Question

the adjacent figure shows the diagram of a picture frame having outer dimension 14 cm ×16 cm and inner dimension and 10cm×12 cm if the width of each section is same find area of each section of the frame​

Answer 1

Answer:

Let the outer quadrilateral in the frame be ABCD

Let the inner quadrilateral be PQRS

Let the width marked be as in the figure,equal

PL=PK=QM=QN=RG=RH=SI=JS

BC=28cm, QM=RH (equal width)

QR=20cm

QM+QR+RH=BC

QM+20+QM=28

2 QM=28-20

QM=4 CM

In the figure,

Area of trapezium QBCR=Area of trapezium APSD

=\frac{1}{2}\times\left(AD+PS\right)\times PK=

2

1

×(AD+PS)×PK= =\frac{1}{2}\left(28+20\right)\times4=

2

1

(28+20)×4

=48\times248×2

=96\ cm^296 cm

2

Area of trapezium SRCD =Area of trapezium ABQP

\frac{1}{2}\times\left(CD+RS\right)\times RH=\frac{1}{2}\left(24+16\right)\times4

2

1

×(CD+RS)×RH=

2

1

(24+16)×4

=\frac{1}{2}\times32\times4=

2

1

×32×4

=64\ cm^2=64 cm

2

Area of frame=2{(area of trapezium QBCR)+(area of trapezium SRCD)}

=2(96+64)

=2(160)

=320\ cm^2=320 cm

2