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Marianna [84]
3 years ago
11

A wire is in the shape of a rectangle of length 14cm. Find its breadth if the same wire is re- bent into a square of side 13cm.

Also find out their areas and write which one encloses more area, the rectangle or square.
Mathematics
2 answers:
lara [203]3 years ago
7 0
Length= 40cm

Breadth= 22cm

Perimeter of the rectangle= Length of the wire

=2(l+b)= 2(40cm + 22cm)

=2\times62cm = 124cm

Now, the wire is rebent into a square.

Perimeter= 124cm

4\times side=124

Therefore side= \frac{124}{4}cm= 31cm

So, the measure of each side= 31cm

Area of rectangular shape= l\times b

=40cm\times22cm

=880cm^2

Area of a square shape= \left(Side\right)^2

=\left(31\right)^2=961cm^2

Since 961cm^2\ >\ 880cm^2

Hence, the square encloses more area.
Alborosie3 years ago
6 0

Answer:

See below.

Step-by-step explanation:

Area by square = 169

Area by rectangle = 14 x 12 = 168

So more for square hope it helps Alphonsa

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(This is a "show your work" type of assignment)
joja [24]

Answer:

Part 1) It is not possible to apply the Pythagorean Theorem to find the missing side, because is not a right triangle

Part 2) I would use the law of the cosine, because to apply the law of sine I would need to know the value of angle C or angle B

Part 3) The length of side CB is 369\ units

Part 4) The measure of angle C is 35\°

Part 5) The measure of angle B is 88\°

Step-by-step explanation:

Part 1) Can you use Pythagorean Theorem to find the missing side?

we know that

The Pythagorean Theorem is used in a right triangle

The triangle ABC of the figure is not a right triangle

therefore

It is not possible to apply the Pythagorean Theorem to find the missing side

Part 2) Would you use Law of Sine or Law of Cosine to find the length of side CB?    

I would use the law of the cosine, because to apply the law of sine I would need to know the value of angle C or angle B

Part 3) Find the length of side CB. Show your work and round your answer to the nearest whole number

Applying the law of cosine

BC^{2}=AB^{2}+AC^{2}-2(AB)(AC)cos(A)

substitute the values

BC^{2}=255^{2}+440^{2}-2(255)(440)cos(57\°)

BC^{2}=136,408

BC=369\ units

Part 4) Find the measure of angle C. (Hint: use the other “Law”). Round your answer to the nearest whole degree

Applying the law of sines

\frac{AB}{sin(C)}=\frac{CB}{sin(A)}

substitute the values

\frac{255}{sin(C)}=\frac{369}{sin(57\°)}

sin(C)=255*sin(57\°)/369=0.5796

C=arcsin(0.5796)=35\°

Part 5) What is the measure of angle B? Show your work and round your answer to the nearest whole degree.

Remember that

the sum of the interior angles of a triangle must be equal to 180 degrees

sp

substitute the values and solve for <B

57\°+

4 0
3 years ago
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