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3 years ago

3. The figure below is a composite shape that consists of a semicircle and an equilateral triangle (triangle

Mathematics

1 answer:

Ivenika [448]3 years ago

8 0

Answer:

See explanation below

Step-by-step explanation:

Area = Area of triangle + Area of the semi circle

Area of triangle = 1/2bh

Area of triangle = 1/2 *(5/2)(4)

Area of triangle = 1/2(2.5)*4

Area of triangle = 2 * 2.5

Area of triangle = 5cm^2

Area of semicircle = πr²/2

r = 2.5cm

Area of semicircle =(3.14)(2.5)²/2

Area of semicircle = = 3.14 * 4.5

Area of semicircle = 9.8125cm^2

Area pf the figure = 5+9.8125

Area pf the figure = 14.8125cm^2

For the perimeter

Perimeter of the semicircle = 2πr

Perimeter of the semicircle = 2(3.14)(2.5)

Perimeter of the semicircle = 15.7cm

Perimeter of the semicircle = = 15.7 + 5 + 5

Perimeter of the semicircle = 25.7cm

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Any 10th grader solve it <br>for 50 points

kkurt [141]

Answer:

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0$ is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.

Step-by-step explanation:

Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.

First term of given arithmetic progression is A

and common difference is D

ie., $a_{1}=A$ and common difference=D

The nth term can be written as

$a_{n}=A+(n-1)D$

pth term of given arithmetic progression is a

$a_{p}=A+(p-1)D=a$

qth term of given arithmetic progression is b

$a_{q}=A+(q-1)D=b$ and

rth term of given arithmetic progression is c

$a_{r}=A+(r-1)D=c$

We have to prove that

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)=0$

Now to prove LHS=RHS

Now take LHS

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)$

$=\frac{A+(p-1)D}{p}\times (q-r)+\frac{A+(q-1)D}{q}\times (r-p)+\frac{A+(r-1)D}{r}\times (p-q)$

$=\frac{A+pD-D}{p}\times (q-r)+\frac{A+qD-D}{q}\times (r-p)+\frac{A+rD-D}{r}\times (p-q)$

$=\frac{Aq+pqD-Dq-Ar-prD+rD}{p}+\frac{Ar+rqD-Dr-Ap-pqD+pD}{q}+\frac{Ap+prD-Dp-Aq-qrD+qD}{r}$

$=\frac{[Aq+pqD-Dq-Ar-prD+rD]\times qr+[Ar+rqD-Dr-Ap-pqD+pD]\times pr+[Ap+prD-Dp-Aq-qrD+qD]\times pq}{pqr}$

$=\frac{Arq^{2}+pq^{2} rD-Dq^{2} r-Aqr^{2}-pqr^{2} D+qr^{2} D+Apr^{2}+pr^{2} qD-pDr^{2} -Ap^{2}r-p^{2} rqD+p^{2} rD+Ap^{2} q+p^{2} qrD-Dp^{2} q-Aq^{2} p-q^{2} prD+q^{2}pD}{pqr}$

$=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2}-pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}$

$=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2} -pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}$

$\neq 0$

ie., $RHS\neq 0$

Therefore $LHS\neq RHS$

ie., $\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0$

Hence proved

5 0

3 years ago

Working alone, Darryl can mop a warehouse in 11 hours. Elisa can mop the same warehouse in 9

Harrizon [31]

Answer:

≈ 5 hours

Step-by-step explanation:

<u>Time needed if they work together:</u>

1 / (1/11 + 1/9) =
1/ (20 / 99) =
99/20 ≈ 5 hours

3 0

2 years ago

PLEASE HELP ME!!!

crimeas [40]

B=4676+10043 c=b-2160 answer equals b+c
I hope this is helpful I am trying not to say the exact answer but to teach you instead

4 0

3 years ago

Owen is signing up for a gym membership with a one-time fee to join and then a

Katena32 [7]

The cost of the monthly fee is 50$.. if the joining fee is 100.. take 200 and subtract that fee since it is included in the total. You’re left with 100. Note, the 200 is calculated for a 2 month period, so split the remaining 100 ,from the total 200, into 2. You’re left with 50 each month.

T=(C-100)/2 (T being 1 month) (C being the total cost/200)

Hopefully

3 0

3 years ago

In which number is the value represented by the digit 9 equal to 110

Tresset [83]

Answer:

Step-by-step explanation:

a+b/2xh

aksfsaowcapa

8 0

3 years ago