MCQ Questions for Class 9 Maths Heron'S Formula Quiz 2

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

0%
Remains the same
0%
Becomes doubled
0%
Becomes three times
0%
Becomes four times

Explanation

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

$\therefore A_1=\sqrt{16s(s-a)(s-b)(s-c)}$

$\therefore A_1=4A$

If every side of a triangle is doubled, then the area of the new triangle is 'K' times the area of the old one. The value of K is

0%
2
0%
3
0%
$\sqrt{2}$
0%
4

Explanation

By using Heron's formula,

$A=\sqrt{s(s-a)(s-b)(s-c)}$
Now,

$a=2a$ ,

$b=2b$ ,

$c=2c$

$\Rightarrow s=2s$

$\therefore A_1=\sqrt{2s(2s-2a)(2s-2b)(2s-2c)}$

$A_1=\sqrt{16s(s-a)(s-b)(s-c)}$

$A_1=4\sqrt{s(s-a)(s-b)(s-c)}$

$A_1=4A$

The area of a triangle whose sides are

$4\ cm, 13\ cm$ and

$15\ cm$ , is

0%
$\sqrt{420}\ {cm}^2$
0%
$24\ {cm}^2$
0%
$48\ {cm}^2$
0%
$56\ {cm}^2$

Explanation

The sides of triangle are

$4\ cm, 13\ cm, 15\ cm$
Using Heron's formula,
Area of triangle

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

$s = \displaystyle \frac{a+b+c}{2}$

$\therefore s= \displaystyle \frac{4+13+15}{2} = 16\ cm$

$\therefore$ Area

$=\sqrt{16 (16-4)(16-13)(16-15)}$

$\therefore$ Area

$=\sqrt{16\times 12\times 3\times 1}$

$= 4\times6$

$= 24\ {cm}^2$

Hence, option B.

The area of a triangle whose sides are

$13 cm, 14 cm$ and

$15 cm$ is:

0%
$42 cm^2$
0%
$86 cm^2$
0%
$84 cm^2$
0%
$100 cm^2$

Explanation

Let

$a=13cm$ ,

$b=14cm$ and

$c=15cm$

Now,

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

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

$=21cm$

Area:

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

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

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

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

$=2\times 2\times 3\times 7$

$=84c{ m }^{ 2 }$

A quadrilateral

$ABCD$ has

$\angle C =90,AB=9 cm,BC=8cm,CD=6cm$ and

$AD=7 cm.$ How much area does it occupy?

0%
$6 \sqrt {26} + 24$
0%
$48$
0%
$12 \sqrt {26}$
0%
none of these

Explanation

It is given that

$∠C = 90º, AB = 9$ cm,

$BC = 8$ cm,

$CD = 6$ cm and

$AD = 7$ cm

$BD$ is joined.

$ΔBCD$ ,

By applying Pythagoras theorem,

$BD^2 = BC^2 + CD^2$

$⇒ BD^2 = 8^2 + 4^2$

$⇒ BD^2 = 64+16=80$

$BD=4\sqrt { 5 }$ cm

Area of

$ΔBCD$

$=\dfrac { 1 }{ 2 } \times 8\times 6=24$ cm

$^2$

Now, semi perimeter of

$ΔABD$

$=\dfrac { 1 }{ 2 } (7+9+4\sqrt { 5 } )=\dfrac { 16+4\sqrt { 5 } }{ 2 } =8+2\sqrt { 5 }$ cm

$^2$

Using heron's formula,

Area of

$ΔABD$ is:

$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 8+2\sqrt { 5 } (8+2\sqrt { 5 } -4\sqrt { 5 } )(8+2\sqrt { 5 } -9)(8+2\sqrt { 5 } -7) }$

$=\sqrt { (8+2\sqrt { 5 } )(8-2\sqrt { 5 } )(2\sqrt { 5 } -1)(2\sqrt { 5 } +1) } =\sqrt { [\left( 8 \right) ^{ 2 }-\left( 2\sqrt { 5 } \right) ^{ 2 }][\left( 2\sqrt { 5 } \right) ^{ 2 }-\left( 1 \right) ^{ 2 }] } =\sqrt { [64-20][20-1] } =\sqrt { 44\times 19 }$

$=\sqrt { 936 } =6\sqrt { 26 }$ cm

$^2$

Area of quadrilateral

$ABCD$ = Area of

$ΔBCD +$ Area of

$ΔABD$

$6\sqrt { 26 } +24$ cm

$^2$ .

