Chapter 12 Heron's Formula Extra Questions

Explanation: 

 Set the triangle's height equal to h.

Due to the triangle's isosceles shape,

Let height = base + h

The answer is that the triangle's area is equal to 8 cm2 times 21 bases and 8 heights.

⟹ 21×h×h=8\s⟹h 2=16

⟹h=4cm

Height + Base = 4 cm

The triangle's right angles make it

Hypotenuse 2 equals Base 2 + Height 2.

Hypotenuse 2 = (4 + 4)

Hypertensive 2 = 32

Since Hypotenuse=32, Option A is the appropriate response.

Explanation: 

The right answer is D.

Justification for the ideal selection:

Determine the triangle's equilateral side.

We already know that the equilateral triangle's three sides are equal.

Thus let the equilateral triangle's side be.

Now put the three sides together to find the triangle's perimeter.

the equilateral triangle's perimeter is 60 m.

So,

a+a+a=60

=>3a=60

=>a=20 m

Calculate the equilateral triangle's area.

We are aware that the equilateral triangle's area can be expressed as: Where is the equilateral triangle's side?

The area will now be, and (d) is the best choice.

A= √3/4(20) 2 (∵ a=20 m)

=>A=√3/4(400)

=>A=100√32

Hence, the appropriate choice is (d)

Explanation: 

Given that a triangle's three sides are 56 cm, 60 cm, and 52 cm, respectively.

So, the triangle's semi-perimeter is s=a+b+c2=56+60+522=1682=84cm.

Triangle area equals s(s-a)(s-b)(s-c) [according to Heron's formula]

=√84(84−56)(84−60)(84−52) 

=√84×28×24×32 

=√4×7×3×4×7×4×2×3×4×4×2

 =√(4)6×(7)2×(3)2 =(4)3×7×3=1344cm2

Consequently, the triangle's area is 1344 cm2.

Explanation:

The right response is A 5.196 cm2

A given equilateral triangle has a side of 2 √ 3 cm.

Equilateral triangle area 

= 34(side)2

=√ 34(2+√3)2 + √34(2+4)3

= 3√3 = 3x1.732 = 5.196 cm2.

Consequently, 5.196 cm2 is the area of an equilateral triangle.

Explanation:

C 6 cm is the right answer choice.

Given that the area of an equilateral triangle is =9√3 cm2,

=√34(Side)2

⇒√34(Side)2=9√3

⇒(Side)2=36

∴Side=6cm

the following formula is used to get its side measurements: side = 6 cm (using the positive square root since the side is always positive).

Consequently, an equilateral triangle has a 6 cm length.

Explanation: 

The right answer is B 24 cm.

Given that the area of an equilateral triangle is equal to 16 √cm2,

the area of the triangle is equal to √34 (sides)2, ⇒√34(Side)2=16√3⇒(Side)2=64

Given that side is always positive, side equals 8 cm (positive square root).

An equilateral triangle has a perimeter of 3 and a side of 3 by 8 that is 24 cm.

Consequently, an equilateral triangle has a 24 cm perimeter.

Explanation: 

Consider the triangle ABC, which has sides of 35 cm, 54 cm, and 61 cm.

The triangle's semi-perimeter is now 35+54+612=1502=75cm [semi-perimeter, s=a+b+c2].

Area of ABC = √s(s-a)(s-b)(s-c)

= √75 (75- 35) (75- 54) (75 -61) 

= √75 x40 x21 x14

= √25x 3 x4x 2x5x 7x 3x 7x 2

= 5 2 2 3 7 √5 = 420 cm2

Additionally, the Area of ABC=12ABAltitude 1235CD=4205 CD=4202535 CD=245

Consequently, the longest altitude length is 24 cm.

Explanation: 

The right answer is A 15 c m 2

Assume that the dimensions of ABC are AB = AC = 4 cm and BC = 2 cm.

AB2=AD2+BD2[by Pythagoras theorem] in right-angled

 ( 4 ) 2 = A D 2 + 1

⇒A D  2= 16 − 1

⇒A D 2 = 15

∴Given that the square root of length is always positive, AD is equal to 15 cm.

