Q: In the given figure ∠QPR = 90°, QR = 26 cm, PM = 6 cm, MR = 8 cm and ∠PMR = 90°, find the area of ΔPQR.

Given in the question, QR = 26 cm, PM = 6 cm, MR = 8 cm According to question, in Δ PMR (Pythagoras Theorem) PR = √PM2 + MR2 = √36 + 64 = √ 100 = 10 cm According to question, in Δ PQR (Pythagoras Theorem) PQ = √QR2 - PR2 = √262 - 102 = √576 - 100 = √ 476 = 24 ∴ area of triangle ∆PQR = Base length x Height / 2 ⇒ area of triangle ∆PQR = PR x PQ / 2 ⇒ area of triangle ∆PQR = 10 x 24 / 2 = 10 x 12 ⇒ area of triangle ∆PQR = 120

Q: If ABCD is a parallelogram and AC and BD be its diagonals, then:

In ||gm ABCD , we have Since diagonals of parallelogram(||gm ) bisect each other, ∴ M will be the mid-point of each of the diagonal AC and BD ∴ In ΔABC AB2 + BC2 = 2(AM2 + MB2) ......... ( 1 ) [Appolonius Theorem] In ΔADC AD2 + CD2 = 2(AM2 + DM2) ......... ( 2 ) = 2(AM2 + MB2) [∴ DM = BM] Adding equations ( 1 ) and ( 2 ) AB2 + BC2 + CD2 + DA2 = 2(AM2 + MB2 + 2(AM2 + MB2 = 4AM2 + 4MB2 = (2AM)2 + (2MB)2= AC2+ BD2. { ∴ AM = MC , MB = MD }

Q: In center of a triangle lies in the interior of :

We know that In center of every triangle ( acute , obtuse and right angle ) lies in its interior. Hence required answer will be any triangle .

Q: In the given figure, which of the following is true :

From above figure , we can see that Join B and D and produce BD to E Then, p + q = β and s + t = x We know that , Exterior angle = Sum of two interior angle Now, s = p + α (Exterior angle of a Δ prop.) Similarly, t = q + γ (Exterior angle of a Δ prop.) Adding, s + t = p + q + α + γ Putting s + t = x and p + q = β , Hence x = α + β + γ

Q: In the given figure, the side BC of a Δ ABC is produced on both sides. Then ∠ 1 + ∠ 2 is equal to :

From the given figure, In ΔABC We know that , Exterior angle = Sum of two interior angles ∠1 = ∠A + ∠5 ................. ( 1 ) and ∠2 = ∠A + ∠6 ................. ( 2 ) Adding equations ( 1 ) and ( 2 ). ∠1 + ∠2 = 2∠A + ∠5 + ∠6 = 2∠A + (180° − ∠A) = ∠A + 180° The given question can be restated as the sum of two exterior angles exceeds ∠A of the ΔABC by 2 right angles.

Q: If two diameters of a circle intersect each other at right angles, then the quadrilateral formed by joining here end points is a :

From given figure , we have In quadrilateral ABCD , we can see that Diagonal AB = Diagonal CD ......... ( ⅰ ) AO = OB ......... ( ⅱ ) DO = CO ......... ( ⅲ ) ∠DOB = 90° ......... ( ⅳ ) From ( ⅰ ) , ( ⅱ ) , ( ⅲ ) and ( ⅳ ) , we get From above , it is clear that In quadrilateral ABCD , diagonals are equal and they bisect each other at 90° . ⇒ Quadrilateral ABCD is square.

Question 1

In ΔABC, the angle bisectors of ∠B and ∠C meet at O. If ∠A = 70°, then ∠BOC is equal to:

Accepted Answer

Question 2

The complement of an angle exceeds the angle by 60°. Then the angle is equal to:

Accepted Answer

Let, the angle be A ⇒ Its complement angle = 90° – A According to the question (90 – A) = A + 60° ⇒ 90 - 60 = A + A ⇒ 2A = 30° ⇒ A = 15°

Question 3

In the given figure, AB || CD and AC || BD. If ∠EAC = 40°, ∠FDG = 55°, ∠HAB = x; then the value of x is:

Accepted Answer

∠DCK = ∠FDG = 55° (corr. ∠s) ∴ ∠ACE = ∠DCK = 55° (vert. opp. ∠s) So, ∠AEC = 180° – (40° + 55°) = 85° ∴ ∠HAB = ∠AEC = 85° (corr. ∠s) Hence, x = 85°.

Question 4

In the given figure, OP bisect ∠BOC and OQ bisects ∠AOC. Then ∠POQ is equal to :

Accepted Answer

Since OP bisects ∠BOC, ∴ ∠BOC = 2∠POC Again, OQ bisects ∠AOC, ∴ ∠AOC = 2∠QOC Since ray OC stands on line AB, ∴, ∠AOC + ∠BOC = 180° ⇒ 2∠QOC + 2∠POC = 180° ⇒ 2∠QOC + ∠POC = 180° ⇒ ∠QOC + ∠POC = 90° ⇒ ∠POQ = 90°. The above sum can also be restated as follows; The angle between the bisectors of a linear pair of angles is a right angle.

