MCQ Questions for Class 9 Maths Heron'S Formula Quiz 5

The sides of a triangle measures

$13\,cm , 14\,cm$ and

$15\,cm.$ Its area is

0%
$84 \,cm^2$
0%
$91\,cm^2$
0%
$168 \,cm^2$
0%
$182 \,cm^2$

Explanation

Let

$a, b, c$ be the sides of the triangle

Then,

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

$= \dfrac{13 + 14 + 15}{2}$

$= \dfrac{42}{2}$

$= 21$

Area of triangle

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

$= \sqrt{(21 \times (21 - 13) \times (21 - 14) \times (21 - 15)}$

$= \sqrt{(21 \times (8) \times (7) \times (6)}$

$= \sqrt{(3 \times 7 \times 2 \times 2 \times 2 \times 7 \times 2 \times 3)}$

$= \sqrt{(3^2 \times 7^2 \times 2^2 \times 2^2)}$
By cancelling square and square root we get,

$= (3 \times 7 \times 2 \times 2)$

$= 84\ cm^2$

The area of an equilateral triangle is

$4 \sqrt{3} \,cm^2.$ The length of each of its sides is

0%
$3 \,cm$
0%
$4 \,cm$
0%
$2 \sqrt{3} \,cm$
0%
$\dfrac{1}{2} \sqrt{3} \,cm$

Explanation

From the question is given that,
Area of an equilateral triangle is

$(4 \sqrt{3})$
Area of the equilateral triangle

$= ((\sqrt{3}) / 4) \times (sides)^2$ sq. units

$Area= ((\sqrt{3} / 4) \times (sides)^2$

$= (sides)^2 = (4 \sqrt{3}) / (4 / (\sqrt{3})$

$= (sides)^2 = (4 \sqrt{3} \times 4) / \sqrt{3}$

$= (sides)^2 = 4 \times 4$

$= Sides = \sqrt{} 16$

$= Sides = 4 \,cm$
Hence, the length of each side of the triangle is

$4 \,cm$

Write true or false and justify your answer. The are of isosceles triangle is

$5/4 \sqrt{11} cm^{2}$ , if the perimeter is

$11\ cm$ and the base is

$5 cm$ .

0%
True
0%
False

Explanation

Justification:

According to the question,

Perimeter =

$11cm$

And base,

$a = 5$

As the triangle is isosceles,

$b = c$

Perimeter =

$11$

$\implies a + b + c = 11$

$\implies 5 + b + b = 11$

$\implies 5 + 2b = 11$

$\implies 2b = 6$

$\implies b = 3$

So, we have,

$a = 5, b = 3, c = 3$

$s = (a + b + c)/2$

$\implies s = (5 + 3 + 3)/2 = 11/2$

Area of triangle =

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

$=\sqrt{\dfrac{11}{2}(\dfrac{11}{2}-5)(\dfrac{11}{2}-3)(\dfrac{11}{2}-3)}$

$=\sqrt{\dfrac{11}{2}(\dfrac{1}{2})(\dfrac{5}{2})(\dfrac{5}{2})}$

$\implies$ Area of triangle =

$(5\sqrt{11})/4 cm^{2}$

Hence the statement

$\text{ The area of the isosceles triangle is }(5\sqrt{11})/4 cm^{2},\text{ if the perimeter is }11\ cm\text{ and the base is }5\ cm\text{ is True}$ .

Each side of an equilateral triangle is

$8 \,cm$ long. Its area is:

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$32 \,cm^2$
0%
$64 \,cm^2$
0%
$16 \sqrt{3} \,cm^2$
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$16 \sqrt{2} \,cm^2$

Explanation

Area of the equilateral triangle

$= \dfrac{\sqrt 3}{ 4} \times (sides)^2$

$= \dfrac{\sqrt 3} {4} \times (8)^2$

$= \dfrac{\sqrt 3}{ 4} \times 64$

$= \sqrt 3 \times 16$

$= 16 \sqrt 3 \,cm^2$

In an isosceles triangle

$PQR$ , the altitude

$PS = 6$

$cm$ and

$PQ = PR = 10$

$cm$ . Find the area of

$\triangle PSR$ .

