Herons Formula Test

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Herons Formula Test - 25

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Question 1
1 / -0

Find the area of the triangle by Heron's formula.

A
$$12m^2$$

B
$$24m^2$$

C
$$40m^2$$

D
$$6m^2$$

Solution

We use heron's formula to find the area of the triangle.

Let us find the semi perimeter:

$$S=\dfrac { 5+3+4 }{ 2 } =\dfrac { 12 }{ 2 } =6$$ cm

Now area of the triangle is:

$$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 6(6-5)(6-3)(6-4) } =\sqrt { 6\times 1\times 3\times 2 } =\sqrt { 36 } =6$$ m$$^2$$

Hence, the area of the triangle is $$6$$ m$$^2$$.
Question 2
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The area of a triangle by Heron's formula is

A
$$\sqrt{(s-a)(s-b)(s-c)}$$

B
$$\sqrt{s(s-a)(s-b)(s-c)}$$

C
$$\sqrt{(s+a)(s-b)(s-c)}$$

D
$$\sqrt{(s+a)(s+c)(s+c)}$$

Solution

Using Herons formula, we first have to find the semi perimeter with sides $$a,b$$ and $$c$$:

$$s=\dfrac { a+b+c }{ 2 }$$

And then area of the triangle is:

$$A=\sqrt { s(s-a)(s-b)(s-c) }$$

Hence, area of a triangle by heron's formula is $$A=\sqrt { s(s-a)(s-b)(s-c) }$$.
Question 3
1 / -0

Find the area of a triangle using Heron's formula

A
$$170 cm^2$$

B
$$160 cm^2$$

C
$$180 cm^2$$

D
$$200 cm^2$$

Solution

We find the area of triangle using herons formula:

$$S=\dfrac { 9+40+41 }{ 2 } =\dfrac { 90 }{ 2 } =45$$ cm

Now area is:

$$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 45(45-9)(45-40)(45-41) } =\sqrt { 45\times 36\times 5\times 4 } =\sqrt { 32400 } =180$$ cm$$^2$$

Hence, area of triangle is $$180$$ cm$$^2$$.
Question 4
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Find the area of an equilateral triangle with side $$10cm$$.

A
$$25\sqrt{3}cm^2$$

B
$$24\sqrt{3}cm^2$$

C
$$26\sqrt{3}cm^2$$

D
None of the above

Solution

Formula: Are of an equilateral triangle having side length $$a$$ is given by, $$\dfrac{\sqrt 3}{4}a^2$$

Here it is given that, side length $$, a=10$$ cm
Thus its area $$=\dfrac{\sqrt 3}{4}\times 10^2$$ cm$$^2=25\sqrt 3$$ cm$$^2$$
Question 5
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Use Heron's formula to find the area of a triangle of lengths 3,3 and 4

A
$$5\sqrt{2}$$

B
$$2\sqrt{5}$$

C
$$2\sqrt{7}$$

D
$$7\sqrt{2}$$
Question 6
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Use Heron's formula to find the area of a triangle of lengths $$5 cm, 9 cm$$ and $$12 cm$$.

A
$$\sqrt{26}\ cm^2$$

B
$$2\sqrt{26}\ cm^2$$

C
$$3\sqrt{26}\ cm^2$$

D
$$4\sqrt{26}\ cm^2$$

Solution

We will find the area of triangle with lengths $$5$$ cm, $$9$$ cm and $$12$$ cm using heron's formula:

$$S=\dfrac { 5+9+12 }{ 2 } =\dfrac { 26 }{ 2 } =13$$ units

Now area is:

$$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 13(13-5)(13-9)(13-12) } =\sqrt { 13\times 8\times 4\times 1 } =\sqrt { 416 } =4\sqrt { 26 }$$ sq.units

Hence, area of triangle is $$4\sqrt { 26 } cm^2$$
Question 7
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A land ABCD is in the shape of a Quadrilateral has $$\angle C = 90, AB = 9m, BC = 12m, CD = 5m$$ and $$AD = 8m$$. How much area does it occupy?

A
$$30+6\sqrt{5}m^2$$

B
$$30+6\sqrt{35}m^2$$

C
$$30+6\sqrt{125}m^2$$

D
None of the above

Solution

$$BD = \sqrt{12^2+5^2} = \sqrt{144+25} = \sqrt{169}=13m$$
Area of $$\triangle BDC = \dfrac{1}{2} \times 12\times 5m^2 = 6\times 5=30m^2$$
Area of \triangle ABD
$$S=\dfrac{9+8+13}{2} = \dfrac{17+13}{2} = \dfrac{30}{2} = 15$$
$$=\sqrt{15\times (15-9)(15-8)(15-13)}$$
$$=\sqrt{15\times 6\times 7\times 2}$$
$$=\sqrt {3\times 5\times 3\times 2\times 7\times 2}$$
$$=3\times 2\sqrt{35}=6\sqrt{35}$$
Total area $$= 30+6\sqrt{35}m^2$$
Question 8
1 / -0

The sides of a triangle are $$5y, 6y $$ and $$9y$$. find an expression for its area $$A$$.

