Herons Formula Test

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Herons Formula Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

Find the area of triangle of sides $$6cm ,8cm ,10cm$$

A
$$24cm^2$$

B
$$12cm^2$$

C
$$30cm^2$$

D
$$none$$

Solution

The sides are $$6cm,8cm,10cm$$
$$s=\dfrac {6+8+10}2=\dfrac {24}2=12$$
The area is given as $$\sqrt {s(s-a)(s-b)(s-c)}\\\sqrt {12(6)(4)(2)}\\\sqrt {576}=24cm^2$$
Question 2
1 / -0

Find the area of a triangle whose sides are respectively $$150\ cm$$, $$120\ cm$$, and $$200\ cm.$$

A
$$8545.77\text{ sq. cm}$$

B
$$8966.57\text{ sq. cm}$$

C
$$9257.66\text{ sq. cm}$$

D
$$5627.66 \text{ sq. cm}$$

Solution

Let $$a=150\ cm,\ b=120\ cm$$ and $$c=200\ cm.$$
$$s=\dfrac{a+b+c}{2}$$
$$\Rightarrow s=\dfrac{150+120+200}{2}$$
$$\Rightarrow s=235\ cm$$

Now, by using the heron's formula,
Area $$=\sqrt{s(s-a)(s-b)(s-c)}$$

$$=\sqrt{235(235-150)(235-120)(235-200)}$$

$$=\sqrt{235(85)(115)(35)}$$

$$=\sqrt{80399375}$$

$$=8966.57\text{ sq. cm}$$
Question 3
1 / -0

If every side of a triangle is doubled, the area of the new triangle is $$K$$ times the area of the old one. Find the value of $$K$$.

A
$$\sqrt{2}$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Let, the sides of the original triangle be $$a,b$$ and $$c$$
so, the semi-perimeter, $$s=\dfrac{a+b+c}{2}$$
Original area $$=\sqrt{s(s-a)(s-b)(s-c)}$$

Now every side is doubled
So, the new sides are $$2a, 2b$$ and $$ 2c$$
new semi-perimeter, $$s'=\dfrac{2a+2b+2c}{2}=2\left(\dfrac{a+b+c}{2}\right)=2s$$

New area $$=\sqrt{s'(s'-2a)(s'-2b)(s'-2c)}$$
$$=\sqrt{(2s)(2s-2a)(2s-2b)(2s-2c)}$$
$$=\sqrt{2.2.2.2s(s-a)(s-b)(s-c)}$$
$$=4\sqrt{s(s-a)(s-b)(s-c)}$$
$$=4\times Original\ Area$$

Hence, the value of $$K$$ is $$4$$.
Question 4
1 / -0

If the edges of a triangular board are $$15$$ cm, $$36$$cm and $$39$$cm, then the cost of painting its one of the faces at the rate of $$5$$ paise per cm$$^2$$ is

A
Rs.$$1212$$

B
Rs.$$13.5$$

C
Rs.$$1350$$

D
Rs$$11.6$$

Solution

Semi-perimeter, $$s=\dfrac{15+36+39}{2}$$
$$=45$$
From Heron's formula
$$\Delta =\sqrt{s(s-a)(s-b)(s-c)}$$
$$=\sqrt{45\times 30\times 9\times 6}$$
$$=9\times 5\times 6=30\times 9=270$$cm

given that cost is 5 paisa per $$cm^2$$
so, for 270 $$cm^2$$ cost in rs will be:
Cost$$=\dfrac{270\times 5}{100}$$Rs. $$=\dfrac{1350}{100}$$
$$\Rightarrow (B)$$.
Question 5
1 / -0

In the given figure, the area of the $$\triangle {ABC}$$ is

A
$$13.24{cm}^{2}$$

B
$$12.28{cm}^{2}$$

C
$$11.32{cm}^{2}$$

D
$$15.37{cm}^{2}$$

Solution

We know that the area of the triangle using heron's formula is given by
Area $$= \sqrt{s(s-a)(s-b)(s-c)} $$

So,
$$ S = \dfrac{8+4+11}{2} = \dfrac{23}{2} = 11.5 $$

Area $$ = \sqrt{11.5(11.5-11)(11.5-8)(11.5-4)}$$
$$ =\sqrt{11.5\times 0.5\times 3.5\times 7.5}$$
$$ = 12.28\text{ cm}^{2}$$
Question 6
1 / -0

If the sides of a triangle are $$\dfrac{y}{z} + \dfrac{z}{x},\dfrac{z}{x} + \dfrac{x}{y}$$ and $$\dfrac{x}{y} + \dfrac{y}{z}$$ then its area is

A
xyz

B
$$\sqrt {\dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x}} $$

C
$$\sqrt {xyz} $$

D
$$\dfrac{{{x^2}{y^2}{z^2}}}{2}$$
Question 7
1 / -0

Three sides of a triangular field are 20m, 21m, and 29m long, respectively. The area of the field is :

A
$$215{ m }^{ 2 }$$

B
$$230{ m }^{ 2 }$$

C
$$210{ m }^{ 2 }$$

D
None of these

Solution

Given: Sides of the triangular field; $$a=20m, b=21m, c=39m$$
Now, Semiperimeter $$S = \dfrac{a+b+c}{2} = \dfrac{20+21+29}{2} = 35m$$
$$Area = \sqrt{s\times (s-a)\times (s-b)\times (s-c)}$$
$$= \sqrt{35\times (35-20)\times (35-21)\times (35-29)}$$
$$= \sqrt{35\times 15\times 14\times 6}$$
$$=210 sq.m$$
Question 8
1 / -0

