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CBSE Questions for Class 9 Maths Heron'S Formula Quiz 3 - MCQExams.com
CBSE
Class 9 Maths
Heron'S Formula
Quiz 3
What is the area of a triangle whose sides are $$3\ \mathrm{cm}$$ , $$5\,\mathrm{ cm}$$ and $$4\,\mathrm{ cm}$$?
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$$6 \ \displaystyle \mathrm{cm^{2}}$$
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$$7.5 \ \displaystyle \mathrm{cm^2}$$
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$$\displaystyle 5\sqrt{2} \,\mathrm{cm^{2}}$$
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none
Explanation
Given $$a=3\,\mathrm{cm},b=5\,\mathrm{cm},c=4\,\mathrm{cm}$$
Now $$s=\dfrac{a+b+c}{2}$$
Hence $$s=\dfrac{3+5+4}{2}=6$$
Now using Heron's Formula
$$\Delta=\sqrt{s(s-a)(s-b)(s-c)}$$
$$\Delta=\sqrt{6(6-3)(6-5)(6-4)}$$
$$\Delta=6\,\mathrm{cm^2}$$
$$\therefore$$ Area of the triangle $$= 6 \mathrm {cm^2}$$
Find the area of the parallelogram with sides $$10 cm$$ and $$12 cm$$ and one of the diagonals as $$14 cm. $$
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$$48 \sqrt{6}$$ $$cm^2$$
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$$24 \sqrt{6}$$ $$cm^2$$
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$$40 \sqrt{6}$$ $$cm^2$$
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$$20 \sqrt{6}$$ $$cm^2$$
Explanation
Let $$ABCD$$ be the given parallelogram.
Area of parallelogram $$ABCD = 2$$ x (area of triangle $$ABC$$)
Now $$a = 10$$ cm,
$$b = 12$$ cm
and
$$c = 14$$ cm
$$S=\dfrac { 10+12+14 }{ 2 } =\dfrac { 36 }{ 2 } =18$$ cm$$^2$$
Area of triangle $$ABC$$:
$$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 18(18-10)(18-12)(18-14) } =\sqrt { 18\times 8\times 6\times 4 } =\sqrt { 3456 } =24\sqrt { 6 }$$
Area of parallelogram $$ABCD$$:
$$2\times 24\sqrt { 6 } =48\sqrt { 6 }$$
cm$$^2$$
Hence, area of parallelogram is
$$48\sqrt { 6 }$$
cm$$^2$$.
The perimeter of a triangular field is 144 m and the ratio of the sides is 3 : 4 : 5 The area of the field is
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$$864\ \displaystyle m^{2}$$
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$$468\ \displaystyle m^{2}$$
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$$824\ \displaystyle m^{2}$$
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none
Explanation
Let the side of the triangle $$a,b$$ and $$c$$ be $$3x,4x,5x$$
Perimeter $$=144$$ m
$$\therefore 3x+4x+5x=144$$
$$\Rightarrow 12x=144$$
$$\Rightarrow x=12$$
Then$$ a=12\times 3=36m$$
$$b=4\times 12=48 m$$
$$c=5\times 12=60$$
Then $$s=\dfrac{a+b+c}{2}=\dfrac{36+48+60}{2}=\dfrac{144}{2}=72$$
$$\therefore $$ Area of the triangle$$=\sqrt{s(s-a)(s-b)(s-c)}$$
$$\Rightarrow \sqrt{72(72-36)(72-48)(72-60)}$$
$$\Rightarrow \sqrt{72\times 36\times 24\times 12}=864 \ m^2$$
Find the area of a triangle with sides having length $$40, 24$$ and $$32 m$$
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$$384 m^2$$
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$$364 m^2$$
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$$374 m^2$$
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None of the above
Explanation
We use heron's formula to find the area of the triangle.
Let us find the semi perimeter:
$$S=\dfrac { 40+24+32 }{ 2 } =\dfrac { 96 }{ 2 } =48$$
Now area of the triangle is:
$$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 48(48-40)(48-24)(48-32) } =\sqrt { 48\times 8\times 24\times 16 } =\sqrt { 147456 } =384$$ m$$^2$$
Hence, the area of the triangle is $$384$$
m$$^2$$.
