find the area of ☐ABCD.

Answer:

Area of BDC = 12×13×60=390 m2
Area of BAD = 12×AB×AD=12×40×9=180 m2
Area of ☐ABCD = Ar of BDC + Ar of BAD
= 390 + 180 
= 570 m2

 



Page No 102:

Question 1:

Find the areas of given plots. (All measures are in metres.)
(1) 


(2) 

Answer:

(1)
Ar of PQA = 12×QA×PA=12×30×50=750 m2
Ar of PBT = 12×PB×BT = 12×60×30=900 m2
Ar of SBT = 12×SB×BT=12×90×30=1350 m2
Ar of RCS = 12×RC×CS=12×25×60=750 m2
Ar of QRCA = 12QA+RC×AC=12×50+25×60=2250 m
Ar os PQRST = Ar of PQA + Ar of PBT + Ar of SBT + Ar of RCS + Ar of QRCA
= 750 + 900 + 1350 + 750 + 2250
= 6000 m2

(2) Disclaimer: The information given is insufficient to solve the question. 



Page No 104:

Question 1:

Radii of the circles are given below, find their areas.
(1) 28 cm
(2) 10.5 cm
(3) 17.5 cm

Answer:

(1) 28 cm
Ar=πr2=π×282=227×28×28=2464 sq cm

(2) 10.5 cm
Ar=πr2=π×10.52=227×10.5×10.5=346.5 sq cm

(3) 17.5 cm
Ar=πr2=π×17.52=227×17.5×17.5=962.5 sq cm

Page No 104:

Question 2:

Areas of some circles are given below find their diameters.
(1) 176 sq cm
(2) 394.24 sq cm
(3) 12474 sq cm

Answer:

(1) 176 sq cm
Ar=πr2176=227×r2r2=176×722=56r=56d=2r=256

(2) 394.24 sq cm
Ar=πr2394.24=227×r2r2=394.24×722=125.44r=11.2d=2r=2×11.2=22.4 cm

(3) 12474 sq cm
Ar=πr212474=227×r2r2=12474×722=3969r=63d=2r=2×63=126 cm
 

Page No 104:

Question 3:

Diameter of the circular garden is 42 m. There is a 3.5 m wide road around the garden. Find the area of the road.

Answer:

Diameter of the garden(d) = 42 m
Radius, = 21 m
Diameter of the garden including the road = 42 + 3.5 + 3.5 = 49 m
Radius of the garden with the road = 24.5 m
Area of road = 
Ar of garden with road-Ar of garden=πR2-πr2=πR2-r2=22724.52-212=227×600.25-441=227×159.25=500.5 m2

Page No 104:

Question 4:

Find the area of the circle if its circumfence is 88 cm.

Answer:

Circumference = 88 cm
2πr=88r=88×72×22=14 cm
Area of the circle=πr2=227×142=616 cm2



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