Area Theorems — Exercise 15

In the given figure, BD : DC = 3 : 5, then area of △ ABD : area of △ ACD is :

1. 5 : 3

2. 3 : 5

3. 25 : 9

4. 9 : 25

Answer

In the given figure,

△ ABD and △ ACD has same vertex A and bases along the same straight line BC.

∴ $\dfrac{\text{Area of △ ABD}}{\text{Area of △ ACD}} = \dfrac{BD}{DC} = \dfrac{3}{5}$

∴ area of △ ABD : area of △ ACD = 3 : 5.

Hence, Option 2 is the correct option.

A median of a triangle divides it into two :

1. triangles of equal areas

2. congruent triangles

3. triangles of areas in the ratio 2 : 1

4. right triangles

Answer

A median of a triangle divides it into two triangles of equal areas.

Hence, Option 1 is the correct option.

In the given figure, AB is parallel to DC and AB ≠ DC, the area of △ AOD is equal to area of triangle :

1. AOB

2. COD

3. ACB

4. BOC

Answer

We know that,

The area of triangles lying on same base and between same parallel lines are equal.

From figure,

△ ADC and △ BCD lie on same base CD and between same parallel lines AB and DC.

∴ Area of △ ADC = Area of △ BCD

Subtracting area of △ COD from both sides, we get :

⇒ Area of △ ADC - Area of △ COD = Area of △ BCD - Area of △ COD

⇒ Area of △ AOD = Area of △ BOC.

Hence, option 4 is the correct option.

In the given figure, AF // BE and PQ // RS, FC and ED are perpendiculars to RS. The area of parallelogram ABEF is equal to :

1. rect. CDEF

2. quad. CBEF

3. 2 × △ACF

4. 2 × △EBD

Answer

We know that,

Parallelogram on equal bases and between the same parallel are equal in area.

A rectangle is also a parallelogram.

From figure,

Parallelogram ABEF and rectangle CDEF lie on same base FE and between same parallel lines PQ and RS.

∴ Area of parallelogram ABEF = Area of rectangle CDEF.

Hence, Option 1 is the correct option.

In the given figure, D is mid-point of side BC, the area of triangle BEA is equal to area of triangle :

1. BED

2. CED

3. CEA

4. ACD

Answer

In △ ABC,

Since, D is the mid-point of side BC.

∴ AD is the median of BC.

∴ Area of △ ABD = Area of △ ACD (As, median of a triangle divides it into two triangles of equal areas.) ...........(1)

In △ EBC,

Since, D is the mid-point of side BC.

∴ ED is the median of BC.

∴ Area of △ EBD = Area of △ ECD (As, median of a triangle divides it into two triangles of equal areas.) .........(2)

Subtracting equation (2) from (1), we get :

⇒ Area of △ ABD - Area of △ EBD = Area of △ ACD - Area of △ ECD

⇒ Area of △ BEA = Area of △ CEA.

Hence, Option 3 is the correct option.

In the given figure, if area of triangle ADE is 60 cm²; state, giving reason, the area of :

(i) parallelogram ABED;

(ii) rectangle ABCF;

(iii) triangle ABE.

Answer

(i) We know that,

Area of triangle is half that of a parallelogram, on the same base and between the same parallels.

From figure,

△ ADE and parallelogram ABED lies on same base DE and between same parallel lines AB and DE.

∴ Area of △ ADE = $\dfrac{1}{2}$ Area of parallelogram ABED

⇒ Area of parallelogram ABED = 2 × Area of △ ADE

⇒ Area of parallelogram ABED = 2 × 60 = 120 cm².

Hence, area of parallelogram ABED = 120 cm².

(ii) We know that,

Area of parallelogram is equal to area of a rectangle on the same base and between same parallel lines.

From figure,

Parallelogram ABED and rectangle ABCF lie on lie on same base AB and between same parallel lines AB and CD.

∴ Area of rectangle ABCF = Area of parallelogram ABED = 120 cm².

Hence, area of rectangle ABCF = 120 cm².

(iii) We know that,

Area of a triangle is half that of a parallelogram on the same base and between the same parallels.

