In the RT triangle ABC, ∠ C = 90 °, ab = radical 10, AC: BC = 2:1, calculate the perimeter and area of RT triangle

In the RT triangle ABC, ∠ C = 90 °, ab = radical 10, AC: BC = 2:1, calculate the perimeter and area of RT triangle

Root number 2 + 2 times root number 2 + root number 10

1. The hypotenuse ab of the RT triangle ABC = 4 times the root sign 2 + 2 right angle side AC = 2 times the root sign 2 + 4. Find the area and perimeter of the triangle ABC 1. The hypotenuse ab of the RT triangle ABC = 4 times the root sign 2 + 2 right angle side AC = 2 times the root sign 2 + 4. Find the area and perimeter of the triangle ABC? 2. The three sides of a triangle are the root 20 cm. The root 12 cm. The root 32 cm. Find the perimeter and area of the triangle?

Please add a comma in the sentence later, you almost read it wrong
This is simple:
1. AB = 4 √ 2 + 2; AC = 2 √ 2 + 4; BC = 2 √ 2; (Pythagorean theorem, the square of an oblique edge = the sum of two bisections)
Area s = 4 + 4 √ 2; perimeter C = 8 √ 2 + 6;
2. This is the RT triangle, because the sum of the two short sides is equal to the square of the hypotenuse
Perimeter C = 2 √ 5 + 2 √ 3 + 4 √ 2; area s = 2 √ 15

In the RT triangle ABC, if the angle c is equal to 90 degrees, cos = 2 / 2 root sign 3, then the value of Tan 2 / B is

In the RT triangle ABC, if the angle c is equal to 90 degrees, CoSb = the root of 2 / 3, then the value of B in tan2 is
Cosb2 fraction B
=sinB/(1+cosB)
=2-radical 3

In the triangle ABC, Sina = 1 / 2, tanb = radical 3. Then Cosa tanb =? I have to process... Thank you

TanB= root 3
∠ B=60
sinA=1/2
∠ a = 30 or 150 (omit, 150 + 60 > 180)
Cosa tanb = root 3 / 2-radical 3
=-Radical 3 / 2

In RT △ ABC, CD is the height on the hypotenuse AB, ab = 8ac = 4, root sign 3, then ad =? Ninth grade people's education press Volume 2

RT△ABC∽RT△ACD
AC/AB=AD/AC
AD=AC^2/AB=48 /8=6

As shown in the figure, in the RT triangle ABC, the angle ABC = 90 °, the point D is the midpoint of the hypotenuse AB, and the perpendicular foot of de ⊥ AC is e. if de = 2, CD is equal to 2, the length of be is

Because the point D is the middle point of the hypotenuse AB, so ad = DB CD = AB / 2 because CD = 2, radical 5, so AB = 4, because De is perpendicular to AC, so De is parallel to BC, so de = BC / 2 AE = CE = AC / 2 because de = 2, so BC = 4 in right triangle ACB

As shown in the figure, in RT △ ABC, ∠ C = 90 °, ad bisection ∠ cab, CD = radical 3, BD = 2, radical 3, find the length of ab No more pictures

Let the length of AC be a
According to the title,
∠C=90°,CD=√3,BD=2√3
AD=√(3+a²)
BA=√(27+a²)
Theorem of bisector from angle
BD：DC=BA：CA
So,
2=√(27+a²)/a
The solution
A=3
Therefore, Ba = √ (27 + a?) = 3 √ 3

In a right triangle ABC, the angle ACB is equal to 90 degrees, AC is equal to BC and is equal to 2 times the root sign 2

Friend, we can only give you a complete answer to the question!

As shown in the figure, in the right triangle ABC, the angle ACB = 90 degrees, D is the midpoint of side AB, be is vertical CD, the perpendicular foot is point E, known AC = 15, cosa = 3 / 5, find (1) the length of the line CD and (2) the value of sindbe

(2) s ⊿ BCD = 1 / 2S ⊿ ABC = 75, i.e. 1 / 2 × CD × be = 75,

As shown in the figure △ ABC is a right triangle, ∠ ABC = 90 ° CD ⊥ AB in D

1. It is proved that if CD is perpendicular to AB and E is the midpoint of AC, then de = AC / 2 = AE, ∠ a = ∠ ade = ∠ FDB;
And ∠ a = ∠ DCG (all are the residual angles of ∠ ECD), so ∠ FDB = ∠ DCB
If [F=] [F], then ⊿ FDB ⊿ FCD, FD/FC=FB/FD, FD? =FBxFC
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