Can you find the missing side lengths of the triangle? | (Justify) |#math #maths | #geometry

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Can you find the missing side lengths of the triangle? | (Justify) |#math #maths | #geometry

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[Alishbavlogs-bmip]

very good sharing bro thanks for sharing❤ stay connected keep it up🎉🎉🎉🎉🎉🎉🎉🎉

[himo]

a=√[8²-5²]=√39
8h/2=5√39/2 8h=5√39 h=5√39/8
q=5*5/8=25/8
p=8-25/8=64/8-25/8=39/8

[unknownidentity]

Let's find the missing side lengths:
.
..
...
....


Let's have a look at the interior angles of all three right triangles ABC, ACD and BCD:

ABC: ∠ACB=90° ∠ABC=α ∠BAC=β
ACD: ∠ADC=90° ∠ACD=α ∠CAD=β
BCD: ∠BDC=90° ∠CBD=α ∠BCD=β

Therefore all these triangles are similar and we can conclude:

a:b:c = h:q:b = p:h:a

Additionally we can apply the Pythagorean theorem:

a² + b² = c²
a² + 5² = 8²
a² + 25 = 64
a² = 39
⇒ a = √39

b/c = h/a
⇒ h = a*b/c = 5√39/8

a/c = p/a
⇒ p = a²/c = 39/8

b/c = q/b
⇒ q = b²/c = 25/8

Best regards from Germany

[aljawad]

Nice! I used the Ratios behind the trigonometric identities (sin, cos & tan in terms of “opposite”, “adjacent” and “hypotenuse”) of the two similar triangles to reach the answers.

[prossvay]

In Right triangle ABC
b^2=q(c)
5^2=q(8)
So q=25/8
a^2=P(c)=(8-25/8)(8)
So a=√39

h=√975/64=5√39/8
So P=8-25/8=39/8.❤❤❤ Thanks Sir

[allanflippin]

I accidentally stumbled onto a quicker way to find the length of q. Notice that triangle ABC is similar to ACD. That means AB/CA = CA/q. AB = 8 and CA = 5. So 8/5 = 5/q. Cross multiplying and simplifying, q = 25/8. As you did, Pythagorean formula finds a = sqrt(AB^2 - CA^2). Or a = sqrt(39). As you did, c = 8 - q or c = 64/8 - 25/8 = 39/8. h can be found by again comparing similar triangles. ABC is similar to CBD so AB/CA = CB/CD. That means 8/5 = a/h. Cross multiplying and solving for h: h = 5a/8. a = sqrt 39, so h = 5*sqrt(39)/8.

[BBMathTutorials]

Very useful & informative video👍🏼thanks a lot🙋🏻‍♂️

[skverma]

Sir, by using Pythagoras theorem we can get BC directly.
It is sq.root of 64-25=39
That is BC is root 39.
We can easily find other sides using similarity of triangles.

[thewolfdoctor]

Alternative way to get pq = SQR(h): Since ABC is a right triangle, use Thales theorem to see that BA is the diameter of a circle and C is a point on the circle. So intersecting chords get you p*q = h*h

[yalchingedikgedik]

Thanks Prof. For your explaining this difficult situation
That’s very nice
Good luck with glades
❤❤❤

[quigonkenny]

Triangle ∆ACB:
AC² + CB² = BA²
5² + a² = 8²
a² = 64 - 25 = 39
a = √39 <--- [a]

As ∠BDC = ∠ACB = 90° and ∠B is common, triangle ∆BDC is similar to ∆ACB. As ∠CDA = ∠ACB = 90° and ∠A is common, triangle ∆CDA is also similar to ∆ACB.