Question 5

If two parallel lines are intersected by a transversal, then the bisectors of the two pairs of interior angles enclose a:

Accepted Answer

Question 6

In the given figure, ∠B = ∠C = 55° and ∠D = 25°. Then:

Accepted Answer

According to question, ∠B = ∠C = 55° , ∠D = 25° We can say , AB = AC ( ∴ ∠B = ∠C = 55° ) In Triangle ABC, ∠A + ∠B + ∠C = 180° ⇒ ∠A + 55° + 55° = 180° ⇒ ∠A + 110° = 180° ⇒ ∠A = 180° - 110° ⇒ ∠A = 70° ..........................(1) As per given figure, ∠ACD + ∠ACB = 180° ( ∠ACB = ∠C = 55°) ⇒ ∠ACD + 55° = 180° ⇒ ∠ACD = 180° - 55° ⇒ ∠ACD = 125° ....................... (2) Now in Triangle ACD, ∠CAD + ∠ACD + ∠CDA = 180° ⇒ ∠CAD + 125° + 25° = 180° ⇒ ∠CAD + 150° = 180° ⇒ ∠CAD = 30° ...........................(3) ( In a Δ, greater angle has longer side opposite to it ) From the equation (1) , (2) and (3); ∠B < ∠A and ∠CAD > ∠D ; ∴ BC > CA and CA < CD

Question 7

In the given figure, AM ⊥ BC and AN is the bisector of ∠A. What is the measure of ∠MAN.

Accepted Answer

Question 8

In the given figure ∠QPR = 90°, QR = 26 cm, PM = 6 cm, MR = 8 cm and ∠PMR = 90°, find the area of ΔPQR.

Accepted Answer

Given in the question, QR = 26 cm, PM = 6 cm, MR = 8 cm According to question, in Δ PMR (Pythagoras Theorem) PR = √PM2 + MR2 = √36 + 64 = √ 100 = 10 cm According to question, in Δ PQR (Pythagoras Theorem) PQ = √QR2 - PR2 = √262 - 102 = √576 - 100 = √ 476 = 24 ∴ area of triangle ∆PQR = Base length x Height / 2 ⇒ area of triangle ∆PQR = PR x PQ / 2 ⇒ area of triangle ∆PQR = 10 x 24 / 2 = 10 x 12 ⇒ area of triangle ∆PQR = 120

Question 9

In the given figure, find the length of BD.

Accepted Answer

Question 10

In, ΔABC, D and E are the mid-points of AB and AC respectively. Find the ratio of the areas of ΔADE and ΔABC.

Accepted Answer

Question 11

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.

Accepted Answer

Question 12

In a right angled Δ ABC, Right angled at A, AD ⊥ BC. Then :

Accepted Answer

Question 13

If ABCD is a parallelogram and AC and BD be its diagonals, then:

Accepted Answer

In ||gm ABCD , we have Since diagonals of parallelogram(||gm ) bisect each other, ∴ M will be the mid-point of each of the diagonal AC and BD ∴ In ΔABC AB2 + BC2 = 2(AM2 + MB2) ......... ( 1 ) [Appolonius Theorem] In ΔADC AD2 + CD2 = 2(AM2 + DM2) ......... ( 2 ) = 2(AM2 + MB2) [∴ DM = BM] Adding equations ( 1 ) and ( 2 ) AB2 + BC2 + CD2 + DA2 = 2(AM2 + MB2 + 2(AM2 + MB2 = 4AM2 + 4MB2 = (2AM)2 + (2MB)2= AC2+ BD2. { ∴ AM = MC , MB = MD }

Question 14

In the given figure, DE || BC and DE : BC = 3 : 5 the ratio of the areas of Δ ADE and the trapezium BCED.

Accepted Answer

Question 15

In a Δ ABC, AD intersects ∠ A and BC. If BC = a, AC = b and AB = c, Then :

Accepted Answer

Question 16

In the given figure DE||BC and AD : DB = 5 : 4, Then ar(∆DFE) ar(∆CFB)

Accepted Answer

Question 17

In center of a triangle lies in the interior of :

Accepted Answer

We know that In center of every triangle ( acute , obtuse and right angle ) lies in its interior. Hence required answer will be any triangle .

Question 18

In the given figure, which of the following is true :

Accepted Answer

From above figure , we can see that Join B and D and produce BD to E Then, p + q = β and s + t = x We know that , Exterior angle = Sum of two interior angle Now, s = p + α (Exterior angle of a Δ prop.) Similarly, t = q + γ (Exterior angle of a Δ prop.) Adding, s + t = p + q + α + γ Putting s + t = x and p + q = β , Hence x = α + β + γ

Question 19

In the given figure, the side BC of a Δ ABC is produced on both sides. Then ∠ 1 + ∠ 2 is equal to :

Accepted Answer

From the given figure, In ΔABC We know that , Exterior angle = Sum of two interior angles ∠1 = ∠A + ∠5 ................. ( 1 ) and ∠2 = ∠A + ∠6 ................. ( 2 ) Adding equations ( 1 ) and ( 2 ). ∠1 + ∠2 = 2∠A + ∠5 + ∠6 = 2∠A + (180° − ∠A) = ∠A + 180° The given question can be restated as the sum of two exterior angles exceeds ∠A of the ΔABC by 2 right angles.

Question 20

If two diameters of a circle intersect each other at right angles, then the quadrilateral formed by joining here end points is a :

Accepted Answer

From given figure , we have In quadrilateral ABCD , we can see that Diagonal AB = Diagonal CD ......... ( ⅰ ) AO = OB ......... ( ⅱ ) DO = CO ......... ( ⅲ ) ∠DOB = 90° ......... ( ⅳ ) From ( ⅰ ) , ( ⅱ ) , ( ⅲ ) and ( ⅳ ) , we get From above , it is clear that In quadrilateral ABCD , diagonals are equal and they bisect each other at 90° . ⇒ Quadrilateral ABCD is square.

Plane Geometry Questions