Herons Formula Test

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Herons Formula Test - 23

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Question 1
1 / -0

The adjacent sides of a parallelogram are 8 cm and 9 cm. The shorter diagonal is 13 cm. What is its area?

A
$$72\, cm^2$$

B
$$12\sqrt {35}\, cm^2$$

C
$$24\sqrt {35}\, cm^2$$

D
$$150\, cm^2$$

Solution

As shown in the diagram, area of triangle ABC = 1/2 area of the parallelogram ABCD. By Heron's formula, area of triangle
$$=\, \sqrt {S(S - a)(S - b)(S - c)}$$
Here S = $$\displaystyle \frac {9 + 8 + 13}{2}\, =\, 15$$
$$\sqrt{180\, \times\, 7}\, =\, \sqrt{1260}\, =\, \sqrt{6\, \times\, 210}$$
$$\sqrt {6\, \times\, 7\, \times\, 30}\, =\, 6\sqrt{35}$$ sq cm.
Hence, area of ABCD
$$=\, 2\, \times\,6\sqrt{35}\, =\, 12\, \sqrt{35}$$ sq cm
Question 2
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The sides of a triangle are $$5 cm$$, $$12 cm$$ and $$13 cm$$, Then its area is

A
$$0.0024 m^2$$

B
$$0.0026 m^2$$

C
$$0.003 m^2$$

D
$$0.0015 m^2$$

Solution

Let $$a=5$$ cm, $$b=12$$ cm, $$c=13$$ cm

Semiperimeter $$s=\dfrac{5+12+13}{2}\Rightarrow 15$$
Area of $$\triangle=\sqrt{s(s-a)(s-b)(s-c)}$$
Area of $$\triangle=\sqrt{15(15-5)(15-12)(15-13)}$$
Area of $$\triangle=\sqrt{15\times 10\times 3\times 2}$$
Area of $$\triangle=\sqrt{900}$$
Area of $$\triangle=30\ cm^2\Rightarrow \dfrac{30}{100\times 100}\Rightarrow 0.003\ m^2.$$
Question 3
1 / -0

The area of a triangle, whose sides are $$4 cm$$, $$13 cm$$ and $$15cm$$, is

A
$$\sqrt{420} cm^2$$

B
$$24 cm^2$$

C
$$48 cm^2$$

D
$$56 cm^2$$

Solution

Let $$a=4$$ cm, $$b=13$$ cm, $$c=15$$ cm

Semiperimeter $$s=\dfrac{4+15+13}{2}\Rightarrow 16$$
Area of $$\triangle=\sqrt{s(s-a)(s-b)(s-c)}$$
Area of $$\triangle=\sqrt{16(16-4)(16-13)(16-15)}$$
Area of $$\triangle=\sqrt{16\times 12\times 3\times 1}$$
Area of $$\triangle=\sqrt{576}$$
Area of $$\triangle=24\ cm^2.$$
Question 4
1 / -0

ABCD is a given field whose sides are indicated. If $$\angle$$DAB = 90$$^o$$, then area of the field is ............

A
180 m$$^2$$

B
126 m$$^2$$

C
306 m$$^2$$

D
300 m$$^2$$

Solution

IN $$\bigtriangleup DBC$$ $$\angle DAB=90^{0}$$
Then $$DB^{2}=(40)^{2}+(9)^{2}=1600+81=1681$$
OR DB=41 m
Area of $$\bigtriangleup DBC=\sqrt{42(42-41)(42-28)(42-15)}=\sqrt{15876}=126$$ $$m^{2}$$
Area of$$\bigtriangleup ADB=\frac{1}{2}(40\times 9)=\frac{360}{2}=180$$$$m^{2}$$
Area of ABCD=$$\bigtriangleup DBC+\bigtriangleup ADB$$
Area of ABCD=$$126+180=306m^{2}$$
Question 5
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The base of an isosceles triangle is $$16$$ cm and its perimeter is $$36$$ cm. The area of the triangle is ........