∴ Area of triangle ABE = $\dfrac{1}{2}$ Area of parallelogram ABED

= $\dfrac{1}{2} \times 120$ = 60 cm².

Hence, area of triangle ABE = 60 cm².

The given figure shows a rectangle ABDC and a parallelogram ABEF; drawn on opposite sides of AB. Prove that :

(i) quadrilateral CDEF is a parallelogram

(ii) Area of quad. CDEF = Area of rect. ABDC + Area of // gm ABEF

Answer

(i) Given,

ABCD is a rectangle.

∴ AB = CD (Opposite sides of rectangle are equal) ...............(1)

Given,

ABEF is a rectangle.

∴ AB = FE (Opposite sides of parallelogram are equal) ...............(2)

From equations (1) and (2), we get :

⇒ CD = FE.

From figure,

CD || FE

Since, one pair of opposite sides of quadrilateral CDEF are equal and parallel.

∴ CDEF is a parallelogram.

Hence, proved that CDEF is a parallelogram.

(ii) In △ AFC and △ BED,

⇒ AF = BE (Opposite sides of parallelogram ABEF)

⇒ AC = BD (Opposite sides of rectangle ABCD)

⇒ CF = ED (Opposite sides of parallelogram CDEF)

∴ △ AFC ≅ △ BED (By S.S.S. axiom)

We know that,

Area of congruent triangle are equal.

∴ Area of △ AFC = Area of △ BED ..........(1)

From figure,

⇒ Area of quadrilateral CDEF = Area of △ ACF + Area of ACDPEF

⇒ Area of quadrilateral CDEF = Area of △ BED + Area of ACDPEF [From equation (1)]

⇒ Area of quadrilateral CDEF = Area of CDBEPA

⇒ Area of quadrilateral CDEF = Area of rect. ABDC + Area of // gm ABEF.

Hence, proved that Area of quadrilateral CDEF = Area of rect. ABDC + Area of // gm ABEF.

In the given figure, diagonals PR and QS of the parallelogram PQRS intersect at point O and LM is parallel to PS. Show that :

(i) 2 Area (△ POS) = Area (// gm PMLS)

(ii) Area (△ POS) + Area (△ QOR) = $\dfrac{1}{2}$ Area (//gm PQRS)

(iii) Area (△ POS) + Area (△ QOR) = Area (△ POQ) + Area (△ SOR)

Answer

(i) We know that,

Area of a triangle is half that of a parallelogram on the same base and between the same parallels.

From figure,

Triangle POS and parallelogram PMLS lie on same base PS and between same parallel lines PS and ML.

∴ Area of triangle POS = $\dfrac{1}{2}$ Area of parallelogram PMLS

⇒ Area of parallelogram PMLS = 2 (Area of triangle POS)

Hence, proved that 2 area (△ POS) = area (// gm PMLS).

(ii) We know that,

Area of a triangle is half that of a parallelogram on the same base and between the same parallels.

From figure,

Triangle POS and parallelogram PMLS lie on same base PS and between same parallel lines PS and ML.

∴ Area of triangle POS = $\dfrac{1}{2}$ Area of parallelogram PMLS ..........(1)

Triangle QOR and parallelogram MQRL lie on same base QR and between same parallel lines QR and ML.

∴ Area of triangle QOR = $\dfrac{1}{2}$ Area of parallelogram MQRL ..........(2)

Adding equations (1) and (2), we get :

⇒ Area of triangle POS + Area of triangle QOR = $\dfrac{1}{2}$ Area of parallelogram PMLS + $\dfrac{1}{2}$ Area of parallelogram MQRL

⇒ Area of triangle POS + Area of triangle QOR = $\dfrac{1}{2}$ (Area of parallelogram PMLS + Area of parallelogram MQRL)

⇒ Area of triangle POS + Area of triangle QOR = $\dfrac{1}{2}$ Area of parallelogram PQRS.

Hence, proved that area (△ POS) + area (△ QOR) = $\dfrac{1}{2}$ area (//gm PQRS).

(iii) We know that,

In a parallelogram diagonals bisect each other.

