Herons Formula Test

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Herons Formula Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

The area of an equilateral triangle with side $$2 \sqrt{3}$$ cm is

A
$$5.196 cm^2$$

B
$$0.866 cm^2$$

C
$$3.496 cm^2$$

D
$$1.732 cm^2$$

Solution

Area of the triangle$$=\dfrac{\sqrt{3}(2\sqrt{3})^2}{4}$$
$$=3\sqrt{3}=3(1.732)=5.196cm^2$$
Question 2
1 / -0

The cost of levelling the ground in the form of a triangle having the sides 51 m, 37 m and 20 m at the rate of Rs $$3$$ per $$m^2$$ is

A
Rs. 918

B
Rs. 818

C
Rs. 718

D
Rs. 618

Solution

$$s=\dfrac{51+37+20}{2}=\dfrac{108}{2}=54$$
Area of the triangle$$=\sqrt{54(54-51)(54-37)(54-20)}$$
$$=\sqrt{54(3)(17)(34)}=\sqrt{93636}$$
Area $$=$$ $$306 m^2$$.
Cost of levelling at $$Rs. 3$$ per $$m^2=306(3)= $$Rs. $$918$$
Question 3
1 / -0

In a triangle, the sides are given as 11 cm, 12 cm and 13 cm. The length of the altitude corresponding to the side having length 12 cm is

A
10.25 cm

B
11.25 cm

C
12.25 cm

D
9.25 cm

Solution

Sides of triangle are $$11,12,13$$
Area of triangle = $$\sqrt{s(s-a)(s-b)(s-c)}$$
$$s = \dfrac{a + b+ c}{2}$$
$$s = \dfrac{11 + 12 + 13}{2}$$
$$s = 18$$
Area of triangle = $$\sqrt{18(7) (6) (5)}$$
Area of triangle = $$\sqrt{18(7) (6) (5)} = 61.5$$

Area of triangle = $$\dfrac{1}{2} (base) (altitude)$$
61.5 = $$\dfrac{1}{2} \times 12 \times (altitude)$$
Altitude = $$10.25$$
Question 4
1 / -0

The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is

A
$$\sqrt{15} cm^2$$

B
$$\sqrt{\dfrac{15}{2}}cm^2$$

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

D
$$4 \sqrt{15}cm^2$$

Solution

$$s=\dfrac{4+4+2}{2}=5$$

Area of the triangle
$$\Delta =\sqrt{s(s-a)(s-b)(s-c)}\\=\sqrt{5(5-4)(5-4)(5-2)}=\sqrt{15}cm^2$$
Question 5
1 / -0

Find the cost of laying grass in a triangular field of sides $$50\>m, 65\>m$$ and $$65\> m$$ at the rate of Rs $$7$$ per $$m^2$$.

A
Rs $$10500$$

B
Rs $$11375$$

C
Rs $$10000$$

D
Rs $$21000$$

Solution

Sides of the triangle are $$a=50\> m, b=65\> m, c=65\> m$$
Area of triangle, by Heron's formula $$=\sqrt{s(s-a)(s-b)(s-c)}$$
where,
$$s= \dfrac{a+b+c}{2}$$
$$s = \dfrac{50 + 65 + 65}{2}$$
$$s = 90$$
Area of triangle = $$\sqrt{90(40) (25)(25)}$$
Area of triangle = $$1500\>m^2$$

Cost of laying grass $$=$$ Area $$\times 7$$
Cost of laying grass $$=1500 \times 7$$
Cost of laying grass $$=$$ Rs $$10500 $$
Question 6
1 / -0

Find the area of a triangle whose two sides are $$18$$ cm and $$10$$ cm and the perimeter is $$42$$ cm.

A
$$83.67$$ cm$$^2$$

B
$$46.12$$ cm $$^2$$

C
$$77.15$$ cm$$^2$$

D
$$69.69$$ cm$$^2$$

Solution

Let $$a$$, $$b$$ and $$c$$ be the sides of triangle.
Perimeter $$ =a+b+c $$$$=42 \text{ cm}$$
$$a=18$$, $$b=10$$
As $$a+b+c=42$$
$$\therefore$$ $$c=42-b-a = 42 - 18-10$$
$$\therefore c=14$$
Area of triangle $$ = \sqrt{s (s -a) (s -b) ( s -c)}$$
where, $$ s=\dfrac{a+b+c}{2}=21 \text{ cm}$$
$$\therefore $$ Area of triangle $$=\sqrt {21(21-18)(21-10)(21-14)}$$
$$\therefore$$ Area of triangle $$=69.69 \text{ cm}^2$$
Question 7
1 / -0

The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then the area of the triangle is

A
$$1322 cm^2$$

B
$$1311 cm^2$$

C
$$1344 cm^2$$

D
$$1392 cm^2$$

Solution

$$s=\dfrac{a+b+c}{2}$$
By Heron's formula,Area of the triangle=$$\sqrt{s(s-a)(s-b)(s-c)}$$
Where a,b,c are sides of the triangle.
$$s=\dfrac{56+60+52}{2}=84$$
Area of the triangle$$=\sqrt{84(84-56)(84-60)(84-52)}=\sqrt{1806336}=1344cm^2$$
Question 8
1 / -0

