Plane Geometry
 If angles of measure (5y + 62°) and (22° + y) are supplementary, then value of y is :

View Hint View Answer Discuss in Forum
As we know that Sum of two supplementary angles = 180°
∴ ( 5y + 62° ) + ( 22° + y ) = 180°
⇒ 6y + 84° = 180°
⇒ 6y = 180° – 84° = 96°Correct Option: A
As we know that Sum of two supplementary angles = 180°
∴ ( 5y + 62° ) + ( 22° + y ) = 180°
⇒ 6y + 84° = 180°
⇒ 6y = 180° – 84° = 96°∴ y = 96 = 16° 6
 In, ΔABC, D and E are the midpoints of AB and AC respectively. Find the ratio of the areas of ΔADE and ΔABC.

View Hint View Answer Discuss in Forum
Clearly DE  BC (by converse of BPT)
∴ ΔADE ∼ ABC (∠A = ∠A and ∠ADE = ∠B)∴ Area (∆ADE) = AD^{2} (Area Theorem) Area (∆ABC) AB^{2} Correct Option: B
Clearly DE  BC (by converse of BPT) ∴ ΔADE ∼ ABC (∠A = ∠A and ∠ADE = ∠B)
∴ area (∆ADE) = AD^{2} (Area Theorem) area (∆ABC) AB^{2} = AD^{2} = 1 (∴AB = 2AD) (2AD)^{2} 4
 A vertical stick 12 cm long casts a shadow 8 cm long on the ground. At the same time a tower casts the shadow 40 m long on the ground. Find the height of the tower.

View Hint View Answer Discuss in Forum
Draw a figure as per given question,
In ΔACB and PCQ
∠C = ∠C (common)
∠ABC = ∠PQC (each 90°)
∴ ΔACB ∼ ΔPC (AA Similarity)
Correct Option: C
In ΔACB and PCQ
∠C = ∠C (common)
∠ABC = ∠PQC (each 90°)
∴ ΔACB ∼ ΔPC (AA Similarity)∴ AB = BC PQ QC h = 4000 12 8
h = 60m
 In the given figure, find the length of BD.

View Hint View Answer Discuss in Forum
In Δs ADE and ΔABC
∠A = ∠A [common]
∠ADE = ∠ACB = x° (Given)
∴ ΔADE ∼ ΔACB ( AA Similarly)Correct Option: A
In ΔsADE and ΔABC
∠A = ∠A [common]
∠ADE = ∠ACB = x°(Given)
∴ ΔADE ∼ ΔACB (AA Similarly)AD = AE (corresponding side of ⁓ ∆s are proportional) AC AB 6 = 9 13 AB AB = 39 = 19.5 cm 2
Hence BD = AB  AD = 19.5  6 = 13.5 cm.
 ABCD is a quadrilateral inscribed in a circle with centre O. If ∠COD = 120° and ∠BAC = 30°, then ∠BCD is :

View Hint View Answer Discuss in Forum
As per the given in question , we draw a figure of a quadrilateral ABCD inscribed in a circle with centre O
Given , ∠COD = 120°
∠BAC = 30°∠CAD = 1 × ∠COD 2 ∠CAD = 1 × 120° = 60° 2
Correct Option: B
As per the given in question , we draw a figure of a quadrilateral ABCD inscribed in a circle with centre O
Given , ∠COD = 120°
∠BAC = 30°∠CAD = 1 × ∠COD 2 ∠CAD = 1 × 120° = 60° 2
∴ ∠BAD = 90°
∴ ∠BCD = 180°  ∠BAD
∴ ∠BCD = 180° – 90° = 90°