Triangle geometry question?
2016-07-31 9:46 am
"A right-angled triangle has height x and base y. Its area is 504cm2 and its perimeter is 144 cm. The value of x is greater than the value of y. What is the value of x?"
回答 (11)
2016-07-31 10:35 am
A right-angled triangle has height x and base y. Its area is 504 cm² and its perimeter is 144 cm. The value of x is greater than the value of y. What is the value of x?
Solution :
By Pythagorean theorem, hypotenuse = √(x² + y²) cm
Area of the triangle (cm²) :
(1/2)xy = 504
xy = 1008 ...... [1]
Perimeter of the triangle (cm) :
x + y + √(x² + y²) = 144
√(x² + y²) = 144 - x - y
[√(x² + y²)]² = (144 - x - y)²
x² + y² = 20736 + x² + y² + 2xy - 288x - 288y
288x + 288y = 2xy + 20736
144x + 144y = xy + 10368 ...... [2]
Subsitute [1] into [2] :
144x + 144y = 1008 + 10368
144x + 144y = 11376
x + y = 79
x = 79 - y ...... [3]
Substitute [3] into [1] :
(79 - y)y = 1008
y² - 79 + 1008 = 0
(y - 16)(y - 63) = 0
y = 16 or y = 63
Substitute the value of y into [3] :
When y = 16 :
x = 79 - 16
x = 63
When y = 63 :
x = 79 - 63
x = 16 (rejected, for x should be greater than y)
Answer : x = 63
Solution :
By Pythagorean theorem, hypotenuse = √(x² + y²) cm
Area of the triangle (cm²) :
(1/2)xy = 504
xy = 1008 ...... [1]
Perimeter of the triangle (cm) :
x + y + √(x² + y²) = 144
√(x² + y²) = 144 - x - y
[√(x² + y²)]² = (144 - x - y)²
x² + y² = 20736 + x² + y² + 2xy - 288x - 288y
288x + 288y = 2xy + 20736
144x + 144y = xy + 10368 ...... [2]
Subsitute [1] into [2] :
144x + 144y = 1008 + 10368
144x + 144y = 11376
x + y = 79
x = 79 - y ...... [3]
Substitute [3] into [1] :
(79 - y)y = 1008
y² - 79 + 1008 = 0
(y - 16)(y - 63) = 0
y = 16 or y = 63
Substitute the value of y into [3] :
When y = 16 :
x = 79 - 16
x = 63
When y = 63 :
x = 79 - 63
x = 16 (rejected, for x should be greater than y)
Answer : x = 63
2016-08-02 8:43 pm
X=463
2016-08-02 1:28 pm
xy = 1008
x + y + h = 144
h = sq rt of (x squared + y squared)
Create two simultaneous equations and solve as a quadratic.
x + y + h = 144
h = sq rt of (x squared + y squared)
Create two simultaneous equations and solve as a quadratic.
2016-07-31 10:28 am
i) Radius of incircle = area/semi perimeter
= 504/72 = 7 cm
ii) If ABC is the triangle with C = 90 deg; O is the incenter; D, E & F are the contact points of the circle with sides, then
OD = OE = CD = CE = 7 cm.
So, BD = y - 7 = BF; AE = x - 7 = AF
[Tangents from same exterior point to the same circle are equal in measure]
{Refer sketch in -
https://www.bing.com/images/search?q=image+of+a+right+triangle+with+a+circle+inscribed+in+it&view=detailv2&&id=0246FF159F9CB9573B2F845239C332D98B48C647&selectedIndex=9&ccid=306yI8n8&simid=608032602731252068&thid=OIP.Mdf4eb223c9fc97349eec046f7979fef9o0&ajaxhist=0}
So perimeter = AF + AE + BD + BF + CD + CE
= 2x + 2y - 14 = 144
==> x + y = 79 ----- (1)
Area of triangle = (1/2)*x*y = 504
==> y = 1008/x ------ (2)
Substituting in (1) & simplifying,
x^2 - 79x + 1008 = 0
==> (x - 63)(x - 16) = 0
So x = 63 cm, being x > y
= 504/72 = 7 cm
ii) If ABC is the triangle with C = 90 deg; O is the incenter; D, E & F are the contact points of the circle with sides, then
OD = OE = CD = CE = 7 cm.
So, BD = y - 7 = BF; AE = x - 7 = AF
[Tangents from same exterior point to the same circle are equal in measure]
{Refer sketch in -
https://www.bing.com/images/search?q=image+of+a+right+triangle+with+a+circle+inscribed+in+it&view=detailv2&&id=0246FF159F9CB9573B2F845239C332D98B48C647&selectedIndex=9&ccid=306yI8n8&simid=608032602731252068&thid=OIP.Mdf4eb223c9fc97349eec046f7979fef9o0&ajaxhist=0}
So perimeter = AF + AE + BD + BF + CD + CE
= 2x + 2y - 14 = 144
==> x + y = 79 ----- (1)
Area of triangle = (1/2)*x*y = 504
==> y = 1008/x ------ (2)
Substituting in (1) & simplifying,
x^2 - 79x + 1008 = 0
==> (x - 63)(x - 16) = 0
So x = 63 cm, being x > y
2016-07-31 10:25 am
xy/2=504 and x+y+root(x^2+y^2)=144. Solve these two.
Alternatively guess the answer and x=63, y=16.
Alternatively guess the answer and x=63, y=16.
2016-07-31 10:29 am
OK, just give me some time to show working. This is one hell of a question, I'll get back to you okay :)
The answers are :
x = 63, y = 16 and the hypotenuse = 65
Hope this helps in the meantime :)
The answers are :
x = 63, y = 16 and the hypotenuse = 65
Hope this helps in the meantime :)
2016-11-03 10:09 am
xy = 1008
x + y + h = 144
h = sq rt of (x squared + y squared)
create two simultaneous equations and solve as a quadratic...
x + y + h = 144
h = sq rt of (x squared + y squared)
create two simultaneous equations and solve as a quadratic...
2016-10-02 4:08 pm
3 cone = 1 cylinder
use this
use this
2016-08-04 12:27 am
X463
2016-08-03 5:29 pm
3 cone = 1 cylinder
use this
use this
2016-08-01 3:08 am
×463
收錄日期: 2021-04-20 16:31:05
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160731014602AABqqX3