Herons Formula Test

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Herons Formula Test - 31

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Weekly Quiz Competition

Question 1
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Find the area of a quadrilateral whose sides are $$12, 5, 6$$ and $$15$$. The angle between the first two sides is $$90$$. (Use Heron's formula)

A
$$20 +$$ $$2\sqrt{2805}$$

B
$$30 +$$ $$2\sqrt{805}$$

C
$$30 +$$ $$2\sqrt{374}$$

D
$$30 +$$ $$2\sqrt{1805}$$

Solution

Consider $$ABCD$$ is a quadrilateral where,
$$AB = 12, BC = 5, CD = 6, DA = 15$$ and $$\angle ABC = 90^o$$

Area of $$ABCD = $$ Area of $$\Delta ABC + $$Area of $$\Delta ACD$$
In $$\Delta$$ ABC, $$\angle B = 90^o$$
Apply Pythagoras theorem in $$\Delta ABC$$
Therefore, $$AC^{2} = AB^{2}+BC^{2}= 12^{2}+5^{2}$$
So, $$AC = 13$$

Area of $$\Delta ABC = \dfrac{1}{2}\times AB \times BC = \dfrac{1}{2}\times12\times 5 = 30 m^{2}$$

In $$\Delta ACD$$, let $$s$$ be the semiperimeter,
$$S = \dfrac{6 + 15 + 13}{2} = 17$$m
Applying Heron's formula,

Area of $$\Delta ACD$$ = $$\sqrt{S(S-a)(S-b)(S-c)}$$ = $$\sqrt{17(17-13)(17-15)(17-6)}$$

= $$\sqrt{17(4)(2)(11)} = 2\sqrt{374}$$

Hence, Area of quadrilateral $$ABCD = 30 + 2\sqrt{374}$$

So, option C is correct.
Question 2
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The perimeter of a triangle is $$540$$ m and its sides are in the ratio $$12 : 25 : 17.$$ Find the area of the triangle.

A
$$5,000\:m^{2}$$

B
$$6,000\:m^{2}$$

C
$$9,000\:m^{2}$$

D
$$8,000\:m^{2}$$

Solution

$$\textbf{Step 1 : Find the sides of triagnle}$$
$$\text{Let the sides of the triangle be 12a, 25a, 17a }$$
$$\text{We know that perimeter of the triangle = Sum of all sides}$$
$$\Rightarrow\text{ 12a + 25a + 17a = 54a}$$
$$\text{Given, perimeter of the triangle = 540 m }$$
$$\Rightarrow\text{54a = 540 m}$$
$$\text{ a = 10 m}$$
$$\text{So, the lengths of the sides of triangle are }$$
$$\text{12a = 120 m}$$
$$\text{25a = 250 m}$$
$$\text{17a = 170 m}$$
$$\textbf{Step 2 : Find the area of triangle using Heron's formula}$$
$$\text{We can use Heron's formula to get the area of triangle}$$
$$\text{Area of triangle with sides with sides a, b, c and semi perimeter s}$$
$$\text{are } \sqrt { s(s-a)(s-b)(s-c)} \text{ respectively}$$.
$$\text{and s =}$$ $$\dfrac{a+b+c}{2}$$
$$\text{For triangle with sides 120 m, 250 m and 170 m, }$$
$$\text{s =}$$ $$\dfrac { 120 + 250 + 170 }{ 2 }$$
$$\text{s = 270 m}$$
$$\text{Substituting the sides 120 m, 250 m and 170 m in the Heron's formula, we get}$$
$$\Rightarrow A= \sqrt { 270(270-120)(270-250)(270-170) } $$
$$=\sqrt { 270\times 150\times 20\times 100 } $$
$$=\sqrt { 9\times30\times30\times5\times20\times20\times5} $$
$$= 3\times 30\times 5 \times 20$$
$$= 9000{m}^{2}$$
$$\textbf{Hence , area of triangle is 9000}$$ $$m^2$$
Question 3
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The triangular side - wall of a flyover have been used for advertisements, The sides of the walls are 122 m, 22m and 120 m. The advertisements requires rent of Rs 5000 per $$\displaystyle m^{2}$$ per year. A company hired one of its walls for 3 months. How much rent did it pay ?

