CTET Solved Paper MCQ - REET Feburary 2016
27
As, we know that
∠AOC+ ∠BOC = 180° (Linear Pair)
⇒ 50° + ∠BOC = 180°
⇒ ∠BOC = 180° - 50°
= 130°
∠AOC+ ∠BOC = 180° (Linear Pair)
⇒ 50° + ∠BOC = 180°
⇒ ∠BOC = 180° - 50°
= 130°
28
As, ∠ACB = 180° - (∠ABC+ ∠BAC)
= 180° - (50°+ 30°) = 100° (Sum of angles of a triangle)
∴ ∠ACD = 180° - 100° = 80° (Linear Pair)
∴ ∠AED = ∠ACD + ∠EDC
= 80° + 25° = 105° (Exterior Angle Theorem).
= 180° - (50°+ 30°) = 100° (Sum of angles of a triangle)
∴ ∠ACD = 180° - 100° = 80° (Linear Pair)
∴ ∠AED = ∠ACD + ∠EDC
= 80° + 25° = 105° (Exterior Angle Theorem).
Sum of length of two sides of a triangle is always greater than the third side.
30
The diagonal of a rectangle field is 17 m and its perimeter is 46 m. What is the area of the field?
Given, 2(l + b) = 46
l + b = 23 ... (i)
and l2 + b2 = (17)2 = 289 ... (ii) (By Pythagoras Theorem)
Now, (l + b)2 = 232
⇒ l2 + b2 + 2lb = 529
⇒ 289 + 2lb = 529
⇒ 2lb = 529 - 289 = 240
⇒ lb = 120
⇒ Area = lb =120 sq m.
l + b = 23 ... (i)
and l2 + b2 = (17)2 = 289 ... (ii) (By Pythagoras Theorem)
Now, (l + b)2 = 232
⇒ l2 + b2 + 2lb = 529
⇒ 289 + 2lb = 529
⇒ 2lb = 529 - 289 = 240
⇒ lb = 120
⇒ Area = lb =120 sq m.
