![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
Junior high school function formula and Pythagorean theorem, the best picture, I am doing mold, thank you for your help I hope I can give you the picture above
Let the opposite sides of right angles a and B be a and B, and the hypotenuse be c. sine: the ratio of the opposite side of an acute angle to the hypotenuse. Sina = A / C cosine: the ratio of the adjacent side of an acute angle to the hypotenuse. Cosa = B / C tangent: the ratio of the opposite side of an acute angle to the adjacent side. Tana = A / b cotangent: the ratio of the adjacent side of an acute angle to the hypotenuse. COTA = B / a Pythagorean Theorem
Pythagorean theorem in junior high school
CD is the high of △ ABC and has CD & sup2; = ad · dB. It is proved that △ ABC is a right triangle
Because: CD is the high value of △ ABC
So: CB & sup2; = CD & sup2; + DB & sup2;, AC & sup2; = ad & sup2; + CD & sup2;;
Because: ab & sup2; = (AD + dB) & sup2; = ad & sup2; + 2ad · DB + DB & sup2;
Because: CD & sup2; = ad · dB
So: ab & sup2; = (AD + dB) & sup2; = ad & sup2; + 2ad · DB + DB & sup2; = ad & sup2; + 2CD & sup2; + DB & sup2; = (AD & sup2; + CD & sup2;) + (CD & sup2; + DB & sup2;) = CB & sup2; + AC & sup2;
So: ab & sup2; = CB & sup2; + AC & sup2;
According to Pythagorean theorem, △ ABC is proved to be a right triangle
Pythagorean theorem in junior high school
In the triangle ABC, ab = AC = 17cm, BC = 16cm, the height ad =? Triangle ABC area =?
The only difficulty is to find the area 96 of triangle ABC with AD = 7.25
Because AB = AC
So BD = CD = BC / 2 = 8
Using Pythagorean theorem: ad = (AB ^ 2-bd ^ 2) ^ (1 / 2)
=(17^2-4^2)^(1/2)
=15
S△ABC=1/2*BC*AD=1/2×16×15=120
RELATED INFORMATIONS
- 1. On the Pythagorean theorem in junior high school It is known that the two right angle sides of the RT triangle ABC are a and B, the hypotenuse is C, and the height of the hypotenuse is h. try to judge the shape of the triangle with H, C + H, and a + B as sides Please explain why,
- 2. In the triangle B2 + C2 BC = A2, find a (ABC followed by square)
- 3. As shown in the figure, it is known that △ ABC and △ DEB are isosceles right triangles, ∠ ACB = ∠ EDB = 90 °, point E is on side AC, CB and ED intersect at point F
- 4. As shown in the picture, OD.OE They are the bisectors of ∠ AOC and ∠ BOC, ∠ AOD = 40 ° and ∠ BOE = 25 ° respectively. To find the degree of ∠ AOB is not only the answer, but also the process plus the answer
- 5. As shown in the figure, OE bisects ∠ BOC, OD bisects ∠ AOC, ∠ BOE = 20 ° and ∠ AOD = 40 ° to calculate the degree of ∠ doe
- 6. As shown in the figure, the bisectors of ∠ CBD and ∠ BCE at the two outer angles of △ ABC intersect at point O, ∠ a = 40 ° to calculate the degree of ∠ BOC
- 7. In the triangle ABC, a = 3, B = root 3, a = 60 degree solution triangle
- 8. (1) In triangle ABC, a = 2, B = 2 √ 3, B = 60 ° are known to solve triangle ABC (2) In triangle ABC, we know a = √ 3, B = √ 2, B = 45 ° to solve triangle ABC
- 9. In the triangle ABC, a = 18, B = 16, a = 150, solve the triangle
- 10. The length of the hypotenuse of a right triangle is 10 cm, and the sum of the two right sides is 12 cm? Pythagorean theorem should be used
- 11. It is known that Y1 is positively proportional to X (scale coefficient is K1) and Y2 is inversely proportional to X (scale coefficient is K2). If the image of function y = Y1 + Y2 passes through points (1,2), (2,12), then the value of 8k1 + 5k2 is______ ．
- 12. Help out on the inverse proportion function and Pythagorean theorem of the problem it! A little basic on the line, but not too simple!
- 13. As shown in the figure, in △ ABC, points D and E are on AB and AC respectively, de ‖ BC, ∠ CED = ∠ BDC. (1) prove: △ DCE ∽ CBD; (2) if BC = 2CD, s △ ade = 1, find the value of s △ ABC
- 14. In the geometry shown in the figure, EA vertical plane ABC, DB vertical plane ABC, AC vertical plane BC and AC = BC = BD = 2ae. M is the midpoint of ab Finding the tangent of the angle between de and plane EMC
- 15. In the triangle ABC, ab = BC, the point D E F is the midpoint on the side of BC AC AB respectively
- 16. As shown in Figure 1, in △ ABC, points e.d.f ab.bc.ca If ad bisects ∠ BAC, then the quadrilateral AEDF is a diamond?
- 17. As shown in the figure, in the quadrilateral ABCD, ∠ bad = ∠ ABC, ad = BC, AC intersects with point o at BD. let's say that the triangle OAB is an isosceles triangle
- 18. As shown in the figure, it is known that ad, be and CF are respectively the heights of the three sides of △ ABC, h is the perpendicular, and the extension line of ad intersects the circumscribed circle of △ ABC at point G
- 19. In △ ABC, ab = AC, be = CF, EF intersect BC with D
- 20. As shown in the figure, the triangle ABC is an equilateral triangle, and D is the midpoint on the side ab. it is known that the area of the triangle BDE is 5 square centimeters. Find the area of the equilateral triangle ABC