MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 9 Maths Heron'S Formula Quiz 4 - MCQExams.com
CBSE
Class 9 Maths
Heron'S Formula
Quiz 4
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are $$15 cm, 14 cm, 13 cm$$, and the parallelogram stands on the base of $$15$$ cm. Find the height of the parallelogram.
Report Question
0%
$$5.9 cm$$
0%
$$6.8 cm$$
0%
$$6.5 cm$$
0%
$$5.6 cm$$
Explanation
Area of triangle = $$\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } $$
$$S=\dfrac { a+b+c }{ 2 } =\dfrac { 15+14+13 }{ 2 } =21cm$$
$$\therefore \quad $$ area of $$\triangle =\sqrt { 21\left( 21-15 \right) \left( 21-14 \right) \left( 21-13 \right) } =\sqrt { 21\times 6\times 7\times 8 } $$
= $$\sqrt { 3\times 7\times 3\times 2\times 7\times 4\times 2 } =2\times 2\times 3\times 7$$
= $${ 84cm }^{ 2 }$$
Now, area of $$\parallel $$gm = area of $$\triangle $$ = $$b\times h=84{ cm }^{ 2 }$$
$$\Rightarrow \quad 15\times h=84\Rightarrow h=\dfrac { 84 }{ 15 } =5.6cm$$
$$\therefore $$ Height of
$$\parallel $$gm = $$5.6cm$$
What is the area of a triangle whose sides are $$13\ cm, 14\ cm$$ and $$15\ cm$$?
Report Question
0%
$$84\ sq.cm$$
0%
$$64\ sq.cm$$
0%
$$825\ sq.cm$$
0%
$$105\ sq.cm$$
Explanation
$$S=\dfrac { a+b+c }{ 2 } =\dfrac { 13+14+15 }{ 2 } =21cm$$
area of $$\triangle =\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } =\sqrt { 21\left( 21-13 \right) \left( 21-14 \right) \left( 21-15 \right) } $$
$$=\sqrt { 21\times 8\times 7\times 6 } =\sqrt { 3\times 7\times 2\times 2\times 2\times 7\times 2\times 3 } =2\times 2\times 3\times 7$$
$$={ 84cm }^{ 2 }$$
Find the number of trees that can be planted in a triangular ground having sides $$51m, 37m$$ and $$20m$$, when each tree occupies $$6m^2$$ of space:
Report Question
0%
$$54$$
0%
$$51$$
0%
$$82$$
0%
$$43$$
Explanation
Area of $$\triangle =\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } ,\quad s=\dfrac { a+b+c }{ 2 } =\dfrac { 51+37+20 }{ 2 } =54m$$
$$\therefore \quad $$ Area of triangle ground $$=\sqrt { 54\left( 54-51 \right) \left( 54-37 \right) \left( 54-20 \right) } $$
$$=\sqrt { 54\times 3\times 17\times 34 } =306{ m }^{ 2 }$$
Number of trees $$=\dfrac { 306 }{ 6 } =51$$
The perimeter of a triangular field is $$420$$ m and its sides are in the ratio $$6: 7: 8$$ Find area of the triangular field.