∴ OS = OQ

In △ PQS,

O is the mid-point of QS.

∴ OP is the median.

We know that,

Median of a triangle divides it into two triangles of equal areas.

∴ Area of △ POS = Area of △ POQ .........(3)

In △ QSR,

O is the mid-point of QS.

∴ OR is the median.

We know that,

Median of a triangle divides it into two triangles of equal areas.

∴ Area of △ QOR = Area of △ SOR .........(4)

Adding equations (3) and (4), we get :

⇒ Area of △ POS + Area of △ QOR = Area of △ POQ + Area of △ SOR.

Hence, proved that area (△ POS) + area (△ QOR) = area (△ POQ) + area (△ SOR).

In parallelogram ABCD, P is a point on side AB and Q is a point on side BC. Prove that :

(i) △ CPD and △ AQD are equal in area.

(ii) Area (△ AQD) = Area (△ APD) + Area (△ CPB)

Answer

We know that,

Area of a triangle is half that of a parallelogram on the same base and between the same parallels.

(i) From figure,

△ CPD and || gm ABCD are on the same base CD and between the same parallel lines AB and CD.

∴ Area of triangle CPD = $\dfrac{1}{2}$ Area of parallelogram ABCD ........(1)

△ AQD and || gm ABCD are on the same base AD and between the same parallel lines AD and BC.

∴ Area of triangle AQD = $\dfrac{1}{2}$ Area of parallelogram ABCD ........(2)

From equations (1) and (2), we get :

Area of triangle CPD = Area of triangle AQD.

Hence, proved that area of triangle CPD = area of triangle AQD.

(ii) From part (i), we get :

⇒ Area of △ CPD = $\dfrac{1}{2}$ Area of parallelogram ABCD

∴ Area of || gm ABCD - Area of △ CPD = $\dfrac{1}{2}$ Area of parallelogram ABCD ........(3)

From figure,

⇒ Area of || gm ABCD - Area of △ CPD = Area of △ APD + Area of △ CPB ............(4)

From equations (3) and (4), we get :

⇒ Area of △ APD + Area of △ CPB = $\dfrac{1}{2}$ Area of || gm ABCD ........(5)

Since,

⇒ Area of △ ADQ = $\dfrac{1}{2}$ Area of || gm ABCD [From eqn. 2]

Substituting value of $\dfrac{1}{2}$ Area of || gm ABCD from above equation in equation (5), we get :

⇒ Area of △ APD + Area of △ CPB = Area of △ ADQ.

Hence, proved that area (△ AQD) = area (△ APD) + area (△ CPB).

In the given figure, M and N are the mid-points of the sides DC and AB respectively of the parallelogram ABCD.

If the area of parallelogram ABCD is 48 cm²;

(i) state the area of the triangle BEC.

(ii) name the parallelogram which is equal in area to the triangle BEC.

Answer

(i) We know that,

Area of a triangle is half that of a parallelogram on the same base and between the same parallels.

Since, triangle BEC and parallelogram ABCD are on the same base BC and between the same parallels AE and BC.

∴ Area of △ BEC = $\dfrac{1}{2}$ Area of || gm ABCD = $\dfrac{1}{2} \times 48$ = 24 cm².

Hence, the area of the triangle BEC = 24 cm².

(ii) Since, M and N are the mid-points of the sides DC and AB respectively.

∴ Area of || gm ANMD = Area of || gm NBCM = $\dfrac{1}{2}$ Area of || gm ABCD .........(1)

From part (i), we get :

Area of △ BEC = $\dfrac{1}{2}$ Area of || gm ABCD ........(2)

From equation (1) and (2), we get :

Area of || gm ANMD = Area of || gm NBCM = Area of △ BEC.

Hence, parallelograms ANMD and NBCM are equal in area to triangle BEC.

In the following figure, CE is drawn parallel to diagonal DB of the quadrilateral ABCD which meets AB produced at point E.

Prove that △ ADE and quadrilateral ABCD are equal in area.

Answer

We know that,

Triangles on the same base and between the same parallel lines are equal in area.

△ BDE and △ BDC lie on the same base BD and along the same parallel lines DB and CE.

