Let's drop two heights from the upper vertexes down to the larger base. They will be h each. Let segments we got by dropping heights be a (the left one) and b (the right one) and middle which is equal to the smaller base and that is 10. Then we got: h²=13²-a² On the other hand: h²=14²-b² So: 169-a²=196-b². And also: a+b+10=25 After few changes, we can make a system: b²-a²=27 a+b=15 Let's solve it. (b-a)(b+a)=27 a+b=15 15(b-a)=27 b-a=1,8 We got a new easier system: b-a=1,8 a+b=15 Let's add equations to each other: 2b=16,8 b=8,4=42/5 a=6,6=33/5 Let's find h: h²=196-(42/5)² 196-1764/25=h² 3136/25=h² h is a height so h>0 => h=56/5=11,2 S=(10+25)11,2/2=35•5,6=196 Answer: 196 square units.
Simple is : Draw 2, 90^ triangles and one rectangle from this shape. Area of 90^ triangle= 1/2 ab Area of rectangle = ab ; a is length, b is height. Add all.
Cool problem. Used different strategy. Created one triangle by taking the left and right sides of the trapezoid. Hence, a triangle with sides 13, 14, and 15. Used Heron's formula to calculate area of triangle. Took that value and calculated height of triangle, which is the height of the trapezoid. I dumped that value into the area formula for a trapezoid. Solution.
In short, it could be solved like this: The height comes from the solution of the equation 13^2-x^2=14^2-(15^2-x^2)^2 because the height is expressed by the Pythagorean theorem in two forms in two right triangles. The solution gives x=6,6. Then apply T.P. in one of the triangles and the height h=11.2 comes out. In the end S=[(25+10)×11,2]÷2= 196cm^2
I took the long way, but got that answer too. I made a regular trapezoid (13 on both left and right) and then the full right triangle had a base of x for 13 and x+x1 for 14. It was fun.
f(h)=√(13^2-h^2)+√(14^2-h^2)-15=0 f'(h)=-h/√(13^2-h^2)-h/√(14^2-h^2) Newton-Raphsin iteration: Start with h=10 and iterate h←h-f(h)/f'(h) until h=11.200... The area is then (1/2)(10+25)11.2=196 sq units.
In method 2, it would be more interesting if you had found the area of the parallelogram and, after adding it to the area of the triangle, finally, we would have the area of the trapezoid: [ABPD]=h×BP=56/5 × 10=112, Therefore, [ABCD]=[ABPD]+[BCP] *[ABCD]=84+112=196*
Seems to me like you're just adding additional unnecessary steps. You use the area of the triangle to get the value of h, but once you have that it's unnecessary to find the area of the parallelogram, since you can use h to directly find the area of the whole trapezoid.
@@quigonkennyUnfortunately, you didn't understand what I meant, in the second method it would be more interesting not to use the formula for the area of the trapezoid, as it is easier to find the area of the parallelogram.
Let height of trapezium be h. Draw AE and BF both equal to h , so that DEFC is a straight line. Area of triangle ADE = 0.5h x DE. Area of triangle BFC = 0.5 h x FC . Area of AEFB (the rectangle) is 10h. hxh= 13x13- DExDE= 14x14- FCxFC by Pythagors'theorem in both triangles FCxFC -DExDE = 14x14-13x13 = 196-169 =27 (FC+DE)(FC-DE)=27 FC+DE+10= 25 so FC-DE = 27/ 15 = 9/5 FC+DE=15 FC-DE=9/5 =1.8 2xFC = (15x5+9)/5 FC=8.4 DE= 6.6 13x13- 6.6x6.6= (13+6.6)(13-6.6)= 19.6 x 6.4= 196 x64/(10x10) = h x h. So h= 14x 8/10 = 112/10 = 11.2 Area of this trapezium = 11.2 (10 + 4.2 + 3.3) = 11.2 x 17.5 = 2.8 x 70 = 196 square units .
