Mathematics Solutions Solutions for Class 8 Maths Chapter 16 Area are provided here with simple step-by-step explanations. These solutions for Area are extremely popular among Class 8 students for Maths Area Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 8 Maths Chapter 16 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 Maths are prepared by experts and are 100% accurate.

#### Question 1:

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

Area of a parallelogram =

#### Question 2:

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

Area of a parallelogram = 29.6 sq cm
Area of a parallelogram = $\mathrm{Base}×\mathrm{Height}$

Height = 3.7 cm

#### Question 3:

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

Area of a parallelogram = $\mathrm{Base}×\mathrm{Height}$
$⇒83.2=b×6.4\phantom{\rule{0ex}{0ex}}⇒b=13$
Length of base = 13 cm

#### Question 1:

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

Area of rhombus =

#### Question 2:

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

Area of rhombus =

#### 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 ?

Perimeter of the rhombus = 100 cm

Thus, each side of the rhombus = 25 cm.
Diagonals of a rhombus bisect each other at 90$°$.
So, AO = OC =
In $△$AOB
We apply Pythagoras theorem,

So, DB =
Area of rhombus =

#### Question 4:

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

Let the other diagonal be d cm
Area of a rhombus =
$⇒240=\frac{1}{2}×\left(30×d\right)\phantom{\rule{0ex}{0ex}}⇒d=\frac{240×2}{30}=16$
AC = 30 cm
DB = 16 cm
Diagonals of a rhombus bisect at right angles.
In $△$AOB,

Thus, the side of the rhombus = 17 cm
Perimeter = $4×17=68$ cm

#### Question 1:

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

Draw perpendicular from c to line AB. Name the point E.
CE = AD = 8 cm
EB =
Area of rectangle AECD =
Area of Triangle BEC =
Area of ☐ ABCD = Area of AECD + Area of triangle BEC = 72 + 16 = 88 cm2

#### 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.

Area of trapezium =

#### 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

Draw a perpendicular from Q to line MR. Where it meets the line MR name it point N.
MN = PQ = 7 cm
In $△$PMS,

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 =

#### Question 1:

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

a = 45 cm
b = 39 cm
c = 42 cm
$s=\frac{a+b+c}{2}=\frac{45+39+42}{2}=63$

#### Question 2:

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

Join PR.
In triangle PSR,
Applying Pthygoras theorem,

In triangle PQR,

Area of triangle PSR =
Area of PQRS = Area of triangle PSR + Area of triangle PQR
= 270 + 420
= 690 m2

#### Question 3:

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

Area of $△$BDC =
Area of $△$BAD =
Area of ☐ABCD = Ar of $△$BDC + Ar of $△$BAD
= 390 + 180
= 570 m2

#### Question 1:

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

(2)

(1)
Ar of $△$PQA = $\frac{1}{2}×\mathrm{QA}×\mathrm{PA}=\frac{1}{2}×30×50=750$ ${\mathrm{m}}^{2}$
Ar of $△$PBT = $\frac{1}{2}×\mathrm{PB}×\mathrm{BT}$ =
Ar of $△$SBT =
Ar of $△$RCS =
Ar of QRCA = $\frac{1}{2}\left(\mathrm{QA}+\mathrm{RC}\right)×\mathrm{AC}=\frac{1}{2}×\left(50+25\right)×60=2250$ 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.

#### Question 1:

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

(1) 28 cm

(2) 10.5 cm

(3) 17.5 cm

#### 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

(1) 176 sq cm
$\mathrm{Ar}=\mathrm{\pi }{r}^{2}\phantom{\rule{0ex}{0ex}}⇒176=\frac{22}{7}×{r}^{2}\phantom{\rule{0ex}{0ex}}⇒{r}^{2}=\frac{176×7}{22}=56\phantom{\rule{0ex}{0ex}}⇒r=\sqrt{56}\phantom{\rule{0ex}{0ex}}d=2r=2\sqrt{56}$

(2) 394.24 sq cm

(3) 12474 sq cm

#### 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.

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 =

#### Question 4:

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