∴ Area of △ BDE = Area of △ BDC ..........(1)

From figure,

⇒ Area of △ ADE = Area of △ ADB + Area of △ BDE

⇒ Area of △ ADE = Area of △ ADB + Area of △ BDC [From equation (1)]

⇒ Area of △ ADE = Area of quadrilateral ABCD.

Hence, proved that △ ADE and quadrilateral ABCD are equal in area.

ABCD is a parallelogram, a line through A cuts DC at point P and BC produced at Q. Prove that triangle BCP is equal in area to triangle DPQ.

Answer

We know that,

Area of a triangle is half that of a parallelogram on the same base and between the same parallels.

Since, triangle APB and parallelogram ABCD are on the same base AB and between the same parallels AB and DC.

∴ Area of △ APB = $\dfrac{1}{2}$ Area of || gm ABCD ........(1)

Since, triangle ADQ and parallelogram ABCD are on the same base AD and between the same parallels AD and BQ.

∴ Area of △ ADQ = $\dfrac{1}{2}$ Area of || gm ABCD ........(2)

Adding equations (1) and (2), we get :

⇒ Area of △ APB + Area of △ ADQ = $\dfrac{1}{2}$ Area of || gm ABCD + $\dfrac{1}{2}$ Area of || gm ABCD

⇒ Area of △ APB + Area of △ ADQ = Area of || gm ABCD ...........(3)

From figure,

⇒ Area of △ APB + Area of △ ADQ = Area of quadrilateral ADQB - Area of △ BPQ ..............(4)

From equation (3) and (4), we get :

⇒ Area of quadrilateral ADQB - Area of △ BPQ = Area of || gm ABCD

⇒ Area of quadrilateral ADQB - Area of △ BPQ = Area of quadrilateral ADQB - Area of △ DCQ

⇒ Area of △ BPQ = Area of △ DCQ

⇒ Area of △ BPQ - Area of △ PCQ = Area of △ DCQ - Area of △ PCQ

⇒ Area of △ BCP = Area of △ DPQ.

Hence, proved that area of △ BCP = area of △ DPQ.

The given figure shows a pentagon ABCDE. EG drawn parallel to DA meets BA produced at G and CF drawn parallel to DB meets AB produced at F. Prove that the area of pentagon ABCDE is equal to the area of triangle GDF.

Answer

We know that,

Triangles on the same base and between the same parallel lines are equal in area.

Since, triangle EDG and EGA lie on the same base EG and between the same parallel lines EG and DA.

∴ Area of △ EDG = Area of △ EGA

Subtracting △ EOG from both sides, we get :

⇒ Area of △ EDG - Area of △ EOG = Area of △ EGA - Area of △ EOG

⇒ Area of △ EOD = Area of △ GOA .........(1)

Since, triangle FDC and FBC lie on the same base FC and between the same parallel lines DB and CF.

∴ Area of △ FDC = Area of △ FBC

Subtracting △ FPC from both sides, we get :

⇒ Area of △ FDC - Area of △ FPC = Area of △ FBC - Area of △ FPC

⇒ Area of △ DPC = Area of △ BPF .........(2)

From figure,

⇒ Area of △ GDF = Area of △ GOA + Area of △ BPF + Area of pentagon ABPDO

⇒ Area of △ GDF = Area of △ EOD + Area of △ DPC + Area of pentagon ABPDO

⇒ Area of △ GDF = Area of pentagon ABCDE.

Hence, proved that area of pentagon ABCDE is equal to the area of triangle GDF.

In the given figure, AP is parallel to BC, BP is parallel to CQ. Prove that the areas of triangles ABC and BQP are equal.

Answer

We know that,

Triangles on the same base and between the same parallel lines are equal in area.

Since, triangle ABC and BPC lie on the same base BC and between the same parallel lines AP and BC.

∴ Area of △ ABC = Area of △ BPC .........(1)

Since, triangle BPC and BQP lie on the same base BP and between the same parallel lines BP and CQ.

∴ Area of △ BPC = Area of △ BQP .........(2)

From equations (1) and (2), we get :

Area of △ ABC = Area of △ BQP.