Draw AE || BC that cuts BC at E Area ( Trapezium ABCD) / area ( ∆ ADE) = ( AB + CD)/DE = ( AB + CD) / ( CD - AB) = ( 25 + 10)/( 25 - 10) = 7/3 Area ( ∆ ADE) of sides 13, 14, 15 = √ ( 21 * 8 * 7 * 6) sqr unit = 7 * 3 * 4 sqr unit = 84 sqr unit Area of Trapezium ABCD = (7/3) * 84 sqr unit = 196 sqr unit Corollary : Area of a trapizum with obleque sides a, b and, parallel sides x, y is (x> y) ∆ ( a, b, (x - y)) * ( x + y)/( x - y) Here ∆ ( a, b, c = (x - y)) = √ ( s * ( s - a) ( s - b)( s - c)) s = ( a + b + c)/2
Draw AE, where E is the point on DC where AE amd BC are parallel. This creates parallelogram ABCE, with sides of 14 and 10, and triangle ∆AED with sides of 13, 14, and 25-10 = 15. Let ∠EDA = θ. By the law of cosines: cos(θ) = (AD²+ED²-AE²)/2(AD)ED cos(θ) = (13²+15²-14²)/2(13)(15) cos(θ) = (169+225-196)/390 cos(θ) = 198/390 = 33/65 sin²(θ) = 1 - cos²(θ) = 1 - (33/65)² sin²(θ) = 1- 1089/4225 = 3136/4225 sin(θ) = √(3136/4225) = 56/65 (1/2)(ED)h = (1/2)AD(ED)sin(θ) h = 13(56/65) = 56/5 Trapezoid ABCD: [ABCD] = h(a+b)/2 = (56/5)(10+25)/2 [ABCD] = (56/5)(35)/2 = 28(7) = 196 sq units
@@MathBoostero importante é que você resolveu! Também, é importante mostrar outras soluções! Matemática, sempre tem outras formas de chegar ao resultado.
he applied: To calculate the area of a triangle using its semi-perimeter, you can use Heron's formula. Here’s how it works: Calculate the semi-perimeter (s): 𝑠 = 𝑎 + 𝑏 + 𝑐 2 s= 2 a+b+c where 𝑎 a, 𝑏 b, and 𝑐 c are the lengths of the sides of the triangle. Calculate the area (A) using Heron's formula: 𝐴 = 𝑠 ( 𝑠 − 𝑎 ) ( 𝑠 − 𝑏 ) ( 𝑠 − 𝑐 ) A= s(s−a)(s−b)(s−c)
Let's drop two heights from the upper vertexes down to the larger base. They will be h each.
Let segments we got by dropping heights be a (the left one) and b (the right one) and middle which is equal to the smaller base and that is 10.
Then we got:
h²=13²-a²
On the other hand:
h²=14²-b²
So:
169-a²=196-b².
And also:
a+b+10=25
After few changes, we can make a system:
b²-a²=27
a+b=15
Let's solve it.
(b-a)(b+a)=27
a+b=15
15(b-a)=27
b-a=1,8
We got a new easier system:
b-a=1,8
a+b=15
Let's add equations to each other:
2b=16,8
b=8,4=42/5
a=6,6=33/5
Let's find h:
h²=196-(42/5)²
196-1764/25=h²
3136/25=h²
h is a height so h>0 => h=56/5=11,2
S=(10+25)11,2/2=35•5,6=196
Answer: 196 square units.
Simple is :
Draw 2, 90^ triangles and one rectangle from this shape.
Area of 90^ triangle= 1/2 ab
Area of rectangle = ab
; a is length, b is height.
Add all.
Cool problem. Used different strategy. Created one triangle by taking the left and right sides of the trapezoid. Hence, a triangle with sides 13, 14, and 15. Used Heron's formula to calculate area of triangle. Took that value and calculated height of triangle, which is the height of the trapezoid. I dumped that value into the area formula for a trapezoid. Solution.
