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
1 / -0

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
1 / -0

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
1 / -0

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
1 / -0

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
1 / -0

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
1 / -0

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
1 / -0

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
1 / -0

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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Question 1

6+21cm26+\sqrt{21}cm^26+21​cm2

6+221cm26+2\sqrt{21}cm^26+221​cm2

6+27cm26+2\sqrt{7}cm^26+27​cm2

None of the above

Solution

Question 2

1201201203\sqrt{3}3​ m2m^{2}m2

1501501503\sqrt{3}3​ m2m^{2}m2

2002002003\sqrt{3}3​ m2m^{2}m2

3003003003\sqrt{3}3​ m2m^{2}m2

Solution

Question 3

631\sqrt{31}31​ m2m^{2}m2

6231\sqrt{231}231​ m2m^{2}m2

621\sqrt{21}21​ m2m^{2}m2

3231\sqrt{231}231​ m2m^{2}m2

Solution

Question 4

0.00240.00240.0024 m2m^2m2

0.00260.00260.0026 m2m^2m2

0.0030.0030.003 m2m^2m2

0.00150.00150.0015 m2m^2m2

Solution

Question 5

5.9cm5.9 cm5.9cm

6.8cm6.8 cm6.8cm

6.5cm6.5 cm6.5cm

5.6cm5.6 cm5.6cm

Solution

Question 6

84 sq.cm84\ sq.cm84 sq.cm

64 sq.cm64\ sq.cm64 sq.cm

825 sq.cm825\ sq.cm825 sq.cm

105 sq.cm105\ sq.cm105 sq.cm

Solution

Question 7

545454

515151

828282

434343

Solution

Question 8

66 cm26\sqrt {6}\ cm^{2}66​ cm2

63 cm26\sqrt {3}\ cm^{2}63​ cm2

62 cm26\sqrt {2}\ cm^{2}62​ cm2

96 cm29\sqrt {6}\,cm^{2}96​cm2

Solution

Question 9

233223\sqrt{32}2332​ sq. cm

253525\sqrt{35}2535​ sq. cm

243524\sqrt{35}2435​ sq. cm

None of these

Solution

Question 10

348348348 sq. m

315315315 sq. m

306306306 sq. m

None of these

Solution

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

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

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

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

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

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

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

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

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

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

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

$0.0024$ $m^2$

$0.0026$ $m^2$

$0.003$ $m^2$

$0.0015$ $m^2$

$5.9 cm$

$6.8 cm$

$6.5 cm$

$5.6 cm$

$84\ sq.cm$

$64\ sq.cm$

$825\ sq.cm$

$105\ sq.cm$

$54$

$51$

$82$

$43$

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

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

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

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

$23\sqrt{32}$ sq. cm

$25\sqrt{35}$ sq. cm

$24\sqrt{35}$ sq. cm

$348$ sq. m

$315$ sq. m

$306$ sq. m