Report Question
0%
$$2100$$ sq m
0%
$$2100\sqrt{15}$$ sq m
0%
$$2000$$ sq m
0%
$$2000\sqrt{15}$$ sq m
Explanation
$$\textbf{Step - 1: Finding the sides}$$
$$\text{Given that the sides are in the ratio }6:7:8$$
$$\implies a=6x,b=7x,c=8x$$
$$\text{Perimeter}=a+b+c$$
$$\implies 420=6x+7x+8x=21x$$
$$\implies x=20$$
$$\implies a=120,b=140,c=160$$
$$\textbf{Step - 2: Calculating area}$$
$$s=\dfrac{a+b+c}2=210$$
$$\implies \text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$$
$$\implies \text{Area} = \sqrt{210\times90\times70\times50}$$
$$\implies \text{Area}=2100\sqrt{15}$$
$$\textbf{Thus, the area of the given field is }\mathbf{2100\sqrt{15}}\textbf{ sq. m.}$$
In $$\triangle ABC, AB = 6\ cm, BC = 7\ cm$$ and $$AC = 5\ cm$$. Find the area of $$\triangle ABC$$
Report Question
0%
$$6\sqrt {6}\ cm^{2}$$
0%
$$6\sqrt {3}\ cm^{2}$$
0%
$$6\sqrt {2}\ cm^{2}$$
0%
$$9\sqrt {6}\,cm^{2}$$
Explanation
Using Heron's formula,
Area of a triangle
$$=\sqrt{s(s-a)(s-b)(s-c)}$$
where
$$s=$$ semiperimeterof the triangle
$$a=$$ length of side $$BC$$
$$b=$$ length of side $$AC$$
$$c=$$ length of side $$AB$$
In $$\triangle ABC,\ a=7,\ b=5,\ c=6$$
Semiperemeter of $$\triangle ABC=s=\dfrac{a+b+c}{2}$$
$$=\dfrac{7+5+6}{2}$$
$$=9$$
Hence, the area of $$\triangle ABC=\sqrt{9\times (9-7)\times (9-5)\times (9-6)}$$
$$=\sqrt{9\times 3\times 2\times 4}$$
$$={6\sqrt{6}\ cm^{2}}$$
Find the area of the parallelogram with sides $$11 cm$$ and $$13 cm$$ and one of the diagonals as $$16 cm. $$
Report Question
0%
$$23\sqrt{32}$$ sq. cm
0%
$$25\sqrt{35}$$ sq. cm
0%
$$24\sqrt{35}$$ sq. cm
0%
None of these
Explanation
ar $$\parallel $$gm ABCD = $$2\times ar\triangle ABC$$
ar $$\triangle $$ABC = $$\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } ,\quad s=\dfrac { a+b+c }{ 2 } =\dfrac { 16+13+11 }{ 2 } =20cm$$
$$\therefore $$ ar
$$\triangle $$ABC =
$$\sqrt { 20\left( 20-16 \right) \left( 20-13 \right) \left( 20-11 \right) } =\sqrt { 20\times 4\times 7\times 9 } =\sqrt { 5\times 4\times 4\times 7\times 9 } =4\times 3\sqrt { 35 } =12\sqrt { 35 } { cm }^{ 2 }$$
$$\therefore $$ ar
$$\triangle $$ABCD =
$$2\times 12\sqrt { 35 } =24\sqrt { 35 } { cm }^{ 2 }$$
Find the area of the quadrilateral whose sides are $$9 m ,40 m, 28 m, 15 m$$. The angle between first two sides is $$90^o$$.
Report Question
0%
$$348$$ sq. m
0%
$$315$$ sq. m
0%
$$306$$ sq. m
0%
None of these
Explanation
Area of quadrilateral $$ABCD$$ = ar $$[\triangle ABC]$$ + ar $$[\triangle ACD]$$
In $$\triangle ABC$$ :
$$ar\ [\triangle ABC] = \dfrac { 1 }{ 2 } \times BC\times AB\\=\dfrac { 1 }{ 2 } \times 40m\times 9m=180{ m }^{ 2 }$$
Using pythagoras theorem,
$$AC= \sqrt { { AB }^{ 2 }+{ BC }^{ 2 } } \\ =\sqrt { { \left( 9 \right) }^{ 2 }+{ \left( 40 \right) }^{ 2 } }\\ =\sqrt { 1681 } =41m$$
Using Heron's formula,
$$ar [\triangle ACD] = \sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } \\\quad s=\dfrac { a+b+c }{ 2 } \\\ \ \ \quad =\dfrac { 41+28+15 }{ 2 } \\\quad \ \ \ =\dfrac { 84 }{ 2 } =42m$$
$$ ar [\triangle ACD ]= \sqrt { 42\left( 42-41 \right) \left( 42-28 \right) \left( 42-15 \right) }\\ \quad \quad \quad \ \ \ \quad =\sqrt { 42\times 1\times 14\times 27 } =126{ m }^{ 2 }$$
