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a = 2, b = 4, c = 4

s = (a + b + c)/2

⇒ s = (2 + 4 + 4)/2 = 10/2 = 5.

Area(Δ) = √s(s-a)(s-b)(s-c)

⇒ Area(Δ) = √5(5-2)(5-4)(5-4)

⇒ Area(Δ) = √5×3×1×1

⇒ Area(Δ) = √15 cm²

Hence, the correct option is A.

Let a, b and c are the sides of triangle and s is 

the semi-perimeter, then its area is given by:

A = \(\sqrt{s(s-a)(s-b)(s-c)}\)where s = \(\frac{a+b+c}2\)[Heron’s Formula]

s = \(\frac{a+b+c}2\) = \(\frac{3+4+5}2\) = 6

A = \(\sqrt{6(6-3)(6-4)(6-5)}\)

A = \(\sqrt{6\times3\times2\times1}\) = 6 cm²

B. 24 cm

Given, area of an equilateral triangle = 16√3 cm²

Area of an equilateral triangle = √3/4 (side)²

√3/4(Side)² = 16√3 

=> (Side)² = 64

=> Side = 8 cm [taking positive square root because side is always positive]

Perimeter of an equilateral triangle = 3 x Side

= 3 x 8 

= 24 cm

Hence, the perimeter of an equilateral triangle is 24 cm.

D. 6 cm

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

∴ Area of an equilateral triangle = √3/4(Side)²

=> √3/4 (Side)² = 9√3

=> (Side)² = 36

∴ Side = 6 cm  [taking positive square root because side is always positive]

Hence, the length of an equilateral triangle is 6 cm.

(C) 1344 cm²

Explanation:

According to the question,

Sides of a triangle,

a = 56, b = 60, c = 52

s = (a + b + c)/2

⇒ s = (56 + 60 + 52)/2

= 168/2 = 84.

Area of triangle = √s(s-a)(s-b)(s-c)

= √84(84-56)(84-60)(84-52)

= √84×28×24×32

= 1344cm²

Hence, Options C is the correct answer.

Sides of triangle are in ratio: 3 : 5 : 7

a = 3x, b = 5x, c = 7x

Since the perimeter of a triangle is given by:

a+b+c = perimeter

3x +5x+7x = 300

x = \(\frac{300}{15}\) = 20

x = 20

Therefore sides of the triangle are:

a = 3x = 3 x 20 = 60,

b = 5x = 5 x 20 = 100,

c = 7x = 7 x 20 = 140

When a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = \(\sqrt{s(s-a)(s-b)(s-c)}\) where s = \(\frac{a+b+c}{2}\)[Heron’s Formula]

s = \(\frac{a+b+c}{2}\) = \(\frac{60+100+140}{2}\) = 150

A = \(\sqrt{150(150-60)(150-100)(150-140)}\)

A = \(\sqrt{150\times 90\times50 \times10}\) = \(\sqrt{15\times 9\times5 \times10000}\) 

= \(\sqrt{15\times 3\times3 \times5 \times10000}\) = \(\sqrt{15\times 3\times15 \times10000}\) 

= \(15\times 100\sqrt{3}\) = \(1500\sqrt{3}\) m²

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = \(\sqrt{s(s-a)(s-b)(s-c)}\)where s = \(\frac{a+b+c}{2}\)[Heron’s Formula]

a = 35, b = 54, c = 61

s = \(\frac{a+b+c}{2}\) = \(\frac{35+54+61}{2}\) = 75

A = \(\sqrt{75(75-35)(75-54)(75-61)}\)

A = \(\sqrt{75\times 40 \times21 \times 14}\) = 939.15 cm²

Altitude on side 35 cm:

Area of triangle = \(\frac{1}2\)(Base x Altitude)

939.15 = \(\frac{1}2\)(35 x Altitude)

Altitude = 53.66 cm 

Altitude on side 54 cm:

Area of triangle =\(\frac{1}2\)(Base x Altitude)

939.15 =\(\frac{1}2\)(54 x Altitude)

Altitude = 34.78 cm 

Altitude on side 61 cm: 

Area of triangle =\(\frac{1}2\)(Base x Altitude)

939.15 =\(\frac{1}2\)(61 x Altitude)

Altitude = 30.79 cm

Therefore smallest Altitude is: 30.79 cm

Given that the sides are in ratio of 6:7:8

Let the sides of the triangle be 6x, 7x and 8x.

