# Chapter – 12 : Theorems On Area | Chapter Solution Class 9

 Book Name : Ganit Prakash Subject : Mathematics (Maths) Class : 9 (Madhyamik/WB) Publisher : Prof. Nabanita Chatterjee Chapter Name : Theorems On Area (12th Chapter)

## Let I do – 12.1

1. If any triangle and rectangle are on the same base and between the same parallel then let us prove logically that the area of triangle is half of the area in the shape of rectangular region.

Solution: Given:

\triangle A B C and Rectangle ABCD are on the same base AB and between the same parallel lines AB and CD.

R.T.P : \triangle A B C=\frac{1}{2} \text { Rectangle } A B C D

Construction :- Joint A and C.

Proof: By construction ABCD is a rectangle and AC is one of its diagonal.

\therefore \triangle A B C=2_2^1 \text { Rectangle } A B C D[\because A B\|C D \ A D\| B C] (Proved)

2. If any triangle and any parallelogram are on the same base and between same parallels. Let us prove logically that the area of the triangular field is half of the area of in the shape of parallelogram field.

Solution: Given \triangle A B C and parallelogram ABDE are on the same base \mathrm{AB} and between the same parallels AB and ED.

R.T.P : \therefore \triangle \mathrm{ABC}=\frac{1}{2} Parallelogram\mathrm{ABDE}

Construction: The straight line through the point A drawn parallel to BC intersects DC produced at the point F.

Proof : By construction ABCF is a parallelogram and AC is one of its diagonals.

\therefore \triangle \mathrm{ABC}=\frac{1}{2} \text { Parallelogram ABDE } \\

\therefore \triangle \mathrm{ABC}=\frac{1}{2} \text { Parallelogram } \mathrm{ABDE} \text { (Proved) }

\therefore The two parallelograms are on the same base and between the same parallels.

## Let’s prove – 12

1. P and Q are the mid points of sides AB and DC of parallelogram ABCD; Let’s prove that the area of the quadrilateral field =\frac{1}{2} \times area of parallelogram field.

Solution: Given:

ABCD is a parallelogram in which P and Q are the mid points of sides AB and DC.

P, Q is drawn, Join A, Q and P, C

R.T.P : Area of quadrilateral field \mathrm{APCQ}=\frac{1}{2} \times area of Parallelogram ABCD

Proof : \because A B\|C D \ B C\| A D

\therefore B C\|A D\| P Q[\because P \ Q are the mid-points of A B and \mathrm{CD}]

\therefore A P Q D and B C Q P are two parallelograms.

In \triangle \mathrm{APQ} and Parallelogram APQD are on the same parallel line \mathrm{AP} \ and \ \mathrm{QD}.

\therefore \triangle \mathrm{APQ}=\frac{1}{2} \text { Parallelogram } \mathrm{APQD}

Similarly, \triangle \mathrm{CPQ}=\frac{1}{2} Parallelogram \mathrm{BCQP}

\therefore \quad \triangle \mathrm{APQ}+\triangle \mathrm{CPQ}= \\

1 \text { Parallelogram APQP }+\frac{1}{2} \text { Parallelogram BCQP }

\therefore \text { Quadrilateral } \mathrm{APCQ}=\frac{1}{2} \text { Prallelogram ABCD (Proved) }

2. The distance between two sides AB and DC of a rhombus ABCD is PQ and distance between sides AD and BC is \mathbf{R S}; Let’s prove that PQ = RS.

Solution : Given:

ABCD is a rhombus.

The distance between two sides AB and DC is PQ and distance between sides AD and BC is R S respectively. PQ and RS are drawn.

R.T.P : PQ = RS.

Construction: Join A,P ; P,B; A,S and S,D

Proof: In \triangle \mathrm{APB} and rhombus \mathrm{ABCD}.

AB is same base and between the same parallel lines AB and \mathrm{CD}.

\therefore \triangle \mathrm{APB}=\frac{1}{2} \text { Rhombus } \mathrm{ABCD}

Again,

In \triangle ASD and rhombus ABCD are on the same base AD and between the same parallel lines AD and BC.

\therefore \triangle \mathrm{ASD}=\frac{1}{2} \text { Rhombus } \mathrm{ABCD} \\

\therefore \triangle \text { Area of } \mathrm{APB}=\text { Area of } \triangle \mathrm{ASD} \\

\text { or } \frac{1}{2} \times \mathrm{AB} \times \mathrm{PQ}=\frac{1}{2} \times \mathrm{AD} \times \mathrm{RS} \\

\therefore P Q=R S[\because A B=A D] \text { (Proved) } \\

3. P and Q are the mid-points of sides AB and DC of parallelogram ABCD respectively. Let’s prove that PBQD is a parallelogram and \triangle \mathrm{PBC}=\frac{1}{2} parallelogram PQBD.

