Question

Find the area of the given hexagon ABCDEF in which each one
of BJ, CL, EM and FK is perpendicular to AD and it is being
given that AJ = 6 cm, AK = 10 cm, AL = 18 cm, AM = 21 cm,
AD = 27 cm, BJ = 5 cm, CL = 6 cm, EM = 4 cm and FK = 6 cm.​

Answer 1

The area of hexagon ABCDEF is  205 cm².  

Step-by-step explanation:

Required Formulas:

• Area of triangle = ½  × (base) × (height)
• Area of trapezium = ½  × (sum of parallel sides) × height

It is given that,

In a hexagon ABCDEF, we have

BJ ⊥ AD

CL ⊥ AD

EM ⊥ AD

FK ⊥ AD

And

AJ = 6 cm, AK = 10 cm, AL = 18 cm, AM = 21 cm, AD = 27 cm, BJ = 5 cm, CL = 6 cm, EM = 4 cm and FK = 6 cm

Referring to the figure attached above, we get

LD = AD - AL = 27 - 18 = 9 cm

LM = AM - AL = 21 - 18 = 3 cm

DM = LD - LM = 9 - 3 = 6 cm

JL = AL - AJ = 18 - 6 = 12 cm

KM = AM – AK = 21 – 10 = 11 cm

Also we can see that,

Area (Hexagon ABCDEF) is given by,

= area ( ABJ) + area ( CLD) + area ( DME) + area ( AFK) + area (Trap. BCLJ) + area (Trap. EMKF) …… (i)

Now,

Area ( ABJ) = ½  × (BJ) × (AJ) = ½  × (6) × (5) = 15 cm² ….. (ii)

Area ( CLD) = ½ × (CL) × (LD) = ½  × (6) × (9) = 27 cm² …….. (iii)

Area ( DME) = ½  × (EM) × (DM) = ½  × (4) × (6) = 12 cm²  …….. (iv)

Area ( AFK) = ½  × (AK) × (FK) = ½ × (10) × (6) = 30 cm²…….. (v)

Area (Trap. BCLJ) = ½  × (BJ + CL) × JL = ½  × (5 + 6) × 12 = 66 cm² …… (vi)

Area (Trap. EMKF) = ½  × (EM + KF) × KM = ½  × (4 + 6) × 11 = 55 cm²…… (vii)

Thus, substituting values from (ii), (iii), (iv), (v), (vi), (vii) in (i), we get

The area of hexagon ABCDEF as

= 15 + 27 + 12 + 30 + 66 + 55  

= 205 cm².

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Also View:

A regular hexagon is inscribed in a circle. If the area of the hexagon is 24root3 cm2 find the area of the circle.

https://brainly.in/question/1027794

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Answer 2

Answer:

205 sq. cm

Step-by-step explanation:

Area of  ABCDEF = area of ABJ + area of CLD + area of DME + area of AFK + area of Trap. BCLJ + area of Trap. EMKF

Area of  ABJ = ½  × BJ × AJ = ½  × 6 × 5 = 15 cm²

Area of CLD= ½ × CL × LD = ½  × 6 × 9 = 27 cm²

Area of DME = ½  × EM × DM = ½  x 4 × 6= 12 cm²

Area of AFK = ½  × AK × FK = ½ × 10 × 6 = 30 cm²

Area of Trap. BCLJ = ½  × (BJ + CL) × JL = ½  × (5 + 6) × 12 = 66 cm²

Area of Trap. EMKF = ½  × (EM + KF) × KM = ½  × (4 + 6) × 11 = 55 cm²

The area of hexagon ABCDEF = 15 + 27 + 12 + 30 + 66 + 55

= 205 cm².