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In triangle ABC, AB = AC = 5 and BC = 8
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15 Jul 2020, 22:50
15 Jul 2020, 22:50
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In triangle ABC, AB = AC = 5 and BC = 8. What is the area of the triangle?
A. 10
B. 12
C. 15
D. 24
E. 30
A. 10
B. 12
C. 15
D. 24
E. 30
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Re: In triangle ABC, AB = AC = 5 and BC = 8
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16 Jul 2020, 04:29
16 Jul 2020, 04:29
1
Refer Attached Image
In triangle ABC, AB = AC = 5, that means its an isosceles triangle. This question is based on a property of isosceles triangle.
"In an Isosceles Triangle the perpendicular from the vertex to the non equal side is a perpendicular bisector to the non-equal side"
In the attached figure, AD which is a perpendicular draw from A to BC will touch BC at its mid-point, which is D. => BD = DC = 8/2 = 4
To find the Area of ABC we need to find the height of ABC which is AD.
Using Pythagoras theorem on ADB we get
\(AB^2\) =\( BD^2\) + \(AD^2\)
=> \(5^2\) =\( 4^2\) + \(AD^2\)
=> \(AD^2\) = 25 - 16 = 9 = \(3^2\)
=> AD = 3
So, Area of ABC = \(\frac{1}{2}\)BC * AD = \(\frac{1}{2}\) * 8 * 3 = 12
So, Answer will be B
Hope it helps!
In triangle ABC, AB = AC = 5, that means its an isosceles triangle. This question is based on a property of isosceles triangle.
"In an Isosceles Triangle the perpendicular from the vertex to the non equal side is a perpendicular bisector to the non-equal side"
In the attached figure, AD which is a perpendicular draw from A to BC will touch BC at its mid-point, which is D. => BD = DC = 8/2 = 4
To find the Area of ABC we need to find the height of ABC which is AD.
Using Pythagoras theorem on ADB we get
\(AB^2\) =\( BD^2\) + \(AD^2\)
=> \(5^2\) =\( 4^2\) + \(AD^2\)
=> \(AD^2\) = 25 - 16 = 9 = \(3^2\)
=> AD = 3
So, Area of ABC = \(\frac{1}{2}\)BC * AD = \(\frac{1}{2}\) * 8 * 3 = 12
So, Answer will be B
Hope it helps!
Attachments
Image.jpg [ 42.46 KiB | Viewed 3880 times ]
Re: In triangle ABC, AB = AC = 5 and BC = 8
[#permalink]
07 Jan 2025, 21:30
07 Jan 2025, 21:30
1
Going by Heron's formula, A= sqrt(s*(s-a)(s-b)(s-c)) where a, b, and c are sides and s is the semi perimeter;
Here s = (5+5+8)/2 = 9
Therefore, A= sqrt(9*4*4*1) = 12.
Here s = (5+5+8)/2 = 9
Therefore, A= sqrt(9*4*4*1) = 12.