A
1750000

B
1600000

C
1650000

D
None of these

Solution

Sides of a triangular wall is $$122m, 22m, 120m.$$
Hence, area of triangle=$$\sqrt{s(s-122)(s-22)(s-120)}$$ and

$$s=\dfrac{122+22+120}{2}$$.
$$\therefore s=132$$

$$\therefore$$Area of triangular wall$$=1320m^2$$
Since, The advertisements yeild an earning of Rs 5000 per $$\displaystyle m^{2}$$ per year.

Hence, Earning per 3 months$$= \dfrac{5000}{4}$$$$=1250$$ per $$m^2$$
$$\therefore$$Rent paid$$=1250(1320)=1650000Rs.$$
Question 4
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Using Heron's formula find the area of a quadrilateral whose sides are $$3\ cm, 4\ cm, 4\ cm$$, and $$5\ cm$$. The angle between the first two sides is $$90^0$$.

A
$$14 cm^{2}$$

B
$$16 cm^{2}$$

C
$$18 cm^{2}$$

D
$$(6 + 6\sqrt{7})cm^{2}$$

Solution

In $$\Box ABCD$$,
$$AB = 3\ cm, BC = 4\ cm, CD = 4\ cm, DA = 5\ cm$$ and $$\angle ABC = 90^o$$

Area of $$ABCD$$ = Area of $$\Delta ABC$$ + Area of $$\Delta ACD$$

In $$\Delta ABC, \angle B = 90^o$$
Therefore, $$AC^{2} = AB^{2}+BC^{2}= 3^{2}+4^{2}$$
So, $$AC = 5$$

Area of $$\Delta ABC = \dfrac{1}{2}\times AB \times AC = \dfrac{1}{2}\times3\times 4 = 6 cm^{2}$$

In $$\Delta ACD, s = \dfrac{4 + 5 + 5}{2} = 7$$m

Applying Heron's formula,
Area of $$\Delta ACD = \sqrt{S(S-a)(S-b)(S-c)}$$ = $$\sqrt{7(7-3)(7-4)(7-4)}$$
= $$\sqrt{7(4)(3)(3)} = 6\sqrt{7}$$

Hence, Area of $$\Box ABCD = (6 + 6\sqrt{7})cm^{2}$$

So, option D is correct.
Question 5
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A traffic signal board, indicating 'SCHOOLAHEAD', is an equilateral triangle with side $$a$$. Find the area of the signal board, using Heron's formula. If its perimeter is $$180 cm$$, what will be the area of the signal board?

A
$$900\sqrt 3 cm^2$$

B
$$300\sqrt 3 cm^2$$

C
$$600\sqrt 3 cm^2$$

D
$$900\sqrt 2 cm^2$$

Solution

Area of triangle $$=\sqrt { S(S-a)(S-b)(S-c) } $$

Perimeter $$=180 cm $$

Semi-perimeter, $$S=\cfrac { a+b+c }{ 2 } =\cfrac { 180 }{ 2 } =90\ cm\quad $$

Given triangle is an equilateral triangle, therefore $$a=b=c$$

Perimeter $$=3a$$

$$180=3a\Rightarrow a=60$$
$$a=b=c=60\ cm$$

Area of triangle
$$=\sqrt { S(S-a)(S-b)(S-c) } $$

$$ =\sqrt { 90(90-60)(90-60)(90-60) } $$

$$ =\sqrt { 90\times 30\times 30\times 30 } $$

$$ =\sqrt { 9\times 3\times 3\times 3\times { (10) }^{ 4 } } $$

$$ =3\times 3\times 10^{ 2 }\sqrt { 3 } $$

$$ =900\sqrt { 3 } cm^2$$
Question 6
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Find the area of a quadrilateral whose sides are $$3\ cm, 4\ cm, 2\ cm$$, and $$5\ cm$$. The angle between the first two sides is $$90^o$$. (Use Heron's formula)

A
$$14 cm^{2}$$

B
$$16 cm^{2}$$

C
$$ (6+ \sqrt{24} ) cm^{2}$$

D
$$20 cm^{2}$$

Solution

In quadrilateral $$ABCD$$,
$$AB = 3\ cm, BC = 4\ cm, CD = 2\ cm, DA = 5\ cm$$ and $$\angle ABC = 90^0 $$

