Herons Formula Test

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Herons Formula Test - 21

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Question 1
1 / -0

If every side of a triangle is doubled, then the area of the new triangle is 'K' times the area of the old one. The value of K is

A
2

B
3

C
$$\sqrt{2}$$

D
4

Solution

By using Heron's formula,
$$A=\sqrt{s(s-a)(s-b)(s-c)}$$
Now,
$$a=2a$$, $$b=2b$$, $$c=2c$$
$$\Rightarrow s=2s$$
$$\therefore A_1=\sqrt{2s(2s-2a)(2s-2b)(2s-2c)}$$
$$A_1=\sqrt{16s(s-a)(s-b)(s-c)}$$
$$A_1=4\sqrt{s(s-a)(s-b)(s-c)}$$
$$A_1=4A$$
Question 2
1 / -0

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

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

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

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

D
$$56\ {cm}^2$$

Solution

The sides of triangle are $$4\ cm, 13\ cm, 15\ cm$$
Using Heron's formula,
Area of triangle $$=\sqrt{s(s-a)(s-b)(s-c)}$$
where $$s = \displaystyle \frac{a+b+c}{2}$$

$$\therefore s= \displaystyle \frac{4+13+15}{2} = 16\ cm$$
$$\therefore $$ Area $$=\sqrt{16 (16-4)(16-13)(16-15)}$$
$$\therefore$$ Area $$=\sqrt{16\times 12\times 3\times 1}$$
$$= 4\times6$$
$$= 24\ {cm}^2$$

Hence, option B.
Question 3
1 / -0

A quadrilateral $$ABCD$$ has $$\angle C =90,AB=9 cm,BC=8cm,CD=6cm$$ and $$AD=7 cm.$$ How much area does it occupy?

A
$$6 \sqrt {26} + 24$$

B
$$48$$

C
$$12 \sqrt {26}$$

D
none of these

Solution

It is given that $$∠C = 90º, AB = 9$$ cm, $$BC = 8$$ cm, $$CD = 6$$ cm and $$AD = 7$$ cm
$$BD$$ is joined.

In $$ΔBCD$$,
By applying Pythagoras theorem,

$$BD^2 = BC^2 + CD^2$$
$$⇒ BD^2 = 8^2 + 4^2$$
$$⇒ BD^2 = 64+16=80$$
$$BD=4\sqrt { 5 }$$ cm

Area of $$ΔBCD$$$$=\dfrac { 1 }{ 2 } \times 8\times 6=24$$ cm$$^2$$

Now, semi perimeter of $$ΔABD$$$$=\dfrac { 1 }{ 2 } (7+9+4\sqrt { 5 } )=\dfrac { 16+4\sqrt { 5 } }{ 2 } =8+2\sqrt { 5 }$$ cm$$^2$$

Using heron's formula,

Area of $$ΔABD$$ is:

$$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 8+2\sqrt { 5 } (8+2\sqrt { 5 } -4\sqrt { 5 } )(8+2\sqrt { 5 } -9)(8+2\sqrt { 5 } -7) }$$
$$=\sqrt { (8+2\sqrt { 5 } )(8-2\sqrt { 5 } )(2\sqrt { 5 } -1)(2\sqrt { 5 } +1) } =\sqrt { [\left( 8 \right) ^{ 2 }-\left( 2\sqrt { 5 } \right) ^{ 2 }][\left( 2\sqrt { 5 } \right) ^{ 2 }-\left( 1 \right) ^{ 2 }] } =\sqrt { [64-20][20-1] } =\sqrt { 44\times 19 }$$
$$=\sqrt { 936 } =6\sqrt { 26 }$$ cm$$^2$$

Area of quadrilateral $$ABCD$$ = Area of $$ΔBCD +$$ Area of $$ΔABD$$
= $$6\sqrt { 26 } +24$$ cm$$^2$$.
Question 4
1 / -0

The area of a triangle whose sides are $$13 cm, 14 cm$$ and $$15 cm$$ is:

A
$$42 cm^2$$

B
$$86 cm^2$$

C
$$84 cm^2$$

D
$$100 cm^2$$

Solution

Let $$a=13cm$$, $$b=14cm$$ and $$c=15cm$$
Now, $$s=\cfrac { a+b+c }{ 2 } $$
$$=\cfrac { 13+14+15 }{ 2 } $$
$$=21cm$$
Area:
$$\triangle =\sqrt { s(s-a)(s-b)(s-c) } $$
$$=\sqrt { 21(21-13)(21-14)(21-15) } $$
$$=\sqrt { 21\times 8\times 7\times 6 } $$
$$=\sqrt { 7\times 3\times 2\times 2\times 2\times 7\times 2\times 3 } $$
$$=2\times 2\times 3\times 7$$
$$=84c{ m }^{ 2 }$$
Question 5
1 / -0

The length of sides of a triangle are in the ratio.$$17:12:25$$ and its semi perimeter is $$270 cm$$. Find the area of the triangle.

