Areas of Parallelograms & Triangles

Given,

BD = 8 cm

AB = DC = 6 cm.

∠DBA = ∠BDC = 90°

Thus, BD is perpendicular to both AB and DC.

∴ AB ∥ DC

Since one pair of opposite sides are equal and parallel, thus ABCD is a parallelogram.

We know that,

Area of ∥gm = Base × Height

Area of ∥gm ABCD = AB × BD

= 6 × 8

= 48 cm².

Hence, proved that ABCD is a parallelogram and area of ∥gm ABCD is 48 cm².

Given,

AB = 16 cm

BF = 8 cm

DE = 6 cm

We know that,

Area of ∥gm = Base × Height

Area of ∥gm ABCD = AB × DE

= 16 × 6

= 96 cm².

Considering side AD,

Area of ∥gm ABCD = Base × Height

96 = AD × BF

96 = AD × 8

AD = $\frac{96}{8}$

AD = 12 cm.

Hence, length of AD is 12 cm.

We know that,

Area of rhombus = $\frac{1}{2}$ × Product of diagonals

= $\frac{1}{2}$ × 18 × 24

= 9 × 24

= 216 cm².

Hence, area of rhombus is 216 cm².

We know that,

Area of trapezium = $\frac{1}{2}$ × (sum of parallel sides) × Height

= $\frac{1}{2}$ × (10 + 8) × 6

= $\frac{1}{2}$ × (18) × 6

= 9 × 6

= 54 cm².

Hence, area of trapezium is 54 cm².

Given,

ABCD be a parallelogram in which P and Q are mid-points of AB and CD respectively.

Let us construct DG ⊥ AB and let DG = h, where h is the altitude on side AB.

Area of ||gm ABCD = base × height = AB × h

Area of ||gm APQD = AP × h = $\frac{AB}{2}$ × h ......(1) [Since P is the mid-point of AB]

The perpendicular distance between two parallel lines is always same everywhere, DG = QR = h.

Area of ||gm PBCQ = PB × h = $\frac{AB}{2}$ × h ......(2) [Since P is the mid-point of AB]

From (1) and (2)

Area of ||gm APQD = Area of ||gm PBCQ.

Hence proved, that the line segment joining the mid-points of a pair of opposite sides of a parallelogram divides it into two equal parallelograms.

(i) ar (||gm ABEF)

||gm ABCD and ||gm ABEF lie on the same base AB and between the same parallels AB and FC.

∴ ar(||gm ABEF) = ar(||gm ABCD) = 90 cm².

Hence, ar(||gm ABEF) = 90 cm².

(ii) Triangle ABD and parallelogram ABCD are on the same base AB and between the same parallels AB and CD, then area of triangle is equal to half of the area of the parallelogram.

ar (△ABD) = $\frac{1}{2}$ ar(||gm ABCD)

= $\frac{1}{2} \times 90$

= 45 cm².

Hence, ar (△ABD) = 45 cm².

(iii) Triangle BEF and a parallelogram ABEF are on the same base EF and between the same parallels AB and EF, then area of triangle is equal to half of the area of the parallelogram.

ar (△BEF) = $\frac{1}{2}$ ar(||gm ABEF)

= $\frac{1}{2} \times 90$

= 45 cm².

Hence, ar (△BEF) = 45 cm².

(i) Triangle ABC and a parallelogram ABCD are on the same base AB and between the same parallels AB and CD, then area of triangle is equal to half of the area of the parallelogram.

ar (△ABD) = $\frac{1}{2}$ ar(||gm ABCD)

64 = $\frac{1}{2}$ ar(||gm ABCD)

ar (||gm ABCD) = 64(2)

ar (||gm ABCD) = 128 cm².

Hence, ar(||gm ABCD) = 128 cm².

(ii) ||gm ABCD and rectangle ABEF lie on the same base AB and between the same parallels AB and ED.

∴ar (||gm ABCD) = ar (rect. ABEF)

ar (rect. ABEF) = 128 cm².

Hence, ar(rect. ABEF) = 128 cm².

Triangles △ ACD and △ ACP on the same base AC and between same parallels AC and DP.

∴ ar (△ACD) = ar (△ACP)

Add the common ar(△ABC) to both sides

ar (△ACD) + ar(△ABC) = ar (△ACP) + ar(△ABC)

From figure,

∴ ar (quad.ABCD) = ar (△ABP)

Hence, proved that ar (△ABP) = ar (quad. ABCD).

We know that,

Area of triangle = $\frac{1}{2}$ × Base × Height

Area of triangle ABD = $\frac{1}{2}$ × BD × AL

Area of triangle CBD = $\frac{1}{2}$ × BD × CM

We know that,

Area of quadrilateral ABCD = Area of △ ABD + Area of △ CBD

= $\frac{1}{2}$ × BD × AL + $\frac{1}{2}$ × BD × CM

= $\frac{1}{2}$ × BD × (AL + CM).

