**Problem 1 : **

Find the area of the square shown below.

**Problem 2 : **

Find the perimeter of the rectangle shown below.

**Problem 3 : **

Find the area of the quadrilateral shown below.

**Problem 4 : **

Find the area of the parallelogram shown below.

**Problem 5 : **

The length of each side of the square inscribed into the circle shown below is 4 cm. Find the perimeter of the circle.

**Problem 6 : **

Find the area of the figure shown below.

**Problem 7 : **

A regular pentagon is inscribed in a circle with radius 1 unit. Find the area of the pentagon.

**Problem 8 :**

Find the area of the regular hexagon shown below.

**Problem 1 : **

Find the area of the square shown below.

**Solution : **

In the square shown above, the length of the diagonal is 8 cm.

When the length of a diagonal is given, formula for area of a square :

= 1/2 ⋅ d^{2}

Substitute 8 for d.

= 1/2 ⋅ 8^{2}

Simplify.

= 1/2 ⋅ 64

= 32

So, the area of the square shown above is is 32 square cm.

**Problem 2 : **

Find the perimeter of the rectangle shown below.

**Solution : **

In the above rectangle, width is 12 cm and the diagonal measures 13 cm.

To find the perimeter of a rectangle, we need to know the length and width.

In the figure shown above, consider the right triangle ABC.

By Pythagorean Theorem, we have

AB^{2} + BC^{2} = AC^{2}

Substitute.

12^{2} + l^{2} = 13^{2}

Simplify and solve for l.

144 + l^{2} = 169

Subtract 144 from each side.

l^{2} = 25

Find positive square root on both sides.

√l^{2} = √25

l = 5

Therefore, the length of the rectangle is 5 cm.

Formula for perimeter of a rectangle :

= 2(l + w)

Substitute 5 for l and 12 for w.

= 2(5 + 12)

= 2(17)

= 34

So, the perimeter of the rectangle shown above is 34 cm.

**Problem 3 : **

Find the area of the quadrilateral shown below.

**Solution : **

Formula for area of a quadrilateral :

= 1/2 ⋅ d ⋅ (h_{1} + h_{2})

Substitute 3 for h_{1} and 6 for h_{2} and 10 for d.

= 1/2 ⋅ 10 ⋅ (3 + 6)

= 1/2 ⋅ 10 ⋅ 9

= 45

So, the area of the quadrilateral shown above is 45 square cm.

**Problem 4 : **

Find the area of the parallelogram shown below.

**Solution : **

Formula for area of a parallelogram :

= b ⋅ h

Substitute 7 for h and 5 for b.

= 7 ⋅ 5

= 35

So, the area of the quadrilateral shown above is 35 square inches.

**Problem 5 : **

The length of each side of the square inscribed into the circle shown below is 4 cm. Find the perimeter of the circle.

**Solution : **

To find the perimeter of the circle, we need to know its radius.

Let us find the radius of the circle.

In the figure shown above, consider the right triangle OBC.

By Pythagorean Theorem, we have

OB^{2} = OC^{2} + CB^{2}

Substitute.

OB^{2} = 2^{2} + 2^{2}

Simplify and solve for OB.

OB^{2} = 4 + 4

OB^{2} = 8

Find positive square root on both sides.

√OB^{2} = √8

OB = 2√2

Therefore, the radius of the circle is 2√2 cm.

Formula for perimeter of a circle is

= ** **2πr

Substitute 2√2 for r.

= ** **2π(2√2)

= 4√2π cm

So, the perimeter of the circle shown above is 4√2π cm.

**Problem 6 : **

Find the area of the figure shown below.

**Solution :**

We can find the area of the figure shown above by dividing the figure into rectangle and triangle as shown below.

In the figure shown above, consider the right triangle CDE.

By Pythagorean Theorem, we have

CD^{2} + DE^{2} = CE^{2}

Substitute.

CD^{2} + 5^{2} = 13^{2}

Simplify and solve for CD.

CD^{2} + 25 = 169

Subtract 25 from each side.

CD^{2} = 144

Find positive square root on both sides.

√CD^{2} = √144

CD = 12

Therefore, the width of the rectangle is 12 cm.

And also, the base of the triangle is 12 cm.

**Area of the rectangle ABCD :**

Area = l ⋅ w

Substitute 6 for l and 12 for w.

Area = 6 ⋅ 12

Area = 72 cm^{2}

**Area of the triangle CDE :**

Area = 1/2 ⋅ b ⋅ h

Substitute 12 for b and 5 for h.

Area = 1/2 ⋅ 12 ⋅ 5

Area = 30 cm^{2}

**Area of the complete figure :**

= Area of the rectangle ABCD + Area of the triangle CDE

= 72 + 30

= 102 cm^{2}

So, the area of the figure shown above is 102 square cm.

**Problem 7 : **

A regular pentagon is inscribed in a circle with radius 1 unit. Find the area of the pentagon.

**Solution : **

Draw a sketch.

To apply formula for the area of a regular pentagon, we must find its apothem and perimeter.

The measure of central ∠ABC is

= 1/5 ⋅ 360°

= 72°

Let us consider the isosceles triangle ΔABC in the above regular pentagon.

The altitude to base AC also bisects ∠ABC and side AC. The measure of ∠DBC, then, is 36°.

In right ΔDBC, we can use trigonometric ratios to find the lengths of the legs.

cos 36° = BD/BC cos 36° = BD/1 cos 36° = BD |
sin 36° = DC/BC sin 36° = DC/1 sin 36° = DC |

So, the pentagon has an apothem of

a = BD = cos 36°

and a perimeter of

P = 5(AC)

P = 5(2 ⋅ DC)

P = 10(DC)

P = 10 ⋅ sin 36°

The area of the pentagon is

A = 1/2 ⋅ apothem ⋅ perimeter of polygon

Substitute.

A = 1/2 ⋅ cos 36° ⋅ 10 ⋅ sin 36°

Simplify.

A = 5 ⋅ cos 36° ⋅ sin 36°

A ≈ 2.38 square units

**Problem 8 :**

Find the area of the regular hexagon shown below.

**Solution : **

To apply formula for the area of a regular hexagon, we must be knowing the apothem and perimeter.

Apothem is already given and we have to find the side length (ED) of the polygon to find its perimeter.

Let us consider the isosceles triangle ΔGED in the above regular pentagon.

Because ΔGED is isosceles, the altitude GH to base ED bisects the side ED.

In right ΔEGH, we can use Pythagorean Theorem to find the length of EH.

EH^{2} + GH^{2 } = EG^{2}

Substitute.

EH^{2} + (10√3)^{2 } = 20^{2}

EH^{2} + 300^{ } = 400

Subtract 300 from each side.

EH^{2}^{ } = 100

Take square root on each side.

EH = 10

So, the hexagon has the perimeter of

P = 6(ED)

P = 6(2 ⋅ EH)

P = 6(2 ⋅ 10)

P = 6(20)

P = 120

The area of the hexagon is

A = 1/2 ⋅ apothem ⋅ perimeter of polygon

Substitute.

A = 1/2 ⋅ 10√3 ⋅ 120

Simplify.

A = 600√3 square units

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