Question

11. Diagram of the adjacent picture frame has outer dimensions = 24 cm x 28 cm
and inner dimensions 16 cm x 20 cm. Find the area of each section of
the frame, if the width of each section is same.
​

Answer 1

The picture frame has 4 sections marked as 1,2,3 and 4

Given that width of each section is same

Let width of section = x cm

Let's first find x

BC = X + 20 cm + X

28 = x + 20 + X

28-20 = 2x

8 = 2x

2x = 8

$x = \frac{8}{2}$

X = 4 cm

So, width of section is 4 cm

Now,

By symmetry,

Area of Section 1 = Area of Section 3

Area of Section 2 = Area of Section 4

So, width of section is 4 cm

Area of section 1

(Attachment 4)

Here,

Parallel sides are AB & EF

Height is 4 cm

Area of section 1 = Area of trapezium

$= \frac{1}{2} \times sum \: of \: parallel \: sides$

$= \frac{1}{2} \times (24 + 16) \times 4$

$= \frac{1}{2} \times 40 \times 4$

$= 20 \times 4$

$= {80}^{2}$

Area of section 2

Here,

(attachment 6 )

C Parallel sides are BC & FG

Height is 4 cm

Area of section 2 = Area of trapezium

$= \frac{1}{2} \times sum \: of \: parallel \: sides \times height \:$

$= \frac{1}{2} \times (28 + 20) \times 4$

$= \frac{1}{2} \times 48 \times 4 = 24 \times 4 = {96}^{2}$

Thus,

Area of section 1 = Area of section 3 = 80 cm²

Area of section 2 = Area of section 4 = 96 cm²

Hence, 80² and 96²

Answer 2

Answer:

The picture frame has 4 sections marked as 1,2,3 and 4

Given that width of each section is same

Let width of section = x cm

Let's first find x

BC = X + 20 cm + X

28 = x + 20 + X

28-20 = 2x

8 = 2x

2x = 8

$x = \frac{8}{2}x= </p><p>2</p><p>8$

X = 4 cm

So, width of section is 4 cm

Now,

By symmetry,

Area of Section 1 = Area of Section 3

Area of Section 2 = Area of Section 4

So, width of section is 4 cm

Area of section 1

(Attachment 4)

Here,

Parallel sides are AB & EF

Height is 4 cm

Area of section 1 = Area of trapezium

$= \frac{1}{2} \times sum \: of \: parallel \: sides= </p><p>2</p><p>1$

×sumofparallelsides

$</p><p>= \frac{1}{2} \times (24 + 16) \times 4=$

2

1

×(24+16)×4

$= \frac{1}{2} \times 40 \times 4=$

2

1

×40×4

= 20 \times 4=20×4

= {80}^{2}=80

2

Area of section 2

Here,

(attachment 6 )

C Parallel sides are BC & FG

Height is 4 cm

Area of section 2 = Area of trapezium

$= \frac{1}{2} \times sum \: of \: parallel \: sides \times height \:= </p><p>21$

×sumofparallelsides×height

$= \frac{1}{2} \times (28 + 20) \times 4= </p><p>2</p><p>1$

×(28+20)×4

$= \frac{1}{2} \times 48 \times 4 = 24 \times 4 = {96}^{2}= </p><p>2</p><p>1$

×48×4=24×4=96

2

Thus,

Area of section 1 = Area of section 3 = 80 cm²

Area of section 2 = Area of section 4 = 96 cm²