Area of triangle = 12(base height); Area of ΔABC = 12x BC xAD 12

=12×2×√15=√15cm2

Alternate Approach

We are aware that the isosceles triangle's area is equal to a4√4b2a2.

where an is the base's length and b is the length of the equal sides.

In this case, the side lengths are b = 4 cm and a = 2 cm.

∴An isosceles triangle's area is 2 √4 4 2 44 =√ 64 42,√ 602 2 √152 = √15 cm2.

Explanation: 

Because a triangle board's edges are 6 cm, 8 cm, and 10 cm, respectively.

Now, a board with a triangle's semi-perimeter,

s=a+b+c2

=6+8+102=242=12cm

Now consider the size of a triangle.=√s(s−a)(a−b)(s−c)  [by Heron's equation]

=√12(12−6)(12−8)(12−10)

=√12×6×4×2

=√(12)2×(2)2

=12×2=24cm2

As a result, the price of painting a 1 cm2 area is equal to Rs. 0.09 An area of 24 cm2 will cost 0.09 times 24 rupees, or Rs. 2.16.

Therefore, Rs 2.16 is needed to paint the triangular board at the rate of 9 paise per cm2.

Explanation: 

Given: The triangle's base is 4 cm

Size equals 6 cm

contain triangles=1/2xBasexheight

=1/2x4x6

=24/2

=12cm2 

Thus, the region is12cm2

Explanation: 

A True

The explanation is the appropriate response.

Due to that,

base=4 cm

height=8cm

size of the triangle =½ xBasex Altitude

=1/2x4x4

=8cm2

Zone of

∆ABC is 8cm2

As a result, the assertion is true.

Explanation:

The right response is True.

Finding the triangle's sides is the first step.

To date, side-by-side isosceles triangles a,b,c

Perimeter = 11 cm

In addition, a=5

Due to the triangle's isosceles shape,b=c

Then, perimeter =a+b+c

=>a+b+c=11

=>5+b+b=11

=>5+2b=11

=>2b=6

=>b=6/2=3

Thus, we have

A = 5cm,

b= 3cm, c=3cm

Step 2:

Locating the location

Let 8 be the semi-perimeter now.

8=(a+b+c)/2

=(5+3+3)/2

=11/2

From Heron's Formula, let's now

size of the triangle =√s(s-a)(s-b)(s-c)

Area=√11/2(11/2-5)(11/2-3)(11/2-3)

       =√11/2x1/2x5/2x5/2

       =5√11/2 cm2

size of the triangle =5√11/2 cm2

This makes the assertion "TRUE"

Explanation:

8 cm is the length of each side.

The supplied statement must be evaluated to see if it is true or untrue.

3/4 (side)2 is the area of an equilateral triangle.

= √3/4 (8)²

= √3/4 (64)

= √3(16)

Size = 16 x 3 cm²

As a result, the assertion is untrue.

Explanation: 

The appropriate response is  96 cm2.

rhombus area equals 2 areas of Δ ABC

The area of a triangle with sides of 10cm, 10cm, and 16 cm is equal to -

S =10+10+162 = 18 cm

Currently, triangular area = √18(18−10)(18−10)(18−16)
= √18(8)(8)(2)
= √(9)(2)(8)(8)(2)
=8×2×3=48

Consequently, rhombic area=2×48=96 cm2

Explanation: 

Given that a parallelogram's base is 10 cm

Height = 3.5 cm

30 cm2 is the parallelogram's area.

The supplied statement must be evaluated to see if it is true or untrue.

Base x height equals the parallelogram's area.

= 10 × 3.5

= 35 cm²

As a result, the assertion is untrue.

Explanation: 

If the area of an equilateral triangle is 43 cm2 with a side of 4 cm, then find the area of a regular hexagon with sides of 4 cm.

Given that the equilateral triangle's area is 43 cm2,

Having said that,

The area of a regular hexagon equals six times the area of an equilateral triangle.

= 6 × 4√3

= 24√3 cm²

Consequently, the regular hexagon's area is 243 cm2.

Explanation: 

True

Assume a triangle has sides of 51, 37, and 20 meters.

Now, the triangle's semi-perimeter is equal to s=a+b+c2=51+37+202=1082=54m. Triangle's area is equal to (s, a, b, and c) using Heron's formula.