Therefore, area of quadrilateral
$$ABCD$$
$$=180 + 126$$
$$=306\ m^{2}$$
Find the area of the parallelogram with sides $$9 cm$$ and $$11 cm$$ and one of the diagonals as $$14 cm. $$
Report Question
0%
$$22\sqrt{15}$$
0%
$$24\sqrt{17}$$
0%
$$21\sqrt{17}$$
0%
None of these
Explanation
Area of $$\parallel $$gm ABCD = 2 $$\times $$ area $$\triangle $$ABC
ar $$\triangle $$ABC = $$\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } ,\quad \& \quad s=\dfrac { a+b+c }{ 2 } =\dfrac { 14+11+19 }{ 2 } =17cm$$
$$\therefore $$ ar
$$\triangle $$ABC =
$$\sqrt { 17\left( 17-14 \right) \left( 17-11 \right) \left( 17-9 \right) } =\sqrt { 17\times 3\times 6\times 8 } =12\sqrt { 17 } { cm }^{ 2 }$$
Therefore, ar
$$\parallel $$gm ABCD = 2
$$\times $$
$$12\sqrt { 17 } =24\sqrt { 17 } { cm }^{ 2 }$$
The parallel sides of a parallelogram are $$60$$m and $$25$$ m and one of the diagonals is $$65$$ m. Find the area of the parallelogram,
Report Question
0%
$$1430$$ sq m
0%
$$1576$$ sq m
0%
$$1500$$ sq m
0%
None of these
Explanation
The area of parallelogram $$ABCD = 2 \times Ar \triangle ABC$$
Now,
$$Ar\ \triangle ABC =\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } \quad \\ s=\dfrac { a+b+c }{ 2 } \\=\dfrac { 65+60+25 }{ 2 } \\=75m$$
Thus,
$$\therefore Ar \triangle ABC = \sqrt { 75\left( 75-65 \right) \left( 75-60 \right) \left( 75-25 \right) } \\=\sqrt { 75\times 10\times 15\times 50 } \\=750{ m }^{ 2 }$$
Thus, area of parallelogram $$=2\times 750\ m^2$$
$$=1500\ m^2$$
Find the area of an equilateral triangle each of whose side is 4 cm long?
Report Question
0%
$$2\sqrt { 3 }$$
0%
$$3\sqrt { 3 }$$
0%
$$5\sqrt { 3 }$$
0%
$$4\sqrt { 3 }$$
Explanation
solution:
$$Area = \dfrac{\sqrt{3}}{4} r^2 $$
$$Area = \dfrac{\sqrt{3}}{4} 4^2 $$
$$Area = \dfrac{\sqrt{3}}{1} 4 $$
hence the correct opt: D
In the given figure, the area of the $$\triangle {ABC}$$ is
Report Question
0%
$$13.24{cm}^{2}$$
0%
$$12.28{cm}^{2}$$
0%
$$11.32{cm}^{2}$$
0%
$$15.37{cm}^{2}$$
Explanation
We know that the area of the triangle using heron's formula is given by
Area $$= \sqrt{s(s-a)(s-b)(s-c)} $$
So,
$$ S = \dfrac{8+4+11}{2} = \dfrac{23}{2} = 11.5 $$
Area $$ = \sqrt{11.5(11.5-11)(11.5-8)(11.5-4)}$$
$$ =\sqrt{11.5\times 0.5\times 3.5\times 7.5}$$
$$ = 12.28\text{ cm}^{2}$$
If the sides of a triangle are $$\dfrac{y}{z} + \dfrac{z}{x},\dfrac{z}{x} + \dfrac{x}{y}$$ and $$\dfrac{x}{y} + \dfrac{y}{z}$$ then its area is
Report Question
0%
xyz
0%
$$\sqrt {\dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x}} $$
0%
$$\sqrt {xyz} $$
0%
$$\dfrac{{{x^2}{y^2}{z^2}}}{2}$$
A triangular park in a city has dimensions 100m X 90m X110m.A contract is given to a company for planting grass in the park at the rate of Rs.4000 per hectare.Find the amount to be paid to the company.(Take $$\sqrt{2}=1.414$$)
Report Question
0%
Rs.$$4532.90$$
0%
Rs.$$4242$$
0%
Rs.$$1696.80$$
0%
Rs.$$1000$$
Explanation
$$Heron's\ Formula\ :$$
Area of a triangle = $$\Delta=\sqrt{s(s-a)(s-b)(s-c)}$$
where $$s$$ is the semi-perimeter and $$a,\ b$$ and $$c$$ are the sides of the triangle.