Solution : Given:

ABCD is a parallelogram in which P and Q are mid-points of AB and DC respectively. PD, BQ and PC are drawn.

R.T.P : (i) PQBD is parallelogram.

(ii) \triangle \mathrm{PBC}=\frac{1}{2} parallelogram PQBD

Proof: \therefore ABCD is a parallelogram

\therefore A B \| D C \\

\therefore P B \| D Q

Again,

\therefore ABCD is a parallelogram.

\therefore A B=D C \\

\therefore \quad \frac{1}{2} A B=\frac{1}{2} D C

\therefore \quad P B=D Q[\therefore \text{P and Q are mid-points of A B and D C}]

Now, P B \| D Q and PB = DQ

\therefore PBQD is parallelogram (i) (Proved)

\therefore \triangle PBC and parallelogram PBQD are on the same base PB between the same parallels PB and DC.

\therefore \triangle \mathrm{PBC}=\frac{1}{2} \text { Parallelogram PBQD (ii) (Proved) }

4. In an isoceles triangle ABC, AB = AC and P is any point on produced side BC. PQ and PR are perpendicular on sides AB and AC from the point P respectively. BS is perpendicular on side AC from point B; Let’s prove that PQ – PR = BS

Solution : Given:

ABC is an isosceles triangle, AB = AC and P is any point on produced side BC. PQ and PR are perpendicular on sides AB and AC from the point P respectively. BS is perpendicular on side AC from point B.

R.T.P: PQ – PR = BS

Construction: A & P are joined.

Proof: In \triangle A B C,

\text { area of } \triangle A B C=\frac{1}{2} \times A C \times B S

In \triangle \mathrm{ACP}

\text { area of } \triangle A C P=\frac{1}{2} \times A C \times P R

In \triangle \mathrm{ABP},

\text { area of } \triangle \mathrm{ABP}=\frac{1}{2} \times A B \times P Q \\

\therefore \text { area of } \triangle \mathrm{ABP}=\text { area of } \triangle \mathrm{ABC}+\text { area of } \triangle \mathrm{ACP} \\

\text { or, } \frac{1}{2} \times A B \times P Q=\frac{1}{2} \times \mathrm{AC} \times \mathrm{BS}+\frac{1}{2} \times \mathrm{AC} \times \mathrm{PR} \\

\text { or, } \frac{1}{2} \times \mathrm{AB} \times \mathrm{PQ}=\frac{1}{2} \times \mathrm{AC}(\mathrm{BS}+\mathrm{PR}) \\

\text { or, } \frac{1}{2} \times \mathrm{AB} \times \mathrm{PQ}=\frac{1}{2} \times \mathrm{AB}(\mathrm{BS}+\mathrm{PR})[\because A B=A C] \\

\text { or, } \mathrm{PQ}=\mathrm{BS}+\mathrm{SR} \\

\therefore P Q-P R=B S \\

\therefore \text { (Proved) }

5. O is any point out side the equilateral triangle ABC and within the angular region on ABC; OP, OQ and OR are the perpendicular distance of AB, BC and CA respectively from the point O. Let us prove that the altitude of the triangle = OP + OQ – OR.

Solution: Given:

ABC is an equilateral triangle. O is a point out side the triangle and within the angular region \mathrm{ABC}. \mathrm{OP}, \mathrm{OQ} and \mathrm{OR} are perpendicular distances of

AB, BC and CA respectively from the point O. Perpedicular AS on BC is drawn. Thus AS is the altitude of \triangle \mathrm{ABC}

Proof : \mathrm{OA}, \mathrm{OB} \ and \ \mathrm{OC} are drawn.

\because \triangle \mathrm{ABC}=\text { Quadrilateral } \mathrm{ABCD}-\triangle \mathrm{AOC}

or, \triangle \mathrm{ABC}=\triangle \mathrm{ABO}+\triangle \mathrm{OBC}-\triangle \mathrm{AOC}

\therefore \frac{1}{2} \mathrm{BC} \times \mathrm{AS}=\frac{1}{2} \times \mathrm{AB} \times \mathrm{OP}+\frac{1}{2} \mathrm{BC} \times \mathrm{OQ}-\frac{1}{2} \mathrm{AC} \times \mathrm{OR} \\

=\frac{1}{2} \times \mathrm{BC} \times \mathrm{OP}+\frac{1}{2} \mathrm{BC} \times \mathrm{OQ}-\frac{1}{2} \mathrm{AC} \times \mathrm{OR} \\

{[\because A B=B C, A C=B C]} \\

=\frac{1}{2} \mathrm{BC}(\mathrm{OP}+\mathrm{OQ}-\mathrm{OR}) \\

\therefore \quad A S=O P+O Q-O R \text { (Proved) } \\

6. A straight line parallel to AB of parallelogram ABCD intersects AD, AC and BC or their produced parts at the points E, F and G respectively. Let’s prove that \triangle \mathrm{AEG}= \triangle \mathrm{AFD}

Solution: Given,

ABCD is a parallelogram, and a straight line parallel to \mathrm{AB} intersects sides \mathrm{AD}, \mathrm{AC} \ and \ \mathrm{BC} or their produced parts at the point E, F and G respectively E G is drawn.

