Question

solve this question ​

Answer 1

$\sf \Huge Solution!!$

$\sf \large Figure\:1$

The length of the rectangle = AB = 18 cm

The breadth of the rectangle = BC = 10 cm

$\sf \to Area_{\sf (Rectangle)}=Length\times Breadth$

$\sf \to Area_{\sf (Rectangle)}=18\:cm\times 10\:cm$

$\sf \to Area_{\sf (Rectangle)}=180\:cm^{2}$

As you can see that there are two triangles formed in the rectangle and the rest area is the shaded portion. So, we have calculated the area of the whole rectangle. We need to find the area of those two triangles and subtract the sum of areas of those two triangles from the area of the rectangle to get the area of the shaded portion.

The equation is like this.

$\sf Area_{\sf (Shaded\: portion)}=Area_{(Rectangle)}-Area_{(∆BCE)}+Area_{(∆EFD)}$

So, let's find out the areas of the two triangles.

∆EFD

Height (FD) = 6 cm

Base (DE) = 10 cm

$\sf Area=\dfrac{1}{2}\times Base\times Height$

$\sf Area=\dfrac{1}{2}\times 10\:cm\times 6\:cm$

Area = 30 cm²

∆BCE

Height (BC) = 10 cm

Base (CE) = DC - DE

Base (CE) = 18 cm - 10 cm

Base (CE) = 8 cm

$\sf Area=\dfrac{1}{2}\times Base\times Height$

$\sf Area=\dfrac{1}{2}\times 8\:cm\times 10\:cm$

Area = 40 cm²

$\sf Area_{\sf (Shaded\: portion)}=Area_{(Rectangle)}-Area_{(∆BCE)}+Area_{(∆EFD)}$

$\sf Area_{\sf (Shaded\: portion)}=180\:cm^{2}-40\:cm^{2}+30\:cm^{2}$

$\boxed{\blue{\sf Area_{\sf (Shaded\: portion)}=110\:cm^{2}}}$

$\sf \large Figure\:2$

The shaded portion is in a square in such a way that it makes 3 triangles which are not shaded. So, we have to find the area of the square. Then area of the 3 triangles to get the area of the shaded portion.

The equation is like this.

$\sf Area_{\sf (Shaded\:portion)}=Area_{(Square)}-Area_{(∆FAE)}+Area_{(∆EBC)}+Area_{(∆CDF)}$

Side of the square = 20 cm

$\sf Area_{(Square)}=(Side)^{2}$

$\sf Area_{(Square)}=(20\:cm)^{2}$

$\sf Area_{(Square)}=400\:cm^{2}$

∆FAE

Height (FA) = 10 cm

Base (AE) = 10 cm

$\sf Area=\dfrac{1}{2}\times Base\times Height$

$\sf Area=\dfrac{1}{2}\times 10\:cm\times 10\:cm$

$\sf Area=50\:cm^{2}$

∆EBC

Height (BC) = 20 cm

Base (EB) = 10 cm

$\sf Area=\dfrac{1}{2}\times Base\times Height$

$\sf Area=\dfrac{1}{2}\times 10\:cm\times 20\:cm$

$\sf Area=100\:cm^{2}$

∆CDF

Height (CD) = 20 cm

Base (DF) = DA - FA

Base (DF) = 20 cm - 10 cm

Base (DF) = 10 cm

$\sf Area=\dfrac{1}{2}\times Base\times Height$

$\sf Area=\dfrac{1}{2}\times 10\:cm\times 20\:cm$

$\sf Area=100\:cm^{2}$

$\sf Area_{\sf (Shaded\:portion)}=Area_{(Square)}-Area_{(∆FAE)}+Area_{(∆EBC)}+Area_{(∆CDF)}$

$\sf Area_{\sf (Shaded\:portion)}=400\:cm^{2}-50\:cm^{2}+100\:cm^{2}+100\:cm^{2}$

$\boxed{\blue{\sf Area_{\sf (Shaded\:portion)}=150\:cm^{2}}}$

Answer 2

Answer:

150 cm

Step-by-step explanation:

The length of the rectangle = AB = 18 cm

The breadth of the rectangle = BC = 10 cm

\sf \to Area_{\sf (Rectangle)}=Length\times Breadth→Area

(Rectangle)

=Length×Breadth

\sf \to Area_{\sf (Rectangle)}=18\:cm\times 10\:cm→Area

(Rectangle)

=18cm×10cm

\sf \to Area_{\sf (Rectangle)}=180\:cm^{2}→Area

(Rectangle)

=180cm

2

As you can see that there are two triangles formed in the rectangle and the rest area is the shaded portion. So, we have calculated the area of the whole rectangle. We need to find the area of those two triangles and subtract the sum of areas of those two triangles from the area of the rectangle to get the area of the shaded portion.