A
$$9000 c$$$$m^2$$

B
$$6000 c$$$$m^2$$

C
$$7000 c$$$$m^2$$

D
$$8000 c$$$$m^2$$

Solution

Let the sides of the triangle be $$17x, 12x$$ and $$25x$$
Semiperimeter (s) $$=\displaystyle \frac{(17x + 12x + 25x)}{ 2}$$
$$= 27x$$
Given that,
$$ 27x = 270x$$
$$= 10$$
So, the sides of the triangle are $$170 cm$$, $$120 cm$$ and $$250 cm$$
Semiperimeter $$= 270 cm$$
Therefore,
Area of the triangle $$=\sqrt {s(s-a)(s-b)(s-c)}$$
$$=\sqrt{270(270-170)(270-120)(270-250)}$$
$$=\sqrt{270(100)(150)(20)}$$
$$=9000 $$ $$cm^2$$
Question 6
1 / -0

The lengths of sides of a triangle are in the ratio $$3:4:5$$ and its perimeter is $$144$$ cm, then find the height corresponding to largest side.

A
$$14.8$$ cm

B
$$28.8$$ cm

C
$$39$$ cm

D
$$16$$ cm

Solution

Given, Side of triangle are in ratio $$3 : 4 : 5$$ and their perimeter $$144 \ cm$$

Let the sides of triangle be $$3x,4x,5x$$

Perimeter $$3x+4x+5x=$$ 144 cm

$$12x=144$$

$$\therefore x=12$$

Then sides of triangle are $$3x$$$$=$$ 3$$\times$$12$$=$$ 36 cm,

$$4x$$$$=$$ 4$$\times$$12$$=$$ 48 cm,

$$5x$$$$=$$ 5$$\times$$12$$=$$ 60 cm.

Now,Semi perimeter, $$s=\dfrac { Sum\quad of\quad sides\quad of\quad triangle }{ 2 }$$

$$ \\ =\dfrac { 36+48+60 }{ 2 } =72\quad cm$$

Using heron's formula, Area of triangle$$=\sqrt { s(s-a)(s-b)(s-c) }$$

$$ \\ =\sqrt { 72(72-36)(72-48)(72-60) } $$

$$\\ =\sqrt { 72\times 36\times 24\times 12 } =864\quad { cm }^{ 2 }$$

Using altitude, area of triangle$$=\dfrac { 1 }{ 2 } \times$$ base $$\times$$ altitude $$=864\quad { cm }^{ 2 }$$

$$=\dfrac { 1 }{ 2 } \times 60\times $$ altitude $$=864$$

$$\therefore$$ altitude $$=\dfrac { 864\times 2 }{ 60 } =\quad 28.8\quad cm$$

So, altitude corresponding to largest side is 28.8 cm.
Question 7
1 / -0

One side of an equilateral triangle is 6 cm. Its area by using Heron's formula is

A
$$9$$ $${cm}^{2}$$

B
$$3$$ $${cm}^{2}$$

C
$$9\sqrt {3}$$ $${cm}^{2}$$

D
$$3\sqrt{8}$$ $${cm}^{2}$$

Solution

$$a = 6 cm$$, $$b = 6 cm$$, $$c = 6 cm$$
$$\therefore s = \displaystyle\frac{a + b + c}{2} = \displaystyle\frac{6 + 6 + 6}{2} = 9 cm$$
$$\therefore$$ Area of the equilateral triangle
$$= \sqrt{s\left(s - a\right) \left(s - b\right) \left(s - c\right)}$$
$$= \sqrt{9\left(9 - 6\right) \left(9 - 6\right) \left(9 - 6\right)}$$
$$= \sqrt{9\left(3\right) \left(3\right) \left(3\right)}$$
$$=9\sqrt{3} {cm}^{2}$$
Question 8
1 / -0

An umbrella is made by stitching $$12$$ triangular pieces of cloth. Each piece measures $$50$$ cm, $$20$$ cm and $$50$$ cm. Find the area of cloth used in it.

A
$$2400\sqrt6$$ c$$m^2$$

B
$$2200\sqrt3$$ c$$m^2$$

C
$$1200\sqrt6$$ c$$m^2$$

D
$$1000\sqrt6$$ c$$m^2$$

Solution

Given, an umbrella is made of 12 triangular pieces each of measure 50 cm, 20 cm, 50 cm.
To find area of cloth required we have to first find the area of one piece of cloth.