$$s=\dfrac{a+b+c}{2}$$
Here, for the triangular park,
$$a=100\ m,\ \ b=90\ m,\ \ c=110\ m$$ and
$$s=\dfrac{100+90+110}{2}=150\ m$$
Area of the triangular park = $$\Delta=\sqrt{150\times50\times60\times40}=3000\sqrt{2}\ m^2=0.3\sqrt{2}\ ha=0.4242\ ha$$
Rate of planting grass = $$Rs.\ 4000\ /ha$$
So, the amount to be paid = $$4000\times0.4242=Rs.\ 1696.80$$. $$[C]$$
Find the area of a triangle whose two sides are $$8$$ cm and $$11$$cm and the perimeter is $$32$$cm.
Report Question
0%
$$8cm^2$$
0%
$$8\sqrt{30}cm^2$$
0%
$$\dfrac{8}{\sqrt{30}}cm^2$$
0%
$$24cm^2$$
Explanation
Let third side be $$x$$
Perimeter $$=8+11+x$$
$$19+x=32$$
$$x=13$$
$$s=\cfrac { 32 }{ 2 } =16$$
Heron's formula
$$A=\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } $$
$$=\sqrt { 16\left( 16-8 \right) \left( 16-11 \right) \left( 16-13 \right) } $$
$$=\sqrt { 16\times 8\times 5\times 3 } $$
$$=8\sqrt { 30 }\ cm^2$$
One side of an equilateral triangle measures $$8$$ cm. Find its area using Heron's formula. What is its altitude
Report Question
0%
$$12\sqrt 3\ \text{cm}^2, 3\sqrt 3$$ cm
0%
$$16\sqrt 3\ \text{cm}^2, 4\sqrt 3$$ cm
0%
$$14\sqrt 3\ \text{cm}^2, \dfrac {7}{2}\sqrt 3$$ cm
0%
$$20\sqrt 3\ \text{cm}^2, 5\sqrt 3$$ cm
Explanation
According to the Heron's formula, Area (A) of the triangle having sides $$a,b,c$$ units is
$$A=\sqrt{s(s-a)(s-b)(s-c)}$$
where
$$s=\cfrac{a+b+c}{2}$$
For the given triangle,
$$a=b=c=8cm$$
$$s=\cfrac{8+8+8}{2}=12$$
$$A=\sqrt{12(12-8)(12-8)(12-8)}$$
$$A=\sqrt{3\times 4\times 4^{3}}$$
$$=4\times 4\sqrt{3}$$
$$=16\sqrt{3}$$
Area of the triangle =$$\cfrac{1}{2}\times base \times altitude=16\sqrt{3}$$
$$\cfrac{1}{2}\times 8 \times$$ altitude$$=16\sqrt{3}$$
Altitude $$=\cfrac{16\sqrt{3}}{4}=4\sqrt{3}$$
The perimeter of a triangle is $$300m$$. If its sides are in the ratio $$3:5:7$$. Find the area of the triangle.
Report Question
0%
$$1500\sqrt{5}m^2$$
0%
$$1500m^2$$
0%
$$1500\sqrt{3}m^2$$
0%
$$1100\sqrt{3}m^2$$
Explanation
Let, 1 side be = $$3x$$
2 side be = $$5x$$
3 side be = $$7x$$
According to question, $$ 3x+5x+7x=300$$
$$\implies x= 20$$
So, sides are $$ 60 ,100, 140 $$
Area = $$\sqrt {s(s-a)(s-b)(s-c)}$$
$$ s=semi-perimeter$$
$$s=150 $$, $$a = 60$$, $$b = 100$$, $$c = 140$$
Area=$$\sqrt {150(150-60)(150-100)(150-140)}$$
$$\implies \sqrt{150(90)(50)(10)}$$
$$\implies1500\sqrt3\ m^2$$
Find the area of a triangle whose sides are 9 cm, 12 cm, and 15 cm.
