Question

The diagram of the adjacent picture frame has outer dimensions 26 cm x 30 cm and inner dimensions 18 cm x 22 cm. Find the area of each section of the frame, if the width of each section is same.

Answer 1

$given$

outer dimension = 26cm * 30 cm

inner dimension = 18cm * 22cm

$to \: find$

to find area

$solution$

Since , it is a rectangle

therefore area = length* breadth

area of outer dimension = l*b = 26 cm*30 cm

= 780 cm²

area of inner dimension = 18 cm* 22 cm

= 396 cm²

total area of each section = (780 - 396)cm²

= 384 cm²

follow me and my fØllØwing ❤️

Answer 2

Answer:

Let

• PQRS represent the outer boundary of the picture frame
• ABCD represent its inner boundary

Width of section PABQ = Width of section RCDS

= ½ (30 - 20) cm

= 4 cm

Width of section PSDA Width of section QBCR

= ½ (26 - 18) cm

= 4 cm

Area of section PQBA = Area of section RCDS

= [ ½ (18 + 26) × 4 ] cm²

= [ ½ (44) × 4 ] cm²

= 88 cm²

Area of section PSDA = Area of section QRCB

= [ ½ (22 + 30) × 4 ] cm²

= [ ½ (52) × 4 ] cm²

= 104 cm²

Hence, the areas of the four sections are 88 cm², 104 cm², 88 cm² and 104 cm²