The equation is like this.

\sf Area_{\sf (Shaded\: portion)}=Area_{(Rectangle)}-Area_{(∆BCE)}+Area_{(∆EFD)}Area

(Shadedportion)

=Area

(Rectangle)

−Area

(∆BCE)

+Area

(∆EFD)

So, let's find out the areas of the two triangles.

∆EFD

Height (FD) = 6 cm

Base (DE) = 10 cm

\sf Area=\dfrac{1}{2}\times Base\times HeightArea=

2

1

×Base×Height

\sf Area=\dfrac{1}{2}\times 10\:cm\times 6\:cmArea=

2

1

×10cm×6cm

Area = 30 cm²

∆BCE

Height (BC) = 10 cm

Base (CE) = DC - DE

Base (CE) = 18 cm - 10 cm

Base (CE) = 8 cm

\sf Area=\dfrac{1}{2}\times Base\times HeightArea=

2

1

×Base×Height

\sf Area=\dfrac{1}{2}\times 8\:cm\times 10\:cmArea=

2

1

×8cm×10cm

Area = 40 cm²

\sf Area_{\sf (Shaded\: portion)}=Area_{(Rectangle)}-Area_{(∆BCE)}+Area_{(∆EFD)}Area

(Shadedportion)

=Area

(Rectangle)

−Area

(∆BCE)

+Area

(∆EFD)

\sf Area_{\sf (Shaded\: portion)}=180\:cm^{2}-40\:cm^{2}+30\:cm^{2}Area

(Shadedportion)

=180cm

2

−40cm

2

+30cm

2

\boxed{\blue{\sf Area_{\sf (Shaded\: portion)}=110\:cm^{2}}}

Area

(Shadedportion)

=110cm

2

\sf \large Figure\:2Figure2

The shaded portion is in a square in such a way that it makes 3 triangles which are not shaded. So, we have to find the area of the square. Then area of the 3 triangles to get the area of the shaded portion.

The equation is like this.

\sf Area_{\sf (Shaded\:portion)}=Area_{(Square)}-Area_{(∆FAE)}+Area_{(∆EBC)}+Area_{(∆CDF)}Area

(Shadedportion)

=Area

(Square)

−Area

(∆FAE)

+Area

(∆EBC)

+Area

(∆CDF)

Side of the square = 20 cm

\sf Area_{(Square)}=(Side)^{2}Area

(Square)

=(Side)

2

\sf Area_{(Square)}=(20\:cm)^{2}Area

(Square)

=(20cm)

2

\sf Area_{(Square)}=400\:cm^{2}Area

(Square)

=400cm

2

∆FAE

Height (FA) = 10 cm

Base (AE) = 10 cm

\sf Area=\dfrac{1}{2}\times Base\times HeightArea=

2

1

×Base×Height

\sf Area=\dfrac{1}{2}\times 10\:cm\times 10\:cmArea=

2

1

×10cm×10cm

\sf Area=50\:cm^{2}Area=50cm

2

∆EBC

Height (BC) = 20 cm

Base (EB) = 10 cm

\sf Area=\dfrac{1}{2}\times Base\times HeightArea=

2

1

×Base×Height

\sf Area=\dfrac{1}{2}\times 10\:cm\times 20\:cmArea=

2

1

×10cm×20cm

\sf Area=100\:cm^{2}Area=100cm

2

∆CDF

Height (CD) = 20 cm

Base (DF) = DA - FA

Base (DF) = 20 cm - 10 cm

Base (DF) = 10 cm

\sf Area=\dfrac{1}{2}\times Base\times HeightArea=

2

1

×Base×Height

\sf Area=\dfrac{1}{2}\times 10\:cm\times 20\:cmArea=

2

1

×10cm×20cm

\sf Area=100\:cm^{2}Area=100cm

2

\sf Area_{\sf (Shaded\:portion)}=Area_{(Square)}-Area_{(∆FAE)}+Area_{(∆EBC)}+Area_{(∆CDF)}Area

(Shadedportion)

=Area

(Square)

−Area

(∆FAE)

+Area

(∆EBC)

+Area

(∆CDF)

\sf Area_{\sf (Shaded\:portion)}=400\:cm^{2}-50\:cm^{2}+100\:cm^{2}+100\:cm^{2}Area

(Shadedportion)

=400cm

2

−50cm

2

+100cm

2

+100cm

2

\boxed{\blue{\sf Area_{\sf (Shaded\:portion)}=150\:cm^{2}}}

Area

(Shadedportion)

=150cm

2