Q: X, Y are the mid-points of opposite sides AB and DC of a parallelogram ABCD. AY and DX are joined intersecting in P; CX and BY are joined intersecting in Q. Then PXQY is a :

From above given figure , Proceeding as in Q. No. 4, we can prove that AXCY is a parallelogram . Similarly, BXDY is a parallelogram. Now, AXCY is a parallelogram . ⇒ AY || CX [∴ Opposite sides of a parallelogram are parallel] ⇒ PY || QX …(1) Also, BXDY is a parallelogram ⇒ DX || BY [∴ Opposite sides of a parallelogram are parallel] ⇒ PX || QY …(2) Thus, in a quadrilateral PXQY, From (i) and (ii) , we have , PY || QX and PX || QY ⇒ PXQY is a parallelogram.

Q: If H is the orthocentre of Δ ABC, then the orthocentre of Δ HBC is (fig. given) :

From above given triangle , It is clear that the altitudes AL, BM and CN of ΔABC intersect at H. Then H is the orthocentre of ΔABC. In ΔABC, HL ⊥ BC and BN ⊥ CH. Thus, the two altitudes HL and BN of ΔHBC, intersects at A. Hence the orthocentre of ΔHBC is M .

Q: A point O in the interior of a rectangular ABCD is joined with each of the vertices A, B, C and D. Then :

As per given figure , we can see that Since the diagonals of a rectangle are equal and bisect each other. Let AC and BD intersect at M. Therefore M is the mid-point of AC and BD and AM = DM From ΔAOC, OA2 + OC2 = 2(AM2 + MO2) [Appollonius Theorem.] .......... ( 1 ) also in ΔODB, OB2 + OD2 = 2(MO2 + DM2) = 2(MO2 + AM2) .......... ( 2 ) From equations ( 1 ) and ( 2 ) . ∴ OA2 + OC2 = OB2 + OD2.

Q: If the sides of a triangle are produced then the sum of the exterior angles i.e., ∠ a + ∠ b + ∠ c is equal to :

From above figure , we have Since every exterior angle is equal to sum of interior opposite angles, So, ∠ a = A + B, ∠ b = B + C and ∠ c = A + C We know that , A + B + C = 180° ∴ ∠a +∠b + ∠c = 2(A + B + C) = 2 × 180° = 360°.

Q: In the given figure, side BC of Δ ABC is produced to form ray BD and CE || BA. Then ∠ ACD is equal to :

From above given figure , we can see that CE || BA and AC is the transversal. ∴ ∠4 = ∠1 (alternate interior ∠s) .......... ( 1 ) Again, CE || BA and BD is the transversal. ∴ ∠5 = ∠2 (corresponding ∠s) ................( 2 ) Adding equations ( 1 ) and ( 2 ) ∴ ∠4 + ∠5 = ∠1 + ∠2 ∴∠ACD = ∠A + ∠B.

Question 1

If the sides of a right triangle are x, x + 1 and x – 1, then the hypotenuse :

Accepted Answer

From given figure , we have Let, (x + 1) be the hypotenuse and perpendicular = x - 1 , base = x. By pythagoras theorem , ∴ (x + 1)2 = x2 + (x + 1)2 ⇒ x = 4 ∴ Hypotenuse = 4 + 1 = 5

Question 2

ABCD is a square, F is mid point of AB and E is a point on BC such that BE is one-third of BC. If area of ∆FBE = 108 m 2, then the length of AC is:

Accepted Answer

Question 3

ABCD is a parallelogram P, Q, R and S are points on sides AB, BC, CD and DA respectively such that AP = DR. If the area of the parallelogram ABCD is 16 cm 2, then the area of the quadrilateral PQRS is:

Accepted Answer

Question 4

The circumcentre of a triangle is always the point of intersection of the :

Accepted Answer

The circumcentre of a triangle is always the point of intersection of the Bisectors .Hence required answer will be bisectors .

Question 5

In the following figure, If BC = 8 cm, AB = 6 cm, AC 9 cm, then DC is equal to :

Accepted Answer

Question 6

In the accompanying figure, AB is one of the diameters of the circle and OC is perpendicular to it through the centre O. If AC is 7√2 cm, then what is the area of the circle in cm 2?

Accepted Answer

Question 7

In a triangle ABC, the length of the sides AB, AC and BC are 3, 5 and 6 cm respectively. If a point D on BC is drawn such that the line AD bisects the angle A internally, then what is the length of BD?

