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2 years ago

2. The perimeter of rectangle is 48cm. If the

Mathematics

1 answer:

natita [175]2 years ago

7 0

lets the length be 3x

width be 5x

perimeter = 48

3x +5x=48

8x = 48

x =48/8

x= 6

so.

3x = 3×6 = 18

5x = 5×6= 30

hope it will be helpful

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The diagonal of the rectangle is 15 cm and one of its sides is 9 cm. find the area of the rectangle.

Tems11 [23]

The area of the rectangle is 108 square centimeters.

Given Information

The following information has been given in the question,

Length of the diagonal of the rectangle, l = 15 cm

Length of one of the sides of the rectangle, a = 9 cm

Let the other side of the rectangle be b.

Finding the Other Side of the Rectangle

According to Pythagoras Theorem,

(hypotenuse)² = (base)² + (perpendicular)²

Applying this theorem in the right-angle triangle formed by the diagonal and 9cm side of the rectangle, we get,

l² = a² + b² [∵ The diagonal makes the hypotenuse]

(15)² = (9)² + b²

b² = (15)² - (9)²

b² = 225 - 81

b² = 144

b = √144

b = 12 cm

Calculating the Area of the Rectangle

Area of the rectangle, A = a × b

A = 9 × 12

A = 108 cm²

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1 year ago

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melomori [17]

Answer: $4\sqrt{3}$

Step-by-step explanation:

For this exercise it is important to remember the following:

$i=\sqrt{-1} \\\\i^2=-1$

Given the following expression:

$-2i\sqrt{-12}$

You can notice that the radicand (the number inside the square root) is negative. Therefore, in order to simplify the expression, you need to follow these steps:

1. Replace $\sqrt{-1}$ with $i$ and simplify:

$(-2i)(i)\sqrt{12}=-2i^2\sqrt{12}=-2(-1)\sqrt{12}=2\sqrt{12}$

2. Descompose 12 into its prime factors:

$12=2*2*3=2^2*3$

3. Substitute into the expression:

$=2\sqrt{2^2*3}$

4. Since $\sqrt[n]{a^n}=a$ , you can simplify it:

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2 years ago

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Zielflug [23.3K]

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Step-by-step explanation:

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1 year ago

The perimeter of a square is 36<br>what is the area of the square?

zmey [24]

Answer: 81 units²

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Therefore, since the perimeter of the given square is 36, we have 36 = 4s and dividing both sides by 4, 9 = s.

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Q.12

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