### Quadrilaterals Class 9th Mathematics AP Board Solution

##### Question 1.ABC is a triangle. D is a point on AB such that and E is a point on AC such that If DE = 2 cm find BC.Answer: Given:- DE=2cm  Formula used:- Line joining midpoints of 2 sides of triangle will behalf of its 3rd side.Solution:-Assume mid points of line AB and AC is M and NThe formation of new triangle will be there Δ AMNAs; ⇒ AC=4AEand ∴ AM=2AEHence;E is the midpoint of AMSimilarlyD is the midpoint of AN→ In Δ AMNIf E,D are midpointsED= MNMN=2EDMN=2×2cm=4cm→ In Δ ABCIf M,N are midpointsAB= MNAB=2MNAB=2×4cm=8cmConclusion:-BC=8cmQuestion 2.ABCD is quadrilateral E, F, G and H are the midpoints of AB, BC, CD and DA respectively. Prove that EFGH is a parallelogram.Answer: Given:- ABCD is quadrilateral E, F, G and H are the midpoints ofAB, BC, CD and DA respectivelyFormula used:- Line joining midpoints of 2 sides of triangleIs parallel to 3rd sideSolution:-BD is diagonal of quadrilateralEH is the line joined by midpoints of triangle ABD,∴EH is parallel to BDGF is line joined by midpoints of side BC&BD of triangle BCD∴GF is parallel to BD⇒ If HE is parallel to BD and BD is parallel to GF∴ It gives HE is parallel to GF⇒ AC is another diagonal of quadrilateralGH is the line joined by midpoints of triangle ADC,∴GH is parallel to ACFE is line joined by midpoints of side BC&AB of triangle ABC∴FE is parallel to AC⇒ If GH is parallel to AC and AC is parallel to FE∴ It gives GH is parallel to FEAs HE||GF and GH||FE∴ EFGH is a parallelogramConclusion:-EFGH is a parallelogramQuestion 3.Show that the figure formed by joining the midpoints of sides of a rhombus successively is a rectangle.Answer: Given:- ABCD is a rhombusE,F,G,H are the mid points of AB,BC,CD,DAFormula used:- Line joining midpoints of 2 sides of triangleIs parallel and half of 3rd sideSolution:-BD is diagonal of rhombusEH is the line joined by midpoints of triangle ABD,∴EH is parallel and half of BDGF is line joined by midpoints of side BC&BD of triangle BCD∴GF is parallel and half BD⇒ If HE is parallel to BD and BD is parallel to GF∴ It gives HE is parallel to GF⇒ If HE is half of BD and GF is also half of BD∴ It gives HE is equal to GF⇒ AC is another diagonal of rhombusGH is the line joined by midpoints of triangle ADC,∴GH is parallel to AC∴GH is half of ACFE is line joined by midpoints of side BC&AB of triangle ABC∴FE is parallel to AC∴ FE is half of AC⇒ If GH is parallel to AC and AC is parallel to FE∴ It gives GH is parallel to FE⇒ If GH is half of AC and FE is also half of AC∴ It gives GH is equal to FEAs diagonals of rhombus intersect at 90°And AC is parallel to FE and GHAll angles of EFGH is 90°All angles are 90° and opposite sides are equal and parallelConclusion:-EFGH is a rectangleQuestion 4.In a parallelogram ABCD, E and F are the midpoints of the sides AB and DC respectively. Show that the line segments AF and EC trisect the diagonal BD. Answer:Given:- ABCD is a parallelogramE and F are the midpoints of the sides AB and DC respectivelyFormula used:- Line drawn through midpoint of one side of triangleParallel to other side , bisect the 3rd side.Solution:-As ABCD is parallelogramAB=CD and AB||CD;⇒ AE=CF and AE||CF [E and F are the midpoints of AB and CD]In quadrilateral AECF⇒ AE=CF and AE||CFAECF is a parallelogram.∵ Quadrilateral having one pair of side equal and parallel are parallelogramIf AECF is a parallelogram∴ AF||CEHence ;PF||CQ and AP||QE [As AF=AP+PF and CE=CQ+QE ]In Δ DQCDF=FC [F is midpoint]PF||CQThen;P is midpoint of DQDP=PQ∵ Line drawn through midpoint of one side of triangle Parallel toother side , bisect the 3rd sideIn Δ APBAE=EB [E is midpoint]AP||QEThen;Q is midpoint of PBPQ=QB∵ Line drawn through midpoint of one side of triangle Parallel toother side , bisect the 3rd sideIf DP=PQ and PQ=QBThen DP=PQ=QBConclusion:-Line segments AF and EC trisect the diagonal BD.Question 5.Show that the line segments joining the midpoints of the opposite sides of a quadrilateral and bisect each other.Answer: Given:- ABCD is a quadrilateralFormula used:- Line joining midpoints of 2 sides of triangleIs parallel and half of 3rd sideSolution:-BD is diagonal of quadrilateralEH is the line joined by midpoints of triangle ABD,∴EH is parallel and half of BDGF is line joined by midpoints of side BC&BD of triangle BCD∴GF is parallel and half BD⇒ If HE is parallel to BD and BD is parallel to GF∴ It gives HE is parallel to GF⇒ If HE is half of BD and GF is also half of BD∴ It gives HE is equal to GF⇒ AC is another diagonal of quadrilateralGH is the line joined by midpoints of triangle ADC,∴GH is parallel to AC∴GH is half of ACFE is line joined by midpoints of side BC&AB of triangle ABC∴FE is parallel to AC∴ FE is half of AC⇒ If GH is parallel to AC and AC is parallel to FE∴ It gives GH is parallel to FE⇒ If GH is half of AC and FE is also half of AC∴ It gives GH is equal to FEIf both opposite sides are parallel and equalThen, the quadrilateral is parallelogramIf EFGH is parallelogramThen their diagonal bisect each otherIf diagonal of parallelogram is the line joining midpoint of opposite sides of quadrilateralThen;Line joining midpoints of opposite sides of quadrilateral bisect each other.Conclusion:-Lines joining midpoints of opposite sides of quadrilateral bisect each otherQuestion 6.ABC is a triangle right angled at C. A line through the midpoint M of hypotenuse AB and Parallel to BC intersects AC at D. Show that(i) D is the midpoint of AC(ii) MD ⊥ AC(iii) CM = MA = AB Answer:Given:- ACB is right angle triangleM is midpoint of ABDM||CBFormula used:- Line drawn through midpoint of one side of triangleParallel to other side , bisect the 3rd side* SAS congruency propertyIf 2 sides and angle between the two sides of boththe triangle are equal, then both triangle are congruentSolution:-(1) In Δ ABCM is midpoint of ABAnd DM||CB∴ D is midpoint of ACAD=DC∵ Line drawn through midpoint of one side of triangleParallel to other side , bisect the 3rd side(2) In Δ ABCDM||CB∠ ADM=∠ ACB [Corresponding angles]∠ ACB =90°∠ ADM=90°(3) In Δ ADM and Δ DMC⇒ DM=DM [Common in both triangles ]⇒ AD=DC [D is the midpoint]As ADC is straight line∠ ADM+∠ MDC=180°∠ MDC=180° - ∠ ADM∠ MDC=90°⇒ ∠ ADM=∠ MDCHence;In Δ ADM ≅ Δ DMC∴ CM=MA⇒ MA= AB [M is midpoint of AB]∴ CM=MA= AB

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