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14 Mar 2019, 01:53
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
72% (02:24) correct
28%
(02:19)
wrong
based on 136
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Triangles ABC and ADC are isosceles with AB = BC and AD = DC. Point D is inside triangle ABC, ∠ABC = 40°, and ∠ADC = 140°. What is the degree measure of ∠BAD?
(A) 20
(B) 30
(C) 40
(D) 50
(E) 60
(A) 20
(B) 30
(C) 40
(D) 50
(E) 60
globaldesi
Joined: 28 Jul 2016
Last visit: 15 Jul 2026
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Concentration: Finance, Human Resources
Schools: ISB '18 (D)
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14 Mar 2019, 21:18
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Figure drawn as per explained.
Now AB = BC hence angles corresponding will be equal hence \(\angle\) BAC= \(180^{\circ}\)-\(40^{\circ}\) =\(\frac{140^{\circ}}{2}\) = \(70^{\circ}\)
for triangle ADC \(180^{\circ}-40^{\circ}\) =\(\frac{40^{\circ}}{2}\)=\(20^{\circ}\)
thus angle ABD = \(70^{\circ}\)-\(20^{\circ}\) =\(50^{\circ}\)
Thus answer D
Now AB = BC hence angles corresponding will be equal hence \(\angle\) BAC= \(180^{\circ}\)-\(40^{\circ}\) =\(\frac{140^{\circ}}{2}\) = \(70^{\circ}\)
for triangle ADC \(180^{\circ}-40^{\circ}\) =\(\frac{40^{\circ}}{2}\)=\(20^{\circ}\)
thus angle ABD = \(70^{\circ}\)-\(20^{\circ}\) =\(50^{\circ}\)
Thus answer D
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General Discussion
Updated on: 14 Mar 2019, 05:30
Originally posted by Archit3110 on 14 Mar 2019, 05:06.
Last edited by Archit3110 on 14 Mar 2019, 05:30, edited 1 time in total.
Last edited by Archit3110 on 14 Mar 2019, 05:30, edited 1 time in total.
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