Perimeter(Δ) = 420

⇒ 6x + 7x + 8x = 420

⇒ 21x = 420

⇒ x = 20

Therefore the sides are 120m, 140m and 160m.

a = 120, b = 140, c = 160

s = (a + b + c)/2

⇒ s = (120 + 140 + 160)/2 = 420/2 = 210.

Area(Δ) = √s(s-a)(s-b)(s-c)

⇒ Area(Δ) = √210(210-120)(210-140)(210-160)

⇒ Area(Δ) = √210×90×70×50

⇒ Area(Δ) = 2100√15 m²

False

Justification:

Area of triangle = ½ × Base × Altitude

= ½ × 4 × 6

= 12cm²

Hence, the statement “the area of a triangle with base 4 cm and height 6 cm is 24 cm²” is False.

Sides of triangle are in ratio: 3 : 4 : 5

a = 3 x, b = 4x, c = 5x

Since the perimeter of a triangle is given by:

a+b+c = perimeter.

3x+4x+5x = 144

x = \(\frac{144}{12}\) = 12

x = 12

Therefore sides of the triangle are:

a = 3x  = 3 x 12 = 36,

b = 4x =4 x 12= 48,

c = 5x = 5 x 12= 60

When a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = \(\sqrt{s(s-a)(s-b)(s-c)}\)where s = \(\frac{a+b+c}2\)[Heron’s Formula]

s = \(\frac{a+b+c}2\) = \(\frac{36+48+60}2\) = 72

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

A = \(\sqrt{72\times 36 \times24 \times12}\) = 864 cm²

Area of triangle = \(\frac{1}2\)(Base x Altitude)

864 = \(\frac{1}2\)(60 x Altitude)

Altitude = 28.8 cm

False. Since in Heron’s formula,

s = (a + b + c)/2 

= 1/2 (perimeter of triangle)

Area of rectangle (ABCD) = BC × CD

⇒ Area of rectangle (ABCD) = 51 × 25 = 1275 cm²

Area of trapezium PQCD = 5/6 × Area of rectangle (ABCD)

⇒ Area of trapezium PQCD = 5/6 × 1275 = 1062.5 cm²

Given that QC:PD = 9:8

Let QC = 9x and PD = 8x

Area (PQCD) = (Sum of parallel sides)/2 x Height

⇒ Area (PQCD) = (9x + 8x)/2 x 25

⇒ 1062.5 = (17x)/2 x 25

⇒ 85 = 17x

⇒ x = 5cm

QC = 9x = 45cm

PD = 8x = 40 cm

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = \(\sqrt{s(s-a)(s-b)(s-c)}\) where s = \(\frac{a+b+c}{2}\)[Heron’s Formula]

a = 15,b = 13, c = 14

a = \(\frac{a+b+c}{2}\) = \(\frac{15+12+14}{2}\) = 21

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

A = \(\sqrt{21\times6 \times8 \times7}\) = 84 cm²

Area of triangle = \(\frac{1}2\)(Base x Altitude)

84 = \(\frac{1}2\) (14 x Altitude)

Altitude = 12 cm

Given: Base of a triangle = 5 cm and altitude = 4 cm 

Area of triangle = 1/2 x base x altitude

= 1/2 x 5 x 4

= 10

Area of triangle is 10 cm².

a = 11, b = 12, c = 13

s = (a + b + c)/2

⇒ s = (11 + 12 + 13)/2 = 36/2 = 18.

Area(Δ) = √s(s-a)(s-b)(s-c)

⇒ Area(Δ) = √18(18-11)(18-12)(18-13)

⇒ Area(Δ) = √18×7×6×5

⇒ Area(Δ) = 6√105 cm² = 61.48 ∼ 61.5 cm²

Area(Δ) = 1/2 × Base × Altitude

⇒ Area(Δ) = 1/2 × 12 × 10.25

⇒ Area(Δ) = 61.5 cm²

Hence the statement is True.

According to the question,

Sides of the triangular field are 50 m, 65 m and 65 m.

Cost of laying grass in a triangular field = Rs 7 per m²

Let a = 50, b = 65, c = 65

s = (a + b + c)/2

⇒ s = (50 + 65 + 65)/2

= 180/2

= 90.