Hence, proved that area of △ ABC = area of △ BQP.

In the figure given alongside, squares ABDE and AFGC are drawn on the side AB and the hypotenuse AC of the right triangle ABC.

If BH is perpendicular to FG, prove that :

(i) △ EAC ≅ △ BAF

(ii) Area of the square ABDE = Area of the rectangle ARHF.

Answer

(i) From figure,

⇒ ∠EAC = ∠EAB + ∠BAC

⇒ ∠EAC = 90° + ∠BAC (As, ABDE is a square and each angle of square equal to 90°) ........(1)

Also,

⇒ ∠BAF = ∠FAC + ∠BAC

⇒ ∠BAF = 90° + ∠BAC (As, AFGC is a square and each angle of square equal to 90°) ............(2)

From equation (1) and (2),

⇒ ∠EAC = ∠BAF

In △ EAC and △ BAF,

⇒ EA = BA (Sides of square ABDE)

⇒ ∠EAC = ∠BAF (Proved above)

⇒ AC = AF (Sides of square AFGC)

∴ △ EAC ≅ △ BAF (By S.A.S. axiom)

Hence, proved that △ EAC ≅ △ BAF.

(ii) From figure,

ABC is a right angled triangle.

⇒ AC² = AB² + BC² [By pythagoras theorem]

⇒ AB² = AC² - BC²

⇒ AB² = (AR + RC)² - (BR² + RC²)

⇒ AB² = AR² + RC² + 2.AR.RC - BR² - RC²

⇒ AB² = AR² + RC² + 2.AR.RC - (AB² - AR²) - RC² [Using pythagoras theorem in △ ABR]

⇒ AB² = AR² + RC² + 2.AR.RC - AB² + AR² - RC²

⇒ AB² + AB² = AR² + AR² + RC² - RC² + 2.AR.RC

⇒ 2AB² = 2AR² + 2.AR.RC

⇒ 2AB² = 2AR(AR + RC)

⇒ AB² = AR(AR + RC)

⇒ AB² = AR.AC

⇒ AB² = AR.AF (As, AC = AF, sides of same sqaure)

∴ Area of square ABDE = Area of rectangle ARFH.

Hence, proved that area of square ABDE = area of rectangle ARFH.

In the following figure, DE is parallel to BC. Show that :

(i) Area (△ ADC) = Area (△ AEB)

(ii) Area (△ BOD) = Area (△ COE)

Answer

We know that,

Triangles on the same base and between the same parallel lines are equal in area.

Since, triangle DEB and DEC lie on the same base DE and between the same parallel lines DE and BC.

∴ Area of △ DEC = Area of △ DEB .........(1)

(i) Adding area of △ ADE in both sides of the equation (1), we get :

⇒ Area of △ DEC + Area of △ ADE = Area of △ DEB + Area of △ ADE

⇒ Area of △ ADC = Area of △ AEB.

Hence, proved that area of △ ADC = area of △ AEB.

(ii) Subtracting area of △ DOE in both sides of the equation (1), we get :

⇒ Area of △ DEC - Area of △ DOE = Area of △ DEB - Area of △ DOE

⇒ Area of △ COE = Area of △ BOD.

Hence, proved that area of △ BOD = area of △ COE.

Show that :

(i) a diagonal divides a parallelogram into two triangles of equal area.

(ii) the ratio of the areas of two triangles of the same height is equal to the ratio of their bases.

(iii) the ratio of the areas of two triangles on the same base is equal to the ratio of their heights.

Answer

(i) Suppose ABCD is the parallelogram and diagonal AC divides it into two triangles ABC and ADC.

In △ ABC and △ ADC,

⇒ AB = CD (Opposite sides of parallelogram are equal)

⇒ AD = BC (Opposite sides of parallelogram are equal)

⇒ AC = AC (Common side)

∴ △ ABC ≅ △ ADC (By S.S.S. axiom)

We know that,

Area of congruent triangle are equal.

∴ Area of △ ABC = Area of △ ADC.

Hence, proved that a diagonal divides a parallelogram into triangles of equal area.