Find the area of a triangle with two equal sides of $$5 cm$$ and remaining one of $$8 cm$$.
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$$9cm^2$$
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$$8cm^2$$
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$$12cm^2$$
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$$16cm^2$$
Explanation
We use heron's formula to find the area of the triangle.
Let us find the semi perimeter:
$$S=\dfrac { 5+5+8 }{ 2 } =\dfrac { 18 }{ 2 } =9$$ cm
Now area of the triangle is:
$$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 9(9-5)(9-5)(9-8) } =\sqrt { 9\times 4\times 4\times 1 } =\sqrt { 144 } =12$$ cm$$^2$$
Hence, the area of the triangle is $$12 cm ^2$$
Find the area of the triangluar field whose sides are $$39\ m, 25\ m$$ and $$56\ m$$ respectively.
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$$\displaystyle420\ m^{2}$$
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$$\displaystyle520\ m^{2}$$
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$$\displaystyle820\ m^{2}$$
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$$\displaystyle320\ m^{2}$$
Explanation
$$\displaystyle s=\left ( 25+39+56 \right )\div 2$$
$$\displaystyle =60m$$
Area $$\displaystyle =\sqrt{s\left ( s-a \right )\left ( s-b \right )\left ( s-c \right )}m^{2}$$
$$\displaystyle =\sqrt{60\left ( 60-39 \right )\left ( 60-25 \right )\left ( 60-56 \right )}$$
$$\displaystyle =\sqrt{60\times 35\times 21\times 4}$$
$$\displaystyle =\sqrt{\left ( 5\times 3\times 4 \right )\times \left ( 5\times 7 \right )\times \left ( 7\times 3 \right )\times 4}$$
$$\displaystyle =\left ( 4\times 3\times 5\times 7 \right )m^{2}=420\ m^{2}$$
The sides of a right triangle are 5cm, 12 cm and 13 cm . Then its area is
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$$0.003{ m }^{ 2 }$$
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$$0.0024{ m }^{ 2 }$$
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$$0.0026{ m }^{ 2 }$$
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$$0.0015{ m }^{ 2 }$$
Explanation
Hypotenuse will be opposite to right angle which is $$13$$ cm
The base and height will be $$5$$ cm and $$12$$ cm.
$$Area\ of\ triangle=\dfrac { 1 }{ 2 } \times \dfrac { 5 }{ 100 } \times \dfrac { 12 }{ 100 }= \dfrac { 30 }{ 10000 } =0.003\ { m }^{ 2 }$$
'
Which of the following is true?
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Area by heron's formula = Area general formula
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Area by heron's formula $$\neq$$ Area general formula
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$$Area = 6cm^2$$
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None of the above
Explanation
General formula of area of triangle is:
$$A=\dfrac { 1 }{ 2 } \times base\times height=A=\dfrac { 1 }{ 2 } \times 3\times 4=6$$ cm$$^2$$
We will now find the area of triangle using herons formula:
$$S=\dfrac { 3+4+5 }{ 2 } =\dfrac { 12 }{ 2 } =6$$ cm
Now area
is:
$$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 6(6-3)(6-4)(6-5) } =\sqrt { 6\times 3\times 2\times 1 } =\sqrt { 36 } =6$$
c
m$$^2$$
Hence, area of triangle
is
$$6$$
c
m$$^2$$ and
Area by Heron's formula$$=$$ Area by general formula$$=6 cm ^2$$
The sides of a triangular plot are in the ratio of $$3:5:7$$ and its perimeter is $$300m$$. Find its area.
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$$1500m^2$$
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$$1500\sqrt{3}m^2$$
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$$1400m^2$$
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$$1400\sqrt{3}m^2$$
Explanation
The sides of a the triangular plot are in the ratio $$3:5:7$$. So, let the sides of the triangle be $$3x$$, $$5x$$ and $$7x$$.