The length of sides of a triangle are in the ratio.

$17:12:25$ and its semi perimeter is

$270 cm$ . Find the area of the triangle.

0%
$9000 c$ $m^2$
0%
$6000 c$ $m^2$
0%
$7000 c$ $m^2$
0%
$8000 c$ $m^2$

Explanation

Let the sides of the triangle be

$17x, 12x$ and

$25x$
Semiperimeter (s)

$=\displaystyle \frac{(17x + 12x + 25x)}{ 2}$

$= 27x$
Given that,

$27x = 270x$

$= 10$
So, the sides of the triangle are

$170 cm$ ,

$120 cm$ and

$250 cm$
Semiperimeter

$= 270 cm$
Therefore,
Area of the triangle

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

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

$=\sqrt{270(100)(150)(20)}$

$=9000$

$cm^2$

The lengths of sides of a triangle are in the ratio

$3:4:5$ and its perimeter is

$144$ cm, then find the height corresponding to largest side.

0%
$14.8$ cm
0%
$28.8$ cm
0%
$39$ cm
0%
$16$ cm

Explanation

Given, Side of triangle are in ratio

$3 : 4 : 5$ and their perimeter

$144 \ cm$

Let the sides of triangle be

$3x,4x,5x$

Perimeter

$3x+4x+5x=$ 144 cm

$12x=144$

$\therefore x=12$

Then sides of triangle are

$3x$

$=$ 3

$\times$ 12

$=$ 36 cm,

$4x$

$=$ 4

$\times$ 12

$=$ 48 cm,

$5x$

$=$ 5

$\times$ 12

$=$ 60 cm.

Now,Semi perimeter,

$s=\dfrac { Sum\quad of\quad sides\quad of\quad triangle }{ 2 }$

$\\ =\dfrac { 36+48+60 }{ 2 } =72\quad cm$

Using heron's formula, Area of triangle

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

$\\ =\sqrt { 72(72-36)(72-48)(72-60) }$

$\\ =\sqrt { 72\times 36\times 24\times 12 } =864\quad { cm }^{ 2 }$

Using altitude, area of triangle

$=\dfrac { 1 }{ 2 } \times$ base

$\times$ altitude

$=864\quad { cm }^{ 2 }$

$=\dfrac { 1 }{ 2 } \times 60\times$ altitude

$=864$

$\therefore$ altitude

$=\dfrac { 864\times 2 }{ 60 } =\quad 28.8\quad cm$

So, altitude corresponding to largest side is 28.8 cm.

One side of an equilateral triangle is 6 cm. Its area by using Heron's formula is

0%
$9$ ${cm}^{2}$
0%
$3$ ${cm}^{2}$
0%
$9\sqrt {3}$ ${cm}^{2}$
0%
$3\sqrt{8}$ ${cm}^{2}$

Explanation

$a = 6 cm$ ,

$b = 6 cm$ ,

$c = 6 cm$

$\therefore s = \displaystyle\frac{a + b + c}{2} = \displaystyle\frac{6 + 6 + 6}{2} = 9 cm$

$\therefore$ Area of the equilateral triangle

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

$= \sqrt{9\left(9 - 6\right) \left(9 - 6\right) \left(9 - 6\right)}$

$= \sqrt{9\left(3\right) \left(3\right) \left(3\right)}$

$=9\sqrt{3} {cm}^{2}$

An umbrella is made by stitching

$12$ triangular pieces of cloth. Each piece measures

$50$ cm,

$20$ cm and

$50$ cm. Find the area of cloth used in it.

0%
$2400\sqrt6$ c $m^2$
0%
$2200\sqrt3$ c $m^2$
0%
$1200\sqrt6$ c $m^2$
0%
$1000\sqrt6$ c $m^2$

Explanation

Given, an umbrella is made of 12 triangular pieces each of measure 50 cm, 20 cm, 50 cm.
To find area of cloth required we have to first find the area of one piece of cloth.

