Ncert solutions

Ncert solutions

Grade 7

Perimeter and Area | Exercise 11.4

Question 10

Q10) In the following figures, find the area of the shaded portions:

Solution 10:

(i) Area of the rectangle= l\times b

=18cm\times\left(6cm+4cm\right)

=18cm\times10cm=180cm^2

Area of the right \Delta BCE=\frac{1}{2}\times b\times h

=\frac{1}{2}\times8\times10=40cm^2

Area of the shaded portion

=180cm^2-70cm^2=110cm^2

(ii) Area of the square PQRS= \left(Side\right)^2

=\left(20\right)^2=400cm^2

Area\ of\ triangle=\ \frac{1}{2}\times b\times h

\frac{1}{2}\times10\times10=50cm^2

Area\ of\ triangle=\ \frac{1}{2}\times b\times h

=\frac{1}{2}\times10\times20=100cm^2

Area\ of\ triangle=\ \frac{1}{2}\times b\times h

=\frac{1}{2}\times10\times20=100cm^2

Area of the three triangles

=50cm^2+100cm^2+100cm^2=250cm^2

Area of the shaded portion

=400cm^2-250cm^2=150cm^2

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subject-cta

Question 10

Q10) In the following figures, find the area of the shaded portions:

Solution 10:

(i) Area of the rectangle= l\times b

=18cm\times\left(6cm+4cm\right)

=18cm\times10cm=180cm^2

Area of the right \Delta BCE=\frac{1}{2}\times b\times h

=\frac{1}{2}\times8\times10=40cm^2

Area of the shaded portion

=180cm^2-70cm^2=110cm^2

(ii) Area of the square PQRS= \left(Side\right)^2

=\left(20\right)^2=400cm^2

Area\ of\ triangle=\ \frac{1}{2}\times b\times h

\frac{1}{2}\times10\times10=50cm^2

Area\ of\ triangle=\ \frac{1}{2}\times b\times h

=\frac{1}{2}\times10\times20=100cm^2

Area\ of\ triangle=\ \frac{1}{2}\times b\times h

=\frac{1}{2}\times10\times20=100cm^2

Area of the three triangles

=50cm^2+100cm^2+100cm^2=250cm^2

Area of the shaded portion

=400cm^2-250cm^2=150cm^2

Still have questions? Our expert teachers can help you out

Book a free class now

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subject-cta
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