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Mathematics 16 Online
OpenStudy (blake57roger):

A right triangle has one leg that measures 12 inches and a hypotenuse that measures 13 inches. Enter the area, in square units, of the triangle.

OpenStudy (bookworm14):

@Agl202 @goformit100 @Nnesha @ParthKohli @TheSmartOne @whpalmer4 Is there any way any of you could help him please? I don't have time at the moment. Thank you so much in advance!

Nnesha (nnesha):

formula to find area of right triangle is \[\large \rm A=\frac{1}{2} b*h\] where b is base and h is height

Nnesha (nnesha):

and it's a right triangle so to find missing side you can use Pythagorean theorem a^2+b^2=c^2 where c is the longest side of right triangle (hypotenuse )

OpenStudy (blake57roger):

25

OpenStudy (blake57roger):

is A^2 + B^2= C^2

Nnesha (nnesha):

what's 25 ??

OpenStudy (blake57roger):

5

Nnesha (nnesha):

correct that's the 3rd side now just use the formula to find area of right triangle

OpenStudy (blake57roger):

30

OpenStudy (blake57roger):

is that the answer

Nnesha (nnesha):

how did you get that ?

OpenStudy (blake57roger):

0.5*5*12

OpenStudy (blake57roger):

1/2*5*12

Nnesha (nnesha):

then yes that's correct

OpenStudy (blake57roger):

thanks

Nnesha (nnesha):

remember it should be in `square ` units and np :=))

OpenStudy (blake57roger):

@Nnesha

OpenStudy (blake57roger):

OpenStudy (whpalmer4):

You need the formula for distance between two points \((x_1,y_1)\) and \((x_2,y_2)\): \[d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\] You know one of the points, and you can find the other by counting marks on the graph from the origin. Doesn't matter which point you call \((x_1,y_1)\) and which point you call \((x_2,y_2)\). This is just the Pythagorean theorem in disguise...the hypotenuse is the distance.

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