Herons Formula Test

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Herons Formula Test - 20

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Question 1
1 / -0

In figure, $$Ar(||gm ABCD)$$ is:

A
$$10 cm$$

B
$$20 cm$$

C
$$10\sqrt{3}cm^2$$

D
$$20\sqrt{3}cm^2$$

Solution

Area of parallelogram ABCD $$=2 \times area \ of \triangle ABC$$
Now,
$$S=\cfrac { a+b+c }{ 2 } =\cfrac { 5+7+8 }{ 2 } =10\ cm$$
Area of $$\triangle ABC=\sqrt { s(s-a)(s-b)(s-c) } $$
$$=\sqrt { 10(10-5)(10-7)(10-8) } \\ =\sqrt { 10\times 5\times 3\times 2 } \\ =\sqrt { 300 } \\ =\sqrt { 3\times 100 } \\ =10\sqrt { 3 } \quad { cm }^{ 2 }$$

Area of parallelogram ABCD$$=2\times \triangle ABC\\ =2\times 10\sqrt { 3 } \\ =20\sqrt { 3 } \ { cm }^{ 2 }$$
Question 2
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Two sides of a triangle are $$16 cm$$ and $$14 cm$$ and its semi-perimeter is $$18 cm$$. Then the third side of the triangle is:

A
$$12 cm$$

B
$$11 cm$$

C
$$10 cm$$

D
$$6 cm$$

Solution

Semiperimeter, $$S=18 \ cm$$

Let the side of the triangle be $$a,b,c$$ such that
$$a=16 \ cm ,\ b=14 \ cm \\$$
$$S=\cfrac { a+b+c }{ 2 } \\ =\cfrac { 16+14+c }{ 2 } \\ 18=\cfrac { 30+c }{ 2 } \\ c=36-30=6 \ cm$$

Therefore, third side of triangle is $$6 \ cm$$
Question 3
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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

When the sides of triangle are given we solve the area of the triangle using Heron's Formula which is
area of a triangle $$=\sqrt { s(s−a)(s−b)(s−c) } $$
where $$s$$ is the semi-perimeter and $$ s=\dfrac{(a+b+c)}{2}$$
Here $$a=b=c$$ ,
Therefore, the formula becomes :
$$Area\,=\,\sqrt { s(s−a)\times(s-a)\times(s-a) } $$

$$\Rightarrow Area\,=\,\sqrt{s(s-a)\times(s-a)^2}$$

$$\Rightarrow Area\,=\,(s-a)\sqrt{s(s-a)}$$

Hence option D is correct.
Question 4
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Heron's formunla 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
Question 5
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
Question 6
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
Question 7
1 / -0

The lengths of four sides and a diagonal of the given quadrilateral are indicated in the diagram. If $$A$$ denotes the area and $$l$$ the length of the other diagonal, then $$A$$ and $$l$$ are respectively

A
$$12\sqrt{6},4\sqrt{6}$$

B
$$12\sqrt{6},5\sqrt{6}$$

C
$$6\sqrt{6},4\sqrt{6}$$

D
$$6\sqrt{6},5\sqrt{6}$$

Solution

For $$\Delta ABC$$, a $$=$$ 6 cm, b $$=$$ 5, c $$=$$ 7 cm

$$\therefore s = \dfrac{6+5+7}{2} = 9$$ cm

$$\therefore$$ Area of $$ \Delta ABC = \sqrt{s(s-a)(s-b)(s-c)}$$

$$= \sqrt{9\times (9-6)(9-5)(9-7)}$$

$$= \sqrt{9\times 3\times 4\times 2}$$

$$= 3\times 2\sqrt{6} = 6\sqrt{6}$$

Thus, area of quadrilateral $$= 2\times$$ Area of $$\Delta ABC$$
$$= 12\sqrt{6}$$ sq cm
From congruency of triangles, it can be proved that $$AE=ED= \dfrac{1}{2}AD$$, and $$ AE \perp BC$$
$$ AE = \dfrac{1}{2}AD = \dfrac{l}{2}$$

Area of $$\Delta ABC = \dfrac{1}{2}\times BC\times AE$$

or $$6\sqrt{6} = \dfrac{1}{2}\times 6\times \dfrac{l}{2}$$

or $$l = 4\sqrt{6}$$
Question 8
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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
$$\sqrt{s(s-a)^3}$$

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

Solution
Question 9
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If the sides of a triangle are doubled, then its area

A
Remains the same

B
Becomes doubled

C
Becomes three times

D
Becomes four times

Solution

Using Heron's formula i.e.
$$A=\sqrt{s(s-a)(s-b)(s-c)}$$
where,
$$s=\displaystyle \frac{a+b+c}{2}$$

Let, $$a_1=2a,b_1=2b,c_1=2c$$
$$\Rightarrow s_1=2s$$
$$\therefore A_1=\sqrt{2s(2s-2a)(2s-2b)(2s-2c)}$$
taking 2 as common
$$\therefore A_1=\sqrt{16s(s-a)(s-b)(s-c)}$$
$$\therefore A_1=4A$$
Question 10
1 / -0

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

A
$$90 m^2$$

B
$$384 m^2$$

C
$$43 m^2$$

D
$$192 m^2$$

Solution

$$S=\dfrac {24+40+32}{2}=\dfrac {96}{2}=48 m$$

Area
$$=\sqrt {48\times (48-24)(48-40)(48-32)}$$

$$=\sqrt {48\times 24\times 8\times 16}$$

$$=\sqrt {24\times 2\times 24\times 8\times 16}$$

$$=24\times 16=384 m^2$$