R.T.P: \triangle \mathrm{AEG}=\triangle \mathrm{AFD}

Construction: We draw IH \|A D\| B C joined A, G.

Proof: \because E G\|A B\| D C \& A D\|I H\| B C

\therefore AIFE, CGFH, BIFG \& DEFH are four parallelograms.

In parallelogram A B C D, \triangle A B C=\triangle A D C

\text{ " " " " " " " "} \mathrm{AIFE}, \quad \triangle \mathrm{AIF}=\triangle \mathrm{AFE} \\

\text{ " " " " " " " "}\mathrm{FGCH} \quad \triangle \mathrm{FGC}=\triangle \mathrm{FHC} \\

\therefore \triangle \mathrm{ABC}-\triangle \mathrm{AIF}-\triangle \mathrm{FGC}=\triangle \mathrm{ADC}-\triangle \mathrm{AEF}-\triangle \mathrm{FHC} \\

\therefore Parallelogram \mathrm{BIFG}= Parallelogram \mathrm{DEFH}

\therefore \angle \mathrm{BIFG}+\angle \mathrm{AIFE}=\angle \mathrm{DEFH}+\angle \mathrm{AIFE}

\therefore \angle \mathrm{EGBA}=\angle \text { DHIA }

In \triangle \mathrm{ADF} and \angle \mathrm{DHIA}, \mathrm{AD} is same base and between the same parallel lines\mathrm{AD} and \mathrm{HI}.

\therefore \triangle \mathrm{ADF}=\frac{1}{2} \times \angle \mathrm{DHIA}

In \triangle \mathrm{AEG} and \angle \mathrm{EGBA}, \mathrm{EG} is same base and between the same parallel lines EG and AB.

\therefore \triangle \mathrm{AEG}=\frac{1}{2} x<\mathrm{EGBA} \\

\therefore \triangle \mathrm{AEG}=\triangle \mathrm{AFD} \quad[\because \angle \mathrm{DHIA}=\angle \mathrm{EGBA}]

7. E is any point on side DC of parallelogram ABCD, produced AE meets produced BC at the point F. Joined D, F. Let’s prove that (i) \triangle A D F= \triangle \mathrm{ABE}, (ii) \triangle \mathrm{DEF}=\triangle \mathrm{BEC}.

Solution : Given:

ABCD is a parallelogram and E is a point on DC. AE produced intersects BC produced at F. DF and BE are drawn.

R.T.P : (i) \triangle \mathrm{ADF}=\triangle \mathrm{ABE}

(ii) \triangle \mathrm{DEF}=\triangle \mathrm{BEC}

Proof: \triangle A D F and parallelogram ABCD are on same base AD and between the same parallels AD and BF

\therefore \triangle A D F=\frac{1}{2} \times \triangle A B C D

\therefore \triangle \mathrm{ABE} \ and \ \angle \mathrm{ABCD} are on the same base A B and between the same parallels AB and DC.

\therefore \triangle A B E=\frac{1}{2} \times \angle A B C D

From (i) and (ii) We get.

Now,

\triangle \mathrm{ADE}+\triangle \mathrm{BEC}=\angle \mathrm{ABCD}-\triangle \mathrm{ABE} \\

\triangle \mathrm{ABCD}-\frac{1}{2} \angle \mathrm{ABCD} \\

{\left[\because \triangle \mathrm{ABE}=\frac{1}{2} \angle \mathrm{ABCD}\right] } \\

=\frac{1}{2} \angle \mathrm{ABCD} \\

\therefore \quad \triangle \mathrm{BEC}=\triangle \mathrm{DEF} \\

\therefore \quad \triangle \mathrm{BEC}=\triangle \mathrm{DEF} (ii) (Proved)

8. Two triangles ABC and ABD with equal area and stand on the opposite side of AB, Let’s prove that A B bisects CD.

Solution:

Given: \triangle A B C \ and\ \triangle A B D are equal in area and they are on the opposite sides of common base AB . CD is drawn. Let CD intersects AB at the point O.

R.T.P: AB bisects CD.

Construction: Perpendiculars CP and DQ are drawn on AB from C and D respectively.

Proof: \mathrm{CP} \| \mathrm{DQ}

Now,

\triangle \mathrm{ABC}=\triangle \mathrm{ABD} \\

\text { or, } =\frac{1}{2} \times \mathrm{AB} \times \mathrm{CP}=\frac{1}{2} \mathrm{AB} \times \mathrm{DQ} \\

\therefore \mathrm{CP}=\mathrm{DQ} \\

\because \mathrm{CP} \| \mathrm{DQ} \text { and } \mathrm{CP}=\mathrm{DQ} \\

\therefore \mathrm{CQDP} \text { is parallelogram, diagonals bisect each other. } \\

\therefore \mathrm{QP} \text { bisects } \mathrm{CD} \text {. (Proved) }

\therefore CQDP is parallelogram, diagonals bisect each other.