=√54(54−51)(54−37)(54−20) =√54×3×17×34

=√9×3×2×3×17×17×2 =3×3×2×17=306m2

Given that leveling costs Rs. 3 per square meter, Leveling per 306 m2 costs 3x306=Rs. 918.

Explanation: 

10.25 cm

Triangle's sides are 11, 12, and 13.

Triangle's area is equal to s(sa)(sb)(sc), where s=a+b+c2s=11+12+132s=18.

Triangle's area is 18(7)(6)(5).

Triangle's area is equal to 18(7)(6)(5)=61.5

Triangle's area equals 12 (base) (altitude).

61.5 equals 12 12 (height)

Height = 10.25

Explanation: 

BC = b = 65m, CA = c = 65m, and AB = a = 50m.

∴trapezoidal field's semi-perimeter, s=a+b+c2 =50+65+652=1802=90m

s=a+b+c2 =50+65+652=1802=90m 

∴ According to Heron's formula, the area of a triangle field is equal to =√s(s−a)(s−b)(s−c).

=√90(90−50)(90−65)(90−65) 

=√90×40×25×25 

=3×2×10×25

 =6×250=1500m2

One square meter of grass installation costs Rs. 7 Putting down grass costs Rs. 10500 for 1500 m2 (7 x 1500).

Therefore, it will cost Rs. 10,500 to install grass on the triangle field.

Explanation: 

Given, flyover side walls with triangular shapes have been used for advertising.

The walls have 13 m, 14 m, and 15 m sides.

The adverts generate an annual revenue per square meter of Rs. 2000.

For six months, a business hired one of its walls.

The company's six-month rent payment must be calculated.

A = 13 m, B = 14 m, and C = 15 m

Area of triangle = s(s - a)(s - b)(s - c) according to Heron's formula.

Where (a + b + c)/2 is the semiperimeter, s is

So, s = (13 + 14 + 15)/2 = 42/2 

s = 21 m

Area = √21(21 - 13)(21 - 14)(21 - 15)

 = √21(8)(7)(6)

 = √7 × 3 × 4 × 2 × 7 × 3 × 2 

= 7 × 3 × 2 × 2 

= 21 × 4

Area = 84 m²

One square meter of advertising costs Rs. 2000 per year.

Advertising expense for 84 m2 per year is 2000 divided by 84, or Rs. 168000.

The sum of the company's six-month rent payments, divided by two, is Rs. 84000.

Consequently, the corporation pays a monthly rent of Rs. 84000.

Explanation: 

Let O serve as the interior point of the equilateral triangle ABC, and let AQ, BR, and CP represent the perpendiculars emanating from point O.

Make an equilateral triangle with m on each side.

Area of an ΔOAB is equal to 12 AB + 12 OP + 12 (base + height)

= 12 a + 14 = 7 a cm2.(i)

Area of OBC=12BCOQ 

=12a10=5cm2(ii)

Area of OAC=12 AC OR 

=12 a 6 = 3 cm2(iii)

The area of an equilateral ΔABC is equal to the Area of (ΔOAB+ΔOBA+ΔOAC) = (7a+5a+3a)=15a cm2.(iv)

Semi-perimeter s=a+a+a2s=3a2cm is what we got.

∴ The area of an equilateral ΔABC is given by Heron's formula as=√ s(s-a)(s-b)(s-c).

= √3a2(3a2-a) (3a2-a) (3a2-a) (3a2-a) 

=√3a2×a2×a2×a2=√34 a2 ⋯(v)

Equations (iv) and (v) yield:

√ 34 a2=15a 

a=154333=6033=203cm.

Equation (v) yields

Area of ABC=√34(20√3)

when a=2032 =√34×400×3

 =300√3cm2

As a result, an equilateral triangle has a 300√3cm2 area.

Explanation: 

The base's ratio to the equal side is 3:2.

Make the sides 3 x 2. Third =3x, please.

the perimeter is 32 given

The perimeter is equal to the total of the sides, as we are aware. Thus, ⇒3x+2x+3x=32

⇒8x=32

⇒x=4

⇒232=16

The sides are therefore 12 cm, 8 cm, and 12 cm.

