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1 year ago

Please help i been asking since yesterday

Mathematics

2 answers:

Advocard [28]1 year ago

4 0

Answer:

To calculate the perimeter of a triangle, add the length of its sides. For example, if a triangle has sides a, b, and c, then the perimeter of that triangle will be P = a + b + c.

svp [43]1 year ago

4 0

Answer:

1) 21.24cm

2)19.8ft

3)22.84mm [2.284cm]

Step-by-step explanation:

To calculate the perimeter of a triangle, add the length of its sides.

1) perimeter = 5.4+8.84+7

=> 5.4+8.84+7 = 21.24,

21.24cm

2) perimeter = 6.6 + 6.6 + 6.6 i.e 6.6* 3

=> 6.6 x 3 = 19.8

19.8ft

3) perimeter = 8.3 + 4.9 + 9.64

=> 8.3 + 4.9 + 9.64 = 22.84

22.84mm or 2.284cm

AREA

1) it's a right triangle,

=> Area of a right triangle = 1/2 × base × height.

=> here, base is a and height is b

=> a = 7, b = 5.4

=> now we multiply it,

=> 7*5.4 = 37.8

=> now multiply it with 1/2,

=> 1/2 x 37.8 = 37.8/2

=> 37.8/2 = 18.9

2) it's an equilateral triangle,

18.8620

the formula is √3 a^2/ 4

=> √3* 6.6^2/ 4

=> 75.4481/4 = 18.8620

3) 20.335

base = a = 8.3

height = b = 4.9

=> 8.3 * 4.9 = 40.68

=> 40.68/2 = 20.335

hope this helps:)

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2 years ago

A triangle has three sides 35cm 54 cm and 61 cm find its area Also find the smallest of its altitude

I am Lyosha [343]

Answer:

Area of given triangle is 939.15cm² and smallest altitude is 30.8cm

<h3>Solution:</h3>

We are given three sides of a triangle, Let the sides be :

( a ) = 35 cm

( b ) = 54 cm

( c ) = 61 cm

We can find the area of the triangle with its three sides using Heron's Formula

<u>Heron's </u><u>Formula</u>

Heron's formula was founded by hero of Alexandria, for finding the area of triangle in terms of the length of its sides. Heron's formula can be written as:

$\sf{ \pmb { \longrightarrow \: \sqrt{s(s - a)(s - b)(s - c)} }}$

where ( s ) :

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

Therefore, for the given triangle first we will calculate ( s )

$\begin {aligned}\quad & \quad \longmapsto \sf s = \dfrac{a + b + c}{2} \\ & \quad \longmapsto \sf s = \dfrac{35 + 54 + 61}{2} \\ & \quad \longmapsto \sf s = \dfrac{150}{2} \\ & \quad \longmapsto \sf s = 75cm \end{aligned}$

Now, Area of triangle will be:

$\begin{aligned}&:\implies \sf\quad \sf \: A = \sqrt{s(s - a)(s - b)(s - c)} \\ &:\implies \sf\quad \sf \: A = \sqrt{75(75 - 35)(75 - 54)(75 - 61)} \\&:\implies \sf\quad \sf \: A = \sqrt{75 \times 40 \times 21 \times 14} \\ &:\implies \sf\quad \sf \: A = \sqrt{5 \times 5 \times 3 \times 3 \times 2 \times 2 \times 7 \times 7 \times 2 \times 2 \times 5} \\ &:\implies \sf\quad \sf \: A =5 \times 3 \times 2 \times 7 \times 2 \sqrt{5} \\ &:\implies \sf\quad \sf \: A =420 \times 2.23 \\ &:\implies \sf\quad \sf \boxed{ \pmb{ \sf A =939.15 {cm}^{2} }} \end{aligned}$

Also, we have to find the smallest altitude, and the smallest altitude will be on the longest side. So,

$\begin{aligned}&:\implies \sf\quad \sf \: Area =939.15 \\ &:\implies \sf\quad \sf \: \dfrac{1}{2} \times b \times h =939.15 \\ &:\implies \sf\quad \sf \: \dfrac{1}{2} \times 61 \times h = 939.15 \\&:\implies \sf\quad \sf \: h =939.15 \times \dfrac{2}{61} \\&:\implies \sf\quad \sf \: h = \dfrac{1818.3}{61} \\ &:\implies \sf\quad \boxed{ \pmb{\sf \: h =30.79 \: (approx)}} \end{aligned}$