Report Question
0%
$$34cm^2$$
0%
$$44cm^2$$
0%
$$54cm^2$$
0%
$$55cm^2$$
Explanation
Given sides of the triangle
$$a=9$$cm
$$b=12$$cm
$$c=15$$cm
The given triangle is a right angle triangle
because $$9^2+12^2=15^2$$
$$\therefore$$Area $$=\dfrac{1}{2}\times 9\times 12=54$$ sq cm
Area $$=54$$ sq cm.
The perimeter of a triangle field is $$450\ m$$ and its sides are in the ratio $$25:17:12$$. Find the area of the triangle.
Report Question
0%
$$2250m^2$$
0%
$$6250m^2$$
0%
$$5250m^2$$
0%
$$6050m^2$$
Explanation
Let, 1 side be = $$25x$$
2 side be = $$17x$$
3 side be = $$12x$$
According to question, $$ 25x+17x+12x=450$$
$$\implies x= \dfrac{450}{54}$$
$$\implies x=\dfrac{25}{3}$$
So, sides are $$ \dfrac{625}{3} , \dfrac{425}{3}, \dfrac{100}{1} $$
Area = $$\sqrt {s(s-a)(s-b)(s-c)}$$
$$s=semi-perimeter$$
$$s= 225 $$,
$$a = \dfrac{625}{3}$$,
$$b = \dfrac{425}{3}$$,
$$c = \dfrac{100}{1}$$
Area=$$\sqrt {225(225-\dfrac{625}{3})(225-\dfrac{425}{3})(225-100)}$$
$$\implies \sqrt{225(\dfrac{50}{3})(\dfrac {250}{3})(125)}$$
$$\implies6250\ m^2$$
Find the area of triangle of sides $$6cm ,8cm ,10cm$$
Report Question
0%
$$24cm^2$$
0%
$$12cm^2$$
0%
$$30cm^2$$
0%
$$none$$
Explanation
The sides are $$6cm,8cm,10cm$$
$$s=\dfrac {6+8+10}2=\dfrac {24}2=12$$
The area is given as $$\sqrt {s(s-a)(s-b)(s-c)}\\\sqrt {12(6)(4)(2)}\\\sqrt {576}=24cm^2$$
Find the area of a triangle whose sides are respectively $$150\ cm$$, $$120\ cm$$, and $$200\ cm.$$
Report Question
0%
$$8545.77\text{ sq. cm}$$
0%
$$8966.57\text{ sq. cm}$$
0%
$$9257.66\text{ sq. cm}$$
0%
$$5627.66 \text{ sq. cm}$$
Explanation
Let $$a=150\ cm,\ b=120\ cm$$ and $$c=200\ cm.$$
$$s=\dfrac{a+b+c}{2}$$
$$\Rightarrow s=\dfrac{150+120+200}{2}$$
$$\Rightarrow s=235\ cm$$
Now, by using the heron's formula,
Area $$=\sqrt{s(s-a)(s-b)(s-c)}$$
$$=\sqrt{235(235-150)(235-120)(235-200)}$$
$$=\sqrt{235(85)(115)(35)}$$
$$=\sqrt{80399375}$$
$$=8966.57\text{ sq. cm}$$
If every side of a triangle is doubled, the area of the new triangle is $$K$$ times the area of the old one. Find the value of $$K$$.