(ii) From figure,

AQ and PQ are the altitude of triangle ABC and PRZ respectively wheres bases of both the triangles are equal i.e. BC = RZ = k (let).

We know that,

⇒ Area of triangle = $\dfrac{1}{2}$ × base × height

⇒ Area of △ ABC = $\dfrac{1}{2}$ × BC × AQ

⇒ Area of △ ABC = $\dfrac{1}{2} \times k \times AQ$ ........(1)

⇒ Area of △ PRZ = $\dfrac{1}{2}$ × RZ × PQ

⇒ Area of △ PRZ = $\dfrac{1}{2} \times k \times PQ$ ........(2)

Dividing equation (2) by (1), we get :

$\Rightarrow \dfrac{\text{Area of △ PRZ}}{\text{Area of △ ABC}} = \dfrac{\dfrac{1}{2} × k × PQ}{\dfrac{1}{2} × k × AQ} = \dfrac{PQ}{AQ}$.

Hence, proved that the ratio of the areas of two triangles of the same height is equal to the ratio of their bases.

(iii) We know that,

⇒ Area of triangle = $\dfrac{1}{2}$ × base × height

From figure,

⇒ Area of △ ABC = $\dfrac{1}{2}$ × AC × BM ........(1)

⇒ Area of △ ADC = $\dfrac{1}{2}$ × AC × DN ........(2)

Dividing equation (2) by (1), we get :

$\Rightarrow \dfrac{\text{Area of △ ADC}}{\text{Area of △ ABC}} = \dfrac{\dfrac{1}{2} × AC × DN}{\dfrac{1}{2} × AC × BM} = \dfrac{DN}{BM}$.

Hence, proved that the ratio of the areas of two triangles on the same base is equal to the ratio of their heights.

In the given figure; AD is median of △ ABC and E is any point on median AD. Prove that Area (△ ABE) = Area (△ ACE).

Answer

We know that,

Median of a triangle divides it into two triangles of equal area.

Given,

AD is the median of △ ABC.

∴ Area of △ ABD = Area of △ ACD ..........(1)

Since, E is a point on median AD.

∴ ED is median of △ EBC.

∴ Area of △ EBD = Area of △ ECD ..........(2)

Subtracting equation (2) from (1), we get :

⇒ Area of △ ABD - Area of △ EBD = Area of △ ACD - Area of △ ECD

⇒ Area of △ ABE = Area of △ ACE.

Hence, proved that area (△ ABE) = area (△ ACE).

In the figure of question 14, if E is the mid point of median AD, then prove that :

Area (△ ABE) = $\dfrac{1}{4}$ Area (△ ABC).

Answer

We know that,

Median of a triangle divides it into two triangles of equal area.

From figure,

AD is the median of Δ ABC, so it will divide Δ ABC into two equal triangles.

∴ Area of Δ ABD = Area of Δ ADC = $\dfrac{1}{2}$ Area of Δ ABC

⇒ Area of Δ ABD = $\dfrac{1}{2}$ Area of Δ ABC .........(1)

In Δ ABD,

Since, E is mid-point of AD,

∴ BE is the median.

∴ BE will divide Δ ABD into two equal triangles.

∴ Area of Δ ABE = Area of Δ BED = $\dfrac{1}{2}$ Area of Δ ABD

⇒ Area of Δ ABE = $\dfrac{1}{2}$ Area of Δ ABD .........(2)

Substituting value of Area of Δ ABD from equation (1) in (2), we get :

⇒ Area of Δ ABE = $\dfrac{1}{2} \times \dfrac{1}{2}$ Area of Δ ABC

⇒ Area of Δ ABE = $\dfrac{1}{4}$ Area of Δ ABC.

Hence, proved that area of Δ ABE = $\dfrac{1}{4}$ area of Δ ABC.

ABCD is a parallelogram. P and Q are the mid-points of sides AB and AD respectively. Prove that area of triangle APQ = $\dfrac{1}{8}$ of the area of parallelogram ABCD.

Answer

From figure,

ABCD is the parallelogram and BD is the diagonal of the parallelogram.