The area of a triangle whose sides are 15 m, 16 m and 17 m is :

A
$$24\sqrt { 4 } { m }^{ 2 }$$

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

C
$$24\sqrt { 21 } { m }^{ 2 }$$

D
None of these

Solution

The sides of the triangle are $$a = 15 \text{ m }, b = 16 \text{ m }, c = 17 \text{ m }$$

Area of the triangle is given by
$$S = \cfrac {a + b + c}{2} $$
$$= \cfrac {15 + 16 + 17}{2} $$
$$= \cfrac {48}{2} $$
$$= 24 \text{ m}$$

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

$$= \sqrt {24 (25 - 15) (24 - 16) (24 - 17)}$$

$$= \sqrt {24\times 9\times 8\times 7}$$

$$= \sqrt {\underline {8}\times \underline {3\times 3}\times 3\times \underline {8}\times 7} $$

$$= 24\sqrt {21} \text{ sq. m.}$$
Question 9
1 / -0

The sides of a triangle are $$7$$ cm, $$9$$ cm and $$
14$$ cm. Its area is

A
$$12\sqrt 5 \ \text{cm}^2$$

B
$$12\sqrt 3 \ \text{cm}^2$$

C
$$24\sqrt 5 \ \text{cm}^2$$

D
$$63 \ \text{cm}^2$$

Solution

Area of triangle = $$\sqrt{S(S-a)(S-b)(S-c)}$$
$$S=\dfrac{a+b+c}{2}$$

Given $$a= 7$$ cm
$$b= 9$$ cm
$$c= 14$$ cm

$$S= \dfrac{7+9+14}{2}$$
$$=\dfrac{30}{2}$$

$$S= 15 cm$$

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

$$=\sqrt{15(15-7)(15-9)(15-14)}\\$$
$$=\sqrt{15.8.6.1}\\$$
$$=\sqrt{720}\\$$
$$=\sqrt{5\times 144}\\$$
$$=\sqrt{12\times 12\times 5}\\$$
$$=12\sqrt{5}cm^{2}\\$$
$$\therefore $$ Area of triangle = $$12.\sqrt{5}\ cm^{2}$$

So, the answer is A. $$12.\sqrt{5}\ cm^{2}$$
Question 10
1 / -0

The sides of a triangle are in the ratio $$3:5:7.$$ If the perimeter of the triangle is $$60$$ cm, then its area

A
$$60\sqrt 3 \ \text{cm}^2$$

B
$$30\sqrt 3\ \text{cm}^2$$

C
$$15\sqrt 3 \ \text{cm}^2$$

D
$$120\ \text{cm}^2$$

Solution

Given sides in the ratio $$3:5:7$$

perimeter of the triangle $$=60cm$$

$$3x+5x+7x=60$$

$$\Rightarrow 15x=60$$

$$\Rightarrow x=60/15$$

$$\Rightarrow x=4$$

$$3x=3\times 4=12$$cm

$$5x=5\times 4=20$$cm

$$7x=7\times 4=28$$cm

Let $$S$$ be the semiperimeter of given triangle.

$$S=\dfrac{1}{2}(a+b+c)$$

$$=\dfrac{1}{2}\cdot 60$$

$$=30$$

Now use formula foe area of triangle

Area $$=\sqrt{30(30-12)(30-20)(30-28)}$$

$$=\sqrt{30\cdot18\cdot10\cdot2}$$

$$=\sqrt{60\cdot180}$$

$$=\sqrt{10800}$$

$$=\sqrt{3600\cdot3}$$

$$=60\sqrt{3}cm^2$$

$$\therefore$$ Area of triangle $$=60\sqrt{3}cm^2$$.

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Herons Formula Test - 28

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Herons Formula Test - 28

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SHARING IS CARING

Question 1

$$24cm^2$$

$$12cm^2$$

$$30cm^2$$

$$none$$

Solution

Question 2

$$8545.77\text{ sq. cm}$$

$$8966.57\text{ sq. cm}$$

$$9257.66\text{ sq. cm}$$

$$5627.66 \text{ sq. cm}$$

Solution

Question 3

$$\sqrt{2}$$

$$2$$

$$3$$

$$4$$

Solution

Question 4

Rs.$$1212$$

Rs.$$13.5$$

Rs.$$1350$$

Rs$$11.6$$

Solution

Question 5

$$13.24{cm}^{2}$$

$$12.28{cm}^{2}$$

$$11.32{cm}^{2}$$

$$15.37{cm}^{2}$$

Solution

Question 6

xyz

$$\sqrt {\dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x}} $$

$$\sqrt {xyz} $$

$$\dfrac{{{x^2}{y^2}{z^2}}}{2}$$

Question 7

$$215{ m }^{ 2 }$$

$$230{ m }^{ 2 }$$

$$210{ m }^{ 2 }$$

None of these

Solution

Question 8

$$24\sqrt { 4 } { m }^{ 2 }$$

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

$$24\sqrt { 21 } { m }^{ 2 }$$

None of these

Solution

Question 9

$$12\sqrt 5 \ \text{cm}^2$$

$$12\sqrt 3 \ \text{cm}^2$$

$$24\sqrt 5 \ \text{cm}^2$$

$$63 \ \text{cm}^2$$

Solution

Question 10

$$60\sqrt 3 \ \text{cm}^2$$

$$30\sqrt 3\ \text{cm}^2$$

$$15\sqrt 3 \ \text{cm}^2$$

$$120\ \text{cm}^2$$

Solution

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