The area of a right triangle is 180 m². The height of the right triangle is 40 m. What is the length of the hypotenuse of the right triangle?

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The area of a right triangle is 180 m². The height of the right triangle is 40 m. What is the length of the hypotenuse of the right triangle?

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Answer 1 · 2018-05-27T13:57:13

Assuming the height is one of the legs, the other leg is $9$ $9$ and the hypotenuse is $41$ $41$ .

Explanation:

It's not clear if the height is one of the legs or the altitude to the hypotenuse.

Either way we have $A = \frac{1}{2} b h$ $A =1/2 b h$ or $b = \frac{2 A}{h} = 2 \cdot \frac{180}{40} = 9$ $b={2A}/h=2cdot(180)/40=9$ m.

If $h = 40$ $h=40$ is the height to the hypotenuse, the hypotenuse is $9$ $9$ m and we're done. Let's assume $h = 40$ $h=40$ is one of the legs so $b = 9$ $b=9$ is the other leg so

$c^{2} = h^{2} + b^{2} = 40^{2} + 9^{2} = 1681$ $c^2= h^2 + b^2 = 40^2+9^2=1681$

$c = 41$ $c = 41$

Answer 2 · 2018-05-27T18:06:57

$41 m$ $41 m$

Explanation:

$In the right △ A B C$ $"In the right"" triangle ABC$

$:.angle B= 90^@$ $"given"$

$:.BC=40 m= "height given"= opposite$

$:."Area of a triangle"=1/2 base xxh=180 m^2$

$:."1/2basexx40=180$

$:.1/2 base=180/40$

$:.1/2 base=4.5 m$

$:.base=9m=AB=adjacent$

$:.(opposite)/(adjacent)=(BC)/(AB)=40/9=tan angle A=4.444444444$

$:.angle A=77^@19'11''$

$:."AC= hypotenuse",(AC)/(BC)=sec 77^@19'11''=4.555555556$

$:.AC=sec 77^@19'11''xx 9=41 m=" hypotenuse"$

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The area of a right triangle is 180 m². The height of the right triangle is 40 m. What is the length of the hypotenuse of the right triangle?

2 Answers

Explanation:

Explanation:

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