Hence, proved that ar (quad. ABCD) = $\frac{1}{2}$ × BD × (AL + CM).

(i) Given,

BD = DC

Thus, ED is the median of triangle EBC.

We know that,

Median ED divides a triangle EBC into two triangles EBD and ECD of equal area.

∴ ar (△EBD) = ar (△EDC)

Hence, proved that ar (△EBD) = ar (△EDC).

(ii) Given,

BD = DC

Thus, AD is the median of triangle ABC.

Median AD divides a triangle ABC into two triangles ABD and ADC of equal area.

ar (△ABD) = ar (△ADC)

ar (△ABE) + ar (△EDB) = ar (△ACE) + ar (△EDC)

ar (△ABE) + ar (△EDB) = ar (△ACE) + ar (△EDB) [ar (△EBD) = ar (△EDC)]

ar (△ABE) + ar (△EDB) - ar (△EDB) = ar (△ACE)

ar (△ABE) = ar (△ACE).

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

Median AD divides triangle ABC into two triangles ABD and ACD of equal area.

ar (△ABD) = $\frac{1}{2}$ ar (△ABC) .....(1)

Since E is the mid-point of AD, BE is the median of ΔABD.

ar (△ABE) = $\frac{1}{2}$ ar (△ABD) .....(2)

Substituting value from equation 1 in equation 2, we get :

ar (△ABE) = $\frac{1}{2} \times \Big(\frac{1}{2} \text{ar (△ABC)}\Big)$

ar (△ABE) = $\frac{1}{4}$ ar (△ABC).

Hence, proved that ar (△ABE) = $\frac{1}{4}$ ar (△ABC).

Since, AD = DE thus, D is the mid-point of AE.

Median BD divides ΔABE into two Δs of equal area.

ar (ΔABD) = ar (ΔEBD) ....(1)

Median CD divides ΔACE into two Δs of equal area.

ar (ΔACD) = ar (ΔECD) ....(2)

Adding the Equations:

ar (ΔABD) + ar (ΔACD) = ar (ΔEBD) + ar (ΔECD)

∴ ar (ΔABC) = ar (ΔBEC)

Hence, proved that ar (ΔBEC) = ar (ΔABC).

In ΔABC,

Median AD divides ΔABC into two Δs of equal area.

∴ ar (ΔABD) = ar (ΔACD) ....(1)

In ΔGBC,

Median GD divides ΔGBC into two Δs of equal area.

∴ ar (ΔGBD) = ar (ΔGCD) ....(2)

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

ar (ΔABD) − ar (ΔGBD) = ar (ΔACD) − ar (ΔGCD)

ar (ΔAGB) = ar (ΔAGC) ...........(3)

In ΔABC,

Median BE divides ΔABC into two Δs of equal area.

∴ ar (ΔABE) = ar (ΔBCE) ....(4)

In ΔGAC,

Median GE divides ΔGAC into two Δs of equal area.

∴ ar (ΔGEA) = ar (ΔGEC) ....(5)

Subtracting equation (5) from equation (4), we get :

ar (ΔABE) − ar (ΔGEA) = ar (ΔBCE) − ar (ΔGEC)

ar (ΔAGB) = ar (ΔBGC) ...........(6)

From equation (3) and (6), we get :

∴ ar (ΔAGB) = ar (ΔAGC) = ar (ΔBGC)

From figure,

ar (ΔAGB) + ar (ΔAGC) + ar (ΔBGC) = ar (ΔABC)

3ar (ΔAGB) = ar (ΔABC)

ar (ΔAGB) = $\frac{1}{3}$ × ar (ΔABC)

Hence, proved that ar (ΔAGB) = ar (ΔAGC) = ar (ΔBGC) = $\frac{1}{3}$ ar (ΔABC).

Given,

2BD = DC

$\frac{BD}{DC} = \frac{1}{2}$

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.

$$\begin{aligned}\Rightarrow \frac{\text{Area of Δ ABD}}{\text{Area of Δ ADC}} = \frac{BD}{DC} \\[1em] \Rightarrow \frac{\text{Area of Δ ABD}}{\text{Area of Δ ADC}} = \frac{1}{2} \\[1em] \Rightarrow \text{Area of Δ ADC} = 2 \times \text{Area of Δ ABD}.\end{aligned}$$

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 = $\frac{1}{3}$ ar (ΔABC).

Hence, proved that Area of Δ ABD = $\frac{1}{3}$ ar (ΔABC).