\therefore \quad QP bisects CD. (Proved)

9. D is mid-point af side BC of triangle ABC, Parallelogram CDEF stands between side BC and parallel to BC through the point A. Let’s prove that \triangle \mathrm{ABC} = parallelogram CDEF.

Solution :

Given: ABC is a triangle in which D is the mid-point of side BC and parallelogram CDEF stands between side BC and parallel to BC through A.

R.T.P :\triangle \mathrm{ABC}= parallelogram \mathrm{CDEF}

Construction: A and D are joined.

Proof: \triangle A D C  \ and \ \angle C D E F are situated on the same base DC and lies between same parallel lines DC and AF.

\therefore \triangle \mathrm{ADC}=\frac{1}{2} \times \angle \mathrm{CDEF}

again,

\because \mathrm{AD} is the median of \triangle \mathrm{ABC}

\therefore \triangle \mathrm{ADC}=\frac{1}{2} \triangle \mathrm{ABC} \\

\therefore \triangle \mathrm{ABC}=\angle \mathrm{CDEF} \quad \text { (Proved) }

10. P is any point on diagonal BD of parallelogram ABCD. Let’s prove that \triangle A P D=\triangle A B P

Solution: Given:

ABCD is a parallelogram. P is any point on diagonal BD.

R.T.P : \triangle \mathrm{APD}=\triangle \mathrm{ABP}

Construction: Perpendiculars are drawn on BD from the point A and C which intersectsBD at E and Frespectively.

Proof: In \triangle \mathrm{ADE} \ and \ \triangle \mathrm{CBF},

<\mathrm{AED}=<\mathrm{CFB} \text { (each are right angle) } \\

Now, \quad \because \triangle A P D and \triangle A B P are situated on same base

BD and altitude

[\because \mathrm{AE}=\mathrm{CF}]

\therefore \triangle \mathrm{APD}=\triangle \mathrm{ABP} (Proved)

11. AD and BE are the medians of triangle ABC. Let’s prove that \triangle \mathrm{ACD}=\triangle \mathrm{BCE}

Solution : Given:

ABC is a triangle in which AD and BE are the medians of \triangle \mathrm{ABC} which meets at O.

R.T.P: \triangle A C D=\triangle B C E

Construction: Join D, E such that D E \| A B

Proof: \because B E is the median of \triangle A B C

\therefore \quad \triangle \mathrm{BCE}=\frac{1}{2} \times \triangle \mathrm{ABC}

Again,

\because \quad C F is median

\text { of } \triangle A B C

\therefore \triangle A C D=\frac{1}{2} \times \triangle A B C

From, (i) & (ii)

\therefore \triangle A C D=\triangle B C E(Proved)

12. A line parallel to BC of triangle ABC intersects sides AB and AC at the points P and Q respectively. CQ and BQ intersect each other at the point X. Let’s prove that

(i) \triangle \mathrm{BPQ}=\triangle \mathrm{CPQ}

(ii) \triangle B C P=\triangle B C Q

(iii) \triangle \mathrm{ACP}=\triangle \mathrm{ABQ}

(iv) \triangle \mathrm{BXP}=\triangle \mathrm{CXQ}

Solution: Given:

ABC is a triangle and PQ parallel to BC which intersects AB and AC at P and Q respectively. CP and BQ intersect each other at X.

R.T.P : (i) \triangle \mathrm{BPQ}=\triangle \mathrm{CPQ}

(ii) \triangle B C P=\triangle B C Q

(iii) \triangle \mathrm{ACP}=\triangle \mathrm{ABQ}

(iv) \triangle \mathrm{BXP}=\triangle \mathrm{CXQ}

Proof: \triangle B P Q \ and \ \triangle C P Q are on the same base PQ and between the same parallels PQ and BC.

\therefore \triangle \mathrm{BPQ}=\triangle \mathrm{CPQ}………(i) (Proved)

\therefore \quad \triangle B C P and \triangle B C Q are on the same base BC and between the same parallels BC and PQ.

\therefore \triangle B C P=\triangle B C Q \\

\text { Now, } \triangle A B Q=\triangle B P Q+\triangle A P Q \\

=\triangle C P Q+\triangle A P Q[\because \triangle B A Q=\triangle C P Q]……….(ii) (Proved)

=\triangle \mathrm{ACP} \\

\therefore \triangle \mathrm{BPQ}=\triangle \mathrm{CPQ} .............\text { (iii) (Proved) } \\

\therefore \triangle \mathrm{BXP}+\triangle \mathrm{PXQ}=\triangle \mathrm{CXQ}+\triangle \mathrm{PXQ} \\

\therefore \triangle \mathrm{BXP}=\triangle \mathrm{CXQ}…………. (iv)(Proved).