Area of triangle = √s(s-a)(s-b)(s-c)

= √90(90-50)(90-65)(90-65)

= √90×40×25×25

= 1500m²

Cost of laying grass = Area of triangle × Cost per m²

= 1500×7

= Rs.10500

To find the area of rhombus, we divide it into two triangles.

As all the sides of a rhombus are equal, we have for a triangle

a = 10, b = 10, c = 16

s = (a + b + c)/2

⇒ s = (10 + 10 + 16)/2 = 36/2 = 18.

Area(Δ) = √s(s-a)(s-b)(s-c)

⇒ Area(Δ) = √18(18-10)(18-10)(18-16)

⇒ Area(Δ) = √18×8×8×2

⇒ Area(Δ) = 48cm²

As the sides of the other triangle are also same, so their areas will also be equal.

Area(Rhombus) = Area(Δ) + Area(Δ)

⇒ Area(Rhombus) = 48 + 48 = 96cm²

Hence, the statement is True.

Using Pythagorous theorem:

(Hypothesis)² = (Base)² + (Altitude)²

(13)² = 5² + (Altitude)² 

(Altitude)² = 169 – 25 = 144 

Altitude = 12 cm 

Area of triangle = \(\frac{1}2\)(5 x 12)

= 30 cm²

According to the question,

The lengths of the perpendiculars are 14 cm, 10 cm and 6 cm.

We know that,

Area of an equilateral triangle of side a = √3/4 a²

We divide the triangle into three triangles,

Area of triangle = (1/2 × a × 14) + (1/2 × a × 10) + (1/2 × a × 6)

√3/4 a² = ½ × a × (14 + 10 + 6)

√3/4 a² = ½ × a × 30

a = 60/√3

= 20√3

Area of the triangle = √3/4 a²

=√3/4 (20√3)²

= 300√3 cm²

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = \(\sqrt{s(s-a)(s-b)(s-c)}\) where s = \(\frac{a+b+c}{2}\) [Heron’s Formula]

a = 9, b = 12, c = 15

s = \(\frac{a+b+c}{2}\) = \(\frac{9+12+15}{2}\) = 18

A = \(\sqrt{18(18-9)(18-12)(18-15)}\)

A = \(\sqrt{18 \times9 \times6 \times3}\) = 54 cm²

If two opposite faces of a regular solid have the same shape and the same size, the solid is said to be of uniform cross-section.

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by

A. \(\sqrt{s(s-a)(s-b)(s-c)}\) where s = \(\frac{a+b+c}{2}\)[Heron’s Formula]

a = 150, b = 120, c = 200

s = \(\frac{a+b+c}{2}\) = \(\)\(\frac{100+120+200}{2}\) = 235

A = \(\sqrt{235(235-150)(235-120)(235-200)}\)

A = \(\sqrt{235 \times85 \times115 \times35}\) = \(25\sqrt{47 \times17 \times23 \times7}\) = \(25\sqrt{128369}\) cm²

Each side of an equilateral triangle = a = 4 cm

Formula for Area of an equilateral triangle = (√3/4 ) × a² 

= (√3/4) × 4² 

= 4√3 

Area of an equilateral triangle is 4√3 cm².

For area of one triangle,

a = 25, b = 17, c = 26

s = (a + b + c)/2

⇒ s = (25 + 17 + 26)/2 = 68/2 = 34

Area(Δ) = √s(s-a)(s-b)(s-c)

⇒ Area(Δ) = √34(34-25)(34-17)(34-26)

⇒ Area(Δ) = √34×9×17×8

⇒ Area(Δ) = 204 cm²

Area of design = 8 × Area(Δ)

∴ Area of design = 8 × 204 = 1632 cm²

Area of remaining tile = Area of rectangle – Area of design

⇒ Area of remaining tile = (50×70) – 1632

⇒ Area of remaining tile = 3500 – 1632 = 1868 cm²

Let the smaller side be = x cm

Then, larger side = (x + 4) cm

And, third side = (2x-6) cm

Given that,

Perimeter = 50 cm

⇒ x + (x + 4) + (2x-6) = 50

⇒ 4x-2 = 50

⇒ 4x = 52

⇒ x = 13

Therefore, smaller side = 13cm

Larger side = x + 4 = 13 + 4 = 17cm

Third side = 2x-6 = 2×13 – 6 = 26-6 = 20cm

To find area of triangle,

Let a = 13, b = 17, c = 20

s = (a + b + c)/2

⇒ s = (13 + 17 + 20)/2 = 50/2 = 25.