0%
$20$ $cm^2$
0%
$22$ $cm^2$
0%
$24$ $cm^2$
0%
None of these

Write true or false and justify your answer. The are of the equilateral triangle is

$20\sqrt{3}cm^{2}$ whose each side is

$8\ cm$ .

0%
True
0%
False

Explanation

Justification:

Area of an equilateral triangle of side

$a = \dfrac{\sqrt{3}}{4}a^{2}$

According to the question,

Area of a triangle =

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

$\implies \dfrac{\sqrt{3}}{4}a^{2} =20\sqrt{3}$

$\implies a^2 = 20 \times 4$

$\implies a^2 = 80$

$\implies a = 4\sqrt{5} cm$

Hence, the statement

$\text{the area of the equilateral triangle is }20\sqrt{3}\ cm^2\text{ whose each side is }8 cm\text{ is False}$ .

$AD = 4$

$cm$ and

$AB = 5$

$cm$ in the given figure, then find the area of

$\triangle ADB$ .

0%
$4$ $cm^2$
0%
$6$ $cm^2$
0%
$8$ $cm^2$
0%
$10$ $cm^2$

In the given figure, if

$PL = 2 QL$ , then what will be the area of

$\triangle PLQ$ ?

0%
$28.8$ square units
0%
$26.8$ square units
0%
$28.3$ square units
0%
$26.3$ square units

$\triangle ABC$ is a right isosceles triangle with hypotenuse

$AC= 10$

$cm$ and right angle at

$B$ , then the twice of its area is ____.

0%
$20$ $cm^2$
0%
$30$ $cm^2$
0%
$40$ $cm^2$
0%
$50$ $cm^2$

Riya has a piece of cloth in the shape of equilateral triangle whose all altitudes are equal to

$6$

$cm$ . Find its area. (if each of it's sides are

$4 \sqrt 3$ cm)

0%
$4\sqrt3$ $cm^2$
0%
$8\sqrt3$ $cm^2$
0%
$12\sqrt3$ $cm^2$
0%
$16\sqrt3$ $cm^2$

Find the area of

$\triangle ABD$

0%
$12 \ cm^2$
0%
$24 \ cm^2$
0%
$28 \ cm^2$
0%
none of these

In the given figure, it is given that

$O$ is a mid point of

$AB$ and

$CD$ ,

$AB = 24$ units and

$CD = 26$ units. Find the value of

$ar( \triangle OBC) + 2 ar( \triangle OAD).$

0%
$30$ $units^2$
0%
$60$ $units^2$
0%
$90$ $units^2$
0%
None of these

Diya has a lunch box in the shape of right isosceles triangle of hypotenuse

$30$

$cm$ , find its area.

0%
$150$ $cm^2$
0%
$160$ $cm^2$
0%
$170$ $cm^2$
0%
$225$ $cm^2$

Ankita has a piece of cake in the shape of an equilateral triangle whose all altitudes are equal to

$8$

$cm$ . Then the half of its area is ____. (if each side is equal to

$\dfrac {27 \sqrt 3}{2}$ )

0%
$27\sqrt3$ $cm^2$
0%
$28\sqrt3$ $cm^2$
0%
$29\sqrt3$ $cm^2$
0%
$30\sqrt3$ $cm^2$

Find area of triangle ADC.

0%
$15$ $cm^2$
0%
$20$ $cm^2$
0%
$30$ $cm^2$
0%
$40$ $cm^2$

In given figure,

$AC= 5$

$cm$ ,

$BD = 2$

$cm$ and

$CD = 2BD$ . Find the area of

$\triangle ABC$ .

0%
$9$ $cm^2$
0%
$18$ $cm^2$
0%
$27$ $cm^2$
0%
None of these

Find the area of triangle ABD.

0%
$30$ $cm^2$
0%
$18$ $cm^2$
0%
$6$ $cm^2$
0%
$12$ $cm^2$

Find

$ar(\triangle ABD).$

0%
$6$ $cm^2$
0%
$70$ $cm^2$
0%
$80$ $cm^2$
0%
$90$ $cm^2$

How much paper of blue shade is needed to make a kite given in the figure, if

$AB = BC = 10$

$cm$ and

$BO = 8$

$cm$ ?

