Herons Formula Test

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Herons Formula Test - 26

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Question 1
1 / -0

Calculate the area of quadrilateral $$ABCD$$.

A
$$6+\sqrt{21}cm^2$$

B
$$6+2\sqrt{21}cm^2$$

C
$$6+2\sqrt{7}cm^2$$

D
None of the above

Solution

Area of $$\triangle ABC$$
$$S=\dfrac{5+4+3}{2}=\dfrac{9+3}{2} = \dfrac{12}{2} = 6cm^2$$
$$=\sqrt{s(s-a)(s-b)(s-c)}$$
$$=\sqrt{6\times (6-5)(6-4)(6-3)}$$
$$=\sqrt{6\times 1\times 2\times 3}$$
$$=\sqrt{3\times 2\times 3\times 2}$$
$$=6cm^2$$
Area of $$\triangle ABC$$
$$s=\dfrac{5+5+4}{2} = \dfrac{14}{2} = 7cm^2$$
$$=\sqrt{s(s-a)(s-b)(s-c)}$$
$$=\sqrt{7\times 2\times 2\times 3}$$
$$=2\sqrt{21}$$
Total Area $$=(6+2\sqrt{21}) cm^2$$
Question 2
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The parallel sides of a parallelogram are $$40$$ and $$15$$ m and one of the diagonals is $$35$$ m. Find the area of the parallelogram, if the distance between the parallel sides is $$50$$ m. (Use Heron's formula)

A
$$120$$$$\sqrt{3}$$ $$m^{2}$$

B
$$150$$$$\sqrt{3}$$ $$m^{2}$$

C
$$200$$$$\sqrt{3}$$ $$m^{2}$$

D
$$300$$$$\sqrt{3}$$ $$m^{2}$$

Solution

In $$\parallel$$ gm $$ABCD, AB = CD = 40\ m$$ and $$AD = BC = 15\ m$$, Diagonal, $$AC = 35\ m$$
In $$\Delta ABC$$, $$s= \dfrac{40 + 15 + 35}{2}$$ $$= 45\ m$$
Area of $$\parallel$$ gm = $$2$$ x Area of $$\Delta ABC$$
= $$2\times \sqrt{S(S-a)(S-b)(S-c)}$$
= $$2\times \sqrt{45(45-40)(45-15)(45-35)}$$
= $$2\times \sqrt{45(5)(30)(10)}$$
= $$300\sqrt{3}$$ $$m^{2}$$
Question 3
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The parallel sides of a parallelogram are $$20$$ and $$10$$ m and one of the diagonals is $$12$$ m. Find the area of the parallelogram. (Use Heron's formula)

A
6$$\sqrt{31}$$ $$m^{2}$$

B
6$$\sqrt{231}$$ $$m^{2}$$

C
6$$\sqrt{21}$$ $$m^{2}$$

D
3$$\sqrt{231}$$ $$m^{2}$$

Solution

In $$\parallel$$ gm ABCD, AB = CD = 20 m and AD = BC = 10m, Diagonal, AC = 12 m
In $$\Delta$$ ABC, $$s= \dfrac{20 + 10 + 12}{2}$$ = 21 m
Area of $$\parallel$$ gm = 2 x Area $$\Delta $$ ABC
= 2$$\sqrt{S(S-a)(S-b)(S-c)}$$
= 2$$\sqrt{21(21-20)(21-10)(21-12)}$$
= 2$$\sqrt{21(1)(11)(9)}$$
= 6$$\sqrt{231}$$ $$m^{2}$$
Question 4
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The sides of a triangle are $$5\ cm, 12\ cm$$ and $$13\ cm$$, what is its area?

A
$$0.0024$$ $$m^2$$

B
$$0.0026$$ $$m^2$$

C
$$0.003$$ $$m^2$$

D
$$0.0015$$ $$m^2$$

Solution

Area of $$\triangle =\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } ,\quad \quad s=\dfrac { a+b+c }{ 2 } =\dfrac { 13+12+5 }{ 2 } =15cm$$

$$\therefore $$ ar of $$\triangle =\sqrt { 15\left( 15-13 \right) \left( 15-12 \right) \left( 15-5 \right) } =\sqrt { 15\times 2\times 3\times 10 } =30{ cm }^{ 2 }$$

Therefore area of triangle $$={ 30cm }^{ 2 }=0.003{ m }^{ 2 }\left[ \because \quad 1{ m }^{ 2 }=10000{ cm }^{ 2 } \right] $$
Question 5
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A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are $$15 cm, 14 cm, 13 cm$$, and the parallelogram stands on the base of $$15$$ cm. Find the height of the parallelogram.