Also it is given that the perimeter of the triangle is $$300$$ m therefore,
$$3x+5x+7x=300$$
$$15x=300$$
$$x=20$$
Therefore, the sides of the triangle are $$60,100$$ and $$140$$.
Now using herons formula:
$$S=\dfrac { 60+100+140 }{ 2 } =\dfrac { 300 }{ 2 } =150$$ m
Area of
the triangle is:
$$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 150(150-60)(150-100)(150-140) } =\sqrt { 150\times 90\times 50\times 10 } =\sqrt { 6750000 }$$
$$=1500\sqrt { 3 }$$
m$$^2$$
Hence,
area of the triangular plot is
$$1500\sqrt { 3 }$$
m$$^2$$
.
$$s$$ in Heron's formula is ___ .
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area of a triangle
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perimeter of a triangle
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Semi-perimeter of a triangle
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None of the above
Area of the quadrilateral ABCD is
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$$307 m^2$$
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$$305 m^2$$
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$$306 m^2$$
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$$304 m^2$$
Explanation
If $$ABCD$$ is a quadrilateral, $$AB=9$$ m,
$$AD=28$$ m,
$$BC=40$$ m and
$$CD=15$$ m and
$$\angle ABC=90^{ 0 }$$.
Area of $$ABCD$$= Area of triangle $$ABC+$$
Area of triangle $$ACD$$
In
$$\triangle ABC$$,
$$\angle B=90^{ 0 }$$, therefore,
$$AC^2=AB^2+BC^2=9^2+40^2=81+1600=1681$$
$$\Rightarrow AC=\sqrt { 1681 } =41$$ m
Area of
$$\triangle ABC$$ is:
$$A=\dfrac { 1 }{ 2 } \times AB\times AC==\dfrac { 1 }{ 2 } \times 9\times 40=180$$ m$$^2$$
In
$$\triangle ACD$$,
$$S=\dfrac { 28+15+41 }{ 2 } =\dfrac { 84 }{ 2 } =42$$ m
Now area of
$$\triangle ACD$$ is:
$$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 42(42-28)(42-15)(42-41) } =\sqrt { 42\times 14\times 27\times 1 } =\sqrt { 15876 } =126$$
m$$^2$$
Hence, area of quadrilateral $$ABCD$$ is $$180+126=306$$
m$$^2$$.
Which of the following is / are true?
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Attitude is required for calculating area by heron's formula
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Attitude is not required for calculating area by heron's formula
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Semi-perimeter is required for calculating area by heron's formula
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Semi-perimeter is not required for calculating area by heron's formula
Explanation
By Heron's Formula, $$Area=\sqrt {s(s-a)(s-b)(s-c)}$$ ,
where $$s=semi-perimeter$$ and $$a,b,c$$ are $$side\ lengths$$
Semi-perimeter$$=\dfrac{sum\ of\ side\ lengths}{2}$$
So only the $$semi-perimeter$$ and the $$side\ lengths$$ are required.
Find the area of a triangle, two of its sides are $$8 cm$$ and $$11 cm$$ and the perimeter is $$32 cm$$.
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$$8\sqrt{30}cm^2$$
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$$15\sqrt{2}cm^2$$
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$$4\sqrt{30}cm^2$$
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$$5\sqrt{31}cm^2$$
Explanation
Since the perimeter of the triangle with sides $$a$$, $$b$$ and $$c$$ is $$a+b+c$$, therefore, perimeter of triangle with sides $$8$$ cm,
$$11$$ cm and
$$c$$ cm with perimeter $$32$$ cm
is:
$$32=8+11+c$$
$$c=32-19=13$$
We use heron's formula to find the area of the triangle.
Let us find the semi perimeter:
$$S=\dfrac { 8+11+13 }{ 2 } =\dfrac { 32 }{ 2 } =16$$ cm
Now area of the triangle is:
$$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 16(16-8)(16-11)(16-13) } =\sqrt { 16\times 8\times 5\times 3 } =\sqrt { 1920 } =8\sqrt { 30 }$$
cm$$^2$$
Hence, the area of the triangle is
$$8\sqrt { 30 }$$
c
m$$^2$$.