13. D is the mid point of BC of triangle ABC and P is any point on BC. Join P, A – through the point D a straight line parallel to line segment PA meet AB at point Q. Let’s prove that,

(ii) \triangle \mathrm{BPQ}=\frac{1}{2} \triangle \triangle \mathrm{ABC}

Solution: Given:

ABC is a triangle in which D is the mid-point of BC and P is any point on BC. Join P, A; A, D and P, Q and A P \| Q D

(ii) \triangle \mathrm{BPQ}=\frac{1}{2} \triangle \mathrm{ABC}

Proof: \triangle A D Q \ and \ \triangle P D Q are on the same base QD and between the same parallel line QD and AP.

\therefore \triangle \mathrm{ADQ}=\triangle \mathrm{PDQ} ………(i) (Proved)

We have,

\triangle \mathrm{BPQ} =\triangle \mathrm{BDQ}+\triangle \mathrm{PDQ} \\

=\triangle \mathrm{ABD} \\

\therefore \triangle \mathrm{BPQ} =\triangle \mathrm{ABD}=\frac{1}{2} \triangle \mathrm{ABC} \\

\therefore \triangle \mathrm{BPQ} =\frac{1}{2} \triangle \mathrm{ABC} \quad \text { (Proved) }

14. In triangle ABC of which AB = AC; perpendiculars through the points B and C on sides AC and A B intersed sides AC and AB at the points E and F. Let’s prove that FE \|BC.

Solution: Given: ABC is a triangle such that AB = AC, B and CFare perpendiculars on AC and AB respectively. FE is drawn.

R.T.P : F E \| B C

Proof : In \triangle \mathrm{ABC}, \mathrm{AB}=\mathrm{AC} \\

\text { i.e, } \quad \angle \mathrm{ECB}=\angle \mathrm{FBC}

Now, In \triangle B E C \ and \ \triangle B F C

<\mathrm{ECB}=\angle \mathrm{FBC}

\angle \mathrm{BEC}=\angle \mathrm{BFC} (\text{each is equal to} \ 90^{\circ} ]

BC is common

\therefore \triangle B B E C \cong \triangle B F C \\

\therefore \triangle B E C=\triangle B F C

Since \triangle B E C \ and \ \triangle B F C are equal and they are on the same base BC. Hence they must be between the same parallels.

\therefore \mathrm{FE} \| B C (Proved)

15. In triangle ABC < ABC = \angle A C B; bisectors of an angle \angle A B C \ and \ \angle A C B intersect the side AC and AB at the points E and F respectively. Let’s prove that, F E \| B C.

Solution :

Given: ABC is a triangle such that \angle \mathrm{ABC}=\angle A C B i.e, AB = AC, BE and CF are the bisector of \angle A B C \ and \ \angle A C B respectively.

R.T.P: F E \| B C

Proof: In \mathrm{ABC}, \angle \mathrm{ACB}=\angle \mathrm{ABC} i.e; \angle \mathrm{ECB}=\angle \mathrm{FBC}

\therefore AB = AC

Now, In \triangle B E C \ and \ \triangle B F C

<\mathrm{ECB}=\angle \mathrm{FBC} \quad \text { [Given] } \\

<\mathrm{EBC}=\angle \mathrm{FCB} \quad \text { [Given] }

BC is common

\therefore \quad \triangle B E C \cong \triangle B F C \\

\therefore \triangle B E C=\triangle B F C \\

Since, \triangle B E C are equal and they are on the same base BC.

Hence, they must be between the same parallels.

\therefore \mathrm{FE} \| \mathrm{BC} (Proved)

16. The shape of two parallelograms ABCD and AEFG of which \angle A is common are equal in area and E lies on AB, Let’s prove that DE \| FC.

Solution : Given: ABCD and AEFG be two parallelograms of which \angle A is common. DE & FC are drawn such that \angle \mathrm{OEBC}=\angle \mathrm{ODGF}

R.T.P : DE \| F C

Proof: In \triangle D O F \ and \ \angle \mathrm{ODFG} are on the same base OD and between the same parallel lines OD and GF.

\therefore \triangle \mathrm{ODF}=\frac{1}{2} \times<\mathrm{ODGF}

In \triangle \mathrm{COE} \ and \ \angle \mathrm{OEBC} are one the same base OE and between the same parallel lines OE and BC

\therefore \triangle \mathrm{OEC}=\frac{1}{2}<\mathrm{OEBC}=\frac{1}{2}<\mathrm{ODGF} \\

\therefore \triangle \mathrm{ODF}=\triangle \mathrm{OEC} \\

\because \triangle \mathrm{ODF} \text { and } \triangle \mathrm{OEC} \text { are equal, }

\because \triangle \mathrm{ODF} and \triangle \mathrm{OEC} are equal, Hence, they must be between the parallel lines.

