All categories

3 years ago

em um triângulo retângulo a hipotenusa mede 15 cm e um dos catetos mede 12 cm.Qual é o perímetro desse retângulo?

Mathematics

1 answer:

ElenaW [278]3 years ago

5 0

The answer to Em um triângulo retângulo a hipotenusa mede 15 cm e um dos catetos mede 12 cm.Qual é o perímetro desse retângulo?
borda a =12 unidades
borda b =9 unidades
borda c =15 unidades
ângulo A =53.1 graus
ângulo B =36.9 graus
área=54 square unidades
Agora temos que
Determinar o comprimento da base do triângulo.
Calcula-se a altura do triângulo.
Multiplicar o comprimento da base por o comprimento da altura.
Multiplique por dois.
Adicionar a unidade de medida apropriada.

You might be interested in

What is 7/10 equivalent to write the answer in fractions

Romashka [77]

7/10 is equivalent to 3.5/5 is one

8 0

3 years ago

3(2c-8) - 10c > 0 can some one explain how to solve this ?

Jet001 [13]

Answer:

c < - 6

Step-by-step explanation:

Given

3(2c - 8) - 10c > 0 ← distribute and simplify left side

6c - 24 - 10c > 0

- 4c - 24 > 0 ( add 24 to both sides )

- 4c > 24

Divide both sides by - 4, reversing the inequality symbol as a consequence of dividing by a negative quantity.

c < - 6

8 0

3 years ago

What's the common factors for 39 and 65?

Ivenika [448]

Factors of 39: 1, 3, 13, and 39.
Factors of 65: 1, 5, 13, and 65.

The common factors for 39 and 65 are 1 and 13.
The greatest common factor is 13.

5 0

3 years ago

Sandra is putting up a fence around her garden, which is in the shape of a square. Each foot of fencing costs $9. If each side o

mestny [16]

Answer:

Sandra will need 32 feet of fencing. This is because the shape of the garden is a square. The length is 8 ft, so the perimeter of the garden is 32.

Step-by-step explanation:

4 0

3 years ago

B) T is due north of C, calculate the bearing of B from C

choli [55]

Answer:

(a) 52°

(b) 322°

Step-by-step explanation:

(a) The details of the circle are;

The diameter of the circle = AOC

The center of the circle = Point O

The point the line AT cuts the circle = Point B

The point the tangent PT touches the circle = Point C

Angle ∠COB = 76°

We have that angle AOB and angle COB are supplementary angles, therefore;

∠AOB + ∠COB = 180°

∠AOB = 180° - ∠COB

∴ ∠AOB = 180° - 76° = 104°

∠AOB = 104°

OA = OB = The radius of the circle

Therefore, ΔAOB = An isosceles triangle

∠OAB = ∠OBA by base angles of an isosceles triangle are equal

∠AOB + ∠OAB + ∠OBA = 180° by angle summation property

∴ ∠AOB + ∠OAB + ∠OBA = ∠AOB + ∠OAB + ∠OAB = ∠AOB + 2×∠OAB = 180°

∠OAB = (180° - ∠AOB)/2

∴ ∠OAB = (180° - 104°)/2 = 38°

∠TAC = ∠OAB = 38° by reflexive property

AOC is perpendicular to tangent PT at point C, by tangent to a circle property, therefore;

∠TCA = 90° and ΔTCA = A right triangle

∠TAC + ∠ATC + ∠TCA = 180° by angle sum property

∠ATC = 180° - (∠TAC + ∠TCA)

∴ ∠ATC = 180° - (38° + 90°) = 52°

Angle ATC = 52°

(b) In ΔABC, ∠ABC = Angle subtended by the diameter = 90°

∴ ΔABC = A right triangle

∠ABC and ∠TBC are supplementary angles, therefore;

∠ABC + ∠TBC = 180°

∠TBC = 180° - ∠ABC

∴ ∠TBC = 180° - 90° = 90°

∠TCB = 180° - (∠TBC + ∠ATC)

∴ ∠TCB = 180° - (90° + 52°) = 38°

The bearing of B from C = (360° - 38°) = 322°.

7 0

2 years ago