Find the area of the triangle by Heron's formula.
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$$12m^2$$
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$$24m^2$$
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$$40m^2$$
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$$6m^2$$
Explanation
We use heron's formula to find the area of the triangle.
Let us find the semi perimeter:
$$S=\dfrac { 5+3+4 }{ 2 } =\dfrac { 12 }{ 2 } =6$$ cm
Now area of the triangle is:
$$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 6(6-5)(6-3)(6-4) } =\sqrt { 6\times 1\times 3\times 2 } =\sqrt { 36 } =6$$ m$$^2$$
Hence, the area of the triangle is $$6$$
m$$^2$$.
The area of a triangle by Heron's formula is
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$$\sqrt{(s-a)(s-b)(s-c)}$$
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$$\sqrt{s(s-a)(s-b)(s-c)}$$
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$$\sqrt{(s+a)(s-b)(s-c)}$$
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$$\sqrt{(s+a)(s+c)(s+c)}$$
Explanation
Using Herons formula, we first have to find the semi perimeter with sides $$a,b$$ and $$c$$:
$$s=\dfrac { a+b+c }{ 2 }$$
And then area of
the triangle is:
$$A=\sqrt { s(s-a)(s-b)(s-c) }$$
Hence, area of a triangle by
heron's formula is
$$A=\sqrt { s(s-a)(s-b)(s-c) }$$.
Find the area of a triangle using Heron's formula
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$$170 cm^2$$
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$$160 cm^2$$
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$$180 cm^2$$
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$$200 cm^2$$
Explanation
We find the area of triangle using herons formula:
$$S=\dfrac { 9+40+41 }{ 2 } =\dfrac { 90 }{ 2 } =45$$ cm
Now area
is:
$$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 45(45-9)(45-40)(45-41) } =\sqrt { 45\times 36\times 5\times 4 } =\sqrt { 32400 } =180$$
c
m$$^2$$
Hence, area of triangle
is
$$180$$
c
m$$^2$$.
Find the area of an equilateral triangle with side $$10cm$$.
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$$25\sqrt{3}cm^2$$
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$$24\sqrt{3}cm^2$$
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$$26\sqrt{3}cm^2$$
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None of the above
Explanation
Formula: Are of an equilateral triangle having side length $$a$$ is given by, $$\dfrac{\sqrt 3}{4}a^2$$
Here it is given that, side length $$, a=10$$ cm
Thus its area $$=\dfrac{\sqrt 3}{4}\times 10^2$$ cm$$^2=25\sqrt 3$$ cm$$^2$$
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
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Rs.$$1212$$
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Rs.$$13.5$$
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Rs.$$1350$$
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Rs$$11.6$$
Explanation
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. $$=\dfrac{1350}{100}$$
$$\Rightarrow (B)$$.
Which of the following is true about the above quadrilateral?
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Area of quadrilateral can be calculated by one general formula of area of triangle
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Area of quadrilateral cannot be calculate by the general formula of area of triangle because of the absence of height
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Height is AD and BC in $$\triangle ADC$$ and $$\triangle ABC$$ respectively.
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Area of Quadrilateral can be easily calculated by the heron's formula.
Explanation
If $$ABCD$$ is a quadrilateral, $$AB=3$$ cm,
$$AD=5$$ cm,
$$BC=4$$ cm and
$$CD=4$$ cm and
$$AC=5$$ cm
.
