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The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rhombus, if (A) PQRS is a rhombus (B) PQRS is a parallelogram (C) diagonals of PQRS are perpendicular (D) diagonals of PQRS are equal.

The correct answer is (D) diagonals of PQRS are equal. Solution We know that ABCD is a rhombus So AB = BC = CD = DA Now, Since D and C are... View Article

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if (A) PQRS is a rectangle (B) PQRS is a parallelogram (C) diagonals of PQRS are perpendicular (D) diagonals of PQRS are equal.

The correct answer is (C) diagonals of PQRS are perpendicular Solution Let the rectangle be ABCD, From the properties of the rectangle,... View Article

ABCD is a rhombus such that ∠ACB = 40º. Then ∠ADB is (A) 40º (B) 45º (C) 50º (D) 60º

The correct answer is (C) 50º Solution According to the given details, ABCD is a rhombus ∠ACB = 40° ∵ ∠ACB = 40°... View Article

A diagonal of a rectangle is inclined to one side of the rectangle at 25º. The acute angle between the diagonals is (A) 55º (B) 50º (C) 40º (D) 25º

The correct answer is (B) 50º Solution According to the given details A diagonal of a rectangle is inclined to one side of the rectangle at... View Article

Three angles of a quadrilateral are 75º, 90º and 75º. The fourth angle is (A) 90º (B) 95º (C) 105º (D) 120º

The correct answer is (D) 120º Solution According to the given details Three angles of a quadrilateral are 75°, 90° and 75°... View Article

Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side.

Given Δ ABC with median AD. To Prove AB + AC > 2AD AB + BC > 2AD BC + AC > 2AD Construction: Extend AD to E such that DE = AD... View Article

ABC is an isosceles triangle in which AC = BC. AD and BE are respectively two altitudes to sides BC and AC. Prove that AE = BD.

Given ABC is an isosceles triangle in which AC = BC. To Prove AE = BD Proof According to given details From Δ ADC and Δ... View Article

ABC and DBC are two triangles on the same base BC such that A and D lie on the opposite sides of BC, AB = AC and DB = DC. Show that AD is the perpendicular bisector of BC.

Given Δ ABC and Δ DBC on the same base BC. Also, AB = AC and BD = DC. To Prove AD is the perpendicular bisector of BC i.e.,... View Article

O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that Δ OCD is an isosceles triangle.

Given A square ABCD and OA = OB = AB. To Prove Δ OCD is an isosceles triangle. Proof In square ABCD, We note that ∠1 and... View Article

ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2 AD.

Given A right angles triangle with AB = AC bisector of ∠A meets BC at D. To Prove BC = 2 AD Proof According to given details From right... View Article

ABCD is a quadrilateral in which AB = BC and AD = CD. Show that BD bisects both the angles ABC and ADC.

Given ABCD is a quadrilateral AB = BC AD = CD To Prove BD bisects both the angles ABC and ADC Proof According to the given details... View Article

P is a point on the bisector of ∠ABC. If the line through P, parallel to BA meet BC at Q, prove that BPQ is an isosceles triangle.

Given BP is the bisector of ∠ABC the line through P, parallel to BA meet BC at Q To Prove BPQ is an isosceles triangle. Proof... View Article

ABC is an isosceles triangle with AB = AC and D is a point on BC such that AD ⊥ BC (Figure. To prove that ∠BAD = ∠CAD, a student proceeded as follows: In Δ ABD and Δ ACD, AB = AC (Given) ∠B = ∠C (because AB = AC) And ∠ADB = ∠ADC Therefore, Δ ABD Δ Δ ACD (AAS) So, ∠BAD = ∠CAD (CPCT) What is the defect in the above arguments? [Hint: Recall how ∠B = ∠C is proved when AB = AC].

Solution: FromΔ ABD and Δ ADC, we note that ∠ADB = ∠ADC According to the given details AB = BC AD = AD [Common... View Article

The image of an object placed at a point A before a plane mirror LM is seen at the point B by an observer at D as shown in Fig. 7.12. Prove that the image is as far behind the mirror as the object is in front of the mirror. [Hint: CN is normal to the mirror. Also, angle of incidence = angle of reflection].

Solution: Let AB intersect LM at O. We have to prove that AO = BO. Now, Accoring to law of reflection ∠i = ∠r ---------(i) [∵Angle... View Article

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Archives

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rhombus, if (A) PQRS is a rhombus (B) PQRS is a parallelogram (C) diagonals of PQRS are perpendicular (D) diagonals of PQRS are equal.

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if (A) PQRS is a rectangle (B) PQRS is a parallelogram (C) diagonals of PQRS are perpendicular (D) diagonals of PQRS are equal.

ABCD is a rhombus such that ∠ACB = 40º. Then ∠ADB is (A) 40º (B) 45º (C) 50º (D) 60º

A diagonal of a rectangle is inclined to one side of the rectangle at 25º. The acute angle between the diagonals is (A) 55º (B) 50º (C) 40º (D) 25º

Three angles of a quadrilateral are 75º, 90º and 75º. The fourth angle is (A) 90º (B) 95º (C) 105º (D) 120º

Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side.

ABC is an isosceles triangle in which AC = BC. AD and BE are respectively two altitudes to sides BC and AC. Prove that AE = BD.

ABC and DBC are two triangles on the same base BC such that A and D lie on the opposite sides of BC, AB = AC and DB = DC. Show that AD is the perpendicular bisector of BC.

O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that Δ OCD is an isosceles triangle.

ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2 AD.

ABCD is a quadrilateral in which AB = BC and AD = CD. Show that BD bisects both the angles ABC and ADC.

P is a point on the bisector of ∠ABC. If the line through P, parallel to BA meet BC at Q, prove that BPQ is an isosceles triangle.

Find all the angles of an equilateral triangle.

Q is a point on the side SR of a Δ PSR such that PQ = PR. Prove that PS > PQ.

In Figure BA ⊥ AC, DE ⊥ DF such that BA = DE and BF = EC. Show that Δ ABC ≅ Δ DEF.

CDE is an equilateral triangle formed on a side CD of a square ABCD (Figure). Show that Δ ADE ≅ Δ BCE.

In Figure D and E are points on side BC of a Δ ABC such that BD = CE and AD = AE. Show that Δ ABD ≅ Δ ACE.

ABC is an isosceles triangle with AB = AC and BD and CE are its two medians. Show that BD = CE.

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