Semi perimeter, s=

$=\frac { Sum\quad of\quad sides\quad of\quad triangle }{ 2 } \\ =\frac { 50+20+50 }{ 2 } =60\quad cm$
Using heron's formula, Area of one triangular piece

$=\sqrt { s(s-a)(s-b)(s-c) } \\ =\sqrt { 60(60-50)(60-20)(60-50) } \\ =\sqrt { 60\times 10\times 40\times 10 } =200\sqrt { 6 } { cm }^{ 2 }$
Area of 12 triangular pieces or area of cloth used

$=$ Area of one triangular piece

$\times$ 12

$=200\sqrt { 6 } { cm }^{ 2 }\times 12\\ =2400\sqrt { 6 } { cm }^{ 2 }$

The area of triangle with sides

$35$ cm,

$66$ cm and

$53$ cm is

0%
$924 \text{ cm}^{2}$
0%
$1336 \text{ cm}^{2}$
0%
$693 \text{ cm}^{2}$
0%
$1848 \text{ cm}^{2}$

Explanation

Hen the sides of a triangle are given then the area is given by Heron's formula.

Area

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

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

$\because s=\dfrac{35+66+53}{2}=77$

$\implies$ Area of

$\displaystyle \Delta =\sqrt{77\times \left ( 77-35 \right )\times \left ( 77-66 \right )\times \left ( 77-53 \right )}$

$=\displaystyle \sqrt{77\times 11\times 42\times 24}$

$= 924 \text{ cm}^{2}$

The sides of a triangle are in the ratio of 13:14:15 and its perimeter is 84 cm. Then the area of the triangle is

0%
$136$ $cm^2$
0%
$236$ $cm^2$
0%
$336$ $cm^2$
0%
$436$ $cm^2$

Explanation

Let the sides of the triangle be $13a, 14a, 15a$ .
Perimeter of the triangle $=$ Sum of all sides $= 13a + 14a + 15a = 42a$
Given, perimeter of the triangle $= 84 cm$

$\therefore 42a = 84 cm$

$a = 2 cm$
So, the sides of the triangle are $13a = 26 cm, 14a = 28 cm, 15a = 30 cm$ .

Area of triangle with sides a, b, and c and $s = \displaystyle \frac { a+b+c }{ 2 }$ is $\sqrt { s(s-a)(s-b)(s-c) }$ .

26 cm, 28 cm and 30 cm $=\sqrt { 42(42-26)(42-28)(42-30) }$

$=\sqrt { 42\times 16\times 14\times 12 }$

$=\sqrt { 6\times 7\times 4\times 4\times 7\times 2 \times 6 \times 2}$

$= 6\times 7\times 4 \times 2 = 336 {cm}^{2}$

In a triangular field having sides 30 m, 72 m and 78 m , the length of the altitude to the side measuring 72 m is

0%
25 m
0%
28 m
0%
30 m
0%
35 m

Explanation

$\displaystyle s=\frac{30+72+78}{2}=\frac{180}{2}=90$

$\displaystyle \therefore$ Area of

$\displaystyle \Delta = \sqrt{90(90-30)(90-72)(90-78)m^{2}}$

$\displaystyle =\sqrt{90\times 60\times 18\times 12m^{2}}=1080m^{2}$
Area of the

$\displaystyle \Delta$ with bae ( 72 m)

$\displaystyle =\frac{1}{2}\times 72\times altitude\, =1080$

$\displaystyle \therefore$ Required length of altitude =

$\displaystyle \frac{1080\times 2}{72}=30m$

The area of a triangle whose sides are 13 cm, 14 cm and 15 cm is:

0%
84 sq cm
0%
64 sq cm
0%
825 sq cm
0%
None of these

Explanation

Area of triangle with sides a, b and c is given by: $\sqrt { s(s-a)(s-b)(s-c) }$ .
where $s = \displaystyle \frac { a+b+c }{ 2 }$

For triangle with sides 13 cm, 14 cm and 15 cm:

$s = \displaystyle \frac { 13+14+15 }{ 2 } \ \text{cm}=21\ \text{ cm}$
$\therefore$ Area of the triangle with sides 13 cm, 14 cm and 15 cm $=\sqrt { 21(21-13)(21-14)(21-15) }$

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

$=\sqrt { 7\times 3\times 2\times 4\times 7\times 2 \times 3 }$

$= 7\times 3\times 2 \times 2 = 84\ \text{cm}^{2}$

Find the area of the triangle having three sides given as

$5 cm, 6 cm$ and

$7 cm$ .

