In right angled ∆ABC, ∠B = 90° if P and Q points on the sides AB and BC respectively then-
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Solution
∆ABQ में → (AQ)2 = (AB)2 + (BQ)2 ----------(1)
∆PBC में → (PC)2 = (BP)2 + (BC)2 ----------(2)
(AQ)2 + (PC)2 = (AB)2 + (BC)2 + (BQ)2 + (BP)2
= (AC)2 + (PQ)2
In ∆PQR, AP = 2 cm, AQ = 3 cm, PR = 10 cm and AB || QR. Find the length of BR ?
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Solution
⇒ PR = 10
⇒ BR = × 3 = 6 cm
In ∆ABC, AB = 5 cm, AC = 7 cm. If AD is the angle bisector of ∠A, then BD : CD is
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Solution
(By the angle bisector theorem)
In the adjoining figure ∠BAC = 60°, and BC = a, AC = b and AB = c, then –
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Solution
cos A =
a2 = b2 + c2 – bc
In a triangle ABC, if AB, BC and AC are the three sides of the triangle, then which of the statement is necessarily true?
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Solution
The sum of the smaller sides cannot be equal to or less than the largest side.
The complement of 65° 50′ is
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Solution
65°50' + x = 90° = 89°60'
x = 89°60' – 65°50' = 24°10'
In the given figure, ∠a is greater than one sixth of right angle, then b is –
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Solution
a > ⇒ a > 15°
a + b = 180°
b < 165°
In the given figure AB || DE. Find the value of a° + b° – c° is
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Solution
c + x = b ⇒ x = b – c
a + (b – c) = 180°
a + b – c = 180°
In the given figure AB || CE and BC || FG. Find the value of x°.
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Solution
∠x = ∠ABC = ∠CDG = 180° – 138° = 42°
In figure AB || CD, ∠ABE = 100° find the value of ∠CDE-
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Solution
Extend CD to M, then
∠DME = ∠ABE = 100°
∠MED = 25°
∴ ∠MDE = 180° – (100 + 25°)
= 55°
∴ ∠CDE = 180° – 55° = 125°