Elementary 2 Questions: Right Triangles, Junior 2 Mathematics Right Triangle Problems

1.with a ravine.

2.With the inverse theorem of the valley.

The hypotenuse midline of a right triangle is equal to half of the hypotenuse.

So the hypotenuse is 2*6 12 long

Therefore, the area of the triangle is 1 2*12*5=30

The midline on the hypotenuse is half the length of the hypotenuse.

So the bevel is 12cm long.

The area is 1 2 * 5 * 12 = 30 cm2

No, no, no, no, no, no.

Vertical. Because of rotation, bq=bp=2, qc=pa=1, the angle pbq is a right angle, the pythagorean theorem gives pq=2 2, and the big is perpendicular because the shed burns pq + qc = pc.

It's too hard not to read and calculate.

PQ Brief Draft QC

Because after PAB rotates 90° around point B to QCB, we know that Pb=QB=2, and Pbq=90°, so Pq =Pb +BQ =8

That is, pq=2 2 and because of the simple nature of rotation qc=pa=1, and pq+qc =8+1=9=pc, so pqc is a right triangle, pqc=90°, so pq qc

Vertical. First of all, bp=bq=2, Li Shi pbq=90°, so Wu Ju pq=2 root number 2

cq=ap=1

PC=3 thus can be cavitated in the triangular PQC, PQ 2+CQ 2=PC2 so liposuction.

Because the right-angled side is 8cm

So CB enlarges 8-6 = 2cm

So the area of the triangle is 8*(6+2) 2=32

Isn't it just 8*8 2=64 I feel that there are a lot of typos in the question, and the description is not very clear, and I don't know if what I understand is right or wrong.

Untie; Because the angle abc = 45 degrees, the angle a is a right angle.

So ac=de=610 (e is the point in the middle of ab) so be=de tan39=610

So cd=

Let the height of the building be x, tan39 = 610 - x divided by 610 = tan39.

1.Over B as the high BH of the AC side

then bah=30° and ac=2a

bh=a2.If there is no picture in this question, there are two situations.

One bac = 120 ° and the other bac = 60 ° Cd are 15 cm and 5 cm long, respectively

1) The question is wrong, it should be pa pb, pa=pb, if so, apbc four points are round, pa=pb, they have the same central angle of the circle, cp is the bisector of acb.

2) AP is the bottom, APC is on the same side as APB, ACP= ABP=45, so APBC is a circle of four points.

So apb=90 ap pb

Because abp=45

So ap=pb

The answer is already there, and I will not repeat .........

According to the sum theorem of the inner angle of the triangle, if the base angle is a, then there is a+a+4a=180°, a=30°

Draw a figure, make an extension line on one waist, draw a height, and get a large right-angled triangle, one angle is 30°, one angle is 60°, the hypotenuse is a, and the right-angled side is one-half of the hypotenuse.

Obviously, the bottom angle is 30 °, and the top angle is 120 °.

The height of the triangle is h=a*sin30=a2

Let the bottom angle be x, then 4x+x+x=180, the top angle is 120, the bottom angle is 30, and the extension line from the vertex of the bottom corner to the waist is high. In the large Rt triangle drawn, h=asin30=a2 is obtained

Because the top angle is 4 times the bottom angle, the bottom angle is 30 degrees, the top angle is 120 degrees, and the height to be made can be a 30° right triangle, the hypotenuse is a, and the shorter right angle side (the height on the waist) is half of the hypotenuse, half of a

One-twelfth one-half square.