Triangle ∆BDC:
BD/CB = CB/BA
p/√39 = √39/8
p = (√39)²/8 = 39/8 <--- [p]

DC/CB = AC/BA
h/√39 = 5/8
h = (5√39)/8 <--- [h]

Triangle ∆CDA:
DA/AC = AC/BA
q/5 = 5/8
q = 25/8 <--- [q]

a = √39, h = (5√39)/8, p = 39/8, q = 25/8

[cyruschang]

The left side of the big triangle = ✓(8^2 - 5^2) = ✓39
black ? = 5 ÷ (8/✓39) = (5✓39)/8
red ? = 5 ÷ (8/5) = 25/8
blue ? = (✓39) ÷ (8/✓39) = 39/8

[ChuzzleFriends]

We are given that AC = b = 5, AB = c = 8, △ACB is right, and an altitude CD is shown, represented by h. Therefore, we can use the Geometric Mean Theorems to find a, h, p, & q (the altitude separates △ACB into two other right triangles, △ADC & △BDC, that are thus both similar to it by the Right Triangle Similarity Theorem).
First, use the Geometric Mean (Leg) Theorem.
(AC)² = AB * AD
5² = 8q
8q = 25
q = 25/8
= 3.125
So, BD = p = 8 - 3.125 = 4.875, or 39/8.
(BC)² = AB * BD
a² = 8 * 4.875
a² = 39
a = √39
≈ 6.24
Then, use the Geometric Mean (Altitude) Theorem.
(CD)² = AD * BD
h² = 3.125 * 4.875
h² = 975/64
h = √(975/64)
= (√975)/(√64)
= [(√25)(√39)]/[√(8²)]
= (5√39)/8
≈ 3.90
Alternatively, we could have used the Pythagorean Theorem after using the Geometric Mean (Leg) Theorem once, but this is the way I solved the problem.
So, the labeled side lengths are as follows:
a = √39 ≈ 6.24
b = 5
c = 8
h = (5√39)/8 ≈ 3.90
p = 39/8 = 4.875
q = 25/8 = 3.125

[robertlynch]

The RED line (q) is (bb/c) where b = 5 and c = 8 so ... 25/8 (3.125) in length.
From there, the black riser (h) is sqrt(b² + q²) = sqrt(5² + 25²/8²) = sqrt( 34.7656 ) = 5.896
Then (a) is pythagorean ... 8² - 5² = 64 - 25 = 39 ... sqrt(39) = 6.245
Lastly blue (p) is (aa/c) = 39/8 = 4.875

Just remember those 3 super identities about right triangles.
If sides are A, B and hypotenuse C then
Inside height is AB/C
A-side bit of C is AA/C. and
B-side bit of C is BB/C

FAST. Memorable. The Swiss-Army-Knife of your toolkit!

[sagarmajumder]

Very good representation and explanations 🎉🎉🎉🎉😊😊😊

[ОльгаСоломашенко-ьы]

a=√64-25=√39. S= a*b/2≈c*h/2. h=a*b/c=5*√39/8. Далее по теореме Пифагора.

[joiceroosita]

Here we have the Formula for this ;

(AC)² = (AD).(AB)

(BC)² = (BD).(BA)

(CD)² = (DB).(DA)

[pralhadraochavan]

You are a God in mathematics

[joeschmo]

I just made a table of the 3 similar triangles of "short side", "long side", and hypotenuse.
From 5 and 8 I immediately got sqrt(39). Then just ratioed the others for 5sqrt(39)/8, 39/8, and 25/8.
Final check was last, that the last 2 above totalled up to 8.
Didn't even have to do any 8-x as a side or anything. Took like 1min to just draw the 3 triangles and label the sides.
✨Magic!✨

[marcgriselhubert]

*In ABC: a^2 = BC^2 AB^2 - AC^2 = 64 - 25 = 39,
so a =sqrt(39).
*2.Area of ABC = a.b = 5.sqrt(39) = BC.DC = 8. h,
so h = (5/8).sqrt(39).
*BDC and CDA are similar (same angles),
so BD/BC = CD/CA, so p/sqrt(39) = ((5/8).sqrt(39))/5,
so p = 39/8.
*q = AD = AB -BD = 8 - (39/8) = 25/8.