Area of $$ABCD$$= Area of triangle $$ACD+$$
Area of triangle $$ACB$$
In
$$\triangle ACD$$,
$$S=\dfrac { 5+5+4 }{ 2 } =\dfrac { 14 }{ 2 } =7$$ cm
Now area of
$$\triangle ACD$$ is:
$$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 7(7-5)(7-5)(7-4) } =\sqrt { 7\times 2\times 2\times 3 } =\sqrt { 84 } =2\sqrt { 21 }$$
c
m$$^2$$
In
$$\triangle ACB$$,
$$S=\dfrac { 3+4+5 }{ 2 } =\dfrac { 12 }{ 2 } =6$$ cm
Now area of
$$\triangle ACB$$ is:
$$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 6(6-3)(6-4)(6-5) } =\sqrt { 6\times 3\times 2\times 1 } =\sqrt { 36 } =6$$
c
m$$^2$$
Hence, area of quadrilateral $$ABCD$$ is
$$(6+2\sqrt { 21 })$$
c
m$$^2$$.
Area of quadrilateral $$ABCD$$ cannot be found by the general formula of area of triangle because the height is not given.
Use Heron's formula to find the area of a triangle of lengths 3,3 and 4
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$$5\sqrt{2}$$
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$$2\sqrt{5}$$
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$$2\sqrt{7}$$
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$$7\sqrt{2}$$
Use Heron's formula to find the area of a triangle of lengths $$5 cm, 9 cm$$ and $$12 cm$$.
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$$\sqrt{26}\ cm^2$$
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$$2\sqrt{26}\ cm^2$$
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$$3\sqrt{26}\ cm^2$$
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$$4\sqrt{26}\ cm^2$$
Explanation
We will find the area of triangle with lengths $$5$$ cm,
$$9$$ cm
and
$$12$$ cm
using heron's formula:
$$S=\dfrac { 5+9+12 }{ 2 } =\dfrac { 26 }{ 2 } =13$$ units
Now area
is:
$$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 13(13-5)(13-9)(13-12) } =\sqrt { 13\times 8\times 4\times 1 } =\sqrt { 416 } =4\sqrt { 26 }$$
sq.units
Hence, area of triangle
is
$$4\sqrt { 26 } cm^2$$
A land ABCD is in the shape of a Quadrilateral has $$\angle C = 90, AB = 9m, BC = 12m, CD = 5m$$ and $$AD = 8m$$. How much area does it occupy?
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$$30+6\sqrt{5}m^2$$
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$$30+6\sqrt{35}m^2$$
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$$30+6\sqrt{125}m^2$$
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None of the above
Explanation
$$BD = \sqrt{12^2+5^2} = \sqrt{144+25} = \sqrt{169}=13m$$
Area of $$\triangle BDC = \dfrac{1}{2} \times 12\times 5m^2 = 6\times 5=30m^2$$
Area of \triangle ABD
$$S=\dfrac{9+8+13}{2} = \dfrac{17+13}{2} = \dfrac{30}{2} = 15$$
$$=\sqrt{15\times (15-9)(15-8)(15-13)}$$
$$=\sqrt{15\times 6\times 7\times 2}$$
$$=\sqrt {3\times 5\times 3\times 2\times 7\times 2}$$
$$=3\times 2\sqrt{35}=6\sqrt{35}$$
Total area $$= 30+6\sqrt{35}m^2$$
The sides of a triangle are $$5y, 6y $$ and $$9y$$. find an expression for its area $$A$$.
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$$10y^2\sqrt{2}$$
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$$5y^2\sqrt{2}$$
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$$20y^2\sqrt{2}$$
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$$100y^2\sqrt{2}$$
Explanation
We will find the area of triangle with sides $$5y$$,
$$6y$$
and
$$9y$$
using heron's formula:
Let $$S$$ be the semi perimeter of the given triangle
$$S=\dfrac { 5y+6y+9y }{ 2 } \\=\dfrac { 20y }{ 2 }\\=10y$$
Now area
is:
$$A=\sqrt { s(s-a)(s-b)(s-c) } \\=\sqrt { 10y\left( 10y-5y \right) \left( 10y-6y \right) \left( 10y-9y \right) } \\=\sqrt { 10y\times 5y\times 4y\times y } \\=\sqrt { 200y^{ 4 } } \\=10y^{ 2 }\sqrt { 2 }$$
Hence, area of triangle
is
$$10y^{ 2 }\sqrt { 2 }$$
sq.units
.