The area of an isosceles triangle whose base is $$'b'$$ and equal sides are of length $$'a'$$ is:

A
$$\dfrac{a}{4}\sqrt{4b^2-a^2}$$

B
$$\dfrac{b}{4}\sqrt{4a^2-b^2}$$

C
$$\dfrac{b}{2}\sqrt{4b^2-a^2}$$

D
$$\dfrac{a}{2}\sqrt{4a^2-b^2}$$

Solution

In an isosceles triangle, the altitude bisects the base.
$$Altitude=\sqrt { (side\quad length\quad of\quad equal\quad sides)^{ 2 }+\left( \cfrac { base }{ 2 } \right) ^{ 2 } } \\ =\sqrt { { a }^{ 2 }-\cfrac { { b }^{ 2 } }{ 4 } } $$
Area of isosceles triangle :
$$=\cfrac { 1 }{ 2 } \times altitude\times base\\ =\cfrac { 1 }{ 2 } \times \sqrt { { a }^{ 2 }-\cfrac { { b }^{ 2 } }{ 4 } } \times b\\ =\cfrac { 1 }{ 4 } \times b\sqrt { 4a^{ 2 }-b^{ 2 } } $$
Question 9
1 / -0

The sides of a triangular plot are in the ratio $$4 : 5 : 6$$ and its perimeter is $$150\ \mathrm{cm}$$. Then the sides are:

A
$$4\ \mathrm{cm}, 5\ \mathrm{cm}, 6\ \mathrm{cm}$$

B
$$40\ \mathrm{ cm}, 50\ \mathrm{ cm}, 60\ \mathrm{ cm}$$

C
$$8\ \mathrm{cm}, 10\ \mathrm{cm}, 12\ \mathrm{cm}$$

D
$$120\ \mathrm{cm}, 150\ \mathrm{cm}, 180\ \mathrm{cm}$$

Solution

Let the sides of the triangle be $$4x,5x,6x$$, respectively.
Perimeter of triangle $$=4x+5x+6x$$
$$4x+5x+6x = 150\ \mathrm{cm}$$
$$15x = 150\ \mathrm{cm}$$
$$\Rightarrow x=10\ \mathrm{cm}$$
$$4x = 4\times 10 \ \mathrm{cm} = 40 \ \mathrm{cm} $$
$$5x = 5\times 10 \ \mathrm{cm} = 50 \ \mathrm{cm} $$
$$6x = 6\times 10 \ \mathrm{cm} = 60 \ \mathrm{cm} $$
Therefore, the sides are $$40\ \mathrm{cm}, \ 50 \ \mathrm{cm}, $$ and $$60 \ \mathrm{cm} $$.
Question 10
1 / -0

Two sides of a triangle are $$13 cm$$ and $$14 cm$$ and its semi perimeter is $$18 cm$$. Then third side of the triangle is :

A
$$12 cm$$

B
$$11 cm$$

C
$$10 cm$$

D
$$9 cm$$

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Herons Formula Test - 19

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Herons Formula Test - 19

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SHARING IS CARING

Question 1

$$5.196 cm^2$$

$$0.866 cm^2$$

$$3.496 cm^2$$

$$1.732 cm^2$$

Solution

Question 2

Rs. 918

Rs. 818

Rs. 718

Rs. 618

Solution

Question 3

10.25 cm

11.25 cm

12.25 cm

9.25 cm

Solution

Question 4

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

$$\sqrt{\dfrac{15}{2}}cm^2$$

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

$$4 \sqrt{15}cm^2$$

Solution

Question 5

Rs $$10500$$

Rs $$11375$$

Rs $$10000$$

Rs $$21000$$

Solution

Question 6

$$83.67$$ cm$$^2$$

$$46.12$$ cm $$^2$$

$$77.15$$ cm$$^2$$

$$69.69$$ cm$$^2$$

Solution

Question 7

$$1322 cm^2$$

$$1311 cm^2$$

$$1344 cm^2$$

$$1392 cm^2$$

Solution

Question 8

$$\dfrac{a}{4}\sqrt{4b^2-a^2}$$

$$\dfrac{b}{4}\sqrt{4a^2-b^2}$$

$$\dfrac{b}{2}\sqrt{4b^2-a^2}$$

$$\dfrac{a}{2}\sqrt{4a^2-b^2}$$

Solution

Question 9

$$4\ \mathrm{cm}, 5\ \mathrm{cm}, 6\ \mathrm{cm}$$

$$40\ \mathrm{ cm}, 50\ \mathrm{ cm}, 60\ \mathrm{ cm}$$

$$8\ \mathrm{cm}, 10\ \mathrm{cm}, 12\ \mathrm{cm}$$

$$120\ \mathrm{cm}, 150\ \mathrm{cm}, 180\ \mathrm{cm}$$

Solution

Question 10

$$12 cm$$

$$11 cm$$

$$10 cm$$

$$9 cm$$

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