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

$\sf{\bold{\blue{\underline{\underline{Given}}}}}$

●Dimension of the outer frame = 24 cm × 28 cm

●Dimension of the inner frame = 16 cm × 20 cm

$\sf{\bold{\blue{\underline{\underline{Find}}}}}$

■ The area of each section of the frame if the width of each section is same.

$\sf{\bold{\blue{\underline{\underline{solution }}}}}$

We know that,

h = Height

a = Area

(Refer to the attachment for your reference.)

I've divided the figure into 4 parts.

Now, here,

Figures (I) and (II) are similar in dimensions.

Even figures (III) and (IV) are similar in dimensions.

We know that,

Area of figure (I) = Area of trapezium

By the formula,

$\underline{\boxed{\red{\sf Area \ of \ trapezium= \dfrac{1}{2} \times (a+b) \times h}}}$

Substituting their values,

$\rightsquigarrow{\dfrac{1}{2} \times (28+20) \times 4}$

$\rightsquigarrow{\dfrac{1}{2} \times 48 \times 4}$

$\rightsquigarrow{96 \ cm^{2}}$

Therefore,

Area of figure (I) = 96 cm²

Area of figure (II) = 96 cm²

We know that,

Area of figure (III) = Area of trapezium

By the formula,

$\boxed{\red{\sf Area \ of \ trapezium= \dfrac{1}{2} \times (a+b) \times h}}$

By substituting,

$\rightsquigarrow{\dfrac{1}{2} \times (24+16) \times 4}$

$\rightsquigarrow{\dfrac{1}{2} \times 40 \times 4}$

$\rightsquigarrow{80 \ cm^{2}}$

Therefore,

Area of figure (III) = 80 cm²

Area of figure (IV) = 80 cm²

Answer 2

Step-by-step explanation:

Height

a = Area

(Refer to the attachment for your reference.)

I've divided the figure into 4 parts.

Now, here,

Figures (I) and (II) are similar in dimensions.

Even figures (III) and (IV) are similar in dimensions.

We know that,

Area of figure (I) = Area of trapezium

By the formula,

\underline{\boxed{\red{\sf Area \ of \ trapezium= \dfrac{1}{2} \times (a+b) \times h}}}

Area of trapezium=

2

1

×(a+b)×h

Substituting their values,

\rightsquigarrow{\dfrac{1}{2} \times (28+20) \times 4}⇝

2

1

×(28+20)×4

\rightsquigarrow{\dfrac{1}{2} \times 48 \times 4}⇝

2

1

×48×4

\rightsquigarrow{96 \ cm^{2}}⇝96 cm

2

Therefore,

Area of figure (I) = 96 cm²

Area of figure (II) = 96 cm²

We know that,

Area of figure (III) = Area of trapezium

By the formula,

\boxed{\red{\sf Area \ of \ trapezium= \dfrac{1}{2} \times (a+b) \times h}}

Area of trapezium=

2

1

×(a+b)×h

By substituting,

\rightsquigarrow{\dfrac{1}{2} \times (24+16) \times 4}⇝

2

1

×(24+16)×4

\rightsquigarrow{\dfrac{1}{2} \times 40 \times 4}⇝

2

1

×40×4

\rightsquigarrow{80 \ cm^{2}}⇝80 cm

2

Therefore,

Area of figure (III) = 80 cm²

Area of figure (IV) = 80 cm²