Calculate the unknown side lengths of Trapezoid | Trapezium | Important Geometry skills explained

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Calculate the unknown side lengths of Trapezoid | Trapezium | Important Geometry skills explained

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[zackhemsey]

Great video . Keep it up
You are the best teacher in the world
Thank you for your important knowledge transmission
Love from england 🤟🤟

[abeonthehill]

I worked out the area of triangle to be 1500 sq units . From 600-1500=4500 …..then a must equal 4500/75=60 . Longest parallel side is 60+40 and a =60 .
So 75(160) /2 =6000 sq units .

It seems that my earlier attempt, i had the hyp of triangle as 80 ! …..i need my glasses re-calibrated i think !
Thanks for sharing ….

[theoyanto]

Did it kinda the same way, just a few less steps... But I realise your method us step by step... Thanks again 👍🏻

[spiderjump]

85sq — 75 sq= 1600= 40sq

Let the lengthy shorter of the parallel lines be a.
Longer line= a+ 40
1/2(2a+40)•75= 6000

[alster]

Very easy. Got it in less than 30 seconds

[murdock]

Awesome, many thanks!
75x + (20)75 = 75(x + 20) = 6000 → x + 20 = 80 → x = 60 🙂

[ramaprasadghosh]

difference in parallel edges
= √( 85^2 - 75^2)
= √(32* 2) = 8
Hereby area
= 75( 8+ 2*smaller of the || side)/2
= 75*4 + 75* smaller of the || side
or
6000/75 = 4 + smaller of the || side
or smaller of the || side = 76 unit
larger of the || side = 80 unit

[quigonkenny]

Let a and b be the lengths of CD and AB respectively:

Trapezoid ABCD:
A = h(a+b)/2
6000 = 75(a+b)/2
a + b = 6000(2/75) = 12000/75
a + b = 160 ---- (1)

Drop a perpendicular from D to AB at E. By observation, EBCD is a rectangle, so EB = CD = a, DE = BC = 75, and AE = b-a.

AE² + ED² = DA²
(b-a)² + 75² = 85²
(b-a)² + 5625 = 7225
(b-a)² = 1600
b - a = √1600 = 40
b = 40 + a ---- (2)

Sub (2) into (1) and solve for a:
a + (b) = 160
a + (40+a) = 160
2a = 160 - 40 = 120
a = 60

And solve for b:
b = 40 + (a) = 40 + 60 = 100

AB = 100
CD = 60

[FrogworfKnight]

Done without calculators and before watching the video.

Trapazoid, unknown bases, side with two right angles 75, side with slant 85
Area 6000

b_1 smaller base
b_2=b_1+x

x^2+75^2=85^2
x^2=85^2-75^2
x^2=(85-75)(85+75)
x^2=(10)(160)=1600
x=40

A=(1/2)(b_1+b_2)h
6000=(1/2)(b_1+b_1+40)(75)
6000=(1/2)(2b_1+40)75
6000/75=b_1+20 //6000->3*2*1000 75->25*3
80=b_1+20
60=b_1
100=b_2

[arnoldfrederickangelo]

For this video, the question is easy but usually your question is so hard

Keep on fire!!!😂

[devondevon]

Could avoid using the Pythagorean formula since 8, 15, and 17 are Pythagorean triplets and given
that 85 and 75 are two legs of that triplet scaled up by 5. Hence multiply 8 by 5 to get 40.

[thanhduyenphanthi]

so i calc b-a=40 then calc the triangle in the trapezoid=1500 so 75a=4500 => a=60; b=a+40=100

[yihan]

You don't actually need to use the trapezoid formula so solve this

First, draw a perpendicular from d to make a right-angled triangle AED

Then solve for AE using the Pythagorean theorem, in which b is the unknown variable

85^2-75^2=1600 (c^2 - a^2 = b^2)

sqrt(1600)=40

AE is thus equal to 40

We then subtract the area of the triangle from the area of the trapezoid, leaving behind the "square" (not really a square but I don't know the word for it)

(40·75)/2=1500
6000-1500=4500


Area of square is thus equal to 4500

We now solve the unknown sides using the formula for area of square;
a*b = area

Let x be the side of unknown length

75x=4500

Divide both sides by 75

x=60

AB = 40+60 = 100
CD = 60

[devondevon]

let B = the longer side
(A + B)/2 (h) = 6000
(A+B)/2 (75) =6000
A + B = 12, 000/75
A+ B = 160 equation P
The Pythagorean theorem will be used to find how much greater B is than A.
Since 85 and 75 are two sides of the Pythagorean triplet 17, 15, and 8 scaled up by 5,
then the other is 40 (or 8 scaled up by 5); hence B is 40 units greater than A.
Hence B= A+ 40
But since A+ B =160, see equation P
then A + (A+40) =160;
2A +40 = 160
2A =120
A =60; hence B =100 Answers

[josephwise]

wouldn't have been easier to start off finding the length of the triangle part of the missing side of the trapezoid. (40). Then find the area of that trapezoid. 1/2*40 * 75 = 1500. Subtract that area from 6000 to get 4500 then find the area of the square of 75x=4500. x=60. x+ length of the missing side of the triangle in step 1 = 100 ?

[MrPaulc]

Good video but felt unnecessarily complicated this time. Add imaginary triangle to the left to form rectangle. 85^2 - 75^2 = 1600, so top part of imaginary triangle is 40 . Therefore, 100-40 (one answer). This gives 20*75 as area of imaginary triangle, so 1500. This makes rectangles area 7500. As one side is 75, other is 100 (answer two).

[shreyesveer]

Draw a perpendicular from D to BC where the perpendicular meets BC at E, this splits the trapezium into a rectangle DCBE and a triangle DAE;
Take DC as x and AB y
Now area of trapezium ABCD is (1/2)*(sum of parallel sides)*height = 6000
=> (1/2)*(x+y)*75 = 6000
Simplifying this we get;
x + y = 160
Now AD = AE + EB
From rectangle DCBE we know BE = DC = x and DE = BC = 75
AB = AE + EB
=> AE = AB - EB = y-x
From the right triangle ADE using Pythagoras theorem we get;
AD^2 = DE^2 + AE^2
=> AE^2 = AD^2 - DE^2
=> (y-x)^2 = 85^2 - 75^2
=> (y-x)^2 = 1600
=> y-x = 40 (can’t be -40 as length in any units can’t be negative)
So x + y = 160 and y - x = 40
Therefore using the above equations we get;
y = 100 and x = 60

[eckhardfriauf]

2:39 The 'b' that you come up with might be confused with the previous 'b' (1:06).
No major mistake, yet not completely elegant.

[chrisdrews]

Good video but having your variables and the equations both use a and b might be confusing to some.

[funhcyx]

75a+75(b-a)/2= 6000
75(a+b)=12000
a+b=160