Find the area of the isosceles trapezoid?

Question

Find the area of the isosceles trapezoid. ( Use the Pythagorean Theorem.)?
atrapezoid is drawn: the samller base is 7 cm, the height is 12 cm, the congruent legs are 13cm each.
my answer was 123.5 I am right?
and my second question is about a kite is drawn: the two smaller congruent side are 5cm the diagonal conecting the vertex point of the kit is cut by the smaller diagonal connecting the two nonvertex point the longer diagonal is broken into two parts that are 10 centimeters and 3 centimeters.
my answer is 80
please some help

from where you have the 17 and in the second you have also 4 now i am more confuse sorry

Answer 1

1. The square block in the middle is 7cm x 12cm = 84 cm^2. Then there are two triangular sections, each with height 12 and hypotenuse 13 -> base 5. (√(13^2 - 12^2) = √25 = 5.) So their combined area is 2 * (bh/2) = 2 * (5 * 12 / 2) = 60 cm^2. So the total area is 84 + 60 = 144 cm^2.

2. The two smaller triangles have height 3 and base 4 (it's a 3-4-5 triangle). The two larger triangles have base 4 (it's the same base as the small triangles) and hypotenuse 10, so the height is √(10^2 - 4^2) = √84 = 2√21.
Hence the total area is 2 * (3*4/2) + 2 * (4 * 2√21 / 2)
= 12 + 8√21
= 48.66 cm^2 (2 d.p.)

Answer 2

If i'm right, you have 2 problems with my solutions.

1. Where did the 17 come from for the large base in the trap.

The iso. trap can be split into 2 right triangles and a rectangle by dropping the two heights from the small base to the large base. This splits the large base into 3 pieces. The middle piece is now a side of the rectangle, opposite the small base of 7. This means that the middle piece of the large base must be 7 as well. As I explained in the last answer, you can solve the rigth triangles for the missing leg, which must be 5. Since both ends of the trap. are congruent right triangles, both of those missing pieces of the large base must be 5. This makes the long base the sum of those three pieces (5 + 7 + 5) = 17.

2. Where did the 4 come from in the kite problem.

I kind of rushed through that one, sorry. The Kike has small congruent sides of 5, and the small end of the long diagonal is 3. Since the diagonals of a kite are always perpendicular, kites can always be broken down into 4 right triangles. You should now have two right triangles with a hypotenuse of 5 and a leg of 3(that leg is shared by both triangles). The pythagorean theorem or pythagorean triple would show you that the missing side of each of these right triangles is 4. (3^2 + x^2 = 5^2). This is where I came up with the two lengths of 4, which are the pieces of the short diagonal which is bisected by the long one. You've now got your given long diagonal of 13(pieces 10 and 3) adn your short diagonal of 8(4 and 4).

If you have any more issues, just email me at [e-mail]

Answer 3

The "extra advice" presented make sparkling the variety you acquire here around the outstanding of each and every shape, yet i'm not sure the variety you acquire here in the time of congruent factors or base of triangle. i'll assume there is distinctive techniques you in all probability did now no longer grant. in spite of the undeniable actuality that, if hypotenuse = 10 and appropriate = 6, then a million/2 base = 8, so base of triangle = sixteen, and base of rectangle = sixteen, precise base of trapezoid = sixteen Drop perpendicular from precise triangle to base of precise triangle this factors us 2 precise triangles with outstanding = 6, base = 8, hypotenuse = 10 Drop perpendicular from precise precise corner of trapezoid to backside base of trapezoid we've precise triangle with outstanding = 6 by way of actuality that base attitude of trapezoid = base attitude of isosceles triangle, this precise triangle is congruent to the two precise triangles shaped from isosceles triangle at precise. we are waiting to variety yet yet another congruent triangle on the left component of trapezoid. precise base of trapezoid = sixteen backside base of trapezoid = 8 + sixteen + 8 = 32 backside rectangle: base = 32, outstanding = 6 -------------------- The 4 shapes are: Isosceles triangle with base = sixteen, outstanding = 6 -----> A = a million/2 * sixteen * 6 = 40 8 Rectangle with base = sixteen and appropriate = 6 -----> A = sixteen * 6 = ninety six Trapezoid with bases = sixteen and 32, outstanding = 6 -----> A = (sixteen+32)/2 * 6 = one hundred 40 4 Rectangle with base = 32 and appropriate = 6 -----> A = 32 * 6 = 192 finished section = 480