Therefore, the triangle's area is equal to 232 (16, 12, 16, 8, 16).

 =16×4×8×4

=322cm².

Explanation: 

ABCD is a parallelogram, therefore

We must determine the parallelogram's area.

Knowing that the area of a parallelogram equals double the area of a triangle BCD

Triangle BCD has

a = 25 cm,

b = 12 cm,

and c = 17 cm.

Using Heron's equation,

Triangle's area is equal to√s (s - a)(s - b)(s - c).

Where (a + b + c)/2 is the semiperimeter, s is

s = (25 + 12 + 17)/2

= 54/2

s = 27 cm as a result.

Triangle BCD's area is equal to √27(27 - 25)(27 - 17)(27 - 12)

= √27(2)(10)(15)

= √9 x3x x2x 5x 2x 5x 3

= 3x 3x 5x 2

= 9x 10

The Triangle BCD area is 90 cm2.

ABCD's area is 2(90).

180 cm2 is the parallelogram's area.

We must measure the distance between vertex A and side DC's altitude.

Let the altitude's length be h cm.

Area of parallelogram = base x height 

180 = 12 × h

h = 180/12

h = 90/6

h = 30/2

h = 15 cm

The altitude is 15 cm as a result.

Explanation: 

Assume that the parallelogram ABCD has the sides AB = CD = 60 m, BC = DA = 40 m, and BD = 80 m.

Area of ABCD (parallelogram's area) = 2.(i) Take into account ΔABD, 

Semi-perimeter of a triangle,

where s=a+b+c2

=AB+BD+DA2 

=60+80+402

=1802 

=90m.

∴ Area of [using Heron's formula] ΔABD=√s(s-a)(s-b)(s-c).

=√90(90−60)(90−80)(90−40) 

=√90×30×10×50 

=100×3√15=300√15m2

The area of the parallelogram ABCD is 2x300√15=600√15m2 according to Eq. (i).

Consequently, the parallelogram's area is 600√15m2.

Explanation: 

First step: Identifying the sides

Because the sides have the ratio 6:7:8

=>a=6x,b=7x,c=8x

Perimeter=a+b+c

420=6x+7x+8x=21x

=>x=20

=>a=120,b=140,c=160

Step - 2: estimating the area

 Areas equal to s=2a+b+c=210,

=>√210x90x70x50,

and 210015 are defined as s√(sa- sb- sc) 

210x90x70x50,

and 210x90x70x50, respectively.

Therefore, the given field has a surface size of 2100√15 sq. m.

Explanation: 

Given that ABCD is a quadrilateral with sides measuring 6 cm, 8 cm, 12 cm, and 14 cm, respectively.

Come on, join AC.

We have ABC with a right angle to B.

The Pythagorean theorem states that AC2=AB2+BC2= 62+82=36+64=100.

[Taking positive square] AC = 10 cm

The area of quadrilateral ABCD is equal to the sum of the areas of  ΔABC+ and  ΔACD.

Now, the triangle's area is equal to 12(base + height)

= 12x6 x 8= 24cm2.

In the equation  ΔACD, AC equals 10 cm, CD equals 12 cm, and DA equals 14 cm.

Now, consider the  ΔACD's semi-perimeter, 

s=a+b+c2=10+12+142=362=18 cm.

Area of Δ ACD=√s(s-a)(s-b)(s-c), according to Herom's formula

=√18(18−10)(18−12)(18−14)

 =√18×8×6×4

=√(3)2×2×4×2×3×2×4

 =3×4×2√3×2=24√6cm2

Quadrilateral ABCD area equals area of ΔABC plus area of ΔACD, or =24+24√6=24(1+√6)cm2

Consequently, the quadrilateral's area is 24(1+6)cm2.

Explanation: 

If ABCD is a rhombus, then AB=BC=CD=DA=x

Perimeter of rhombus = 40 cm,

=>4x = 40 cm, and x = 10 cm,

resulting in AB=BC=CD=DA=10 cm.

In △ABC, S=2a+b+c=210+10+12=16cm 

ar,△ ABCD=2x48=96cm2, and ABCD=16(16-10)(16-10)(16-12)=16x6x6x4=48cm2

The cost to paint the sheet is Rs. 960 [both sides]/596 [both sides].