0%
$24$ $cm^2$
0%
$48$ $cm^2$
0%
$96$ $cm^2$
0%
None of these

Two sides of a plot measure 32 m and 24 m and the angle between them is a perfect right angle The other two sides measure 25 m each and the other three angles are not right angles What is the area of the plot?

0%
$768 \displaystyle m^{2}$
0%
$534 \displaystyle m^{2}$
0%
$696.5 \displaystyle m^{2}$
0%
$684 \displaystyle m^{2}$

Explanation

$arABCD=ar\triangle ADC+ar\triangle ABC$

$ar\triangle ADC=\dfrac { 1 }{ 2 } \times 24\times 32=384{ m }^{ 2 }$

Now,

$AC=\sqrt { { AD }^{ 2 }+{ DC }^{ 2 } }$

$=\sqrt { { 24 }^{ 2 }+{ 32 }^{ 2 } } =\sqrt { 1600 } =40m$

$\therefore \quad ar\triangle ABC=\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } ,\quad S=\dfrac { 40+25+25 }{ 2 } =45m$

$=\sqrt { 45\left( 45-40 \right) \left( 45-25 \right) \left( 45-25 \right) }$

$=\sqrt { 45\times 5\times 20\times 20 } =15\times 20=300{ m }^{ 2 }$

$\therefore \quad ar.ABCD=384+300=684{ m }^{ 2 }$

Find the area of a quadrilateral whose sides are

$12, 5, 6$ and

$15$ . The angle between the first two sides is

$90$ . (Use Heron's formula)

0%
$20 +$ $2\sqrt{2805}$
0%
$30 +$ $2\sqrt{805}$
0%
$30 +$ $2\sqrt{374}$
0%
$30 +$ $2\sqrt{1805}$

Explanation

Consider

$ABCD$ is a quadrilateral where,

$AB = 12, BC = 5, CD = 6, DA = 15$ and

$\angle ABC = 90^o$

Area of

$ABCD =$ Area of

$\Delta ABC +$ Area of

$\Delta ACD$
In

$\Delta$ ABC,

$\angle B = 90^o$
Apply Pythagoras theorem in

$\Delta ABC$

Therefore,

$AC^{2} = AB^{2}+BC^{2}= 12^{2}+5^{2}$
So,

$AC = 13$

Area of

$\Delta ABC = \dfrac{1}{2}\times AB \times BC = \dfrac{1}{2}\times12\times 5 = 30 m^{2}$

$\Delta ACD$ , let

$s$ be the semiperimeter,

$S = \dfrac{6 + 15 + 13}{2} = 17$ m
Applying Heron's formula,

Area of

$\Delta ACD$ =

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

$\sqrt{17(17-13)(17-15)(17-6)}$

$\sqrt{17(4)(2)(11)} = 2\sqrt{374}$

Hence, Area of quadrilateral

$ABCD = 30 + 2\sqrt{374}$

So, option C is correct.

The perimeter of a triangle is

$540$ m and its sides are in the ratio

$12 : 25 : 17.$ Find the area of the triangle.

0%
$5,000\:m^{2}$
0%
$6,000\:m^{2}$
0%
$9,000\:m^{2}$
0%
$8,000\:m^{2}$

Explanation

$\textbf{Step 1 : Find the sides of triagnle}$

$\text{Let the sides of the triangle be 12a, 25a, 17a }$
$\text{We know that perimeter of the triangle = Sum of all sides}$

$\Rightarrow\text{ 12a + 25a + 17a = 54a}$

$\text{Given, perimeter of the triangle = 540 m }$
$\Rightarrow\text{54a = 540 m}$
$\text{ a = 10 m}$
$\text{So, the lengths of the sides of triangle are }$

$\text{12a = 120 m}$

$\text{25a = 250 m}$

$\text{17a = 170 m}$

$\textbf{Step 2 : Find the area of triangle using Heron's formula}$

$\text{We can use Heron's formula to get the area of triangle}$

$\text{Area of triangle with sides with sides a, b, c and semi perimeter s}$

$\text{are } \sqrt { s(s-a)(s-b)(s-c)} \text{ respectively}$ .