Report Question
0%
$$\sqrt{2}$$
0%
$$2$$
0%
$$3$$
0%
$$4$$
Explanation
Let, the sides of the original triangle be $$a,b$$ and $$c$$
so, the semi-perimeter, $$s=\dfrac{a+b+c}{2}$$
Original area $$=\sqrt{s(s-a)(s-b)(s-c)}$$
Now every side is doubled
So, the new sides are $$2a, 2b$$ and $$ 2c$$
new semi-perimeter, $$s'=\dfrac{2a+2b+2c}{2}=2\left(\dfrac{a+b+c}{2}\right)=2s$$
New area $$=\sqrt{s'(s'-2a)(s'-2b)(s'-2c)}$$
$$=\sqrt{(2s)(2s-2a)(2s-2b)(2s-2c)}$$
$$=\sqrt{2.2.2.2s(s-a)(s-b)(s-c)}$$
$$=4\sqrt{s(s-a)(s-b)(s-c)}$$
$$=4\times Original\ Area$$
Hence, the value of $$K$$ is $$4$$.
Use Heron's formula to find the area of a triangle of lengths $$5, 7$$ and $$8$$.
Report Question
0%
$$10\sqrt{3}$$
0%
$$20\sqrt{3}$$
0%
$$2\sqrt{6}$$
0%
$$20\sqrt{6}$$
Three sides of a triangular field are 20m, 21m, and 29m long, respectively. The area of the field is :
Report Question
0%
$$215{ m }^{ 2 }$$
0%
$$230{ m }^{ 2 }$$
0%
$$210{ m }^{ 2 }$$
0%
None of these
Explanation
Given: Sides of the triangular field; $$a=20m, b=21m, c=39m$$
Now, Semiperimeter $$S = \dfrac{a+b+c}{2} = \dfrac{20+21+29}{2} = 35m$$
$$Area = \sqrt{s\times (s-a)\times (s-b)\times (s-c)}$$
$$= \sqrt{35\times (35-20)\times (35-21)\times (35-29)}$$
$$= \sqrt{35\times 15\times 14\times 6}$$
$$=210 sq.m$$
The area of a triangle whose sides are 15 m, 16 m and 17 m is :
Report Question
0%
$$24\sqrt { 4 } { m }^{ 2 }$$
0%
$$24\sqrt { 3 } { m }^{ 2 }$$
0%
$$24\sqrt { 21 } { m }^{ 2 }$$
0%
None of these
Explanation
The sides of the triangle are $$a = 15 \text{ m }, b = 16 \text{ m }, c = 17 \text{ m }$$
Area of the triangle is given by
$$S = \cfrac {a + b + c}{2} $$
$$= \cfrac {15 + 16 + 17}{2} $$
$$= \cfrac {48}{2} $$
$$= 24 \text{ m}$$
Area $$ = \sqrt {S(S - a)(S - b)(S - c)}$$
$$= \sqrt {24 (25 - 15) (24 - 16) (24 - 17)}$$
$$= \sqrt {24\times 9\times 8\times 7}$$
$$= \sqrt {\underline {8}\times \underline {3\times 3}\times 3\times \underline {8}\times 7} $$
$$= 24\sqrt {21} \text{ sq. m.}$$
The sides of a triangle are $$7$$ cm, $$9$$ cm and $$
14$$ cm. Its area is
Report Question
0%
$$12\sqrt 5 \ \text{cm}^2$$
0%
$$12\sqrt 3 \ \text{cm}^2$$
0%
$$24\sqrt 5 \ \text{cm}^2$$
0%
$$63 \ \text{cm}^2$$
Explanation
Area of triangle = $$\sqrt{S(S-a)(S-b)(S-c)}$$
$$S=\dfrac{a+b+c}{2}$$
Given $$a= 7$$ cm
$$b= 9$$ cm
$$c= 14$$ cm
$$S= \dfrac{7+9+14}{2}$$
$$=\dfrac{30}{2}$$
$$S= 15 cm$$
$$\therefore $$ Area of triangle = $$\sqrt{S(S-a)(S-b)(S-c)}$$
$$=\sqrt{15(15-7)(15-9)(15-14)}\\$$
$$=\sqrt{15.8.6.1}\\$$
$$=\sqrt{720}\\$$
$$=\sqrt{5\times 144}\\$$
$$=\sqrt{12\times 12\times 5}\\$$
$$=12\sqrt{5}cm^{2}\\$$
$$\therefore $$ Area of triangle = $$12.\sqrt{5}\ cm^{2}$$
So, the answer is A. $$12.\sqrt{5}\ cm^{2}$$
The sides of a triangle are in the ratio $$3:5:7.$$ If the perimeter of the triangle is $$60$$ cm, then its area
Report Question
0%
$$60\sqrt 3 \ \text{cm}^2$$
0%
$$30\sqrt 3\ \text{cm}^2$$
0%
$$15\sqrt 3 \ \text{cm}^2$$
0%
$$120\ \text{cm}^2$$
Explanation
Given sides in the ratio $$3:5:7$$
perimeter of the triangle $$=60cm$$
$$3x+5x+7x=60$$
$$\Rightarrow 15x=60$$
$$\Rightarrow x=60/15$$
$$\Rightarrow x=4$$
$$3x=3\times 4=12$$cm
$$5x=5\times 4=20$$cm
$$7x=7\times 4=28$$cm
Let $$S$$ be the semiperimeter of given triangle.