(i) From figure,

Let DE be altitude on base AC.

Median divides a triangle into two triangles of equal area.

AD is the median of ∆ABC,

Area of ∆ABD = Area of ∆ADC = $\frac{1}{2}$ Area of ∆ABC .....(1)

It is given that,

⇒ area of ∆ADP : area of ∆ABD = 2 : 3

⇒ area of ∆ADP : area of ∆ADC = 2 : 3

$$\begin{aligned}\Rightarrow \frac{\text{Area of Δ ADP}}{\text{Area of Δ ADC}} = \frac{2}{3} \\[1em] \Rightarrow \frac{\frac{1}{2}\times AP \times DE}{\frac{1}{2}\times AC \times DE} = \frac{2}{3} \\[1em] \Rightarrow \frac{AP}{AC} = \frac{2}{3}.\end{aligned}$$

Let AP = 2x and AC = 3x.

From figure,

PC = AC - AP = 3x - 2x = x.

$\frac{AP}{PC} = \frac{2x}{x} = \frac{2}{1}$.

Hence, AP : PC = 2 : 1.

(ii) We know that,

PC : AC = x : 3x = 1 : 3

So,

$$\begin{aligned}\Rightarrow \frac{\text{Area of Δ PDC}}{\text{Area of Δ ADC}} = \frac{\frac{1}{2}\times PC \times DE}{\frac{1}{2}\times AC \times DE} \\[1em] \Rightarrow \frac{\text{Area of Δ PDC}}{\text{Area of Δ ADC}} = \frac{PC}{AC} \\[1em] \Rightarrow \frac{\text{Area of Δ PDC}}{\text{Area of Δ ADC}} = \frac{x}{3x} \\[1em] \Rightarrow \frac{\text{Area of Δ PDC}}{\text{Area of Δ ADC}} = \frac{1}{3} \text{ .....(2)}\end{aligned}$$

Since, AD is median of ∆ABC so,

area of Δ ADC = $\frac{1}{2}$ area of Δ ABC

Substituting above value in equation (2) we get,

$$\begin{aligned}\Rightarrow \frac{\text{Area of Δ PDC}}{\frac{1}{2} \text{ area of Δ ABC}} = \frac{1}{3} \\[1em] \Rightarrow \frac{\text{Area of Δ PDC}}{\text{ area of Δ ABC}} = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}.\end{aligned}$$

Hence, proved that area of △PDC : area of △ABC = 1 : 6.

Given,

BP : PC = 1 : 2

$\frac{BP}{PC} = \frac{1}{2}$

PC = 2BP

From figure,

BC = BP + PC = BP + 2BP = 3BP.

$\frac{PC}{BC} = \frac{2BP}{3BP} = \frac{2}{3}$.

PC = $\frac{2}{3}$ BC

Let AD be the altitude on BC.

Area of △ABC = $\frac{1}{2}$ × BC × AD ......(1)

$$\begin{aligned}\Rightarrow \text{Area of △APC} = \frac{1}{2} \times PC \times AD \\[1em] \Rightarrow \text{Area of △APC} = \frac{1}{2} \times \frac{2}{3} BC \times AD \text{ ....(2)}\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \frac{\text{Area of Δ APC}}{\text{Area of Δ ABC}} = \frac{\frac{1}{2}\times \frac{2}{3} BC \times AD}{\frac{1}{2}\times BC \times AD} \\[1em] \Rightarrow \frac{\text{Area of Δ APC}}{\text{Area of Δ ABC}} = \frac{2}{3} \\[1em] \Rightarrow \text{Area of Δ APC} = \frac{2}{3} \text{Area of Δ ABC}.....(3)\end{aligned}$$

Given,

PQ : AQ = 2 : 3

Let PQ = 2x and AQ = 3x

From figure,

AP = PQ + AQ = 2x + 3x = 5x.

$\frac{AQ}{AP} = \frac{3x}{5x} = \frac{3}{5}$

AQ = $\frac{3}{5}AP$

Let CE be the altitude on side AP.

Area of △AQC = $\frac{1}{2}$ × AQ × CE .....(4)

Area of △APC = $\frac{1}{2}$ × AP × CE ......(5)