$\text{and s =}$ $\dfrac{a+b+c}{2}$

$\text{For triangle with sides 120 m, 250 m and 170 m, }$

$\text{s =}$ $\dfrac { 120 + 250 + 170 }{ 2 }$

$\text{s = 270 m}$
$\text{Substituting the sides 120 m, 250 m and 170 m in the Heron's formula, we get}$

$\Rightarrow A= \sqrt { 270(270-120)(270-250)(270-170) }$

$=\sqrt { 270\times 150\times 20\times 100 }$

$=\sqrt { 9\times30\times30\times5\times20\times20\times5}$

$= 3\times 30\times 5 \times 20$

$= 9000{m}^{2}$

$\textbf{Hence , area of triangle is 9000}$ $m^2$

The triangular side - wall of a flyover have been used for advertisements, The sides of the walls are 122 m, 22m and 120 m. The advertisements requires rent of Rs 5000 per

$\displaystyle m^{2}$ per year. A company hired one of its walls for 3 months. How much rent did it pay ?

0%
1750000
0%
1600000
0%
1650000
0%
None of these

Explanation

Sides of a triangular wall is

$122m, 22m, 120m.$
Hence, area of triangle=

$\sqrt{s(s-122)(s-22)(s-120)}$ and

$s=\dfrac{122+22+120}{2}$ .

$\therefore s=132$

$\therefore$ Area of triangular wall

$=1320m^2$
Since, The advertisements yeild an earning of Rs 5000 per

$\displaystyle m^{2}$ per year.

Hence, Earning per 3 months

$= \dfrac{5000}{4}$

$=1250$ per

$m^2$

$\therefore$ Rent paid

$=1250(1320)=1650000Rs.$

Using Heron's formula find the area of a quadrilateral whose sides are

$3\ cm, 4\ cm, 4\ cm$ , and

$5\ cm$ . The angle between the first two sides is

$90^0$ .

0%
$14 cm^{2}$
0%
$16 cm^{2}$
0%
$18 cm^{2}$
0%
$(6 + 6\sqrt{7})cm^{2}$

Explanation

$\Box ABCD$ ,

$AB = 3\ cm, BC = 4\ cm, CD = 4\ cm, DA = 5\ cm$ and

$\angle ABC = 90^o$

Area of

$ABCD$ = Area of

$\Delta ABC$ + Area of

$\Delta ACD$

$\Delta ABC, \angle B = 90^o$
Therefore,

$AC^{2} = AB^{2}+BC^{2}= 3^{2}+4^{2}$
So,

$AC = 5$

Area of

$\Delta ABC = \dfrac{1}{2}\times AB \times AC = \dfrac{1}{2}\times3\times 4 = 6 cm^{2}$

$\Delta ACD, s = \dfrac{4 + 5 + 5}{2} = 7$ m

Applying Heron's formula,
Area of

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

$\sqrt{7(7-3)(7-4)(7-4)}$
=

$\sqrt{7(4)(3)(3)} = 6\sqrt{7}$

Hence, Area of

$\Box ABCD = (6 + 6\sqrt{7})cm^{2}$

So, option D is correct.

A traffic signal board, indicating 'SCHOOLAHEAD', is an equilateral triangle with side $a$ . Find the area of the signal board, using Heron's formula. If its perimeter is $180 cm$ , what will be the area of the signal board?