A
$$72 \displaystyle cm^{2} $$

B
$$36\displaystyle cm^{2} $$

C
$$24 \displaystyle cm^{2} $$

D
$$48 \displaystyle cm^{2} $$

Solution

In isosceles $$\triangle ABC$$,
perimeter $$=AB+BC+CA =36$$
Let $$AB = AC =x$$
$$x+x+16 =36$$
$$x= \displaystyle \dfrac{1}{2}\left ( 36-16 \right )cm=10cm$$
In an isosceles triangle, median is the perpendicular bisector.
$$\angle AEB =90^o$$
Now, $$\triangle ABE $$ is a right angle triangle.
Then by pythagoras theorem, we have
$$AE^2+BE^2= AB^2$$
$$\displaystyle h^{2}+8^{2}=10^{2} $$
$$h^2=100-64$$
$$\Rightarrow h=6cm$$
$$\displaystyle ar [\Delta ABC]=\dfrac{1}{2}\times \text{base}\times \text{height }\\=\dfrac{1}{2}\times 16cm \times 6cm\\=48\ cm^{2}$$
Question 6
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The sides of a triangle are $$3$$ cm, $$5$$ cm and $$4$$ cm: Its area is

A
$$6\, cm^2$$

B
$$7.5\, cm^2$$

C
$$5\sqrt 2\, cm^2$$

D
None of these

Solution

Given: the sides of a triangle are $$3$$ cm, $$5$$ cm and $$4$$ cm.
To find out: Area of triangle
Solution:-
The lengths of the sides are $$a=3$$ cm, $$b=5$$ cm, $$c=4$$ cm.
$$\therefore s=\dfrac { a+b+2 }{ 2 } =\dfrac { 3+5+4 }{ 2 } cm=6$$ cm
We use the formula,
$$\Delta =\sqrt { { s\left( s-a \right) }{ \left( s-b \right) }{ \left( s-c \right) } } $$.
$$\therefore \Delta =\sqrt { 6{ \left( 6-3 \right) }{ \left( 6-5 \right) }{ \left( 6-4 \right) } } { cm }^{ 2 }\\ =\sqrt { 6\times 3\times 1\times 2 }$$ sq.cm
$$ =6$$ sq.cm
Question 7
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The area of triangle with sides $$35$$ cm, $$66$$ cm and $$53$$ cm is

A
$$924 \text{ cm}^{2} $$

B
$$1336 \text{ cm}^{2} $$

C
$$693 \text{ cm}^{2} $$

D
$$1848 \text{ cm}^{2} $$

Solution

Hen the sides of a triangle are given then the area is given by Heron's formula.
Area $$=\sqrt{s(s-a)(s-b)(s-c)}$$ where semiperimeter$$(s)=\dfrac{a+b+c}{2}$$
$$\because s=\dfrac{35+66+53}{2}=77 $$
$$\implies$$ Area of $$\displaystyle \Delta =\sqrt{77\times \left ( 77-35 \right )\times \left ( 77-66 \right )\times \left ( 77-53 \right )}$$
$$=\displaystyle \sqrt{77\times 11\times 42\times 24}$$
$$ = 924 \text{ cm}^{2}$$
Question 8
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Find the area of a rhombus with sides 20 cm and length of one of the diagonal as 28 cm.

A
$$56 \sqrt{51}$$ $$cm^2$$

B
$$18 \sqrt{51}$$ $$cm^2$$

C
$$25 \sqrt{51}$$ $$cm^2$$

D
none of these

Solution

Let $$ABCD$$ is a rhombus with diagonals $$AC$$ and $$BD$$ which intersect each other at $$O$$.

$$AC = 28 ⇒ AO = 14$$

Let $$BO = x$$ and $$AB = 20\ cm$$ (given)

By Pythagorean theorem,

$$c ^2 = a ^2 + b ^2$$
$$20 ^2 = 14 ^2 + x ^2$$
$$400 = 196 + x ^2$$
$$x ^2 = 400-196$$
$$x ^2 = 204$$
$$x=2\sqrt { 51 }$$ cm

