Math, 18.10.2019 11:00, sabina3193

Aright angled triangle whose sides are 3cm 4cm 5cm is revolved about the sides containing the right angle in two ways . find the diffrrence in volume of the two cones formed .also find their curved surfaces

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The correct answer was given: daniya0

CASE I ) triangle revolved around 3cm side:
h = 3cm, r = 4cm, l = 5cm
volume = ⅓πr²h
= ⅓ * 22/7 * 16 * 3
= 50.29 cm³
CSA = πrl
= 22/7 * 4 * 5
= 62.86 cm²

CASE II ) triangle revolved around 4cm side:
h = 4cm, r = 3cm, l = 5cm
volume = ⅓πr²h
= ⅓ * 22/7 * 9 * 4
= 37.71 cm³
CSA = πrl
= 22/7 * 3 * 5
= 47.14 cm²

therefore difference in volume = 12.58cm³ and difference in CSA is 15.72cm²

The correct answer was given: sabina3193

The difference in volumes of the two cones so formed is 4 π cm³ and the curved surfaces are 20 π cm² & 15 π cm².

Step-by-step explanation:

Given:

Case : 1

When the right angled triangle is revolved around the side 3 cm

Radius of the cone , r = 4 cm

Height of the cone , h = 3 cm

Slant height of the cone , l = 5 cm

Volume of the cone = ⅓ × πr²h

= ⅓ π × 4² × 3 = 16π cm³

Volume of the cone = 16π cm³

Curved surface area = πrl

= π × 4 × 5 = 20π cm²

Curved surface area = 20π cm²

Case 2 :

When the right angled triangle is revolved around the side 4 cm

Radius of the second cone, r = 3 cm

Height of the cone, h = 4 cm

Slant height of the cone , l = 5 cm

Volume of the cone = ⅓ × πr²h

⅓ × π × 3² × 4 = 12π cm³

Volume of the cone = 12π cm³

Curved surface area = πrl

= π × 3 × 5 = 15 π cm²

Curved surface area = 15 π cm²

Difference of the volumes of two cone = (16 - 12) πcm³ = 4π cm³

Hence, the difference in volumes of the two cones so formed is 4 π cm³ and the curved surfaces are 20 π cm² & 15 π cm².

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The correct answer was given: atharvakarawade1

5^2=3^2+4^2
=25=9+16
Therefore, hypotenuse is 5 cm

The correct answer was given: lighy6155

60cm^2
hope it helps you

The correct answer was given: Mannubhalla93

Step-by-step explanation:

By heroins formula

We get s=13+14+15/2

S= 42/2 =21

Now area= root of s(s-a)(s-b)(s-c)

Root of 21 (21-13)(21-14)(21-15)

Root of 21 (8)(7)(6)

Root of 21 (366)

Root of 7056

84cm^2 is the area

Hope it is useful

The correct answer was given: Firdaus17f

Step-by-step explanation:

I hope it's useful for you

Q.1) Find the area of the triangle whose sides are 13cm,14cm,15cm.

need a correct answer

The correct answer was given: jack626259

Step-by-step explanation:

Use the formula i. e Heron's Formula

S= (a+b+C) /2

and then find by using the formula

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The correct answer was given: rajveer8591

The area of the triangle is 6 centimeter square.

Step-by-step explanation:

Given : Triangle whose sides are 3cm ,4cm,5cm.

To find : The area of the triangle?

Solution :

Formula to find area of triangle is

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

Where, s is the semi perimeter,

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

a=3 ,b=4, c= 5 cm sides of the triangle

Substituting in the formula,

$s=\frac{3+4+5}{2}$

$s=\frac{12}{2}$

s=6

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

$A=\sqrt{6(6 - 3)(6 - 4)(6 - 5)}$

$A= \sqrt{6 \times 3 \times 2 \times 1}$

$A= \sqrt{36}$

A= 6 cm^2

Therefore, The area of the triangle is 6 centimeter square.

The correct answer was given: jack626259

The following Question is a part of Application of Heron's Formula

Your Given:-

$\tt \star a = 13cm \\\\ \tt \ \star b = 14cm \\\\ \tt \star c = 15cm \\\\ \tt Where \ \ a,b \ \ and \ \ c \ \ are sides \ \ of \ \ the \ \ Triangle$

Solution:-

We know that

$\tt Area \ \ of \ \ Triangle =\sqrt{s(s-a)(s-b)(s-c)} \\\\ \tt Where \ \ s \ \ is \ \ the \ \ semi-perimeter \ \ of \ \ Triangle$

So, finding 's' first

$\tt s = \dfrac{a+b+c}{2} \\\\ \tt \Rightarrow s = \dfrac{13+14+15}{2} \\\\ \tt \Rightarrow s = \dfrac{42}{2} \\\\ \tt \Rightarrow s= 21$

Now replacing values of s,a,b and c

$\tt Area = \sqrt{21(21-13)(21-14)(21-15)} \\\\ \tt \Rightarrow Area = \sqrt{21 \times 8 \times 7 \times 6} \\\\ \tt \Rightarrow Area = \sqrt{7\times 3 \times 4 \times 2 \times 7 \times 3 \times 2} \\\\ \tt \Rightarrow Area = 7 \times 3 \times 2 \times 2 \\\\ \tt \Rightarrow Area = 21 \times 4 \\\\ \tt \Rightarrow Area = 84cm^2$