According to this theorem ” two triangles having the same base and lies between the same parallel lines are equal in area” In the above figure; ABC and PBC are the two triangles having same base BC. Also, both the triangles lies between the same parallel lines. To Prove: Area ( ABC ) = Area ( PBC ) Construction: Draw line CD parallel to AB Draw line CR parallel to BP Proof: After the construction we now have two parallelogram ABCD and PBCR. Note that both the parallelogram have the same base and lie between the same parallel lines. So according to parallelogram area theorem we can write ..read more

In this post we will discuss properties and formulas of different quadrilateral in detail. Following quadrilaterals are discussed below; (a) Square (b) Rectangle (c) Parallelogram (d) Rhombus (e) Trapezium Along with the properties, try to remember all the formulas as they would help you solve related questions in examination. Square The square has following properties; (a) All sides are equal (b) All angle measures 90 degree (c) Diagonals bisect each other at 90 degree Given above is the square ABCD in which; (i) All sides are equal AB = BC = CD = DA (ii) All angle measure 90 degree ∠A = ∠B ..read more

In this chapter we will prove that “any quadrilateral with equal opposite sides is actually a parallelogram. Given: Consider quadrilateral ABCD in which opposite sides are equal. AB = CD and AD = BC To Prove: Prove that the quadrilateral is parallelogram. i.e. AB || CD and AD || CB Proof: Consider triangle ABC and CDA; AB = CD ( given ) AC = CA ( common side ) AD = CB ( given ) By SSS congruency, both the triangles are congruent. i.e. \mathtt{\triangle ABC\ \cong \triangle CDA} Since both triangles are congruent, we can write; ∠BAC = ∠DCA and ∠BCA = ∠DAC Since ∠BAC = ∠DCA; This is pos ..read more

In this chapter we will prove that ” in parallelogram, opposite sides are equal “. This is very important property so make sure you remember it for your exam. Given: ABCD is a parallelogram in which opposite sides are parallel to each other. AB || CD and AD || CB To Prove: Prove that opposite sides are equal in length. AB = CD and AD = CB Proof: AB & CD are parallel lines intersected by transversal AC. ∠CAB = ∠ACD ( alternate angle ) ∠CAD = ∠ACB ( alternate angle ) Now consider triangle ABC and ADC; ∠CAB = ∠ACD ( alternate angle ) AC = CA ( common side ) ∠CAD = ∠ACB ( alternate angle ..read more

In this chapter we will prove important theorem of parallelogram. According to the theorem ” the diagonal of parallelogram divides the figure into two congruent triangle. “ Given above is parallelogram in which opposite sides are equal and parallel with AC as diagonal. To Prove: Prove that triangle ACD is congruent to triangle ACB. \triangle ABC\ \cong \triangle ADC Proof: AB & CD are parallel lines with transversal AC. ∠BAC = ∠DCA ( alternate angles ) ∠BCA = ∠DAC ( alternate angles ) Now consider triangle ABC and ADC; ∠BAC = ∠DCA ( alternate angles ) AB = CD ( opposite sides of paral ..read more

In this chapter we will solve questions related to congruent triangles. All the questions are to standard of grade 09 and are provided with detailed solution. Question 01 In the below image, line AC bisects ∠A and ∠C. Prove that AB = AD and CB = CD. Solution Consider triangle ABC and ADC. ∠BAC = ∠DAC ( AC is angle bisector of ∠A ) AC = CA ( common side ) ∠BCA = ∠DCA ( AC bisect ∠C) By ASA congruency, both the triangles are congruent. So, AB = AD and CB = CD. Hence Proved. Question 02 In the below image OA = OB and OQ = OP. Prove that PX = QX Solution OA = OB OP + PA = OQ + QB It’s giv ..read more

In this chapter we will solve questions related to congruent triangles. All the questions are to the standard of grade 09. Question 01 In the below figure AD = BC. Prove that line CD bisects AB. Given: AD = BC ∠OAD = ∠OBC = 90 degree To prove: OA = OB Solution Consider triangle OBC and OAD ∠BOC = ∠DOA ( vertically opposite angle) ∠OBC = ∠OAD = 90 degree BC = DA By AAS congruency, both the triangles are congruent. So OB = OA Hence Proved Question 02 In the below figure parallel lines l & m are intersected by parallel transversal p and q. Prove that triangle ABC and CDA are congruent ..read more

In this chapter we will solve questions related to congruent triangles. All the questions are to the standard of grade 09 math. Question 01 Given below is the triangle in which TR = TS. Also ∠1 = 2 (∠2) and ∠4 = 2 (∠3). Prove that triangle RBT and SAT are congruent. Solution ∠1 = ∠4 (vertically opposite angles) 2 (∠2) = 2 (∠3) ∠2 = ∠3 In triangle angle opposite to equal sides are equal. TR = TS ∠R = ∠S ∠TRB + ∠2 = ∠TSA + ∠3 Since ∠2 & ∠3 are equal, we can cancel each of them from the equation. ∠TRB = ∠TSA Now consider triangle RBT and SAT ∠TRB = ∠TSA RT = ST (given ) ∠T = ∠T (common ..read more

Given below are collection of questions related to congruent triangles. All the questions are to the standard of grade 9. Question 01 In below figure, AB = AD and EA = AC. Prove that line ED is parallel to BC. Solution Consider ▵ABC and ▵ABC; AB = AD ∠BAC = ∠EAD ( vertically opposite angle) EA = AC By SAS congruency, ▵ABC and ▵ABC are congruent to each other. Since both the triangle are congruent, we can write; ∠ABC = ∠EDA Both the angles are also alternate interior angles. Since both of them are equal it means that ED & BC are parallel lines intersected by transversal BD. Hence Proved ..read more

In this chapter we will solve questions related to angle sum property of triangle. All the questions are to the standard of grade 09. Question 01 Given below is the image of triangle PQR in which line QT perpendicular to side PR. Find the measure of angle x and y. Solution Consider triangle QTR. We know that sum of internal angle of triangle measures 180 degree. ∠TQR + ∠QTR +∠x = 180 40 + 90 + ∠x = 180 130 + ∠x = 180 ∠x = 180 – 130 ∠x = 50 degree. Using the exterior angle theorem in triangle PSR; ∠y = ∠SPR + ∠x ∠y = 30 + 50 ∠y = 80 degree Hence, we calculated the value of ∠x and ∠y. Q ..read more