Question

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

Answer 1

Given that, the width of each section is same. Therefore,

IB = BJ = CK = CL = DM = DN = AO = AP

IL = IB + BC + CL

28 = IB + 20 + CL

IB + CL = 28 cm − 20 cm = 8 cm

IB = CL = 4 cm

Hence, IB = BJ = CK = CL = DM = DN = AO = AP = 4 cm

Area of section BEFC = Area of section DGHA

$=\left[\frac{1}{2}\right(20+28\left)\right(4\left)\right]c{m}^2=96c{m}^2$ [$\because$ Area of trapezium $=\frac{1}{2}\left(\text{Sum of parallel side}\right)\left(\text{Height}\right)$]

⇒Area of section ABEH = Area of section CDGF
$=\frac{1}{2}[16+24]\left(4\right)=80c{m}^2$