Herons Formula Test

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Herons Formula Test - 18

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Question 1
1 / -0

Area of an equilateral triangle of side $$a$$ units can be calculated by using the formula :

A
$$\sqrt{s^{2}(s-a)^{2}}$$

B
$$(s-a)\sqrt{s^{2}(s-a)}$$

C
$$\sqrt{s(s-a)^{2}}$$

D
$$(s-a)\sqrt{s(s-a)}$$

Solution

We know that for an equilateral triangle, $$a=b=c$$

By Heron's formula, we get $$\triangle=\sqrt{s(s-a)(s-b)(s-c)}$$

So, area of given equilateral triangle $$=\sqrt { s\left( s-a \right) \left( s-a \right) \left( s-a \right) }$$

$$ =\left( s-a \right) \sqrt { s\left( s-a \right) } $$
Question 2
1 / -0

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

Using Heron's formula i.e.
$$A=\sqrt{s(s-a)(s-b)(s-c)}$$
where,
$$a=5 $$ cm
$$b=12$$ cm
$$c=13$$ cm
$$s=\displaystyle \frac{a+b+c}{2}=\frac {5+12+13}{2}=15$$cm
$$A=\sqrt {15(15-5)(15-12)(15-13)}$$
$$A=\sqrt{15\times 10\times 3\times 2}$$
$$\therefore A=30$$ $$cm^2$$
$$\therefore A=0.003$$ $$m^2$$
Question 3
1 / -0

Area of traingle ABC whose sides are 24m, 40m and 32m is-

A
$$96{m}^{2}$$

B
$$384{m}^{2}$$

C
$$43{m}^{2}$$

D
$$192{m}^{2}$$

Solution

Area of triangle with sides $$a$$, $$b$$ and $$c$$ and $$s$$ $$ = \dfrac { a+b+c }{ 2 } $$ is $$ \sqrt {s(s-a)(s-b)(s-c) } \ $$.
For triangle with sides 24 m, 40 m and 32 m, $$ s = \dfrac { 24+40+32 }{ 2 }=48 m\ $$

Area of the triangle with sides 24 m, 40 m and 32 m $$ =\sqrt {48(48-24)(48-40)(48-32) } =\sqrt { 48\times 24\times 8\times 16 } =\sqrt { 24\times 2\times 24\times 8\times 8\times 2 } = 24\times 8\times 2 = 384 {m}^{2}$$
Hence, option 'B' is correct.
Question 4
1 / -0

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

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

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

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

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

Solution

Given side of triangle is 4 cm , 13 cm and 15 cm
We know that area of triangle=$$\sqrt{s(s-a)(s-b)(s-c)}$$
where $$s=\frac{a+b+c}{2} and a, b, c are three sides
$$s=\frac{4+13+15}{2}= 16$$
Then area of triangle=$$\sqrt{16(16-4)(16-13)(16-15)}=\sqrt{16\times 12\times 3\times 1}=\sqrt{576}=24 cm^{2}$$
Question 5
1 / -0

The sides of a triangle are $$4, 5$$ and $$6$$ cm. The area of the triangle is equal to

A
$$\displaystyle \frac{15}{4} cm^2$$

B
$$\displaystyle \frac{15}{4}\sqrt 7 cm^2$$

C
$$\displaystyle \frac{4}{15} \sqrt 7 cm^2$$

D
None of these

Solution

Given that, a = 4 cm, b = 5 cm, and c = 6 cm
We know $$\displaystyle s = \frac{a + b + c}{2} = \frac{4+ 5 + 6}{2} = \frac{15}{2}$$
$$\therefore$$ Area of the triangle $$= \sqrt{s(s - a) (s - b) (s - c)}$$
$$\displaystyle = \sqrt{\left ( \frac{15}{2} \right ) \left ( \frac{15}{2}-4 \right ) \left ( \frac{15}{2} - 5 \right ) \left ( \frac{15}{2} - 6 \right )}$$
$$= \displaystyle \sqrt{\frac{15}{2} \times \frac{7}{2} \times \frac{5}{2} \times \frac{3}{2}} = \frac{15}{4} \sqrt 7 cm^2$$
Question 6
1 / -0

Heron's formula is :

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

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

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

D
$$\Delta = \sqrt{s(s-a)(s-b)(s-c)}$$, $$2s = a+b+c$$

Solution

When the sides of a triangle are given, we solve the area of the triangle using Heron's Formula which is

Area of a triangle $$=\Delta$$ $$=\sqrt { s(s−a)(s−b)(s−c) } $$ where s is semi-perimeter which equals $$ s=\dfrac{(a+b+c)}{2}$$

Thus, the Heron's formula is $$\Delta=\sqrt { s(s−a)(s−b)(s−c) } $$, $$ 2s={(a+b+c)}$$
Question 7
1 / -0

The sides of a triangle are $$3$$ cm, $$4$$ cm and $$5$$ cm. Its area is .......