A
$$5.9 cm$$

B
$$6.8 cm$$

C
$$6.5 cm$$

D
$$5.6 cm$$

Solution

Area of triangle = $$\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } $$
$$S=\dfrac { a+b+c }{ 2 } =\dfrac { 15+14+13 }{ 2 } =21cm$$
$$\therefore \quad $$ area of $$\triangle =\sqrt { 21\left( 21-15 \right) \left( 21-14 \right) \left( 21-13 \right) } =\sqrt { 21\times 6\times 7\times 8 } $$
= $$\sqrt { 3\times 7\times 3\times 2\times 7\times 4\times 2 } =2\times 2\times 3\times 7$$
= $${ 84cm }^{ 2 }$$
Now, area of $$\parallel $$gm = area of $$\triangle $$ = $$b\times h=84{ cm }^{ 2 }$$
$$\Rightarrow \quad 15\times h=84\Rightarrow h=\dfrac { 84 }{ 15 } =5.6cm$$
$$\therefore $$ Height of $$\parallel $$gm = $$5.6cm$$
Question 6
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What is the area of a triangle whose sides are $$13\ cm, 14\ cm$$ and $$15\ cm$$?

A
$$84\ sq.cm$$

B
$$64\ sq.cm$$

C
$$825\ sq.cm$$

D
$$105\ sq.cm$$

Solution

$$S=\dfrac { a+b+c }{ 2 } =\dfrac { 13+14+15 }{ 2 } =21cm$$
area of $$\triangle =\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } =\sqrt { 21\left( 21-13 \right) \left( 21-14 \right) \left( 21-15 \right) } $$
$$=\sqrt { 21\times 8\times 7\times 6 } =\sqrt { 3\times 7\times 2\times 2\times 2\times 7\times 2\times 3 } =2\times 2\times 3\times 7$$
$$={ 84cm }^{ 2 }$$
Question 7
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Find the number of trees that can be planted in a triangular ground having sides $$51m, 37m$$ and $$20m$$, when each tree occupies $$6m^2$$ of space:

A
$$54$$

B
$$51$$

C
$$82$$

D
$$43$$

Solution

Area of $$\triangle =\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } ,\quad s=\dfrac { a+b+c }{ 2 } =\dfrac { 51+37+20 }{ 2 } =54m$$

$$\therefore \quad $$ Area of triangle ground $$=\sqrt { 54\left( 54-51 \right) \left( 54-37 \right) \left( 54-20 \right) } $$

$$=\sqrt { 54\times 3\times 17\times 34 } =306{ m }^{ 2 }$$
Number of trees $$=\dfrac { 306 }{ 6 } =51$$
Question 8
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In $$\triangle ABC, AB = 6\ cm, BC = 7\ cm$$ and $$AC = 5\ cm$$. Find the area of $$\triangle ABC$$

A
$$6\sqrt {6}\ cm^{2}$$

B
$$6\sqrt {3}\ cm^{2}$$

C
$$6\sqrt {2}\ cm^{2}$$

D
$$9\sqrt {6}\,cm^{2}$$

Solution

Using Heron's formula,

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

where $$s=$$ semiperimeterof the triangle
$$a=$$ length of side $$BC$$
$$b=$$ length of side $$AC$$
$$c=$$ length of side $$AB$$

In $$\triangle ABC,\ a=7,\ b=5,\ c=6$$

Semiperemeter of $$\triangle ABC=s=\dfrac{a+b+c}{2}$$

$$=\dfrac{7+5+6}{2}$$

$$=9$$

Hence, the area of $$\triangle ABC=\sqrt{9\times (9-7)\times (9-5)\times (9-6)}$$

$$=\sqrt{9\times 3\times 2\times 4}$$

$$={6\sqrt{6}\ cm^{2}}$$
Question 9
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Find the area of the parallelogram with sides $$11 cm$$ and $$13 cm$$ and one of the diagonals as $$16 cm. $$