\therefore \mathrm{DE} \| \mathrm{FC} \quad (Proved)

17. ABCD is a parallelogram and ABCE is a quadrilateral. Diagonal AC divides the quadrilateral field ABCE into two equal parts. Let’s prove that AC \| DE.

Solution :

Given: ABCD is a parallelogram and ABCE is a quadrilateral. The diagonal AC bisects the quadrilateral, i.e. \triangle \mathrm{ABC}=\triangle \mathrm{AEC}. \mathrm{DE} is drawn.

R.T.P : AC \| DE

Proof : AC is a diagonal of the parallel-ogram ABCD.

\text { But, } \triangle \mathrm{ABC}=\triangle \mathrm{ABC} \\

\because \quad \triangle \mathrm{ADC} \text { and } \triangle \mathrm{AEC} \text { are equal and they } are on the same base AC.

18. D is the mid-point of side BC of triangle ABC ; P and Q lie on sides BC and BA in such a way that \triangle B P Q=1 \triangle A B C. Let’s prove that, D Q \| P A

Solution :

Given: ABC is a triangle in which D is the mid-point of BC. P and Q are points on the sides BC and BA respectively such that \triangle B P Q=\triangle A B C. DQ and PA are drawn.

R.T.P : D Q \| P A

Proof : D is the mid-point of BC.

\therefore \triangle A B D=\frac{1}{2} \triangle A B C

But \triangle \mathrm{BPQ}=\frac{1}{2} \triangle \mathrm{ABC}

\therefore \triangle A B D=\triangle B P Q

or, \triangle \mathrm{BQD}+\triangle \mathrm{AQD}=\triangle \mathrm{BQD}+\triangle \mathrm{PQD}

\therefore \triangle \mathrm{AQD}=\triangle \mathrm{BPQ}

Since \triangle A Q D \ and \ \triangle P Q D are equal and they are on the same base DQ. Hence, they must be between same parallels.

\therefore D Q \| P A \quad (Proved)

19. Parallelogram ABCD of which mid-points are E, F, G and H of sides AB, BC, CD and D A respectively. Let’s prove that

(i) EFGH is a parallelogram

(ii) Area of the shape of parallelogram EFGH is half of area of the shape of parallelogram ABCD.

Solution : (i) Given:

ABCD is a parallelogram of which mid-points are E, F, G & H of sides AB, BC, CD and DA respectively. EF, FG, GH. & HE are drawn.

R.T.P : EFGH is a parallelogram.

Construction: Join A and C.

Proof: G and H are the mid-points of CD and DA of \triangle A C D

\therefore \mathrm{GH} \| \mathrm{AC} \ and \ \mathrm{GH}\frac{1}{2} \mathrm{AC}

Again,

E and F are the mid-points of AB & BC of \triangle A B C

\therefore E F \| A C \ and \ E F{ }\frac{1}{2} A C

\because \mathrm{GH} \| \mathrm{AC} \ and \ \mathrm{EF} \| A C

\therefore \mathrm{GH} \| \mathrm{EF} \ and \ \mathrm{GH} \frac{1}{2} \mathrm{AC}=\mathrm{EF}

\therefore \mathrm{GH} \| \mathrm{EF} \ and \ \mathrm{GH}=\mathrm{EF}

Hence, EFGH is a parallelogram. (Proved)

(ii) Given: ABCD is a parallelogram of which E, F, G & H are the mid points of AB, BC, CD & AD respectively such that EFGH is a parallelogram.

Construction: Join E & G

Proof: In \triangle H E G \ and \ \angle A E G D are on the same base GE and between the same parallel lines GE and AD.

\therefore \triangle \mathrm{HEG}=\frac{1}{2}<\mathrm{AEGD}

Again, In EFG and EBCG are on the same base GE and between the same parallel lines GE and BC

\therefore \triangle \mathrm{EFG}=\frac{1}{2}<B E G C

\therefore \triangle \mathrm{HEG}+\triangle \mathrm{EFG}=\frac{1}{2}<\mathrm{AEGD}+=\frac{1}{2}<\mathrm{BEGC} \\

\text { or } \angle \mathrm{EFGH}=\frac{1}{2}(\angle \mathrm{AEGD}+\angle \mathrm{BEGC}) \\

\therefore \angle \mathrm{EFGH}=\frac{1}{2} \angle \mathrm{ABCD} \quad \text { (Proved). }

20. A B \| D C of a trapezium ABCD and E is mid-point of BC. Let’s prove that area of tringular field A E D=1 \times 2 area of the shape of trapezium field ABCD

Solution :

Given: ABCD is a trapezium and E is mid-point of BC. AE and DE are drawn.

