Can you find the missing side lengths of the triangle? | (Justify) |
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- Опубліковано 17 жов 2024
- Learn how to find the missing side lengths of the right triangle. Important Geometry and Algebra skills are also explained: similar triangles; Pythagorean Theorem; Euclid's Theorem. Step-by-step tutorial by PreMath.com
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Thanks dear ❤️🙏
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
Can u write it properly???
Excellent!
Thanks for sharing ❤️
very good sharing bro thanks for sharing❤ stay connected keep it up🎉🎉🎉🎉🎉🎉🎉🎉
Thanks for visiting🌹
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.
Thanks for the feedback ❤️
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
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^2=P(q)=(8-25/8)(25/8)=975/64
h=√975/64=5√39/8
So P=8-25/8=39/8.❤❤❤ Thanks Sir
Excellent!
You are very welcome!
Thanks for sharing ❤️
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.
can you explain haw are you writing alpha and beeta with no information, so please explain this
Very useful & informative video👍🏼thanks a lot🙋🏻♂️
Let's find the missing side lengths:
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..
...
....
.....
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
Excellent!
Thanks for sharing ❤️
Thanks Prof. For your explaining this difficult situation
That’s very nice
Good luck with glades
❤❤❤
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.
Thanks for sharing
I did the same.
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
That is a proof of Euclid theorim
Excellent!
Thanks for sharing ❤️
Very good representation and explanations 🎉🎉🎉🎉😊😊😊
Glad you liked it🌹
Thanks for the feedback ❤️
Thank you!
a=√64-25=√39. S= a*b/2≈c*h/2. h=a*b/c=5*√39/8. Далее по теореме Пифагора.
Excellent!
Thanks for sharing ❤️
You are a God in mathematics
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!
Triangles ABC and ACD r equiangular.
8/5=5/DC
DC =25/8
AD=√(25 -625/64)
BD=8 -25/8
Then AB =√[(8-25/8)^2+25 - 625/64)]
*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.
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!✨
Triangle ∆ACB:
AC² + CB² = BA²
5² + a² = 8²
a² = 64 - 25 = 39
a = √39
Excellent!
Thanks for sharing ❤️
Here we have the Formula for this ;
(AC)² = (AD).(AB)
(BC)² = (BD).(BA)
(CD)² = (DB).(DA)
Thanks for sharing ❤️
What is wrong with:
ACD is a right triangle. Set h=4, q=3 then b=5 due to Pythagorean theorem. Then p=8-3=5. Then due to Pythagorean theorem, a**2=(5**2) + (4**2), so a=sqrt(25+16)=sqrt(41)
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
I took a order going through it, but got the same answers. Very satisfying.
I find these problems easier when using trigonometric identities. I also don't like leaving the answers in fraction form or as a multiple of a square root. I prefer the answers in a decimal form. These answers make more sense to me as an engineering technician, who had to find the sides of triangles in an effort to determine dimensions. In this case, the 5÷8 = arcsine beta, which is 38.68°. Now that I know the angle, I can use the trigonometric functions of the angle to determine the other sides of both smaller triangles. The side "a" then would 6.2449 which in the case of needing an answer in inches it would be rounded to 6 1/4 inches. If I was working 5 and 8 feet the answer would be 6 ft 2 15/16 in or 6 ft 3 in. Depending on the accuracy of the dimension required.
I got this one!
My take: (1) Calculate on ABC using Pythagore: a=√39. (2) Calculate on ABC using both formulae for the surface: h=5√39/8. (3) Calculate p and q. More direct, easier to follow IMHO.
Just show the previous comment... Another way would be to use upfront the three similar triangles, not just two. Then explain which ratios are interesting, depending on the available data. Perhaps remind people that there are three ratios each time - a/b, a/c and b/c - and help them those that will lead more directly to the result. You could try to put them all on display, using different colors for numbers and variables, then highlithing the "good ones".
My way of solution ▶
5²= q*c
c= 8
⇒
25= q*8
q= 25/8 length units
p= 8- 25/8
p= 39/8 length units
h²= p*q
h²= (39/8)*(25/8)
h= √975/64
h = √39*25/8
h = 5√39/8
a²+b²= c²
a²+5²= 8²
a= √39
It could be solved easily by similarity😂😂
b = 5, c = 8
p+q = c
===========
a² = c² - b²
a = √(c²-b²)
a = √(8²-5²)
a = √39
===========
a²-p² = h²
b²-q² = h²
a²-p² = b²-q²
a²-b² = p²-q²
39-25 = (p+q)(p-q)
14 = 8(p-q)
14 = 8p-8q
8p-8q = 14
8p+8q = 64 ---> (p+q = 8)8
------------ +
16p = 78
p = 39/8
========
8 = p + q
8 = 39/8 + q
64 = 39 + 8q
25 = 8q
25/8 = q
=========
h = √(a²-p²) or √(b²-q²)
h = √(a²-p²)
h = √(39-39²/8²)
h = √(39*8²/8²-39²/8²)
h = ⅛√(39*8²-39²)
h = ⅛√975
h = ⅛5√39
h = ⅝√39
h = √(b²-q²)
h = √(5²-25²/8²)
h = √(5²*8²/8²-25²/8²)
h = ⅛√(25*64-625)
h = ⅛√975
h = ⅝√39
p = 39/8, q = 25/8, h = ⅝√39 9:13
STE-BY-STEP RESOLUTION PROPOSAL :
01) BC = a = sqrt(39) lin un ~ 6,245 li un
02) AC = b lin un = 5 lin un
03) AB = c lin un = 8 lin un
04) BD = p lin un ~ 4,875 lin un
05) AD = q lin un ~ 3,123 lin un
06) CD = h lin un = ((5 * sqrt(39)) / 8) lin un ~ 3,9 lin un
07) a^2 + b^2 = c^2 ; a^2 + 25 = 64 ; a^2 = 64 - 25 ; a^2 = 39 ; a = sqrt(39)
08) Right Triangle Area (A) = (a * b) / 2 ; A = ((5 * sqrt(39)) / 2) sq un ; A ~ 15,6 sq un
09) 2 * A = (h * c) ; 2 * ((5 * sqrt(39)) / 2) = 8 * h ; h = (5 * sqrt(39)) / 8 ; h ~ 3,9 lin un
10) q / b = h / a ; q / 5 = 3,9 / 6,245 ; q = (5 * 3,9) / 6,245 ; q ~ 3,123 lin un
11) p^2 = a^2 - h^2 ; p^2 = 39 - 975/64 ; p^2 = (2.496 - 975) / 64 ; p^2 = 1.521 / 64 ; p^2 ~ 23,8 ; p = 4,875 lin un
Thus,
OUR BEST ANSWER :
a ~ 6,245 li un
p ~ 4,875 lin un
q ~ 3,123 lin un
h ~ 3,9 lin un
Greetings.