A
$$10y^2\sqrt{2}$$

B
$$5y^2\sqrt{2}$$

C
$$20y^2\sqrt{2}$$

D
$$100y^2\sqrt{2}$$

Solution

We will find the area of triangle with sides $$5y$$, $$6y$$ and $$9y$$ using heron's formula:

Let $$S$$ be the semi perimeter of the given triangle

$$S=\dfrac { 5y+6y+9y }{ 2 } \\=\dfrac { 20y }{ 2 }\\=10y$$

Now area is:

$$A=\sqrt { s(s-a)(s-b)(s-c) } \\=\sqrt { 10y\left( 10y-5y \right) \left( 10y-6y \right) \left( 10y-9y \right) } \\=\sqrt { 10y\times 5y\times 4y\times y } \\=\sqrt { 200y^{ 4 } } \\=10y^{ 2 }\sqrt { 2 }$$

Hence, area of triangle is $$10y^{ 2 }\sqrt { 2 }$$ sq.units.
Question 9
1 / -0

Use Heron's formula to find the area of a triangle of lengths $$7, 8$$ and $$9$$.

A
$${12\sqrt{5}}$$

B
$$11\sqrt{5}$$

C
$$6\sqrt{5}$$

D
None of the above

Solution

We will find the area of triangle with lengths $$7,8$$ and $$9$$ using herons formula:

$$S=\dfrac { 7+8+9 }{ 2 } =\dfrac { 24 }{ 2 } =12$$ units

Now area is:

$$A=\sqrt { s(s-a)(s-b)(s-c) } $$

$$=\sqrt { 12(12-7)(12-8)(12-9) } $$

$$=\sqrt { 12\times 5\times 4\times 3 } $$

$$=\sqrt { 720 } =12\sqrt { 5 }$$ sq.units

Hence, area of triangle is $$12\sqrt { 5 }$$ sq.units.
Question 10
1 / -0

Use Heron's formula to find the area of a triangle of lengths $$4, 5$$ and $$6.$$

A
$$\dfrac{15}{4}\sqrt{7}$$

B
$$\dfrac{13}{4}\sqrt{7}$$

C
$$4\sqrt{7}$$

D
$$3\sqrt{7}$$

Solution

Given: Lengths of a triangle $$=4,$$ $$5,$$ and $$6$$

$$S=\dfrac { 4+5+6 }{ 2 } =\dfrac { 15 }{ 2 }$$ units

Using heron's formula to find the area of the triangle:

$$A=\sqrt { s(s-a)(s-b)(s-c) } $$

$$=\sqrt { \dfrac { 15 }{ 2 } \left( \dfrac { 15 }{ 2 } -4 \right) \left( \dfrac { 15 }{ 2 } -5 \right) \left( \dfrac { 15 }{ 2 } -6 \right) }$$

$$ =\sqrt { \dfrac { 15 }{ 2 } \times \dfrac { 7 }{ 2 } \times \dfrac { 5 }{ 2 } \times \dfrac { 3 }{ 2 } } $$

$$=\sqrt { \dfrac { 1575 }{ 16 } } $$

$$=\dfrac { 15 }{ 4 } \sqrt { 7 }$$

Hence, area of triangle is $$\dfrac { 15 }{ 4 } \sqrt { 7 }$$ sq. units.

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Herons Formula Test - 25

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Herons Formula Test - 25

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SHARING IS CARING

Question 1

$$12m^2$$

$$24m^2$$

$$40m^2$$

$$6m^2$$

Solution

Question 2

$$\sqrt{(s-a)(s-b)(s-c)}$$

$$\sqrt{s(s-a)(s-b)(s-c)}$$

$$\sqrt{(s+a)(s-b)(s-c)}$$

$$\sqrt{(s+a)(s+c)(s+c)}$$

Solution

Question 3

$$170 cm^2$$

$$160 cm^2$$

$$180 cm^2$$

$$200 cm^2$$

Solution

Question 4

$$25\sqrt{3}cm^2$$

$$24\sqrt{3}cm^2$$

$$26\sqrt{3}cm^2$$

None of the above

Solution

Question 5

$$5\sqrt{2}$$

$$2\sqrt{5}$$

$$2\sqrt{7}$$

$$7\sqrt{2}$$

Question 6

$$\sqrt{26}\ cm^2$$

$$2\sqrt{26}\ cm^2$$

$$3\sqrt{26}\ cm^2$$

$$4\sqrt{26}\ cm^2$$

Solution

Question 7

$$30+6\sqrt{5}m^2$$

$$30+6\sqrt{35}m^2$$

$$30+6\sqrt{125}m^2$$

None of the above

Solution

Question 8

$$10y^2\sqrt{2}$$

$$5y^2\sqrt{2}$$

$$20y^2\sqrt{2}$$

$$100y^2\sqrt{2}$$

Solution

Question 9

$${12\sqrt{5}}$$

$$11\sqrt{5}$$

$$6\sqrt{5}$$

None of the above

Solution

Question 10

$$\dfrac{15}{4}\sqrt{7}$$

$$\dfrac{13}{4}\sqrt{7}$$

$$4\sqrt{7}$$

$$3\sqrt{7}$$

Solution

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