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Advocard [28]
3 years ago
7

The area of rectangle ABCD is 72 square inches. A diagonal of rectangle ABCD is 12 inches and the diagonal of rectangle EFGH is

22 inches. Find the area of rectangle EFGH. Round to the nearest square inch if necessary.
Mathematics
1 answer:
luda_lava [24]3 years ago
4 0

Answer:

The area of rectangle EFGH is 242\ in^{2}

Step-by-step explanation:

For this problem I assume that rectangle ABCD and rectangle EFGH are similar

step 1

Find the scale factor

we know that

If two figures are similar, then the ratio of its corresponding sides is proportional, and this ratio is called the scale factor

Let

z ------> the scale factor

The scale factor is the ratio between the diagonals of rectangles

so

z=\frac{22}{12}=\frac{11}{6}

step 2

Find the area of rectangle EFGH

we know that

If two figures are similar, then the ratio of its areas is the scale factor squared

Let

z------> the scale factor

x -----> area of rectangle EFGH

y ----> area of rectangle ABCD

so

z^{2}=\frac{x}{y}

we have

z=\frac{11}{6}

y=72\ in^{2}

substitute and solve for x

(\frac{11}{6})^{2}=\frac{x}{72}

\frac{121}{36}=\frac{x}{72}

x=\frac{121}{36}(72)

x=242\ in^{2}

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Find the area of a square with sides of length 6.5 inches
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andriy [413]

Answer:

i. x² + 7x - 16800 = 0 ii. x = 126.16 km/h or -133.16 km/h iii. 5.01 h

Step-by-step explanation:

i. If he drove at an average speed of x km/h on his journey from City P to City Q formulate an equation in x and show that it reduces to x2 + 7x – 16 800 = 0.

For the first journey from City P to City Q, with Mr Lee moving at an average speed of x km/h, he reaches there in time, t and covers the distance, d = 600 km

So, xt = 600 (1)

On his return journey from City Q to CIty P, his average speed increases by 7 km/h, so it is (x + 7)km/h and his time is 15 minutes less than his first journey. 15 min = 15/60 h = 0.25 h, we have that his time for the journey is (t - 0.25) h. Since the distance covered is the same d = 600 km,

We have (x + 7)(t - 0.25) = 600  (2)

Expanding the brackets, we have

xt - 0.25x + 7t - 0.25(7) = 600

xt - 0.25x + 7t - 1.75 = 600

From (1) t = 600/x and xt = 600

Substituting these into the equation, we have

600 - 0.25x + 7(600/x) - 1.75 = 600

simplifying

-0.25x + 4200/x - 1.75 = 600 - 600

-0.25x + 4200/x - 1.75 = 0

multiplying through by x, we have

-0.25x² + 4200 - 1.75x = 0

dividing through by -0.25, we have

-0.25x²/-0.25 + 4200/-0.25 - 1.75x/-0.25 = 0

x² - 16800 + 7x = 0

re-arranging, we have

x² + 7x - 16800 = 0

ii. Solve the equation x² + 7x - 16 800 = 0, giving both your answers correct to  2 decimal places.

Using the quadratic formula, we solve x² + 7x - 16800 = 0 for x

So, x = \frac{-7 +/-\sqrt{7^{2} - 4 X 1 X -16800} }{2 X 1}\\x = \frac{-7 +/-\sqrt{49 + 67200} }{2} \\x = \frac{-7 +/-\sqrt{67249} }{2} \\x = \frac{-7 +/- 259.32}{2} \\x = \frac{-7 + 259.32}{2} or x = \frac{-7 - 259.32}{2} \\x = 252.32/2 or x=  -266.32/2\\x = 126.16 km/hor x = -133.16 km/h

So, x = 126.16 km/h or -133.16 km/h

iii. Find the time taken for the return journey

The time taken for the return journey is t' = t + 0.25. Now. t = 600/x

Since x cannot be negative, we use x = 126.16 km/h.

So, t = 600/x = 600/126.16 = 4.76 h

t' = t + 0.25

t' = 4.76 + 0.25

t' = 5.01 h

3 0
2 years ago
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