Semi perimeter, s=$$=\frac { Sum\quad of\quad sides\quad of\quad triangle }{ 2 } \\ =\frac { 50+20+50 }{ 2 } =60\quad cm$$
Using heron's formula, Area of one triangular piece$$=\sqrt { s(s-a)(s-b)(s-c) } \\ =\sqrt { 60(60-50)(60-20)(60-50) } \\ =\sqrt { 60\times 10\times 40\times 10 } =200\sqrt { 6 } { cm }^{ 2 }$$
Area of 12 triangular pieces or area of cloth used$$=$$ Area of one triangular piece $$\times$$ 12
$$=200\sqrt { 6 } { cm }^{ 2 }\times 12\\ =2400\sqrt { 6 } { cm }^{ 2 }$$
Question 9
1 / -0

In a triangular field having sides 30 m, 72 m and 78 m , the length of the altitude to the side measuring 72 m is

A
25 m

B
28 m

C
30 m

D
35 m

Solution

$$\displaystyle s=\frac{30+72+78}{2}=\frac{180}{2}=90$$
$$\displaystyle \therefore $$ Area of $$\displaystyle \Delta = \sqrt{90(90-30)(90-72)(90-78)m^{2}}$$
$$\displaystyle =\sqrt{90\times 60\times 18\times 12m^{2}}=1080m^{2}$$
Area of the $$\displaystyle \Delta $$ with bae ( 72 m)
$$\displaystyle =\frac{1}{2}\times 72\times altitude\, =1080$$
$$\displaystyle \therefore $$ Required length of altitude =$$\displaystyle \frac{1080\times 2}{72}=30m$$
Question 10
1 / -0

The sides of a triangle are in the ratio of 13:14:15 and its perimeter is 84 cm. Then the area of the triangle is

A
$$136$$ $$cm^2$$

B
$$236$$ $$cm^2$$

C
$$336$$ $$cm^2$$

D
$$436$$ $$cm^2$$

Solution

Let the sides of the triangle be $$ 13a, 14a, 15a $$.
Perimeter of the triangle $$ = $$ Sum of all sides $$ = 13a + 14a + 15a = 42a$$
Given, perimeter of the triangle $$ = 84 cm $$
$$ \therefore 42a = 84 cm $$
$$ a = 2 cm $$
So, the sides of the triangle are $$ 13a = 26 cm, 14a = 28 cm, 15a = 30 cm $$.
Area of triangle with sides a, b, and c and $$s = \displaystyle \frac { a+b+c }{ 2 } $$ is $$ \sqrt { s(s-a)(s-b)(s-c) } $$.
For triangle with sides 26 cm, 28 cm and 30 cm, $$ s = \displaystyle \frac { 26+28+30 }{ 2 }= 42 cm $$
Area of the triangle with 26 cm, 28 cm and 30 cm $$ =\sqrt { 42(42-26)(42-28)(42-30) }$$
$$ =\sqrt { 42\times 16\times 14\times 12 } $$
$$=\sqrt { 6\times 7\times 4\times 4\times 7\times 2 \times 6 \times 2} $$
$$= 6\times 7\times 4 \times 2 = 336 {cm}^{2}$$

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Herons Formula Test - 21

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Herons Formula Test - 21

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SHARING IS CARING

Question 1

2

3

$$\sqrt{2}$$

4

Solution

Question 2

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

$$24\ {cm}^2$$

$$48\ {cm}^2$$

$$56\ {cm}^2$$

Solution

Question 3

$$6 \sqrt {26} + 24$$

$$48$$

$$12 \sqrt {26}$$

none of these

Solution

Question 4

$$42 cm^2$$

$$86 cm^2$$

$$84 cm^2$$

$$100 cm^2$$

Solution

Question 5

$$9000 c$$$$m^2$$

$$6000 c$$$$m^2$$

$$7000 c$$$$m^2$$

$$8000 c$$$$m^2$$

Solution

Question 6

$$14.8$$ cm

$$28.8$$ cm

$$39$$ cm

$$16$$ cm

Solution

Question 7

$$9$$ $${cm}^{2}$$

$$3$$ $${cm}^{2}$$

$$9\sqrt {3}$$ $${cm}^{2}$$

$$3\sqrt{8}$$ $${cm}^{2}$$

Solution

Question 8

$$2400\sqrt6$$ c$$m^2$$

$$2200\sqrt3$$ c$$m^2$$

$$1200\sqrt6$$ c$$m^2$$

$$1000\sqrt6$$ c$$m^2$$

Solution

Question 9

25 m

28 m

30 m

35 m

Solution

Question 10

$$136$$ $$cm^2$$

$$236$$ $$cm^2$$

$$336$$ $$cm^2$$

$$436$$ $$cm^2$$

Solution

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