Question 1

Two angles are supplementary and the ratio of the angels is 1:4. what is the value of smaller angle?

Accepted Answer

As we know that the angles are supplementary so sum of angles will be 180 degree. Let us assume that the ratio factor is r. According to question, Angles are supplementary and have a ratio of 1:4. r + 4r = 180 ⇒ 5r = 180 ⇒ r = 180/5 ⇒ r = 36

Question 2

The complement of 30°20′ is:

Accepted Answer

complement of 30°20′ = 90° – ( 30°20′ ) = 90° – ( 30° + 20′ ) = (89° – 30°) + (1° – 20′) = 59° + 60′ – 20′ [ ∴ 1° = 60°′] = 59° + 40′ = 59°40′.

Question 3

In the given, AB || CD. Then X is equal to:

Accepted Answer

Through O, draw a line l parallel to both AB and CD. Then ∠1 = 45° (alt. ∠S) and ∠2 = 30° (alt. ∠S) ∴ ∠BOC = ∠1 + ∠2 = 45° + 30° = 75° So, X = 360° – ∠BOC = 360° – 75° = 285° Hence X = 285°.

Question 4

An angle is equal to one-third of its supplement. Its measure is equal to :

Accepted Answer

Question 5

In the adjoining figure, AB || CD, t is the traversal, EG and FG are the bisectors of ∠BEE and ∠DFE respectively, then ∠EGF is equal to :

Accepted Answer

Question 6

Find the measure of an angle, if six times its complement is 12° less than twice its supplement :

Accepted Answer

Let, the measure of the required angled be A°. Then, measure of its complement = ( 90 – A )° measure of its supplement = (180 – A)° According to question, 6(90° – A) = 2(180° – A) –12° ⇒ 540° – 6A = 360° – 2A – 12° ⇒ 4A = 192° ⇒ A = 48°.

Question 7

In fig., AB || CD, ∠a is equal to:

Accepted Answer

CD || AB (Given) Produce RQ to meet AB in S ∠CRS = ∠PSR (at. int. ∠s) But ∠CRS = 55° ∴ ∠PSR = 55° Now in QSP ∠QSP + ∠QPS + ∠PQS = 180° 55° + 38° + ∠SQP = 180° ∴ ∠SQP = 180° – 93° = 87° But angle a and ∠PQS are linear ∠a = 180° – 87° ∠a = 93°

Question 8

In the following figure, ∠B : ∠C = 2 : 3, find ∠B + ∠C.

Accepted Answer

∠DAC = ∠B + ∠C (Exterior angle prop. of a Δ ABC) According to question, 130° = 2A + 3A 5A = 130° A = 26° ∴ ∠B = 52°; ∠C = 78°

Question 9

In a ΔABC, if 2∠A = 3∠B = 6∠C, Then ∠A is equal to:

Accepted Answer

Question 10

A, B, C are the three angles of a Δ. If A − B = 15° and B − C = 30°. Then ∠A is equal to :

Accepted Answer

Since A, B and C are the angles of a Δ, ∴ A + B + C = 180° .......................................... (1) According to question, A – B = 15° ; ⇒ A = B + 15°....................................................(2) B – C = 30°; ⇒ B = C + 30°;...................................................(3) Put the value of B from equation (2) in Equation (1), we will get ∴ A = B + 15° A = C + 30° + 15° A = C + 45° ........................................................(4) From a equation, ∴ A + B + C = 180° ⇒ (C + 45°) + (C + 30°) + C = 180° ⇒ 3C + 45° + 30° = 180° ⇒ 3C = 180° – 75° = 105° ⇒ C = 35° ..........................................................(5) From equation (4) A = C + 45° Put the value of C from equation (5) , we will get ∴ ∠A = 35° + 45° = 80°.

Question 11

The sides AB and AC of ΔABC have been produced to D and E respectively. The bisectors of ∠CBD and ∠BCE meet at O. If ∠A = 40°, then ∠BOC is equal to:

Accepted Answer

Question 12

In the given figure, DE || BC if AD = 1.7 cm, AB = 6.8 cm and AC = 9 cm, find AE.

Accepted Answer

Question 13

If the bisector of an angle of Δ bisects the opposite side, then the Δ is :

Accepted Answer

Question 14

The areas of two similar Δs are 81 cm 2 and 144 cm 2. If the largest side of the smaller Δ is 27 cm, then the largest side of the larger Δ is :

Accepted Answer

Question 15

In the given figure ∠BAD = ∠CAD. AB = 4 cm, AC = 5.2 cm, BD = 3 cm. Find BC.

Accepted Answer

Question 16

A ladder 15 m long reaches a window which is 9 m above the ground on one side of street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 12 m high. What is the width of the street:

Accepted Answer

Draw a figure as per given question, In a Right triangle ADC, Use the (Pythagoras Theorem) AC = √DC2 - AD2 ⇒ AC = √ 152 - 9 2 ⇒ AC = √ 225 - 81 ⇒ AC = √ 144 ⇒ AC = 12 cm In a Right triangle BCE, Use the formula CB = √ CD2 - BE2 ⇒ CB = √15 2 - 122 ⇒ CB = √225 - 144 ⇒ CB = √ 81 ⇒ CB = 9 m ∴ Width of the street (AC + BC) = AB = 12 + 9 = 21 m.

Question 17

D and E are the points on the sides AB and AC respectively of ΔABC such that AD = 8 cm, BD = 12 cm, AE = 6 cm and EC = 9 cm. Then find BC/ DE.

Accepted Answer

Question 18

In an equilateral Δ ABC, if AD ⊥ BC, then which option is true?

Accepted Answer

Question 19

In the given figure, AB || DC, find the value of x.

Accepted Answer

Question 20

In the adjoining figure ∠A + ∠B + ∠C + ∠D + ∠E + ∠F =?

Accepted Answer

In ΔACE ∠A + ∠C + ∠E = 180° Similarly in ΔDFB ∠D + ∠F + ∠B = 180° ∴ (∠A + ∠C + ∠E) + (∠D + ∠F + ∠B) = 360° ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360°

Plane Geometry Questions