Use Heron's formula to find the area of a triangle of lengths $$7, 8$$ and $$9$$.
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$${12\sqrt{5}}$$
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$$11\sqrt{5}$$
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$$6\sqrt{5}$$
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None of the above
Explanation
We will find the area of triangle with lengths $$7,8$$ and $$9$$ using herons formula:
$$S=\dfrac { 7+8+9 }{ 2 } =\dfrac { 24 }{ 2 } =12$$ units
Now area
is:
$$A=\sqrt { s(s-a)(s-b)(s-c) } $$
$$=\sqrt { 12(12-7)(12-8)(12-9) } $$
$$=\sqrt { 12\times 5\times 4\times 3 } $$
$$=\sqrt { 720 } =12\sqrt { 5 }$$
sq.units
Hence, area of triangle
is
$$12\sqrt { 5 }$$
sq.units
.
Use Heron's formula to find the area of a triangle of lengths $$4, 5$$ and $$6.$$
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$$\dfrac{15}{4}\sqrt{7}$$
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$$\dfrac{13}{4}\sqrt{7}$$
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$$4\sqrt{7}$$
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$$3\sqrt{7}$$
Explanation
Given: Lengths of a triangle $$=4,$$
$$5,$$
and
$$6$$
$$S=\dfrac { 4+5+6 }{ 2 } =\dfrac { 15 }{ 2 }$$ units
Using heron's formula to find the area of the triangle:
$$A=\sqrt { s(s-a)(s-b)(s-c) } $$
$$=\sqrt { \dfrac { 15 }{ 2 } \left( \dfrac { 15 }{ 2 } -4 \right) \left( \dfrac { 15 }{ 2 } -5 \right) \left( \dfrac { 15 }{ 2 } -6 \right) }$$
$$ =\sqrt { \dfrac { 15 }{ 2 } \times \dfrac { 7 }{ 2 } \times \dfrac { 5 }{ 2 } \times \dfrac { 3 }{ 2 } } $$
$$=\sqrt { \dfrac { 1575 }{ 16 } } $$
$$=\dfrac { 15 }{ 4 } \sqrt { 7 }$$
Hence, area of triangle
is
$$\dfrac { 15 }{ 4 } \sqrt { 7 }$$
sq. units
.
Calculate the area of quadrilateral $$ABCD$$.
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$$6+\sqrt{21}cm^2$$
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$$6+2\sqrt{21}cm^2$$
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$$6+2\sqrt{7}cm^2$$
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None of the above
Explanation
Area of $$\triangle ABC$$
$$S=\dfrac{5+4+3}{2}=\dfrac{9+3}{2} = \dfrac{12}{2} = 6cm^2$$
$$=\sqrt{s(s-a)(s-b)(s-c)}$$
$$=\sqrt{6\times (6-5)(6-4)(6-3)}$$
$$=\sqrt{6\times 1\times 2\times 3}$$
$$=\sqrt{3\times 2\times 3\times 2}$$
$$=6cm^2$$
Area of $$\triangle ABC$$
$$s=\dfrac{5+5+4}{2} = \dfrac{14}{2} = 7cm^2$$
$$=\sqrt{s(s-a)(s-b)(s-c)}$$
$$=\sqrt{7\times 2\times 2\times 3}$$
$$=2\sqrt{21}$$
Total Area $$=(6+2\sqrt{21}) cm^2$$
Which of the following are true?
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Area of $$\triangle ABC$$ cannot be calculated by Heron's formula.
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Area of $$\triangle ABC$$ can be calculated by Heron's formula.
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Area of $$\triangle ABC$$ cannot be calculated by the general formula of area.
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Area of $$\triangle ABC$$ can be calculated by the general formula of area.