Explanation: 

Utilize the method above to find TR.

∴ The area of the trapezium is equal to 12

(Sum of Parallel Lines) Distance between two points is =12(PS+QR)×TR=12×(12+7)×12

=12×19×12=114m2

Consequently, the trapezium's area is 114m2.

Explanation: 

We are aware that ABCD's square sheet's area is 21x diagonal2.

By changing the value

ABCD's square sheet area is 21x44x44= 968 cm2.

From the illustration, we may deduce that the yellow sheet area equals the sum of regions I and II.

It can be expressed mathematically as follows: Area of yellow sheet = 21x area of square sheet ABCD

By changing the value

Yellow sheet area =21x968=484 cm2.

We can deduce from the figure that Region IV corresponds to the area of the bed sheet.

One way to spell it is as

The red sheet area equals 1/4 of a square sheet of ABCD

By changing the number 1/4x968=242cm2 for the area of the red sheet

Think about ΔAEF

AE = 20 cm, EF = 14 cm, and AF = 20 cm are known values.

If an is 20 cm, b is 14 cm, and c is 20 cm, then

So, we have s=2a+b+c.

Using the values s=220+14+20 as a substitute 

So, we obtain s=27cm.

The area is defined as =s(s-a)(s-b)(s-c).

By changing Area to Area=27(27-20)(27-14)(27-20)

The result is Area=27 x7 x13 x7.

By multiplying, Area = 2139

By changing Area=21x6.25 to Area=131.65cm2, we obtain Area.

The figure shows that the area of the green sheet equals the sum of the areas in regions III and V.

You can write it out as follows: Area of green sheet = 41 x Area of square sheet ABCD + 131.25

Area of green sheet = 41x968+131.25 by substituting

The area of the green sheet calculated at =242 + 131.25

Additionally, the green sheet's area is =373.25 cm2.

As a result, the yellow sheet's area is 484 cm2, the red sheet's area is 242 cm2, and the green sheet's area is 373.25 cm2.

Explanation: 

Given that a triangle's perimeter is 50 cm

A triangle has one side that is 4 cm longer than the other side.

The smaller side is 6 cm smaller than the third side.

We must calculate the triangle's area.

Let x cm be the smaller side.

x + 4 cm on one side

Third side = 6 cm - 2x

The triangle perimeter equals the total of its sides.

50 = x + 4 + x + 2x - 6

50 = 4x - 2

4x = 50 + 2

4x = 52

x = 52/4

x = 13 cm

the smaller side measures 13 cm.

13 + 4 = 17 cm on one side.

Third side: =13(2) – 6= 26–6= 20 cm

Using Heron's equation,

Triangle's area is equal to √(s - a)(s - b)(s - c).

Where (a + b + c)/2 is the semiperimeter, s is

Here, the measurements are a=13 cm, 17 cm, and c=20 cm.

s = (13 + 17 + 20)/2

= 50/2

s = 25 cm as a result.

Area = √25(25 - 13)(25 - 17)(25 - 20)

 = √25(12)(8)(5)

 = √5 × 5 × 4 × 3 × 4 × 2 × 5

 = (5 × 4)√5 × 3 × 2 = 20√30 cm²

Consequently, the triangle has an area of 20 by 30 cm2.

Explanation: 

Think of the smaller side of the trapezium as being x cm.

Consequently, the larger parallel can be expressed as (x + 4) cm.

We are aware that the area of a trapezium is equal to 21 plus the height.

Using the values

475=21(x+(x+4))x19 as a substitute

Further calculations show that

25=21(2x+4).

The result is 50 = 2x + 4.

By subtracting 

2x =50 -42

x becomes 46.

divided by x, 23 cm

hence, we do (x + 4) = 23 + 4 = 27 cm, the larger parallel side.

Two parallel sides are therefore 23 cm and 27 cm long.

Explanation:
Let ABCD, where AB is 15 meters and BC is 40 meters, be the specified rectangular plot.

The front, back, and each of the other sides should each have a minimum of a 3 m wide area, and the other sides should each have a minimum of a 2 m wide space. 