Area of $$ABCD$$ = Area of $$\Delta ABC +$$ Area of $$\Delta ACD$$

In $$\Delta ABC, \angle B = 90^0$$

$$\therefore$$ Area of $$\Delta ABC$$ = $$\dfrac{1}{2}\times AB \times BC = \dfrac{1}{2}\times3\times 4\ cm^2 = 6 cm^{2}$$

In $$\Delta ABC$$, applying pythagorous theorem we get,
$$AC^2=AB^2+BC^2$$
$$\Rightarrow AC^2=3^2+4^2$$
$$\Rightarrow AC^2=9+16$$
$$\Rightarrow AC^2=25$$
$$\Rightarrow AC=5$$
$$\therefore AC=5\ cm$$

In $$\Delta ACD$$, semi-perimeter $$s = \dfrac{2 + 5 + 5}{2}cm = 6$$ $$cm$$

Applying Heron's formula,
Area of $$\Delta\ ACD$$ = $$\sqrt{S(S-a)(S-b)(S-c)}$$ = $$\sqrt{6(6-5)(6-5)(6-2)}$$
= $$\sqrt{6(1)(1)(4)} = \sqrt{24}$$

Hence, Area of $$\Box ABCD =( 6 + \sqrt{24}) cm^2$$

So, option $$C$$ is correct.
Question 7
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Use Heron's formula to find the area of a triangle of lengths $$5, 7$$ and $$8$$.

A
$$10\sqrt{3}$$

B
$$20\sqrt{3}$$

C
$$2\sqrt{6}$$

D
$$20\sqrt{6}$$
Question 8
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Find the area of a field which is in shape of a parallelogram with sides 12 cm and 14 cm and one of the diagonal of length 18 cm.

A
$$16 \sqrt{110}$$ $$cm^2$$

B
$$18 \sqrt{11}$$ $$cm^2$$

C
$$8 \sqrt{10}$$ $$cm^2$$

D
$$6 \sqrt{11}$$ $$cm^2$$

Solution

Solution:
To find:
$$ar(llgm ABCD)=?$$
Solution;
$$ar(\triangle ABD)=\sqrt{s(s-a)(s-b)(s-c)}$$
here, $$a=12, b=14 , c=18 $$
$$s=\cfrac{12+14+18}2=22$$
$$ar(\triangle ABD)=\sqrt{22(22-12)(22-14)(22-18)} cm^2$$

$$=\sqrt{22\times10\times8\times4}\quad cm^2$$

$$=\sqrt{7040}\quad cm^2$$

$$=8\sqrt{110}\quad cm^2$$

Now,
$$ar(llgm ABCD)=2\times ar(\triangle ABD)$$

$$=2\times 8\sqrt{110}$$

$$=16\sqrt{110}\quad cm^2$$

Hence, A is the correct answer.
Question 9
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Two sides of a plot measure 32 m and 24 m and the angle between them is a perfect right angle The other two sides measure 25 m each and the other three angles are not right angles What is the area of the plot?

A
$$768 \displaystyle m^{2}$$

B
$$534 \displaystyle m^{2}$$

C
$$696.5 \displaystyle m^{2}$$

D
$$684 \displaystyle m^{2}$$

Solution

$$arABCD=ar\triangle ADC+ar\triangle ABC$$
$$ar\triangle ADC=\dfrac { 1 }{ 2 } \times 24\times 32=384{ m }^{ 2 }$$
Now, $$AC=\sqrt { { AD }^{ 2 }+{ DC }^{ 2 } } $$
$$=\sqrt { { 24 }^{ 2 }+{ 32 }^{ 2 } } =\sqrt { 1600 } =40m$$
$$\therefore \quad ar\triangle ABC=\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } ,\quad S=\dfrac { 40+25+25 }{ 2 } =45m$$
$$=\sqrt { 45\left( 45-40 \right) \left( 45-25 \right) \left( 45-25 \right) } $$
$$=\sqrt { 45\times 5\times 20\times 20 } =15\times 20=300{ m }^{ 2 }$$
$$\therefore \quad ar.ABCD=384+300=684{ m }^{ 2 }$$
Question 10
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In an equilateral triangle, $$3$$ coins of radii $$1$$ unit each are kept so that they touch each other and also the sides of the triangle. Area of the triangle is?

A
$$4+2\sqrt{3}$$

B
$$6+4\sqrt{3}$$

C
$$12+\displaystyle\frac{7\sqrt{3}}{4}$$

D
$$3+\displaystyle\frac{7\sqrt{3}}{4}$$