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X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (See Figure). Prove that ar (LZY) = ar (MZYX)

Solution: Given According to the given details LX=XY=YN XZ II LM To Prove ar (LZY) = ar (MZYX) Proof ar(LZX)+(XZY)=ar(LZY)... View Article

In Figure PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR = RS and PA || QB || RC. Prove that ar (PQE) = ar (CFD).

Solution: Given According to the given details PA║QB║RC║SD & PQ=QR=RS From the Equal intercept theorem, we know that if... View Article

PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then ar (PAS) = 30 cm2. Write True or False and justify your answer:

The correc answer is (C) Perimeter of ABCD > Perimeter of ABEM Solution From rectangle ABEM, AB = EM ------(eq.1) [sides of rectangle]... View Article

ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm², then ar (ABC) = 24 cm². Write True or False and justify your answer:

The correc answer is (C) DC × DL Solution We know that Area of parallelogram = Base × Corresponding altitude = AB ×... View Article

In Figure parallelogram ABCD and rectangle ABEF are of equal area, then : (A) Perimeter of ABCD = Perimeter of ABEM (B) Perimeter of ABCD < Perimeter of ABEM (C) Perimeter of ABCD > Perimeter of ABEM (D) Perimeter of ABCD = 1/2 (Perimeter of ABEM)

The correc answer is (C) DC × DL Solution Explanation: Area of parallelogram = Base × Corresponding altitude = AB ×... View Article

In Figure the area of parallelogram ABCD is: (A) AB × BM (B) BC × BN (C) DC × DL (D) AD × DL

The correc answer is (D) a rhombus of area 24 cm2 Solution According to the given details Let ABCD be the rectangle. E is the... View Article

In which of the following figures (Figuere), you find two polygons on the same base and between the same parallels?

Solution: The correct answer is (D) Explanation: In diagram (D), the parallelograms, PQRA and BQRS are on the same base QR and... View Article

The median of a triangle divides it into two (A) triangles of equal area (B) congruent triangles (C) right triangles (D) isosceles triangles

The correct answer is (A) triangles of equal area Solution A median of a triangle is a line segment that joins a vertex to the mid-point... View Article

In Figure AB || DE, AB = DE, AC || DF and AC = DF. Prove that BC || EF and BC = EF.

Solution: Given According to the given details AB || DE, AB = DE, AC || DF and AC = DF To Prove BC || EF and BC = EF Proof From... View Article

ABCD is a quadrilateral in which AB || DC and AD = BC. Prove that ∠A = ∠B and ∠C = ∠D.

Given ABCD is a quadrilateral in which AB || DC and AD = BC. To Prove ∠A = ∠B and ∠C = ∠D. Proof According to... View Article

P and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. AQ intersects DP at S and BQ intersects CP at R. Show that PRQS is a parallelogram.

Given P and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. AQ intersects DP at S and BQ intersects CP at R... View Article

A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus.

Given A diagonal of a parallelogram bisects one of its angles. To Prove We have to prove that it is a rhombus Proof Let the... View Article

P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square.

Given P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD.... View Article

P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD such that AC ⊥ BD. Prove that PQRS is a rectangle.

Given P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD such that AC ⊥ BD. To Prove We... View Article

P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. Prove that PQRS is a rhombus.

Given P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. To... View Article

In a parallelogram ABCD, AB = 10 cm and AD = 6 cm. The bisector of ∠A meets DC in E. AE and BC produced meet at F. Find the length of CF.

Given A parallelogram ABCD, AB = 10 cm and AD = 6 cm. The bisector of ∠A meets DC in E. AE and BC produced meet at F. Find out We... View Article

A square is inscribed in an isosceles right triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.

Given ΔABC with ∠A = 90o and ABC is an isosceles triangle, To Prove The vertex of the square opposite the vertex of... View Article

E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. Show that BFDE is a parallelogram.

Given E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF To Prove BFDE is a parallelogram. Proof Construction:... View Article

ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus.

Solution: According to given details ABCD is a rhombus. DE is the altitude on AB then AE = EB. From ΔAED and ΔBED, We note... View Article

The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60º. Find the angles of the parallelogram.

Solution: According to given details ABCD is parallelogram, DP ⊥ AB DQ ⊥ BC. ∠PDQ = 60o In quadrilateral DPBQ, We know... View Article

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