Therefore, $$BO=2\sqrt { 51 }\ cm$$

Diagonal $$BD=2\times 2\sqrt { 51 } =4\sqrt { 51 }\ cm$$

Area = $$\dfrac { 1 }{ 2 }$$ [ product of diagonals]
$$=\dfrac { 1 }{ 2 } \times 28\times 4\sqrt { 51 } =56\sqrt { 51 }$$

Hence, the area of rhombus is $$56\sqrt { 51 }\ cm^2$$.
Question 9
1 / -0

The area of triangle whose sides are 4cm, 13 cm and 15 cm is

A
$${ 576cm }^{ 2 }$$

B
$${ 30cm }^{ 2 }$$

C
$${ 26cm }^{ 2 }$$

D
$${ 24cm }^{ 2 }$$

Solution

$$\dfrac { s }{ 2 } =\dfrac { 4+13+15 }{ 2 } =16cm\\\\ Area\ =\sqrt { s(s-a)(s-b)(s-c) } \\\\ =\sqrt { 16(16-4)(16-13)(16-15) } \\\\ =\sqrt { 16\times 12\times 3\times 1 } ={ 24\ cm }^{ 2 }$$'
Question 10
1 / -0

What is the area of a triangle whose sides are $$3\ \mathrm{cm}$$ , $$5\,\mathrm{ cm}$$ and $$4\,\mathrm{ cm}$$?

A
$$6 \ \displaystyle \mathrm{cm^{2}}$$

B
$$7.5 \ \displaystyle \mathrm{cm^2}$$

C
$$\displaystyle 5\sqrt{2} \,\mathrm{cm^{2}}$$

D
none

Solution

Given $$a=3\,\mathrm{cm},b=5\,\mathrm{cm},c=4\,\mathrm{cm}$$
Now $$s=\dfrac{a+b+c}{2}$$

Hence $$s=\dfrac{3+5+4}{2}=6$$

Now using Heron's Formula

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

$$\Delta=\sqrt{6(6-3)(6-5)(6-4)}$$

$$\Delta=6\,\mathrm{cm^2}$$

$$\therefore$$ Area of the triangle $$= 6 \mathrm {cm^2}$$

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Herons Formula Test - 23

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Herons Formula Test - 23

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SHARING IS CARING

Question 1

$$72\, cm^2$$

$$12\sqrt {35}\, cm^2$$

$$24\sqrt {35}\, cm^2$$

$$150\, cm^2$$

Solution

Question 2

$$0.0024 m^2$$

$$0.0026 m^2$$

$$0.003 m^2$$

$$0.0015 m^2$$

Solution

Question 3

$$\sqrt{420} cm^2$$

$$24 cm^2$$

$$48 cm^2$$

$$56 cm^2$$

Solution

Question 4

180 m$$^2$$

126 m$$^2$$

306 m$$^2$$

300 m$$^2$$

Solution

Question 5

$$72 \displaystyle cm^{2} $$

$$36\displaystyle cm^{2} $$

$$24 \displaystyle cm^{2} $$

$$48 \displaystyle cm^{2} $$

Solution

Question 6

$$6\, cm^2$$

$$7.5\, cm^2$$

$$5\sqrt 2\, cm^2$$

None of these

Solution

Question 7

$$924 \text{ cm}^{2} $$

$$1336 \text{ cm}^{2} $$

$$693 \text{ cm}^{2} $$

$$1848 \text{ cm}^{2} $$

Solution

Question 8

$$56 \sqrt{51}$$ $$cm^2$$

$$18 \sqrt{51}$$ $$cm^2$$

$$25 \sqrt{51}$$ $$cm^2$$

none of these

Solution

Question 9

$${ 576cm }^{ 2 }$$

$${ 30cm }^{ 2 }$$

$${ 26cm }^{ 2 }$$

$${ 24cm }^{ 2 }$$

Solution

Question 10

$$6 \ \displaystyle \mathrm{cm^{2}}$$

$$7.5 \ \displaystyle \mathrm{cm^2}$$

$$\displaystyle 5\sqrt{2} \,\mathrm{cm^{2}}$$

none

Solution

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