0%
$6 \sqrt6$ $cm^2$
0%
$6 \sqrt3$ $cm^2$
0%
$4 \sqrt6$ $cm^2$
0%
None of the above

Explanation

Let,

$a=7cm,\quad b=5cm,\quad c=6cm$

Now by applying Heron's formula :

$s=\cfrac { 7+5+6 }{ 2 }$

$=9cm$

$\therefore area\,of\, \Delta =\sqrt { 9(9-7)(9-5)(9-6) }$

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

$area \,of\,\Delta =2\times 3\sqrt { 6 }$

$=6\sqrt { 6 } c{ m }^{ 2 }$

Hence option A is correct.

The area of a triangle is

$\displaystyle 216cm^{2}$ and its sides are in the ratio 3 : 4 : 5 The perimeter of the triangle is

0%
6 cm
0%
12 cm
0%
36 cm
0%
72 cm

Explanation

Let the sides of the triangle be

$3x cm, 4x cm$ and 5x cm Then

$\displaystyle s=\frac{3x+4x+5x}{2}=\frac{12x}{2}=6x$

$\displaystyle \therefore$ Area of

$\displaystyle \Delta=\sqrt{s\left ( s-a \right )\left ( s-b \right )\left ( s-c \right )}$

$\displaystyle = \sqrt{6x\left ( 6x - 3 \right )\left ( 6x - 4x \right )\left ( 6x - 5x \right )}$

$\displaystyle= \sqrt{6x\times 3x\times 2x\times x=6x^{2}}$
Given

$\displaystyle 6x^{2}=216\Rightarrow x^{2}=\frac{216}{6}=36\Rightarrow x=6$

$\displaystyle \therefore$ The sides of the triangle are

$(3\times 6)cm, (4\times 6)\ cm$ and

$(5\times 6)\ cm$ i.e

$18\ cm, 24\ cm$ and

$30\ cm$
Perimeter of the

$\displaystyle \Delta$ =

$18 cm + 24 cm +30 cm$
= 72 cm

State Heron's formula for area of a triangle.

0%
$\sqrt { s(s+a)(s-b)(s-c) }$ .
0%
$\sqrt { s(s-a)(s+b)(s-c) }$ .
0%
$\sqrt { s(s-a)(s-b)(s-c) }$ .
0%
None of the above

Explanation

The Heron's formula for the area of a triangle is

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

where,

$a,\ b,\ c$ are the three sides and

$s$ is the semi-perimeter of the triangle.

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

The perimeter of a triangle field is

$144 m$ and ratio of the sides is

$3 : 4 : 5$ . Then the area of the field is

0%
$864$ sq m
0%
$764$ sq m
0%
$854$ sq m
0%
$754$ sq m

Explanation

Let the sides of the triangle be $3a, 4a, 5a$ .
Perimeter of the triangle $= 3a + 4a + 5a$

$= 12a$
Given, perimeter of the triangle $= 144 m$

$\Rightarrow 12a = 144 m$

$a = 12 m$
So, the sides of the triangle are

$3a = 36 m,\\ 4a = 48 m, \\5a = 60 m$ .

Area of triangle with sides $a,b$ and $c$ is given as is $\sqrt { s(s-a)(s-b)(s-c) }$ .