Area of triangle = √s(s-a)(s-b)(s-c)

= √25(25-13)(25-17)(25-20)

= √25×12×8×5

= 20√30 cm²

According to the figure,

AC = BD = 44cm

AO = 44/2 = 22cm

BO = 44/2 = 22cm

From ΔAOB,

AB² = AO² + BO²

⇒ AB² = 22² + 22²

⇒ AB² = 2 × 22²

⇒ AB = 22√ 2 cm

Area of square = (Side)²

= (22√2)²

= 968 cm²

Area of each triangle (I, II, III, IV) = Area of square /4

= 968 /4

= 242 cm²

To find area of lower triangle,

Let a = 28, b = 28, c = 14

s = (a + b + c)/2

⇒ s = (28 + 28 + 14)/2 = 70/2 = 35.

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

= √35(35-28)(35-28)(35-14)

= √35×7×7×21

= 49√15 = 189.77 cm²

Therefore,

We get,

Area of Red = Area of IV

= 242 cm²

Area of Yellow = Area of I + Area of II

= 242 + 242

= 484 cm²

Area of Green = Area of III + Area of lower triangle

= 242 + 189.77

= 431.77 cm²

In a rhombus, all sides are equal.

Perimeter of rhombus = 40 cm

Side of rhombus = 40/4 = 10 cm

a = 10, b = 10, c = 12

s = (a + b + c)/2

⇒ s = (10 + 10 + 12)/2 = 32/2 = 16.

Area(ΔBCD) = √s(s-a)(s-b)(s-c)

⇒ Area(ΔBCD) = √16(16-10)(16-10)(16-12)

⇒ Area(ΔBCD) = √16×6×6×4

⇒ Area(ΔBCD) = 48 cm²

As ABCD is a rhombus, Area(ΔBCD) = Area(ΔABD)

⇒ Area of rhombus(ABCD) = Area(ΔBCD) + Area(ΔABD)

⇒ Area of rhombus(ABCD) = 48 + 48

⇒ Area of rhombus(ABCD) = 96cm²

Since the sides of a rhombus are equal there fore each side = \(\frac{perimeter}4\) = \(\frac{22}4\) = 8 m

Let a, b and c are the sides of triangle and s is 

the semi-perimeter, then its area is given by:

A = \(\sqrt{s(s-a)(s-b)(s-c)}\)where s = \(\frac{a+b+c}2\)[Heron’s Formula]

s = \(\frac{a+b+c}2\) = \(\frac{8+8+10}2\) = 13

A = \(\sqrt{13(13-8)(13-8)(13-10)}\)

A = \(\sqrt{13\times5\times5\times3}\) = 31.22 cm²

Hence, area of rhombus ABCD = 2 ×31.22 m² 

= 62.44 m²

Total painting area of rhombus = 62.44 × 2 = 124.88 m² 

Cost of painting of rhombus on both sides = 124.88 × 5 = Rs 624.50

a = 6.5, b = 7, c = 7.5

s = (a + b + c)/2

⇒ s = (6.5 + 7 + 7.5)/2 = 21/2 = 10.5

Area(Δ) = √s(s-a)(s-b)(s-c)

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

⇒ Area(Δ) = √10.5×4×3.5×3

⇒ Area(Δ) = 21 cm²

Area of parallelogram (BCED) = Area (Δ)

⇒ BC × DF = 21

⇒ 7 × DF = 21

⇒ DF = 3 cm

We know that,

Area of parallelogram(ABCD) = Area(ΔBCD) + Area(ΔABD)

For Area (ΔBCD),

We have,

a = 12, b = 17, c = 25

s = (a + b + c)/2

⇒ s = (12 + 17 + 25)/2 = 54/2 = 27.