Draw BE parallel to AD. Then EC =15 . Determine area of triangle BEC . Then determine its height . Now determine the area of Trapezium.
x = √(13^2 - h^2) = 15 - √(14^2 - h^2) → h = 56/5 → area ABCD = 125h
or: ∆ BPC → CP = 15; BP = 13; BC = 14; BPC = θ → cos(θ) = 33/65 →
sin(θ) = 56/65 = h/13 → h = 56/5 → area ABCD = 125h
Very good illustration.thanks
In short, it could be solved like this: The height comes from the solution of the equation 13^2-x^2=14^2-(15^2-x^2)^2 because the height is expressed by the Pythagorean theorem in two forms in two right triangles. The solution gives x=6,6. Then apply T.P. in one of the triangles and the height h=11.2 comes out. In the end S=[(25+10)×11,2]÷2= 196cm^2
Yeah, solved it the same way.
I took the long way, but got that answer too. I made a regular trapezoid (13 on both left and right) and then the full right triangle had a base of x for 13 and x+x1 for 14. It was fun.
f(h)=√(13^2-h^2)+√(14^2-h^2)-15=0
f'(h)=-h/√(13^2-h^2)-h/√(14^2-h^2)
Newton-Raphsin iteration: Start with h=10 and iterate h←h-f(h)/f'(h) until h=11.200...
The area is then (1/2)(10+25)11.2=196 sq units.
In method 2, it would be more interesting if you had found the area of the parallelogram and, after adding it to the area of the triangle, finally, we would have the area of the trapezoid:
[ABPD]=h×BP=56/5 × 10=112, Therefore,
[ABCD]=[ABPD]+[BCP]
*[ABCD]=84+112=196*
Seems to me like you're just adding additional unnecessary steps. You use the area of the triangle to get the value of h, but once you have that it's unnecessary to find the area of the parallelogram, since you can use h to directly find the area of the whole trapezoid.
We need not find the height in the 2nd method. Area of the parallelogram = 84*10*2/15=112. Total area= 84+112=196
@@quigonkennyUnfortunately, you didn't understand what I meant, in the second method it would be more interesting not to use the formula for the area of the trapezoid, as it is easier to find the area of the parallelogram.
@@purnajinananandaavadhuta8605It can be done like this, but for the sake of explanation, the interesting thing is to find the height!
@@quigonkenny Area parallelogram is just 10x56/5 =112, and add to 84. That is much simpler.
DC=a+10+(25-10-a)→ 13²-a² =h²= 14²-(15-a)²→ a=33/5→ h²=13²-(33/5)²→ h=56/5→ Área ABCD =[(10+25)/2]*(56/5)=196 ud².
Gracias y saludos.
Let height of trapezium be h. Draw AE and BF both equal to h , so that DEFC is a straight line.
Area of triangle ADE = 0.5h x DE. Area of triangle BFC = 0.5 h x FC . Area of AEFB (the rectangle) is 10h.
hxh= 13x13- DExDE= 14x14- FCxFC by Pythagors'theorem in both triangles
FCxFC -DExDE = 14x14-13x13 = 196-169 =27
(FC+DE)(FC-DE)=27
FC+DE+10= 25 so FC-DE = 27/ 15 = 9/5
FC+DE=15
FC-DE=9/5 =1.8
2xFC = (15x5+9)/5
FC=8.4
DE= 6.6
13x13- 6.6x6.6= (13+6.6)(13-6.6)= 19.6 x 6.4= 196 x64/(10x10) = h x h. So h= 14x 8/10 = 112/10 = 11.2
Area of this trapezium = 11.2 (10 + 4.2 + 3.3) = 11.2 x 17.5 = 2.8 x 70 = 196 square units .
(13)^2=169 (14)^2=196 {169+196}=365 (10)^2=100 (25)^2=625{100+625}=725 {365+725}=1090/180°ABCD =6 .50ABCD 6.5^10 6.5^2^5 3^2.1^2^1 3^1.2^1 3.2 (ABCD ➖ 3ABCD+2).