Explanation
As $$\Delta ABC$$ is right-angled,
$$AB^2 + BC^2 = AC^2$$ (hypotenuse theorem)
$$AC^2 = 20^2+ 50^2$$
$$=400+2500$$
$$AC^2 = 2900$$
$$AC = \sqrt{2900}cm$$.
a) Area of $$\Delta ABC$$ using general formula $$= \dfrac{1}{2}\times b \times h$$
$$=\dfrac{1}{2} \times 50cm \times 20cm$$
$$=500cm^2$$.
$$\therefore$$ Area of $$\Delta ABC$$ can be calculated using general formula of area.
b) Area of $$\Delta ABC$$ using Heron's formula
Area $$=\sqrt{s(s-a)(s-b)(s-c)}$$ ($$s\to$$ semiperimeter)
Area $$=499.9999 cm^2$$ $$s\to \dfrac{a+b+c}{2} = 61.928$$
$$\therefore$$ Area of $$\Delta ABC$$ can be calculated by Herou's formula
$$\therefore$$ Hence, option (B) and option (D) are correct.
The parallel sides of a parallelogram are $$40$$ and $$15$$ m and one of the diagonals is $$35$$ m. Find the area of the parallelogram, if the distance between the parallel sides is $$50$$ m. (Use Heron's formula)
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$$120$$$$\sqrt{3}$$ $$m^{2}$$
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$$150$$$$\sqrt{3}$$ $$m^{2}$$
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$$200$$$$\sqrt{3}$$ $$m^{2}$$
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$$300$$$$\sqrt{3}$$ $$m^{2}$$
Explanation
In $$\parallel$$ gm $$ABCD, AB = CD = 40\ m$$ and $$AD = BC = 15\ m$$, Diagonal, $$AC = 35\ m$$
In $$\Delta ABC$$, $$s= \dfrac{40 + 15 + 35}{2}$$ $$= 45\ m$$
Area of $$\parallel$$ gm = $$2$$ x Area of $$\Delta ABC$$
= $$2\times \sqrt{S(S-a)(S-b)(S-c)}$$
= $$2\times \sqrt{45(45-40)(45-15)(45-35)}$$
= $$2\times \sqrt{45(5)(30)(10)}$$
= $$300\sqrt{3}$$ $$m^{2}$$
The parallel sides of a parallelogram are $$20$$ and $$10$$ m and one of the diagonals is $$12$$ m. Find the area of the parallelogram. (Use Heron's formula)
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6$$\sqrt{31}$$ $$m^{2}$$
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6$$\sqrt{231}$$ $$m^{2}$$
0%
6$$\sqrt{21}$$ $$m^{2}$$
0%
3$$\sqrt{231}$$ $$m^{2}$$
Explanation
In $$\parallel$$ gm ABCD, AB = CD = 20 m and AD = BC = 10m, Diagonal, AC = 12 m
In $$\Delta$$ ABC, $$s= \dfrac{20 + 10 + 12}{2}$$ = 21 m
Area of $$\parallel$$ gm = 2 x Area $$\Delta $$ ABC
= 2$$\sqrt{S(S-a)(S-b)(S-c)}$$
= 2$$\sqrt{21(21-20)(21-10)(21-12)}$$
= 2$$\sqrt{21(1)(11)(9)}$$
= 6$$\sqrt{231}$$ $$m^{2}$$
The sides of a triangle are $$5\ cm, 12\ cm$$ and $$13\ cm$$, what is its area?
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$$0.0024$$ $$m^2$$
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$$0.0026$$ $$m^2$$
0%
$$0.003$$ $$m^2$$
0%
$$0.0015$$ $$m^2$$
Explanation
Area of $$\triangle =\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } ,\quad \quad s=\dfrac { a+b+c }{ 2 } =\dfrac { 13+12+5 }{ 2 } =15cm$$
$$\therefore $$ ar of $$\triangle =\sqrt { 15\left( 15-13 \right) \left( 15-12 \right) \left( 15-5 \right) } =\sqrt { 15\times 2\times 3\times 10 } =30{ cm }^{ 2 }$$
Therefore area of triangle $$={ 30cm }^{ 2 }=0.003{ m }^{ 2 }\left[ \because \quad 1{ m }^{ 2 }=10000{ cm }^{ 2 } \right] $$
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