S.chand publication New Learning Composite mathematics solution of class 8 Chapter 12 Mensuration Exercise 12B

 Exercise 12B

Find area of the following polygons

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Q1 | Ex-12B |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 1

ΔABE-

Area₁$=\frac{1}{2}\times 12\times 30$

=180cm²

Area₂$=\frac{1}{2}h(a+b)$

$=\frac{1}{2}\times 15\times (30+24)$

=405cm²

Total area=Area₁+Area₂

=180+405=585cm²

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Q2 | Ex-12B |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 2

Sol :

ΔPQU-

Area₁$=\frac{1}{2}\times 40\times 15$

=300 cm²

In rectangle QRTU

Area₂=40×28=1120cm²

In ΔRST-

Area₃$=\frac{1}{2}\times 40 \times 15$

=300cm²

∴Total area=Area₁+Area₂+Area₃

=300+1120+300=1720cm²

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Q3 | Ex-12B |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 3

Sol :

ABCDEF is a hexagonal figure.

∴A regular hexagon can be divided into a equilateral triangle

∴Area of ABCDEF=6×Area of one triangle

$=6\times \left[\frac{\sqrt{3}}{4}\times (sides)^2\right]$

$=6\left[\frac{\sqrt{3}}{4}\times 8^2\right]$

$=6\times \frac{\sqrt{3}}{4}\times 8\times 8$

=6×1.732×2×8

=166.272m²

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Q4 | Ex-12B |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 4

Sol :

EI=5cm ,HI=8cm

EF=6cm ,GH=3cm

Join O and G 

OG=8cm=IH

IOGH is a rectangle

∴EFGO is a trapezium

∴Area of IEFGH=Area of IOGH+Area of EFGO

$=(8×3)+\frac{1}{2}\times (6+8) \times 2$

=24+$\frac{1}{2}\times 14 \times 2$

=24+14

=38 cm²

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Q5 | Ex-12B |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 5

Sol :

ΔRSV-

Area₁$=\frac{1}{2}\times \text{Base}\times \text{Height}$

$=\frac{1}{2}\times 8\times 10$

=40cm²

Trapezium STUV-

Area₁$=\frac{1}{2}\times \text{h}\times (8+14)$

[∵RO=10

RP=17

∴OP=h=17-10=7cm]

$=\frac{1}{2}\times 7 \times 22$

=77

∴Area=Area₁+Area₁

=40+77=117cm²

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Q6 | Ex-12B |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 6

Sol :

Let ,ABCP is a rectangle

∴A₁=(4×10)=40cm²

Let, 

EFQD is a rectangle

∴A₂=(9×8)=72cm²

PC and QD join "O"

∴POQG is a rectangle

∴PO=GQ=15-8=7cm

PG=27-10=17cm

∴Area=7×17=119cm²

Now , ΔODC➝

OD=27-(10+9)=27-19=8cm

OC=PO-PC=7-4=3cm

∴Area$=\frac{1}{2}\times 8\times 3$=12

∴Area of PCDQG➝

A₃=119-12=107cm²

∴Total Area=A₁+A₂+A₃

=40+72+107

=219cm²

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Q7 | Ex-12B |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 7

Find the area of the hexagon ABCDFF in two ways:

(a) By splitting into two congruent triangles and a rectangle.

(b) By splitting into two congruent trapeziums.

Sol :

(a)

⇒ABCDEF is a hexagon, by splitting into two congruent by BF and CE

∴BCEF is rectangle

ABF ,DCE are triangle

ABF , DCE are triangle

∴Area of BCEF=A₁=(7×10)=70cm²

Area of ABF=A₂=$=\frac{1}{2}\times 10\times 4$=20cm²

ΔABF=ΔDCF=20

∴Total area of ABCDEF=70+20+20=110cm²

(b) 

ABCDEF is hexagon splitting by AD into two congruent trapezium

ABCD and ADEF are trapezium 

∴Area of ABCD=A₁$=\frac{1}{2}h(a+b)$

[h=OC$=\frac{10}{2}$=5

a=7 , b=15]

$=\frac{1}{2}5(7+15)$

$=\frac{1}{2}\times 5\times 22$

=55cm²

∴Area of ABCD=Area of ADEF=55cm²

∴Total area of ABCDEF=55+55=110cm²

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Q8 | Ex-12B |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 8

The perimeter of a square and a rectangle hexagon are equal. Find the ratio of the area of the hexagon to the area of the square.

Sol :

Perimeter of a square=4a

Perimeter of a regular hexagon=6b

ATQ,

4a=6b

∴$\frac{a}{b}=\frac{6}{4}$

∴$b=\frac{4a}{6}$

Now,

Area of hexagon : Area of square

$=6\times \frac{\sqrt{3}}{4}\times b^2$ : a²

$=\dfrac{6\times \frac{\sqrt{3}}{4}\times \frac{4a}{6}\times \frac{4a}{6}}{a^2}$

$=\frac{2\sqrt{3}}{3}$

=2√3:3