Are you smart enough? shape help?!?

Question

Find is the area of a right triangle where:

Longest side=10 in

2nd longest side=6 in

shortest side=7.2 in

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Find the area of a trapizod where:

longest side=20 ft

2nd longest=14 ft

short side=12 ft

other short side=11ft

and down the middle=9 1/5 ft

PLEASE HELP 10 POINTS!

Answer 1

i'm not sure but this question should be in geometry not geography

Answer 2

With the sides 10, 7.2 & 6 it cannot be a right(angled) triangle. It can be if instead of 7.2, it is 8.

Area = sqrt[s(s-a)(s-b)(s-c)]

where s= (10+6+7.2)/2 = 11.6

A = sqrt[11.6*1.6*4.4*5.6] = 21.385 sq.inches.

In trapezoid by definition has a pair parallel sides(opposite). The distance (perpendicular) is given as 9.2'; the other sides with which it makes 2 triangles are 11', 12'. Bases of these

b1= sqrt[11^2 - 9.2^2] = 6.03;

b2 = sqrt[12^2 - 9.2^2] = 7.7045.

Area of these triangles are

1/2*9.2*[6.03+7.7] = 63.36.

Area of the central rectangle=14*9.2=128.8

Total area=63.36 + 128.8 = 192.16 sq.feet

Answer 3

The area of a triangle equals one half of base ttimes height.  Because of the right angle you can consider the2 shorter sides to be the base and height. So the area of your triangle is one half of 6 X7.2 or 3X7.2= 21.6 square inches.   As to your trapizod, which may be the same thing as a trapezoid, draw it and draw 2 lines such that it is divided into 2 triangles and a rectangle.  Then calculate the area of each of the 3 figures and add them together.