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4 years ago

Question 64

Mathematics

1 answer:

olga55 [171]4 years ago

7 0

Answer:

987 cm²

Step-by-step explanation:

We are given a figure which consists of the a triangle, a rectangle, and a semicircle.

To find the area of the figure, calculate the area of each shape that makes up the figure as follows:

=>Area of triangle: ½(a*b)

Leg a = 20cm

Leg b = 49-34 = 15cm

Area = ½(20*15) = 300/2

Area = 150 cm²

==>Area of rectangle: L*W

L = 34cm

W = 20cm

Area = 34*20 = 680cm²

==>Area of semicircle: ½πr²

r = ½ of 20cm = 10cm

Area = ½*3.14*10² = ½*3.14*100

Area = 314/2

Area = 157cm²

Area of the figure = 150+680+157 = 987 cm²

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Two altitudes of a triangle have lengths $12$ and $14$. What is the longest possible integer length of the third altitude

nataly862011 [7]

The longest possible altitude of the third altitude (if it is a positive integer) is 83.

According to statement

Let h is the length of third altitude

Let a, b, and c be the sides corresponding to the altitudes of length 12, 14, and h.

From Area of triangle

A = 1/2*B*H

Substitute the values in it

A = 1/2*a*12

a = 2A / 12 -(1)

Then

A = 1/2*b*14

b = 2A / 14 -(2)

Then

A = 1/2*c*h

c = 2A / h -(3)

Now, we will use the triangle inequalities:

a < b+c

2A/12 < 2A/14 + 2A/h

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h<84

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2A/14 < 2A/12 + 2A/h

Solve it and get

h > -84

c < a+b

2A/h < 2A/12 + 2A/14

Solve it and get

h > 6.46

From all the three inequalities we get:

6.46<h<84

So, the longest possible altitude of the third altitude (if it is a positive integer) is 83.

Learn more about TRIANGLE here brainly.com/question/2217700

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2 years ago

What are the solutions of x^2-2x+5=0

sweet [91]

The solution of $x^{2}-2 x+5=0$ are 1 + 2i and 1 – 2i

Solution:

Given, equation is $x^{2}-2 x+5=0$

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Here in $x^{2}-2 x+5=0$ a = 1 ; b = -2 ; c = 5

$b^2 - 4ac = 2^2 - 4(1)(5) = 4 - 20 = -16$

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Now plug in values in eqn 1, we get,

$x=\frac{-(-2) \pm \sqrt{(-2)^{2}-4 \times 1 \times 5}}{2 \times 1}$

On solving we get,

$x=\frac{2 \pm \sqrt{4-20}}{2}$

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$x=\frac{2 \pm \sqrt{16} \times \sqrt{-1}}{2}$

we know that square root of -1 is "i" which is a complex number

$\begin{array}{l}{\mathrm{x}=\frac{2 \pm 4 i}{2}} \\\\ {\mathrm{x}=1 \pm 2 i}\end{array}$

Hence, the roots of the given quadratic equation are 1 + 2i and 1 – 2i

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