$$S=\dfrac{1}{2}(a+b+c)$$
$$=\dfrac{1}{2}\cdot 60$$
$$=30$$
Now use formula foe area of triangle
Area $$=\sqrt{30(30-12)(30-20)(30-28)}$$
$$=\sqrt{30\cdot18\cdot10\cdot2}$$
$$=\sqrt{60\cdot180}$$
$$=\sqrt{10800}$$
$$=\sqrt{3600\cdot3}$$
$$=60\sqrt{3}cm^2$$
$$\therefore$$ Area of triangle $$=60\sqrt{3}cm^2$$.
The percentage increase in the area of a triangle if its each side is doubled, is
Report Question
0%
$$200\% $$
0%
$$400\% $$
0%
$$300\% $$
0%
$$500\% $$
Explanation
If the sides of triangle are 4, 5 and 6 cm. Then the area (in sq cm) of triangle is
Report Question
0%
$$\dfrac{\pi}{4}$$
0%
$$\dfrac{\pi}{4} \sqrt{7}$$
0%
$$\dfrac{4}{15}$$
0%
$$\dfrac{4}{15} \sqrt{7}$$
0%
$$\dfrac{15}{4} \sqrt{7}$$
Explanation
Given, triangle of sides $$a,b,c=4,5,6cm$$
$$\therefore S=\dfrac{a+b+c}{2}$$
$$=\dfrac{4+5+6}{2}=\dfrac{15}{2}$$
Then, area of triangle
$$A=\sqrt{S(S-a)(S-b)(S-c)}$$
$$=\sqrt{\dfrac{15}{2}\left(\dfrac{15}{2}-4\right)\left(\dfrac{15}{2}-5\right)\left(\dfrac{15}{2}-6\right)}$$
$$=\sqrt{\dfrac{15}{2}.\left(\dfrac{7}{2}\right)\left(\dfrac{5}{2}\right)\left(\dfrac{3}{2}\right)}$$
$$=\dfrac{15}{4}\sqrt{7}$$
Find the area of equilateral triangle each of whose sides measures 20 cm.
Report Question
0%
173$${cm^2}$$
0%
150
$${cm^2}$$
0%
120
$${cm^2}$$
0%
153
$${cm^2}$$
Explanation
Area of Equilateral Triangle =$$\dfrac{\sqrt{3}{b^2}}{4}$$
where b is the Side of the Triangle.
Area=
$$\dfrac{\sqrt{3}\times {20^2}}{4}$$
Area=$$173.2{cm^2}$$
Find the area of equilateral triangle (in $${cm^2}$$)each of whose sides measure 18 cm.
Report Question
0%
150
0%
140
0%
180
0%
190
Explanation
Area of Equilateral Triangle =$$\dfrac{\sqrt{3}{a^2}}{4}$$
where $$a$$ is the Side of the Triangle.
Given $$a= 18\ cm$$
Area $$=\dfrac{\sqrt{3}\times {18^2}}{4}$$
Area $$=140.29\ {cm^2}$$
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 9 Maths Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