R.T.P : Area of triangular field AED =\frac{1}{2} x area of the trapezium field ABCD

Constructuion: We draw EF parallel to AB

Proof : In \triangle E F D and trapezium CEFD are on the same base FE and between the same parallel lines FE and CD.

\therefore \triangle \mathrm{AEF}=\frac{1}{2} \times \text { trapazium CEFD }

Again,

In \triangle \mathrm{AEF} \text{and trapezium} \ \mathrm{ABEF} \ \text{are on the same base}\ \mathrm{EF} and between the same parallel lines FE and AB

\therefore \triangle \mathrm{AEF}=\frac{1}{2} \times \text { trapazium AFEB } \\

\therefore \triangle \mathrm{EFD}+\triangle \mathrm{AEF}=\frac{1}{2} \text {  Trapezium CEFD + Trapezium AFEB) } \\

\therefore \triangle \mathrm{AE} = \frac{1}{2} \times \text { Trapeziun ARCO }

## 21. M.C.Q :

(i) D, E and F are mid-points of sides BC, CA and AB respectively of a triangle ABC. If \triangle A B C=16 \mathrm{sq} . cm; then the area of the shape of trapezium FBCE is

(a) 40 \mathrm{sq} . \mathrm{cm}.

(b) 8 sq. cm.

(c) 12 \mathrm{sq} \cdot \mathrm{cm}

(d) 100 \mathrm{sq} \cdot \mathrm{cm}.

Solution :\because \triangle A B C=16 sq.cm

In the figure, We see that,

Four triangles are formed with equal areas since D, E and F are the mid-points of BC, CA and A B respectively.

\therefore \text { Area } \triangle \mathrm{AFE}=\frac{16}{4} \mathrm{Sq} . \mathrm{cm} .=4 \mathrm{sq} \cdot \mathrm{cm} \text {. }

\therefore Area of trapezium \mathrm{FBCE}=4 \times 3 sq. cm. (Ans.)

\therefore (c) is correct option

(ii) A, B, C, D are the mid-points of sides PQ, QR, RS and SP respectively. of parallelogram PQRS. I area of the shape of parallelogram PQ S = 36 sq. cm. then area of ABCD field is

(a) 24 sq. cm.

(b) 18 sq. cm.

(c) 30 sq. cm.

(d) 36 sq. cm.

Solution : We have,

Area of parallelogram \ABCD

\text { Area of Parallelogram }=\frac{1}{2} \times \text { area of parallelogram PQRS } \\

\text { ABCD } \\

=\frac{1}{2} \times 36 \mathrm{Sq} . \mathrm{cm} \\

= 18 \mathrm{sq} \mathrm{cm}.

\therefore (b) is correct option.

(iii) O is any a point inside parallelogram ABCD.

If \triangle A O B+\triangle C O D=16 sq.cm., then area of the shape of parallelogram ABCD is

(a) 8 sq. cm.

(b) 4 sq. cm.

(c) 32 sq. cm.

(d) 64 sq. cm.

Solution : We draw a parallel line EF Such that A B\|C D\| E F

\therefore \triangle A O B=\frac{1}{2} area of parallelogram ABEF

\therefore \triangle C O D=\frac{1}{2} area of parallelogram CDEF

\therefore area of parallelogram A B C D=2(\triangle A O B+\triangle C O D) =2 \times 16 \mathrm{sq} . \mathrm{cm} .=32 \mathrm{sq} . \mathrm{cm}

\therefore (c) is correct option.

(iv) D is the mid-point of side BC of triangle ABC. E is the mid-point of side BD and O is the mid-point of A E; area of triangular field  BOE is

(a) \frac{1}{3} \times  Area \ of \triangle A B C

(b) 1 \times  Area \ of \triangle A B C

(c) \frac{1}{6} \times Area \ of \triangle A B C

(d) 8 \times \ Area \ of \triangle A B C

Solution :

Area of  \triangle \mathrm{BOE} =\frac{1}{2} \triangle \mathrm{ABE} \\

=\frac{1}{2} \times \frac{1}{2} \triangle \mathrm{ABD} \\

=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \triangle A B C \\

=\frac{1}{8} \triangle \mathrm{ABC}

\therefore(d) is correct option.

(v) A parallelogram, a rectangle and a triangle stand on same base and between same parallel and if their area are P, Q and T respectively then

(a) \mathrm{P}=\mathrm{R}=2 \mathrm{~T}

(b) \mathrm{P}=\mathrm{R}=\frac{\mathrm{T}}{2}

(c) 2P = 2R = T

(d) \mathrm{P}=\mathrm{R}=\mathrm{T}

Solution : We know that,

Area of a triangle =\frac{1}{2} Area of parallelogram.

\therefore area of parallelogram = 2 × area of triangle

\therefore (a) is correct option.

