All categories

3 years ago

A rectangle has a diagonal 15 cm and one side 8 cm finde the perimeter p of the rectangle

Mathematics

2 answers:

snow_lady [41]3 years ago

6 0

Consider the right angles triangle with hypotenuse 15 and leg 8 cm formed by drawing the diagonal.

The other leg (x) will be the other side of the rectangle and , by Pythagoras we have:-

15^2 = 8^2 + x^2

x^2 = 225 - 64

x^2 = 161

x = 12.69 cms

So the perimeter p = 2*8 + 2* 12.69

= 16 + 25.38

= 41.38 cms (answer)

Tamiku [17]3 years ago

6 0

Hi there!

The diagonal of a rectangle forms, with 2 adjacent sides, a rectangle triangle. This means that you just have to use Pythagora's theorem to calculate the length of the second side.

The pythagorean theorem states that the square of the hypothenuse (which is in this case the diagonal), is equal to the sum of the square of the other two sides.

Here's what the formula looks like:

"C" being the hypothenuse (diagonal)

"A" & "B" being the two other sides

C² = A² + B²

Replace the values you are given in the problem in the formula:

C² = A² + B²

15² = 8² + B²

Solve this equation by isolating "B" :

15² = 8² + B²

225 = 64 + B²

Subtract 64 from each side of the equation → 225 - 64 = 161

161 = B²

Square root on each side of the equation → √161 = 12.6885775...

12.69 ≈ B

Now that you have the measures of the length and the width of the rectangle, you can calculate the perimeter:

"P" being the perimeter

"L" being the length

"W" being the width

P = (L × 2) + (W × 2)

P = (12.69 × 2) + (8 × 2)

P = 25.38 + 16

P = 41.38

The answer is: The perimeter of the rectangle is 41.38 cm.

There you go! I really hope this helped, if there's anything just let me know! :)

You might be interested in

You are a lifeguard and spot a drowning child 60 meters along the shore and 40 meters from the shore to the child. You run along

sukhopar [10]

Answer:

The lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

Step-by-step explanation:

This is a problem of optimization.

We have to minimize the time it takes for the lifeguard to reach the child.

The time can be calculated by dividing the distance by the speed for each section.

The distance in the shore and in the water depends on when the lifeguard gets in the water. We use the variable x to model this, as seen in the picture attached.

Then, the distance in the shore is d_b=x and the distance swimming can be calculated using the Pithagorean theorem:

$d_s^2=(60-x)^2+40^2=60^2-120x+x^2+40^2=x^2-120x+5200\\\\d_s=\sqrt{x^2-120x+5200}$

Then, the time (speed divided by distance) is:

$t=d_b/v_b+d_s/v_s\\\\t=x/4+\sqrt{x^2-120x+5200}/1.1$

To optimize this function we have to derive and equal to zero:

$\dfrac{dt}{dx}=\dfrac{1}{4}+\dfrac{1}{1.1}(\dfrac{1}{2})\dfrac{2x-120}{\sqrt{x^2-120x+5200}} \\\\\\\dfrac{dt}{dx}=\dfrac{1}{4} +\dfrac{1}{1.1} \dfrac{x-60}{\sqrt{x^2-120x+5200}} =0\\\\\\ \dfrac{x-60}{\sqrt{x^2-120x+5200}} =\dfrac{1.1}{4}=\dfrac{2}{7}\\\\\\ x-60=\dfrac{2}{7}\sqrt{x^2-120x+5200}\\\\\\(x-60)^2=\dfrac{2^2}{7^2}(x^2-120x+5200)\\\\\\(x-60)^2=\dfrac{4}{49}[(x-60)^2+40^2]\\\\\\(1-4/49)(x-60)^2=4*40^2/49=6400/49\\\\(45/49)(x-60)^2=6400/49\\\\45(x-60)^2=6400\\\\$

$x$

As $d_b=x$ , the lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.