Question

find the area of the shaded portions in the following figures:​

Answer 1

Answer:

Step-by-step explanation:

Hey mate here is your answer:

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Given:

→Figure 1 and Figure 2,

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Answer for Figure 1:

→To find the area of the shaded are you should divide the figure into three parts:

→Rectangle,

→Right-angled triangle,

→Rectangle,

→See the Figure 1 given in the attachment.

→Let BDFC and ABGF be 2 rectangles  and EGF be

right-angled triangle.

→To find  area of rectangle BDFC You need length and breadth,

→length=12 cm and breadth= 8 cm.

So area = length × breadth,

12 cm × 8 cm,

→96 cm².

→To find the  area of the rectangle ABEG also we need length and breadth,

→length=6 cm,

→As Points ABD lie on same line and AD=12 cm, DB=8 cm(opposite and equal to side FC),

→So breadth AB = AD-BD,

→12 cm - 8cm,

→4 cm.

So area of the rectangle = length × breadth,

→6 × 4 cm,

→24 cm².

Now we should find the area of the right-angled triangle,

→Area of the right-angled triangle = $\frac{perpendicular\:*\:base}{2}$.

→Perpendicular = EG=4cm(opposite and equal to AB) and base = is also 4 cm .

→4×4/2,

→16/2,

→8 cm is the area.

So total are of the figure = BDFC + ABGE + EGF ,

Substitute the values:

→96  + 24 +8,

→128 cm is the area of figure 1.

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Answer for Figure 2 :

→To find the area of the figure 2 shaded portion we should find the area of the whole figure and area of the  2 isosceles triangle and subtract both the areas.

→ABDC-(AED+BFC),

Fist find area of the ABDC rectangle,

→ length = 12 cm , breadth=10 cm.

→ Area= 12×10,

→120 cm².

Now area of the ADE or BFC( because both have same sides and will have same area and if we find one area then another one will also have same),

→Area of  the isosceles triangle  :

→ $\frac{bh_b}{2}$,

→b= 12 cm = (BC) and $b_h$ = 4 cm  FG .

→$\frac{12\:*\:4}{2}$,

→48/2,

24 cm is the area of one isosceles triangle .

Hence ADE = 24 cm so BFC = 24cm,

So area of the figure 2=

→ABDC-(AED+BFC)

→120-(24+24),

→120-48,

→72 cm is the area of the figure 2.

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Hope it helps you.

Answer 2

Answer:

→To find the area of the shaded are you should divide the figure into three parts:

→Rectangle,

→Right-angled triangle,

→Rectangle,

→See the Figure 1 given in the attachment.

→Let BDFC and ABGF be 2 rectangles and EGF be

right-angled triangle.

→To find area of rectangle BDFC You need length and breadth,

→length=12 cm and breadth= 8 cm.

So area = length × breadth,

12 cm × 8 cm,

→96 cm².

→To find the area of the rectangle ABEG also we need length and breadth,

→length=6 cm,

→As Points ABD lie on same line and AD=12 cm, DB=8 cm(opposite and equal to side FC),

→So breadth AB = AD-BD,

→12 cm - 8cm,

→4 cm.

So area of the rectangle = length × breadth,

→6 × 4 cm,

→24 cm².

Now we should find the area of the right-angled triangle,

→Area of the right-angled triangle = \frac{perpendicular\:*\:base}{2}

2

perpendicular∗base

.

→Perpendicular = EG=4cm(opposite and equal to AB) and base = is also 4 cm .

→4×4/2,

→16/2,

→8 cm is the area.

So total are of the figure = BDFC + ABGE + EGF ,

Substitute the values:

→96 + 24 +8,

→128 cm is the area of figure 1.

→To find the area of the figure 2 shaded portion we should find the area of the whole figure and area of the 2 isosceles triangle and subtract both the areas.

→ABDC-(AED+BFC),

Fist find area of the ABDC rectangle,

→ length = 12 cm , breadth=10 cm.

→ Area= 12×10,

→120 cm².

Now area of the ADE or BFC( because both have same sides and will have same area and if we find one area then another one will also have same),

→Area of the isosceles triangle :

→ \frac{bh_b}{2}

2

bh

b

,

→b= 12 cm = (BC) and b_hb

h

= 4 cm FG .

→\frac{12\:*\:4}{2}

2

12∗4

,

→48/2,

24 cm is the area of one isosceles triangle .

Hence ADE = 24 cm so BFC = 24cm,

So area of the figure 2=

→ABDC-(AED+BFC)

→120-(24+24),

→120-48,

→72 cm is the area of the figure 2.