OD=CL=AP=BK=3m and DN=CM=BJ=AI=2m are the results.

The length of the home with the maximum buildable area is therefore =EF=AB-(AI+BJ)=15m-(2m+2m)=11m.

Similar to this, the width of the largest possible house is equal to EH=AD-(DO+AP)=40m-(3m+3m)=34m.

Therefore, the built-up area of the house is =length×breadth=11m×34m=374m2.

Explanation: 

We draw a line CE that is parallel to line AB in the trapezium ABCD.

We know that DC = AE = 30m.

Consequently, BE = AB - AE = 90 - 30 = 60m.

Using Pythagoras' theorem, (BC)2=(BE)2+(EC)2 for right-angled ΔBEC.

⇒ (100)2=(60)2+(EC)2

⇒ (EC)2=10000−3600

⇒ (EC)2=6400

∴ EC=√6400=80m

Taking a positive square root because length is usually positive, EC=6400=80m

∴ Area of the trapezium ABCD = 12 (Sum of Parallel Sides) 

=12(AB+CD)×EC=12(90+30)×80

=12×(120)×80=4800 m2

The cost to plough a field measuring 1 m2 is Rs. 4 The cost of ploughing a 4800 m2 field is 4800 x 4 = Rs. 19200.

The entire cost of preparing the field is therefore Rs. 19,200.

Explanation: 

We know that ABC is a triangle.

The sides measure AB=7.5 cm, 

AC=6.5 cm, 

and BC=7 cm.

A parallelogram DBCE with the same area as triangle ABC is built on foundation BC.

We must determine the parallelogram's DF height.

According to Heron's formula, 

the area of triangle ABC is =√s(s - a)(s - b)(s - c).

Where (a + b + c)/2 is the semiperimeter, s is

Here, an is equal to 7.5 cm, 

b to 7 cm, and c to 6.5 cm.

s = (7.5 + 7 + 6.5)/2 

= 21/2

s = 10.5 cm as a result.

Area = √10.5(10.5 - 7.5)(10.5 - 7)(10.5 - 6.5) 

= √10.5(3)(3.5)(4) 

= √36.75 × 12 

= √441

Triangle ABC's area is 21 cm2.

Base x height equals the parallelogram's area.

DBCE parallelogram area equals BC minus DF

Given that area of triangle ABC = area of parallelogram DBCE

Therefore, 21 = 7 DF

 DF = 21/7

DF = 3 c

The parallelogram is 3 cm tall as a result.

Explanation: 

QC=45 cm, PD=40 cm

A rectangle with the dimensions AB = 51 cm and BC = 25 cm is ABCD.

Let QC=9x and PD=8x because the ratio of the parallel sides QC and PD is 9:8.

A trapezium's surface area is equal to 12 times its parallel sides' sum and their separation.

Now, the trapezium's PQCD 

area is equal to 12 (9 x + 8 x) x25 cm2

=12x 17 x 25.

Rectangle area equals length times width

Rectangle ABCD's area is= BCxCD=51x25.

Given is that trapezium PQCD area.

The area of the rectangle ABCD is =56.

12×17x×25=56×51×25

 ⇒x=56×51×25×2×117×25=5

As a result, the lengths QC=9x=95=45cm and PD=8x=85=40cm.

Explanation: 

Given that a rectangle tile measures 50 cm by 70 cm.

∴ Tile surface area is = 50 x 70, = 3500 cm2.

One triangle's sides should be

A = 25 cm, b = 17 cm, and c = 26 cm.

At this point, the semi-perimeter

s=a+b+c2=25+17+262=682=34.

∴ The area of a triangle is equal to s(sa), sb, and sc [according to Heron's formula].

=√34×9×17×8

 =√17×2×3×3×17×2×2×2 

=17×3×2×2=204 cm2

Eight triangles have a total area of = 204 x 8 = 1632 cm2.

Now, the design's area is equal to the sum of the eight triangles, =

 1632 cm2.

Additionally, the remaining tile area is equal to the sum of the rectangle's and the design's areas, which is  =3500 - 1632, 

= 1868 m2.

Consequently, the design's overall area is 1632 cm2, leaving 1868 cm2 of the tile unoccupied.