Question

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Answer 1

Given :

AD = 8cm

AH = 6cm

AG = 4cm

AF = 3cm

BF = 2cm

CH = 3cm

EG = 2.5 cm

now,

triangles in the the given fig:-

∆1DEG

∆2CHD

∆3ABF

∆4AEG

for ∆1DEG

base = (AD) - (AG)

8cm - 4cm

(DG) = 4cm

height = GE = 2.5 cm

so,

area∆DEG = 1/2 × B × H

1/2 × 4cm × 2.5 cm

area∆ DEG = 5cm^2. -. (

now,

in ∆ AEG

area∆ AEG = 1/2 × B × H

1/2 × (AG) × ( EG)

= 1/2 × 4cm × 2.5 cm

5 cm^2

now in∆ CHD

= 1/2 × B × H

= 1/2 × (CH ) × (DH) [ DH = AD - AH

DH = 8cm - 6cm

= 2cm]

1/2 × 3cm × 2cm

area∆ CHD = 3cm^2

now,∆ AFB = 1/2 × B × H

= 1/2 × (AF) × ( FB)

= 1/2 × 3cm × 2cm

area∆ AFB = 3cm^2

now , in BFHC

( refer to pic)

Therefore, in ∆FHC

angle(CHF) = 90°

FH = AH - AF

FH = 6cm - 3cm

FH = 3cm

so by using Pythagoras theorem in ∆ FCH

$fc^{2} = fh ^{2} + ch ^{2}$

we get

fc = 3✓2

so,

area∆CHF = 1/2 × B × H

= 1/2 × 3cm × 3cm

= 9/2cm

= 4.5 cm^2

now, let take ∆ BFG

by using Pythagoras theorem we will find BC

so ,

$fc ^{2} = bf^{2} + bc ^{2}$

from this we can find out BC

so BC = ✓14cm

so, area∆FBC

1/2 × Base(BC) + height (BF)

1/2 × ✓14 × 2

area∆BFC = ✓14 cm^2

so now

add all the area if triangle

(∆1 + ∆2 + ∆3 + ∆4 + ∆FHC + ∆ BFC)area

= 20.5 + ✓14 cm^2

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