∴ BD divides || gm ABCD into two triangles of equal area.

∴ Area of Δ ABD = Area of Δ DBC = $\dfrac{1}{2}$ Area of || gm ABCD ..........(1)

Since, P is the mid-point of AB,

∴ DP is the median of Δ ABD.

∴ Area of Δ APD = Area of Δ DPB = $\dfrac{1}{2}$ Area of Δ ABD (Median divides triangle into two triangles of equal area)

⇒ Area of Δ APD = $\dfrac{1}{2}$ Area of Δ ABD

⇒ Area of Δ APD = $\dfrac{1}{2} \times \dfrac{1}{2}$ Area of || gm ABCD [From equation (1)]

⇒ Area of Δ APD = $\dfrac{1}{4}$ Area of || gm ABCD .......(2)

In Δ APD,

Q is the mid-point of AD.

∴ PQ is the median.

∴ Area of Δ APQ = Area of Δ DPQ = $\dfrac{1}{2}$ Area of Δ APD

⇒ Area of Δ APQ = $\dfrac{1}{2}$ Area of Δ APD

⇒ Area of Δ APQ = $\dfrac{1}{2} \times \dfrac{1}{4}$ Area of || gm ABCD [From equation (2)]

⇒ Area of Δ APQ = $\dfrac{1}{8}$ Area of || gm ABCD.

Hence, proved that area of Δ APQ = $\dfrac{1}{8}$ area of || gm ABCD.

The base BC of triangle ABC is divided at D so that BD = $\dfrac{1}{2}$ DC.

Prove that the area of Δ ABD = $\dfrac{1}{3}$ of the area of Δ ABC.

Answer

In △ ABC,

⇒ BD = $\dfrac{1}{2}$ DC

⇒ $\dfrac{BD}{DC} = \dfrac{1}{2}$

We know that,

Ratio of the area of triangles with same vertex and bases along the same line is equal to the ratio of their respective bases.

$\Rightarrow \dfrac{\text{Area of Δ ABD}}{\text{Area of Δ ADC}} = \dfrac{BD}{DC} \\[1em] \Rightarrow \dfrac{\text{Area of Δ ABD}}{\text{Area of Δ ADC}} = \dfrac{1}{2} \\[1em] \Rightarrow \text{Area of Δ ADC} = \text{2 Area of Δ ABD}.$

From figure,

⇒ Area of Δ ABC = Area of Δ ABD + Area of Δ ADC

⇒ Area of Δ ABC = Area of Δ ABD + 2 Area of Δ ABD

⇒ Area of Δ ABC = 3 Area of Δ ABD

⇒ Area of Δ ABD = $\dfrac{1}{3}$ Area of Δ ABC.

Hence, proved that area of Δ ABD = $\dfrac{1}{3}$ area of Δ ABC.

In a parallelogram ABCD, point P lies in DC such that DP : PC = 3 : 2. If area of Δ DPB = 30 sq.cm, find the area of the parallelogram ABCD.

Answer

Given,

DP : PC = 3 : 2

We know that,

Ratio of the area of triangles with same vertex and bases along the same line is equal to the ratio of their respective bases.

$\Rightarrow \dfrac{\text{Area of Δ DPB}}{\text{Area of Δ PCB}} = \dfrac{DP}{PC} \\[1em] \Rightarrow \dfrac{\text{Area of Δ DPB}}{\text{Area of Δ PCB}} = \dfrac{3}{2} \\[1em] \Rightarrow \text{Area of Δ PCB} = \dfrac{2}{3} \times \text{Area of Δ DPB} \\[1em] \Rightarrow \text{Area of Δ PCB} = \dfrac{2}{3} \times 30 \\[1em] \Rightarrow \text{Area of Δ PCB} = 20 \text{ cm}^2.$

From figure,

Area of Δ CDB = Area of Δ PCB + Area of Δ DPB = 20 + 30 = 50 cm².

Since, diagonal divides a || gm into two triangles of equal area.

∴ Area of || gm ABCD = 2 Area of Δ CDB = 2 × 50 = 100 cm².

Hence, area of \\ gm ABCD = 100 cm².