Where

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

For triangle with sides $36 m, 48 m \ and\ 60 m,$

$s = \dfrac { 36+48+60 }{ 2 } \\= 72 m$
Thus

$A =\sqrt { 72(72-36)(72-48)(72-60) } \\=\sqrt { 72\times 36\times 24\times 12 }$

$=\sqrt { 36\times 2\times 36\times 12\times 2\times 12 }$

$= 36\times 12\times 2 \\= 864 {m}^{2}$

The sides of a triangular board are

$13\ metres$ ,

$14\ metres$ and

$15\ metres$ . The cost of painting it at the rate of

$Rs. 8.75$ per

$m^2$ is

0%
$Rs. 688.80$
0%
$Rs. 735$
0%
$Rs. 730.80$
0%
$Rs. 722.50$

Explanation

Area of triangle with sides $a, b,$ and $$c$$ and

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

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

For triangle with sides 13 m, 14 m and 15 m,

$s = \dfrac { 13+14+15 }{ 2 } = 21\ m$

Area of the triangle with $13\ m$ , $14\ m$ and $15\ m$ $=\sqrt { 21(21-13)(21-14)(21-15) }$

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

$=\sqrt { 3\times 7\times 2\times 4\times 7\times 3 \times 2}$

$= 3\times 7\times 2 \times 2 = 88\ {m}^{2}$

$\therefore$ cost of painting it at the rate of Rs $8.75$ per ${ m }^{ 2 } = 84 \times 8.75 =$ Rs. 735

If each side of a triangle is doubled, than find the ratio of area of the new triangle than formed and the given triangle.

0%
$4 : 1$
0%
$1:4$
0%
$2:1$
0%
$1:2$

Explanation

Let

$a, b$ and

$c$ denotes the length of the sides of the triangle.
Area of the triangle,

${ A }_{ 1 }$

$=$

$\sqrt{s(s-a)(s-b)(s-c)}$ , where s is the semi-perimeter of the triangle.
So, semiperimeter

$\displaystyle s = \frac{a+b+c}{2}$
When the sides of the triangle are doubled, we get

${ s }^{ \prime }=\dfrac { 2a+2b+2c }{ 2 } =a+b+c=2s$ , where

${ s }^{ \prime }$ is the semi-perimeter of the new triangle

Now, Area of the new triangle,

${ A }_{ 2 }=\sqrt { { s }^{ \prime }({ s }^{ \prime }-2a)({ s }^{ \prime }-2b)({ s }^{ \prime }-2c) }$

$=\sqrt { 2s(2s-2a)(2s-2b)(2s-2c) }$

$=\sqrt { 16s(s-a)(s-b)(s-c) }$

$=4\sqrt { s(s-a)(s-b)(s-c) } =4{ A }_{ 1 }$

So, ratio of the are of new triangle to the given triangle

$=\dfrac { { A }_{ 2 } }{ { A }_{ 1 } } =\dfrac { 4{ A }_{ 1 } }{ { A }_{ 1 } } =4:1$

For a quadrilateral, with the

$4$ sides given and with none of the angles

$90^o,$

0%
area can be found by using Heron's formula only
0%
sum of all the angles is $360$
0%
diagonals are always equal
0%
none of these

Explanation

For a quadrilateral with

$4$ sides given and none of the angle is

$90°$ , then area of quadrilateral area can found by using Heron's formula only and sum of all the angles is

$360°$

If A is the area of a triangle in

$\displaystyle cm^{2}$ , whose sides are 9 cm, 10 cm and 11 cm,then which one of the following is correct?

0%
$A < 40$ $\displaystyle cm^{2}$
0%
$40$ $\displaystyle cm^{2}$ $< A < 45$ $\displaystyle cm^{2}$
0%
$45$ $\displaystyle cm^{2}$ $< A < 50$ $\displaystyle cm^{2}$
0%
$A > 50$ $\displaystyle cm^{2}$

Explanation

$A=\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } ,s=\cfrac { a+b+c }{ 2 } =\cfrac { 9+10+11 }{ 2 } =15㎝$

$\therefore A=\sqrt { 15\left( 15-9 \right) \left( 15-10 \right) \left( 15-11 \right) } =\sqrt { 15\times 6\times 5\times 4 } =\sqrt { 300\times 6 }$

$=\sqrt { 100\times 9\times 2 } =30\sqrt { 2 } =30\times 1.41=42.30cm^2$

$\therefore 40cm^2<A<45cm^2$ is correct answer

The sides of a triangle are 16 cm, 30 cm and 34 cm. Its area is

0%
$120\, cm^2$
0%
$260\, cm^2$
0%
$240\, cm^2$
0%
$272\, cm^2$

Explanation

Since

$16^2 + 30^2 = 34^2$ , clearly, given triangle is a right angled triangle. Hence
Area

$=\, \displaystyle \frac {1}{2}\, \times\, base\, \times\, height$

$=\, \displaystyle \frac {1}{2}\, \times\, 16\, \times\, 30\, =\, 240\, cm^2$

The adjacent sides of a parallelogram are 8 cm and 9 cm. The shorter diagonal is 13 cm. What is its area?