Accepted Answer

In a triangle ABC, Given :- AB = 3 cm , AC = 5 cm and BC = 6 cm We know that BD : DC = AB : AC BD : DC = 3 : 5 ∴ Divided BC = 6 in the ratio 3 : 5 BC = 3x + 5x = 6 ⇒ 8x = 6 ⇒ x = 6/8 = 0.75 ⇒ BD = 2.25, CD = 3.75.

Question 8

The number of tangents that can be drawn to two non-intersecting circles :

Accepted Answer

We know that when two circle neither touch nor intersect and one lies outside the other, then 4 common tangents can be drawn. For 4 common tangents , we get ∴ r1 + r2 < c1c2 Hence ,The number of tangents = 4

Question 9

X, Y are the mid-points of opposite sides AB and DC of a parallelogram ABCD. AY and DX are joined intersecting in P; CX and BY are joined intersecting in Q. Then PXQY is a :

Accepted Answer

From above given figure , Proceeding as in Q. No. 4, we can prove that AXCY is a parallelogram . Similarly, BXDY is a parallelogram. Now, AXCY is a parallelogram . ⇒ AY || CX [∴ Opposite sides of a parallelogram are parallel] ⇒ PY || QX …(1) Also, BXDY is a parallelogram ⇒ DX || BY [∴ Opposite sides of a parallelogram are parallel] ⇒ PX || QY …(2) Thus, in a quadrilateral PXQY, From (i) and (ii) , we have , PY || QX and PX || QY ⇒ PXQY is a parallelogram.

Question 10

ABCD is a rhombus with ∠ ABC = 56°, then ∠ ACD is equal to :

Accepted Answer

Question 11

The diagonals of a rectangle ABCD meet at 0. If ∠ BOC = 44°, then ∠ OAD is equal to :

Accepted Answer

Question 12

In the given figure, In a Δ ABC , ∠B = ∠ C. If AM is the bisector of ∠ BAC and AN ⊥ BC, then ∠ MAN is equal to :

Accepted Answer

Question 13

If H is the orthocentre of Δ ABC, then the orthocentre of Δ HBC is (fig. given) :

Accepted Answer

From above given triangle , It is clear that the altitudes AL, BM and CN of ΔABC intersect at H. Then H is the orthocentre of ΔABC. In ΔABC, HL ⊥ BC and BN ⊥ CH. Thus, the two altitudes HL and BN of ΔHBC, intersects at A. Hence the orthocentre of ΔHBC is M .

Question 14

A point O in the interior of a rectangular ABCD is joined with each of the vertices A, B, C and D. Then :

Accepted Answer

As per given figure , we can see that Since the diagonals of a rectangle are equal and bisect each other. Let AC and BD intersect at M. Therefore M is the mid-point of AC and BD and AM = DM From ΔAOC, OA2 + OC2 = 2(AM2 + MO2) [Appollonius Theorem.] .......... ( 1 ) also in ΔODB, OB2 + OD2 = 2(MO2 + DM2) = 2(MO2 + AM2) .......... ( 2 ) From equations ( 1 ) and ( 2 ) . ∴ OA2 + OC2 = OB2 + OD2.

Question 15

ABC is a Δ in which AB = AC and D is a point on AC such that BC2 = AC × CD. Then :

Accepted Answer

Question 16

The area of two similar Δs are 121 cm 2 and 81 cm 2 respectively. What is the ratio of their corresponding heights (altitudes):

Accepted Answer

Question 17

In Δ ABC, the median BE intersects AC at E, if BG = 6 cm, where G is the centro-id, then BE is equal to :

Accepted Answer

Question 18

If the sides of a triangle are produced then the sum of the exterior angles i.e., ∠ a + ∠ b + ∠ c is equal to :

Accepted Answer

From above figure , we have Since every exterior angle is equal to sum of interior opposite angles, So, ∠ a = A + B, ∠ b = B + C and ∠ c = A + C We know that , A + B + C = 180° ∴ ∠a +∠b + ∠c = 2(A + B + C) = 2 × 180° = 360°.

Question 19

In the given figure, side BC of Δ ABC is produced to form ray BD and CE || BA. Then ∠ ACD is equal to :

Accepted Answer

From above given figure , we can see that CE || BA and AC is the transversal. ∴ ∠4 = ∠1 (alternate interior ∠s) .......... ( 1 ) Again, CE || BA and BD is the transversal. ∴ ∠5 = ∠2 (corresponding ∠s) ................( 2 ) Adding equations ( 1 ) and ( 2 ) ∴ ∠4 + ∠5 = ∠1 + ∠2 ∴∠ACD = ∠A + ∠B.

Question 20

In fig, AB = AC, D is a point on AC and E on AB such that AD = ED = EC = BC. Then ∠A : ∠B :

Accepted Answer

Plane Geometry Questions