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Right triangle ABC has sides with length x, y and z. If triangle ABC
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30 Apr 2021, 23:50
30 Apr 2021, 23:50
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Right triangle ABC has sides with length x, y and z. If triangle ABC has perimeter 17, and \(x^2 + y^2 + z^2 = 98\), then what is the area of triangle ABC?
A. 12.75
B. 13.25
C. 14.00
D. 14.50
E. 15.25
A. 12.75
B. 13.25
C. 14.00
D. 14.50
E. 15.25
Re: Right triangle ABC has sides with length x, y and z. If triangle ABC
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01 May 2021, 00:51
01 May 2021, 00:51
need help!
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Re: Right triangle ABC has sides with length x, y and z. If triangle ABC
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01 May 2021, 14:44
01 May 2021, 14:44
3
Explanation:
Let the legs of ABC be \(x\) and \(y\) and the \(z\) be the hypotenuse.
In Right triangle, \(x^2 + y^2 = z^2\) .... (1)
We are given:
\(x + y + z = 17\) ... (2) and
\(x^2 + y^2 + z^2 = 98\) ... (3)
Substituting (1) in (3):
\(z^2 + z^2 = 98\)
\(z^2 = 49\)
\(z = 7\)
So, \(x + y + 7 = 17\) i.e. \(x + y = 10\)
On squaring both sides, \(x^2 + y^2 + 2xy = 100\) ... (4)
Substituting (4) in (3):
\(100 - 2xy + 7^2 = 98\)
\(51 = 2xy\)
\(xy = 25.5\)
Area of triangle ABC = \(\frac{1}{2}xy = 12.75\)
Hence, option A
Let the legs of ABC be \(x\) and \(y\) and the \(z\) be the hypotenuse.
In Right triangle, \(x^2 + y^2 = z^2\) .... (1)
We are given:
\(x + y + z = 17\) ... (2) and
\(x^2 + y^2 + z^2 = 98\) ... (3)
Substituting (1) in (3):
\(z^2 + z^2 = 98\)
\(z^2 = 49\)
\(z = 7\)
So, \(x + y + 7 = 17\) i.e. \(x + y = 10\)
On squaring both sides, \(x^2 + y^2 + 2xy = 100\) ... (4)
Substituting (4) in (3):
\(100 - 2xy + 7^2 = 98\)
\(51 = 2xy\)
\(xy = 25.5\)
Area of triangle ABC = \(\frac{1}{2}xy = 12.75\)
Hence, option A