0%
$72\, cm^2$
0%
$12\sqrt {35}\, cm^2$
0%
$24\sqrt {35}\, cm^2$
0%
$150\, cm^2$

Explanation

As shown in the diagram, area of triangle ABC = 1/2 area of the parallelogram ABCD. By Heron's formula, area of triangle

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

$\displaystyle \frac {9 + 8 + 13}{2}\, =\, 15$

$\sqrt{180\, \times\, 7}\, =\, \sqrt{1260}\, =\, \sqrt{6\, \times\, 210}$

$\sqrt {6\, \times\, 7\, \times\, 30}\, =\, 6\sqrt{35}$ sq cm.
Hence, area of ABCD

$=\, 2\, \times\,6\sqrt{35}\, =\, 12\, \sqrt{35}$ sq cm

The sides of a triangle are

$5 cm$ ,

$12 cm$ and

$13 cm$ , Then its area is

0%
$0.0024 m^2$
0%
$0.0026 m^2$
0%
$0.003 m^2$
0%
$0.0015 m^2$

Explanation

Let

$a=5$ cm,

$b=12$ cm,

$c=13$ cm

Semiperimeter

$s=\dfrac{5+12+13}{2}\Rightarrow 15$

Area of

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

Area of

$\triangle=\sqrt{15(15-5)(15-12)(15-13)}$

Area of

$\triangle=\sqrt{15\times 10\times 3\times 2}$

Area of

$\triangle=\sqrt{900}$

Area of

$\triangle=30\ cm^2\Rightarrow \dfrac{30}{100\times 100}\Rightarrow 0.003\ m^2.$

The area of a triangle, whose sides are

$4 cm$ ,

$13 cm$ and

$15cm$ , is

0%
$\sqrt{420} cm^2$
0%
$24 cm^2$
0%
$48 cm^2$
0%
$56 cm^2$

Explanation

Let

$a=4$ cm,

$b=13$ cm,

$c=15$ cm

Semiperimeter

$s=\dfrac{4+15+13}{2}\Rightarrow 16$

Area of

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

Area of

$\triangle=\sqrt{16(16-4)(16-13)(16-15)}$

Area of

$\triangle=\sqrt{16\times 12\times 3\times 1}$

Area of

$\triangle=\sqrt{576}$

Area of

$\triangle=24\ cm^2.$

ABCD is a given field whose sides are indicated. If

$\angle$ DAB = 90

$^o$ , then area of the field is ............

0%
180 m $^2$
0%
126 m $^2$
0%
306 m $^2$
0%
300 m $^2$

Explanation

$\bigtriangleup DBC$

$\angle DAB=90^{0}$
Then

$DB^{2}=(40)^{2}+(9)^{2}=1600+81=1681$
OR DB=41 m
Area of

$\bigtriangleup DBC=\sqrt{42(42-41)(42-28)(42-15)}=\sqrt{15876}=126$

$m^{2}$
Area of

$\bigtriangleup ADB=\frac{1}{2}(40\times 9)=\frac{360}{2}=180$

$m^{2}$
Area of ABCD=

$\bigtriangleup DBC+\bigtriangleup ADB$
Area of ABCD=

$126+180=306m^{2}$

The base of an isosceles triangle is

$16$ cm and its perimeter is

$36$ cm. The area of the triangle is ........