0%
$900\sqrt 3 cm^2$
0%
$300\sqrt 3 cm^2$
0%
$600\sqrt 3 cm^2$
0%
$900\sqrt 2 cm^2$

Explanation

Area of triangle

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

Perimeter

$=180 cm$

Semi-perimeter,

$S=\cfrac { a+b+c }{ 2 } =\cfrac { 180 }{ 2 } =90\ cm\quad$

Given triangle is an equilateral triangle, therefore

$a=b=c$

Perimeter

$=3a$

$180=3a\Rightarrow a=60$

$a=b=c=60\ cm$

Area of triangle

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

$=\sqrt { 90(90-60)(90-60)(90-60) }$

$=\sqrt { 90\times 30\times 30\times 30 }$

$=\sqrt { 9\times 3\times 3\times 3\times { (10) }^{ 4 } }$

$=3\times 3\times 10^{ 2 }\sqrt { 3 }$

$=900\sqrt { 3 } cm^2$

Find the area of a quadrilateral whose sides are

$3\ cm, 4\ cm, 2\ cm$ , and

$5\ cm$ . The angle between the first two sides is

$90^o$ . (Use Heron's formula)

0%
$14 cm^{2}$
0%
$16 cm^{2}$
0%
$(6+ \sqrt{24} ) cm^{2}$
0%
$20 cm^{2}$

Explanation

In quadrilateral

$ABCD$ ,

$AB = 3\ cm, BC = 4\ cm, CD = 2\ cm, DA = 5\ cm$ and

$\angle ABC = 90^0$

Area of

$ABCD$ = Area of

$\Delta ABC +$ Area of

$\Delta ACD$

$\Delta ABC, \angle B = 90^0$

$\therefore$ Area of

$\Delta ABC$ =

$\dfrac{1}{2}\times AB \times BC = \dfrac{1}{2}\times3\times 4\ cm^2 = 6 cm^{2}$

$\Delta ABC$ , applying pythagorous theorem we get,

$AC^2=AB^2+BC^2$

$\Rightarrow AC^2=3^2+4^2$

$\Rightarrow AC^2=9+16$

$\Rightarrow AC^2=25$

$\Rightarrow AC=5$

$\therefore AC=5\ cm$

$\Delta ACD$ , semi-perimeter

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

$cm$

Applying Heron's formula,
Area of

$\Delta\ ACD$ =

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

$\sqrt{6(6-5)(6-5)(6-2)}$
=

$\sqrt{6(1)(1)(4)} = \sqrt{24}$

Hence, Area of

$\Box ABCD =( 6 + \sqrt{24}) cm^2$

So, option

$C$ is correct.

Find the area of a field which is in shape of a parallelogram with sides 12 cm and 14 cm and one of the diagonal of length 18 cm.

0%
$16 \sqrt{110}$ $cm^2$
0%
$18 \sqrt{11}$ $cm^2$
0%
$8 \sqrt{10}$ $cm^2$
0%
$6 \sqrt{11}$ $cm^2$

Explanation

Solution:

To find:

$ar(llgm ABCD)=?$

Solution;

$ar(\triangle ABD)=\sqrt{s(s-a)(s-b)(s-c)}$

here,

$a=12, b=14 , c=18$

$s=\cfrac{12+14+18}2=22$

$ar(\triangle ABD)=\sqrt{22(22-12)(22-14)(22-18)} cm^2$

$=\sqrt{22\times10\times8\times4}\quad cm^2$

$=\sqrt{7040}\quad cm^2$

$=8\sqrt{110}\quad cm^2$

Now,

$ar(llgm ABCD)=2\times ar(\triangle ABD)$

$=2\times 8\sqrt{110}$

$=16\sqrt{110}\quad cm^2$

Hence, A is the correct answer.

In an equilateral triangle,

$3$ coins of radii

$1$ unit each are kept so that they touch each other and also the sides of the triangle. Area of the triangle is?

0%
$4+2\sqrt{3}$
0%
$6+4\sqrt{3}$
0%
$12+\displaystyle\frac{7\sqrt{3}}{4}$
0%
$3+\displaystyle\frac{7\sqrt{3}}{4}$

Find the area of the triangle whose sides are

$84$ metres,

$80$ metres and

$52$ metres.

0%
$@056$ sq.m
0%
$2756$ sq.m
0%
$2016$ sq.m
0%
$2716$ sq.m
0%
NONE OF THESE

The side of an equilateral triangle whose area is

$\sqrt{3}{cm}^{2}$ is:

0%
$1cm$
0%
$2cm$
0%
$3cm$
0%
$4cm$

CBSE Questions for Class 9 Maths Heron'S Formula Quiz 5 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Heron'S Formula Quiz 5 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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