Question

t/f
5.the area of the equilateral triangle is 20√3cm^2 whose each side is 8cm.
6.In a triangle, the sides are given as 11 cm 12 cm 13 cm. the length of the altitude is 10.25cm corresponding to the side having length 12cm.
(plz answer fast it's urgent)​

Answer 1

True/False

[5] The area of the equilateral triangle is 20√3cm^2 whose each side is 8cm.

Given:

• Area of equilateral triangle is 20√3 cm².
• Each side of triangle is 8 cm.

To Check:

• The statement is true or false .

Solution: As we know that area of an equilateral triangle is

★Area of Equilateral triangle=√3(a)²/4★

A/q

$\implies{\rm }$ 20√3 = √3(8)²/4

$\implies{\rm }$ 20√3 = √3 $\times$ 64/4

$\implies{\rm }$ 20√3 = 16√3

Here, LHS ≠ RHS Therefore this is false.

___________________________

[6] In a triangle, the sides are given as 11 cm , 12 cm 13 cm. the length of the altitude is 10.25cm corresponding to the side having length 12cm.

Given:

• Sides of triangle are 11 cm , 12 cm and 13 cm.
• The length of altitude is 10.25 cm corresponding to side 12 cm.

To Check:

• This statement is true or false ?

Solution: Let in ∆ABC, A = 11 cm , B = 12 cm , C = 13 cm & AD = Altitude to side 12 cm.

We have to find the area of ∆ABC , by Heron's Formula.

➟ Semi Perimeter (S) = (a + b + c/2) cm

➟ S = (11 + 12 + 13/2) cm

➟ S = 36/2 = 18 cm

★ Heron's Formula = √S ( s – a ) ( s – b ) ( s – c ) ★

$\implies{\rm }$ Area ∆ABC = √18( 18–11 ) ( 18–12 ) ( 18–13 ) cm²

$\implies{\rm }$ √18 $\times$ 3 $\times$ 6 $\times$ 5 cm²

$\implies{\rm }$ √3780 cm²

$\implies{\rm }$ 61.48 cm²

Now, taking 12 cm as base, area of ∆ABC

➱ Area of ∆ABC = ( 1/2 x Base x Height )

➱ Area of ∆ABC = ( 1/2 x BC x AD )

➱ 61.48 = ( 1/2 x 12 x AD )

➱ 61.48 = 6AD

➱ 61.48/6 = AD

➱ 10.246 or 10.25 cm = AD

Hence, the length of altitude corresponding to side 12 cm is 10.25 cm. This statement is true.

Answer 2

$\\ Given:</p><p> \\ Area of equilateral triangle is \\ 2013 \\ cm2</p><p>. \\ Each side of triangle is 8 cm.</p><p> \\ To Check:</p><p>. \\ The statement is true or false.</p><p> \\ Solution: As we know that \: area of an \: equilateral triangle is</p><p> \: \\ *Area of Equilateral \\ triangle=v3(a)/4*Alq</p><p> \\ 2013 = 73(8)/4</p><p> \\ 20v3 = 73 x 64/4</p><p> \\ 2013 = 1673</p><p> \\ Here, LHS + RHS Therefore \: this is false.</p><p> \\ [6] In a triangle, the sides are \: given as 11 \: cm, 12 cm 13 \: cm. the length of the</p><p> \: altitude is 10.25cm \: corresponding to the</p><p> \: side having length 12cm.</p><p> \\ Given:</p><p> \\ Sides of triangle are 11 cm, \: 12 cm \: and 13 cm</p><p> \\ The length of altitude is 10.25 \: cm</p><p> \: corresponding to side 12 cm.</p><p> \: \\ To Check:</p><p> \\ This statement is true or \: false?</p><p> \\ Solution: Let in AABC, A = 11 \: cm, B = 12 \: cm, C = 13 cm & AD = Altitude to side 12cm.</p><p> \\ We have to find the area of \\ AABC, by</p><p> \\ Heron's Formula.- Semi \: Perimeter (S) = (a + b + c/2) cm</p><p> \\ - S = (11 + 12+ 13/2)cm</p><p> \\ - S = 36/2 = 18 cm</p><p> \\ * Heron's Formula = S(S-a) \\ (s-b)</p><p> \\ (s-C)*</p><p> \\ Area AABC = 718(18-11) (18- \\ 12) (18-13) cm?</p><p>V18 x 3 x 6 x 5 cm?</p><p> \\ V3780 cm</p><p> \\ => 61.48 cm</p><p> </p><p>$