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3 years ago

Use special right triangles to solve for the exact value of x

Mathematics

2 answers:

Savatey [412]3 years ago

6 0

Pythagorean theorem, a^2 + b^2 = c^2
5 squared is 25, 8 squared is 64, square root of 89 is 9.43 thus the hypotenuse is 9.43, correct me if I'm wrong but that is what I got

Vanyuwa [196]3 years ago

3 0

Answer:

The value of x is $\sqrt{89}$

Step-by-step explanation:

Given : A right angled triangle with base = 8 , perpendicular = 5

We have to find the value of x that is we have to find the length of hypotenuse.

Consider the given right angled triangle,

PYTHAGORAS THEOREM states that for a given right angle triangle the sum of square of base and perpendicular is equal to the square of hypotenuse.

Mathematically written as ,

$H^2=B^2+P^2$

Here, B = 8 , P = 5

substitute, we have,

$H^2=8^2+5^2$

Simplify, we have,

$H^2=64+25$

$H^2=89$

Taking square root both side, we have,

$H=\sqrt{89}$

Thus, The value of x is $\sqrt{89}$

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Work out the area of the trapezium ABDE. B 9 cm 6 cm 8 cm →D E

Evgen [1.6K]

Answer:

Area of the trapezium ABDE = 30 cm²

Step-by-step explanation:

Area of a trapezium = $\frac{1}{2}(b_1+b_2)h$

Here, $b_1$ and $b_2$ are the parallel sides of the trapezium

h = Distance between the parallel sides

From the picture attached,

ΔCAE and ΔCBD are the similar triangles.

So by the property of similarity their sides will be proportional.

$\frac{AE}{BD}= \frac{CE}{CD}$

$\frac{9}{6}=\frac{CE}{8}$

CE = $\frac{9\times 8}{6}$

CE = 12 cm

Therefore, DE = CE - CD

DE = 12 - 8 = 4 cm

Now area of trapezium ABDE = $\frac{1}{2}(AE+BD)(DE)$

= $\frac{1}{2}(6+9)4$

= 30 cm²

Therefore, area of the trapezium ABDE = 30 cm²

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3 years ago

One angle of an isosceles triangle measures 62º. Which other angles could be in that

oee [108]

Answer:

62 and 56 or 59 and 59

Step-by-step explanation:

62, 62, 56

62, 59, 59

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An aquarium with a square base has no top. There is a metal frame. Glass costs 8 dollars/m^2 and the frame costs 7 dollars/m. Th

Irina-Kira [14]

Given that the volume of the aquarium is 20m^3.

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Thus, the length of the base of the aquarium is $\sqrt{area \ of \ base} = \sqrt{ \frac{20}{h} }$

The frame is to cover 8 sides with the length equal to the length of the base and 4 sides with the length of the height.

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Therefore, total cost in terms of the height, h, is given by

$C=\sqrt{\frac{62,720}{h}}+28h+\frac{160}{h}+\sqrt{20,480h$

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