(i) DE is perpendicular on side AB from the point D of parallelogram ABCD and B F is perpendicular on side AD from the point B; If AB = 10 cm, AD = 8 cm and DE = 6 cm, let us write how much length of BF is

Solution : Given,

\mathrm{AB}=\mathrm{CD}=10 \mathrm{~cm} \\

\mathrm{DE}=6 \mathrm{~cm}

Area of \triangle A B D=\frac{1}{2} A B \times D E

\therefore \frac{1}{2} \mathrm{AD} \times \mathrm{BF}=\frac{1}{2} \mathrm{AB} \times \mathrm{DE} \\

\text { or, } 8 \times \mathrm{BF}=10 \times 6 \\

\text { or, } \mathrm{BF}=\frac{10 \times 6}{8} \times \frac{15}{2} \\

\therefore \mathrm{BF}=7.5 \mathrm{~cm} \quad \text { (Ans.) }

(ii) The area of the shape of parallelogram ABCD is 100 sq. units. P is mid-point of side BC, let us write how much area of triangular field ABP is

Solution :

We have,

Area of \triangle ABP

=\frac{1}{2} \text { Area of } \triangle \mathrm{ABC} \\

=\frac{1}{2} \times \frac{1}{2} \text { Parallelogram } \mathrm{ABCD} \\

=\frac{1}{2} \times 100 \text { sq.cm. } \\

= 25 \text { sq. units. (Ans.) }

(iii) AD is the median of triangle ABC and P is any point on side AC in such a way that area of \triangle ADP: area of \triangle A B D=2: 3 Let us write the area of \triangle P D C: area of \triangle \mathrm{ABC}

Solution:

We have,

area of \triangle \mathrm{ADP} \text { : area of } \triangle \mathrm{ABD}=2: 3

\text { or, } \frac{\text { area of } \triangle \mathrm{ADP}}{\text { area of } \triangle \mathrm{ABD}}=\frac{2}{3}

Area of \triangle A B D=\frac{3}{2} Area of \triangle A D P

\therefore \quad D is the median of \triangle ABC

\triangle \mathrm{ABD} =\frac{1}{2} \triangle \mathrm{ABC} \\

\triangle \mathrm{ADC} =\frac{1}{2} \triangle \mathrm{ABC} \\

\therefore \triangle \mathrm{ABD} =\triangle \mathrm{ADC} \\

Now,

or, \triangle \mathrm{ADP}+\triangle \mathrm{PDC}=\frac{1}{2} \triangle \mathrm{ABC}

or, \triangle \mathrm{ADP}+\triangle \mathrm{PDC}=\frac{1}{2} \times 2 \triangle \mathrm{ABD} [\therefore \triangle A B D=\triangle A D C]

\therefore \triangle \mathrm{ADP}=2 \triangle \mathrm{PDC} \\

\therefore \quad \frac{\text { area of } \triangle \mathrm{PDC}}{\text { area of } \triangle \mathrm{ABC}}=\frac{\text { area of } \triangle \mathrm{PDC}}{2 \times \text { area of } \triangle \mathrm{ACD}} \\

=\frac{\text { area of } \triangle \mathrm{PDC}}{2 \times(\triangle \mathrm{ADP}+\triangle \mathrm{PDC})}=\frac{\text { area of } \triangle \mathrm{PDC}}{2 \times(2 \triangle \mathrm{ACD}+\triangle \mathrm{PDC})} \\

=\frac{\text { area of } \triangle \mathrm{PDC}}{2 \times 3 \text { area of } \triangle \mathrm{PDC}}=\frac{1}{6} \\

\therefore \text { Area of PDC : Area of } \mathrm{ABC}=1: 6 \text { (Ans.) }

(iv) ABDE is a parallelogram. F is mid-point of side ED. If area of triangular field ABD is 20sq. unit, then let us write how much area of triangular field AEF is.

Solution : \because \triangle ABD = 20 sq. unit.

\therefore \triangle A B D=\triangle A D E=20 \text { sq. unit. }

\therefore \triangle \mathrm{AEF} =\frac{1}{2} \triangle \mathrm{ADE} \\

=\frac{1}{2} \times 20 \text { sq.unit } \\

=10 sq. unit. (Ans.)

(v) PQRS is a parallelogram X and Y are the mid-points of side PQ and SR respectively. Joint diagonal SQ, Let us write the area of the shape of parallelogram field XQRY: area of triangular field QSER

Solution:

\because Parallelogram XQRY and Parallelogram PQRS are on the same base PQ and between the same parallel lines PQ and SR.

\therefore Parallelogram XQRY =\frac{1}{2} \times Parallelogram PQRS………(i)

Again,

\triangle QSR and Parallelogram PQRS are on the same base PQ and between the same parallel lines PQ and SR.

\therefore \triangle QRS=\frac{1}{2} Parallelogram PQRS……. (ii)

\therefore from (i) and (ii)

Parallelogram XQRY = \triangle Q S R

\therefore Parallelogram XQRY: \triangle Q S R=1: 1 (Ans)

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