Question

Question 2....................

Answer 1

Answer:

32 sq.cm

Step-by-step explanation:

4(a)

a=16

4×16

=64.

2(a)

2×16= 32

64-32=32

therefore area is 32cm sq

Answer 2

The area of the shaded region in the given figure is 109.72 cm².

Step-by-step explanation:

It is given that,

ABCD is a square with side 16 cm.

Step 1:

Join the points B and D.

Since each angle in a square is 90° ∴ ∠BAD = 90° and Δ ABD is a right-angled triangle.

Considering the given figure,  

The area of the minor segment BSD is given by,

= [Area of the sector ABSD] – [Area of the right ∆ABD]

= [πr² * (θ/360)] – [½ * AD * AB]

Here r = AD = AB = 16 cm and θ = 90°

= [(22/7)(16)² * (90/360)] – [½ * 16 * 16]

= 201.14 – 128

= 73.14 cm²

Step 2:

Similarly, we can also solve for the area of the minor segment BFD which will also be equal to 73.14 cm².

Therefore,

The total area of both the minor segments is given by,

= 2 * area of one of the minor segments

= 2 * 73.14  

= 146.28 cm²

Step 3:

The area of the square ABCD will be = side² = 16² = 256 cm²

Thus,  

The area of the shaded region in the given figure is given by,

= [Area of the square ABCD] – [Total area of the two minor segments]

= 256 – 146.28

= 109.72 cm²

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