Nice use of the Herons on Method 2
ar=11.2(10+25)/2=112×35/2×10==56×7/2=28×7=196 sq.
ar.=196sq unit.ans
Draw AE || BC that cuts BC at E
Area ( Trapezium ABCD)
/ area ( ∆ ADE)
= ( AB + CD)/DE
= ( AB + CD) / ( CD - AB)
= ( 25 + 10)/( 25 - 10)
= 7/3
Area ( ∆ ADE) of sides 13, 14, 15
= √ ( 21 * 8 * 7 * 6) sqr unit
= 7 * 3 * 4 sqr unit
= 84 sqr unit
Area of Trapezium ABCD
= (7/3) * 84 sqr unit
= 196 sqr unit
Corollary : Area of a trapizum with obleque sides a, b and, parallel sides x, y is (x> y)
∆ ( a, b, (x - y)) * ( x + y)/( x - y)
Here ∆ ( a, b, c = (x - y))
= √ ( s * ( s - a) ( s - b)( s - c))
s = ( a + b + c)/2
Super solution
10+25=35÷2=17.5×11.2=196(100%raite)
10+25/2=17.50, 13+14/2=13.50 17.5*13.50=236.25
99/15=33/5=6.6
AB is not parallel to CD because AD =13, BC=14 if AD=BC, AB and CD parallel
По условию же трапеция
Bhai parallel hogi because AD and BC are transversal line not perpendicular distance between these parallel line.
17'- 6" × 13' - 6".
99/15は33/5と約分可能。
17.5
Trapezoid area=1/2(10+25)(56/5)=196
2 parellal sides divided by 2 into other 2 sides divided by 2 and if multiplied how ?
196
Se complica con un desarrollo tan extenso y es más fácil que la tabla del uno.
189.4
x = 99/15 = 33/5
what abt nos 13, 17,, 19 , to divide by these nos
Draw AE, where E is the point on DC where AE amd BC are parallel. This creates parallelogram ABCE, with sides of 14 and 10, and triangle ∆AED with sides of 13, 14, and 25-10 = 15.
Let ∠EDA = θ. By the law of cosines:
cos(θ) = (AD²+ED²-AE²)/2(AD)ED
cos(θ) = (13²+15²-14²)/2(13)(15)
cos(θ) = (169+225-196)/390
cos(θ) = 198/390 = 33/65
sin²(θ) = 1 - cos²(θ) = 1 - (33/65)²
sin²(θ) = 1- 1089/4225 = 3136/4225
sin(θ) = √(3136/4225) = 56/65
(1/2)(ED)h = (1/2)AD(ED)sin(θ)
h = 13(56/65) = 56/5
Trapezoid ABCD:
[ABCD] = h(a+b)/2 = (56/5)(10+25)/2
[ABCD] = (56/5)(35)/2 = 28(7) = 196 sq units
Thanks but so easy
Nice problem
Ans.210 sq.ft. Area
99/15 = 33/5
Would have made calculatioj much simpler. I don't know why you chose not to simplify it further...
I just didn't noticed it at that time.
@@MathBoostero importante é que você resolveu! Também, é importante mostrar outras soluções! Matemática, sempre tem outras formas de chegar ao resultado.
hard to think this one,,i cant pass this,1+1 =2. that's all ,,, in army when there is war you cannot think that,,
195.6519
I imagined method 2 right away. BUT, somewhere, I went astray and got the answer wrong. Oh well.
المساحة ١٩٦ وحدة مربعة
Area= 196
alt.11.2
Ca is 25-x
Sorry QC is 25-X
Nein falsch = 236
Why you guys not use S=
S mean surface, A means area but they are actually the same just because the difference of habit.习惯不同而已
Your method is very lengthy. After getting x = 6.6 apply pythagorus theorem to get h= 11.2 and apply formula of trapezium you will get answer 198.
Please I won't JHs2 and 3
👍👍👍👍
2nd Method
13.58min
What is S
he applied: To calculate the area of a triangle using its semi-perimeter, you can use Heron's formula. Here’s how it works:
Calculate the semi-perimeter (s):
𝑠
=
𝑎
+
𝑏
+
𝑐
2
s=
2
a+b+c
where
𝑎
a,
𝑏
b, and
𝑐
c are the lengths of the sides of the triangle.
Calculate the area (A) using Heron's formula:
𝐴
=
𝑠
(
𝑠
−
𝑎
)
(
𝑠
−
𝑏
)
(
𝑠
−
𝑐
)
A=
s(s−a)(s−b)(s−c)
It is not trengle
25 or 15 ??
what us yesquaire? Awful pronunciation and explanation
196.437