0%
$72 \displaystyle cm^{2}$
0%
$36\displaystyle cm^{2}$
0%
$24 \displaystyle cm^{2}$
0%
$48 \displaystyle cm^{2}$

Explanation

In isosceles

$\triangle ABC$ ,

perimeter

$=AB+BC+CA =36$

Let

$AB = AC =x$

$x+x+16 =36$

$x= \displaystyle \dfrac{1}{2}\left ( 36-16 \right )cm=10cm$

In an isosceles triangle, median is the perpendicular bisector.

$\angle AEB =90^o$
Now,

$\triangle ABE$ is a right angle triangle.

Then by pythagoras theorem, we have

$AE^2+BE^2= AB^2$

$\displaystyle h^{2}+8^{2}=10^{2}$

$h^2=100-64$

$\Rightarrow h=6cm$

$\displaystyle ar [\Delta ABC]=\dfrac{1}{2}\times \text{base}\times \text{height }\\=\dfrac{1}{2}\times 16cm \times 6cm\\=48\ cm^{2}$

1129394_345138_ans_3178bd1014d046d894c2ad1ab170436e.png

The sides of a triangle are

$3$ cm,

$5$ cm and

$4$ cm: Its area is

0%
$6\, cm^2$
0%
$7.5\, cm^2$
0%
$5\sqrt 2\, cm^2$
0%
None of these

Explanation

Given: the sides of a triangle are

$3$ cm,

$5$ cm and

$4$ cm.

To find out: Area of triangle

Solution:-

The lengths of the sides are

$a=3$ cm,

$b=5$ cm,

$c=4$ cm.

$\therefore s=\dfrac { a+b+2 }{ 2 } =\dfrac { 3+5+4 }{ 2 } cm=6$ cm

We use the formula,

$\Delta =\sqrt { { s\left( s-a \right) }{ \left( s-b \right) }{ \left( s-c \right) } }$ .

$\therefore \Delta =\sqrt { 6{ \left( 6-3 \right) }{ \left( 6-5 \right) }{ \left( 6-4 \right) } } { cm }^{ 2 }\\ =\sqrt { 6\times 3\times 1\times 2 }$ sq.cm

$=6$ sq.cm

Find the area of a rhombus with sides 20 cm and length of one of the diagonal as 28 cm.

0%
$56 \sqrt{51}$ $cm^2$
0%
$18 \sqrt{51}$ $cm^2$
0%
$25 \sqrt{51}$ $cm^2$
0%
none of these

Explanation

Let

$ABCD$ is a rhombus with diagonals

$AC$ and

$BD$ which intersect each other at

$O$ .

$AC = 28 ⇒ AO = 14$

Let

$BO = x$ and

$AB = 20\ cm$ (given)

By Pythagorean theorem,

$c ^2 = a ^2 + b ^2$

$20 ^2 = 14 ^2 + x ^2$

$400 = 196 + x ^2$

$x ^2 = 400-196$

$x ^2 = 204$

$x=2\sqrt { 51 }$ cm

Therefore,

$BO=2\sqrt { 51 }\ cm$

Diagonal

$BD=2\times 2\sqrt { 51 } =4\sqrt { 51 }\ cm$

Area =

$\dfrac { 1 }{ 2 }$ [ product of diagonals]

$=\dfrac { 1 }{ 2 } \times 28\times 4\sqrt { 51 } =56\sqrt { 51 }$

Hence, the area of rhombus is

$56\sqrt { 51 }\ cm^2$ .

The area of triangle whose sides are 4cm, 13 cm and 15 cm is

0%
${ 576cm }^{ 2 }$
0%
${ 30cm }^{ 2 }$
0%
${ 26cm }^{ 2 }$
0%
${ 24cm }^{ 2 }$

Explanation

$\dfrac { s }{ 2 } =\dfrac { 4+13+15 }{ 2 } =16cm\\\\ Area\ =\sqrt { s(s-a)(s-b)(s-c) } \\\\ =\sqrt { 16(16-4)(16-13)(16-15) } \\\\ =\sqrt { 16\times 12\times 3\times 1 } ={ 24\ cm }^{ 2 }$ '

CBSE Questions for Class 9 Maths Heron'S Formula Quiz 2 - MCQExams.com

0:0:1

Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Heron'S Formula Quiz 2 - MCQExams.com

0:0:1

Practice Class 9 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.