A
$$23\sqrt{32}$$ sq. cm

B
$$25\sqrt{35}$$ sq. cm

C
$$24\sqrt{35}$$ sq. cm

D
None of these

Solution

ar $$\parallel $$gm ABCD = $$2\times ar\triangle ABC$$
ar $$\triangle $$ABC = $$\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } ,\quad s=\dfrac { a+b+c }{ 2 } =\dfrac { 16+13+11 }{ 2 } =20cm$$
$$\therefore $$ ar $$\triangle $$ABC = $$\sqrt { 20\left( 20-16 \right) \left( 20-13 \right) \left( 20-11 \right) } =\sqrt { 20\times 4\times 7\times 9 } =\sqrt { 5\times 4\times 4\times 7\times 9 } =4\times 3\sqrt { 35 } =12\sqrt { 35 } { cm }^{ 2 }$$
$$\therefore $$ ar $$\triangle $$ABCD = $$2\times 12\sqrt { 35 } =24\sqrt { 35 } { cm }^{ 2 }$$
Question 10
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Find the area of the quadrilateral whose sides are $$9 m ,40 m, 28 m, 15 m$$. The angle between first two sides is $$90^o$$.

A
$$348$$ sq. m

B
$$315$$ sq. m

C
$$306$$ sq. m

D
None of these

Solution

Area of quadrilateral $$ABCD$$ = ar $$[\triangle ABC]$$ + ar $$[\triangle ACD]$$
In $$\triangle ABC$$ :
$$ar\ [\triangle ABC] = \dfrac { 1 }{ 2 } \times BC\times AB\\=\dfrac { 1 }{ 2 } \times 40m\times 9m=180{ m }^{ 2 }$$

Using pythagoras theorem,
$$AC= \sqrt { { AB }^{ 2 }+{ BC }^{ 2 } } \\ =\sqrt { { \left( 9 \right) }^{ 2 }+{ \left( 40 \right) }^{ 2 } }\\ =\sqrt { 1681 } =41m$$

Using Heron's formula,
$$ar [\triangle ACD] = \sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } \\\quad s=\dfrac { a+b+c }{ 2 } \\\ \ \ \quad =\dfrac { 41+28+15 }{ 2 } \\\quad \ \ \ =\dfrac { 84 }{ 2 } =42m$$

$$ ar [\triangle ACD ]= \sqrt { 42\left( 42-41 \right) \left( 42-28 \right) \left( 42-15 \right) }\\ \quad \quad \quad \ \ \ \quad =\sqrt { 42\times 1\times 14\times 27 } =126{ m }^{ 2 }$$
Therefore, area of quadrilateral $$ABCD$$
$$=180 + 126$$
$$=306\ m^{2}$$

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Herons Formula Test - 26

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Herons Formula Test - 26

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SHARING IS CARING

Question 1

$$6+\sqrt{21}cm^2$$

$$6+2\sqrt{21}cm^2$$

$$6+2\sqrt{7}cm^2$$

None of the above

Solution

Question 2

$$120$$$$\sqrt{3}$$ $$m^{2}$$

$$150$$$$\sqrt{3}$$ $$m^{2}$$

$$200$$$$\sqrt{3}$$ $$m^{2}$$

$$300$$$$\sqrt{3}$$ $$m^{2}$$

Solution

Question 3

6$$\sqrt{31}$$ $$m^{2}$$

6$$\sqrt{231}$$ $$m^{2}$$

6$$\sqrt{21}$$ $$m^{2}$$

3$$\sqrt{231}$$ $$m^{2}$$

Solution

Question 4

$$0.0024$$ $$m^2$$

$$0.0026$$ $$m^2$$

$$0.003$$ $$m^2$$

$$0.0015$$ $$m^2$$

Solution

Question 5

$$5.9 cm$$

$$6.8 cm$$

$$6.5 cm$$

$$5.6 cm$$

Solution

Question 6

$$84\ sq.cm$$

$$64\ sq.cm$$

$$825\ sq.cm$$

$$105\ sq.cm$$

Solution

Question 7

$$54$$

$$51$$

$$82$$

$$43$$

Solution

Question 8

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

$$6\sqrt {3}\ cm^{2}$$

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

$$9\sqrt {6}\,cm^{2}$$

Solution

Question 9

$$23\sqrt{32}$$ sq. cm

$$25\sqrt{35}$$ sq. cm

$$24\sqrt{35}$$ sq. cm

None of these

Solution

Question 10

$$348$$ sq. m

$$315$$ sq. m

$$306$$ sq. m

None of these

Solution

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