A
$$12 cm^2$$

B
$$15 cm^2$$

C
$$20 cm^2$$

D
$$6 cm^2$$

Solution

By Heron's formula,
Area of the triangle $$= \sqrt{s(s - a)(s - b)(s -c )}$$

We know, $$s = \displaystyle \frac{3 + 4 + 5}{2} = 6$$

$$\therefore $$ Area of triangle $$ = \sqrt{6(6 - 3)(6 - 4)(6 - 5)}$$

$$ = \sqrt{6 \times 3 \times 2 \times1} = 6 cm^2$$
Question 8
1 / -0

Heron's formula is :

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

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

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

D
$$\Delta =\sqrt{s(s-a)(s-b)(s-c)}, 2s=a+b+c$$

Solution

$${\textbf{Step -1: A triangle having side lengths as }}\mathbf{a,}{\textbf{ }}\mathbf b{\textbf{ and }}\mathbf c.{\text{ }}$$
$${\text{Area of triangle }} = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} $$
$${\text{Where, s is the semi perimeter of the triangle;}}$$
$${\text{that is, }}s = \dfrac{{a + b + c}}{2}$$
$${\text{Thus, the Heron's formula is}}$$ $$\vartriangle = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} $$ $${\text{Where, }}2s = a + b + c$$
$${\textbf{Hence, option D is correct. }}$$
Question 9
1 / -0

The sides of a $$\Delta$$ are $$7 cm, 24 cm$$ and $$25 cm$$. Find its area.

A
$$168 cm^2$$

B
$$84 cm^2$$

C
$$87.5 cm^2$$

D
$$300 cm^2$$

Solution

Let $$a=7$$cm $$b=24$$cm $$c=25$$cm
$$S=\cfrac { a+b+c }{ 2 } \\ =\cfrac { 7+24+25 }{ 2 } \\ =28\ cm$$
Now, area of $$\triangle =\sqrt { S(S-a)(S-b)(S-c) } $$
$$=\sqrt { 28(28-7)(28-24)(28-25) } \\ =\sqrt { 24\times 24\times 4\times 3 } \\ =\sqrt { 4\times 7\times 7\times 3\times 4\times 3 } \\ =4\times 7\times 3\\ =84\ { cm }^{ 2 }$$
Question 10
1 / -0

If s is the semi-perimeter of a $$\Delta A B C$$ whose sides are a,b,c then s=.............?

A
$$a + b + c$$

B
$$\dfrac { a + b + c } { 2 }$$

C
$$\dfrac { a + b + c } { 3 }$$

D
$$\dfrac { a + b + c } { 4 }$$

Solution

semiperimeter of a triangle is given as half of the perimeter which means half of the sum of the lengths of all three sides
Hence, in given question $$\triangle ABC$$
semiperimeter $$=s=\cfrac{a+b+c}{2}$$

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Re-Attempt Weekly Quiz Competition

Herons Formula Test - 18

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Herons Formula Test - 18

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SHARING IS CARING

Question 1

$$\sqrt{s^{2}(s-a)^{2}}$$

$$(s-a)\sqrt{s^{2}(s-a)}$$

$$\sqrt{s(s-a)^{2}}$$

$$(s-a)\sqrt{s(s-a)}$$

Solution

Question 2

$$0.0024 m^2$$

$$0.0026 m^2$$

$$0.003 m^2$$

$$0.0015 m^2$$

Solution

Question 3

$$96{m}^{2}$$

$$384{m}^{2}$$

$$43{m}^{2}$$

$$192{m}^{2}$$

Solution

Question 4

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

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

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

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

Solution

Question 5

$$\displaystyle \frac{15}{4} cm^2$$

$$\displaystyle \frac{15}{4}\sqrt 7 cm^2$$

$$\displaystyle \frac{4}{15} \sqrt 7 cm^2$$

None of these

Solution

Question 6

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

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

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

$$\Delta = \sqrt{s(s-a)(s-b)(s-c)}$$, $$2s = a+b+c$$

Solution

Question 7

$$12 cm^2$$

$$15 cm^2$$

$$20 cm^2$$

$$6 cm^2$$

Solution

Question 8

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

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

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

$$\Delta =\sqrt{s(s-a)(s-b)(s-c)}, 2s=a+b+c$$

Solution

Question 9

$$168 cm^2$$

$$84 cm^2$$

$$87.5 cm^2$$

$$300 cm^2$$

Solution

Question 10

$$a + b + c$$

$$\dfrac { a + b + c } { 2 }$$

$$\dfrac { a + b + c } { 3 }$$

$$\dfrac { a + b + c } { 4 }$$

Solution

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