Area of (ΔBCD) = √s(s-a)(s-b)(s-c)

= √27(27-12)(27-17)(27-25)

= √27×15×10×2

= 90 cm²

Since, ABCD is a parallelogram,

Area(ΔBCD) = Area(ΔABD)

Area of parallelogram(ABCD) = Area(ΔBCD) + Area(ΔABD)

= 90 + 90

= 180 cm²

Let altitude from A be = x

Also, Area of parallelogram(ABCD) = CD × (Altitude from A)

⇒ 180 = 12 × x

⇒ x = 15 cm

Let a, b and c are the sides of blade and s is

the semi-perimeter, then its area is given by:

A = \(\sqrt{s(s-a)(s-b)(s-c)}\)where s = \(\frac{a+b+c}2\)[Heron’s Formula]

s = \(\frac{a+b+c}2\) = \(\frac{5+5+1}2\) = 5.5

A = \(\sqrt{5.5(5.5-5)(5.5-5)(5.5-1)}\)

A = \(\sqrt{5.5\times0.5\times0.5\times4.5}\) = 2.49 cm²

Hence, total area of both the blades = 2 × 2.49 = 4.98 cm²

Let a, b and c are the sides of triangle and s is t

he semi - perimeter, then its area is given by:

A = \(\sqrt{s(s-a)(s-b)(s-c)}\)where s = \(\frac{a+b+c}2\)[Heron’s Formula]

s = \(\frac{a+b+c}2\) = \(\frac{13+14+15}2\) = 21

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

A = \(\sqrt{21\times8\times7\times6}\) = 84 cm²

Therefore area of ∆ = \(\frac{1}2\)(Base x Altitude)

84 × 2 = 14× Altitude 

Altitude = 12 cm

Sides of triangle are in ratio: 3 : 4 : 5

a = 3x, b = 4x, c = 5x

Since the perimeter of a triangle is given by:

a+b+c = perimeter

3x+4x+5x = 144

x = \(\frac{144}{12}\)= 12

x = 12

Therefore sides of the triangle are:

a = 3x = 3 x 12 = 36,

b = 4x = 4 x 12 = 48,

c = 5x = 5 x 12 = 60

When a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

\(A = \sqrt{s(s-a)(s-b)(s-c)}\) where s  = \(\frac{a+b+c}2\)[Heron’s Formula]

s = \(\frac{a+b+c}2\) = \(\frac{36+48+60}2\) = 72

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

A = \(\sqrt{72\times36\times24\times12}\) = 864 cm²

Let the third side of the triangle is x

Since the perimeter of a triangle is given by:

a+b+c = perimeter

78+50+x = 240

x = 240 - 128

x = 112^dm

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = \(\sqrt{s(s-a)(s-b)(s-c)}\)where s = \(\frac{a+b+c}{2}\)[Heron’s Formula]

a = 78, b = 50, c = 112

s = \(\frac{a+b+c}{2}\) = \(\frac{78+50+112}{2}\) = 120

A = \(\sqrt{120(120-78)(120-50)(120-112)}\)

A = \(\sqrt{120\times42 \times70 \times8}\) = 1680 dm²

Area of triangle = \(\frac{1}2\)(Base x Altitude)

1680 = \(\frac{1}2\)(50 x Altitude)

Altitude = 67.2 dm

Let a, b and c are the sides of triangular strips and s is

the semi-perimeter, then its area is given by:

A = \(\sqrt{s(s-a)(s-b)(s-c)}\)where s =\(\frac{a+b+c}2\)[Heron’s Formula]

s = \(\frac{a+b+c}2\)= \(\frac{25+25+14}2\) = 32

A = \(\sqrt{32(32-25)(32-25)(32-14}\)

A =\(\sqrt{32\times7\times7\times18}\) = 168 cm²

Hence, total area of 5 Nos of triangular strips of one type = 5 × 168 = 840 cm²

(A) √32 cm

Explanation:

Let height of triangle = h

As the triangle is isosceles,

Let base = height = h

According to the question,

Area of triangle = 8cm²

⇒ ½ × Base × Height = 8

⇒ ½ × h × h = 8

⇒ h² = 16

⇒ h = 4cm

Base = Height = 4cm

Since the triangle is right angled,

Hypotenuse² = Base² + Height²

⇒ Hypotenuse² = 4² + 4²

⇒ Hypotenuse² = 32

⇒ Hypotenuse = √32

Hence, Options A is the correct answer.