Dividing equation (4) by equation (5) we get,

$$\begin{aligned}\Rightarrow \frac{\text{Area of Δ AQC}}{\text{Area of Δ APC}} = \frac{\frac{1}{2} \times AQ \times CE}{\frac{1}{2}\times AP \times CE} \\[1em] \Rightarrow \frac{\text{Area of Δ AQC}}{\text{Area of Δ APC}} = \frac{\frac{1}{2}\times \frac{3}{5} AP \times CE}{\frac{1}{2}\times AP \times CE} \\[1em] \Rightarrow \frac{\text{Area of Δ AQC}}{\text{Area of Δ APC}} = \frac{3}{5} \\[1em] \Rightarrow \text{Area of Δ AQC} = \frac{3}{5} \text{Area of Δ APC} \\[1em] \Rightarrow \text{Area of Δ AQC} = \frac{3}{5} \times \frac{2}{3} \text{Area of Δ ABC} \\[1em] \Rightarrow \text{Area of Δ AQC} = \frac{2}{5} \text{Area of Δ ABC} \\[1em] \Rightarrow \text{Area of Δ AQC}:\text{Area of Δ ABC} = 2 : 5.\end{aligned}$$

Hence, proved that ar (ΔAQC) : ar (ΔABC) = 2 : 5.

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

Area of ∆CPD = $\frac{1}{2}$ Area of ||gm ABCD ......(1)

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

Area of ∆AQD = $\frac{1}{2}$ Area of ||gm ABCD ......(2)

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

Area of ∆CPD = Area of ∆AQD.

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

(i) We know that,

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

∆BCD and ∆BCE are on the same base BC and between the same || lines DE and BC.

⇒ Area of ∆BCD = Area of ∆BCE

Subtracting area of ∆BCD and ∆BCE from area of ∆ABC

⇒ Area of ∆ABC - Area of ∆BCD = Area of ∆ABC - Area of ∆BCE

⇒ Area of ∆ACD = Area of ∆ABE.

Hence proved, that area of ∆ACD = area of ∆ABE.

(ii) We know that,

⇒ Area of ∆BCD = Area of ∆BCE

Subtracting area of ∆OBC from above equation we get,

⇒ Area of ∆BCD - Area of ∆OBC = Area of ∆BCE - Area of ∆OBC

⇒ Area of ∆OBD = Area of ∆OCE.

Hence proved, that area of ∆OBD = area of ∆OCE.

We know that,

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

∆APD and ||gm ABCD are on the same base AD and between the same ∥ lines AD and BC,

Area of ∆APD = $\frac{1}{2}$ Area of ||gm ABCD ....(1)

From figure,

Area of ||gm ABCD = Area of ∆APD + Area of ∆ABP + Area of ∆DPC

Dividing the above equation by 2 we get,

$$\begin{aligned}\frac{\text{Area of ||gm ABCD}}{2} = \frac{\text{Area of ∆APD}}{2} + \frac{\text{Area of ∆ABP}}{2} + \frac{\text{Area of ∆DPC}}{2} \\[1em] \text{Area of ∆APD} = \frac{\text{Area of ∆APD}}{2} + \frac{\text{Area of ∆ABP}}{2} + \frac{\text{Area of ∆DPC}}{2} \\[1em] \text{Area of ∆APD} - \frac{\text{Area of ∆APD}}{2} = \frac{\text{Area of ∆ABP}}{2} + \frac{\text{Area of ∆DPC}}{2} \\[1em] \frac{\text{Area of ∆APD}}{2} = \frac{\text{Area of ∆ABP}}{2} + \frac{\text{Area of ∆DPC}}{2} \\[1em] \text{Area of ∆APD} = \text{Area of ∆ABP} + \text{Area of ∆DPC}.\end{aligned}$$

Hence, proved that Area of ∆ABP + Area of ∆DPC = Area of ∆APD.

Join AP and AQ.

We know that,

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

Since, triangle ABP and BPC lie on the same base BP and between the same parallel lines BP and AC.

∴ Area of △ ABP = Area of △ BPC

Subtracting △ BOP from both sides, we get :

⇒ Area of △ ABP - Area of △ BOP = Area of △ BPC - Area of △ BOP

⇒ Area of △ AOB = Area of △ POC .......(1)

Since, triangle AEQ and EDQ lie on the same base EQ and between the same parallel lines EQ and AD.

∴ Area of △ AEQ = Area of △ EDQ

Subtracting △EXQ from both sides, we get :

⇒ Area of △ AEQ - Area of △ EXQ = Area of △ EDQ - Area of △ EXQ

⇒ Area of △ AXE = Area of △ DQX .......(2)

From figure,

⇒ Area of △ APQ = Area of △ POC + Area of △ DQX + Area of pentagon CDXAO

⇒ Area of △ APQ = Area of △ AOB + Area of △ AXE + Area of pentagon CDXAO

⇒ Area of △ APQ = Area of pentagon ABCDE.

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

Parallelogram ABCD and Parallelogram AEFB are on opposite sides of the common base AB.

Let h₁ be the distance between parallel lines AB and CD, h₂ be the distance between parallel lines AB and EF.

Area of ||gm = Base × Height

ar (∥ gm ABCD) = AB × h₁

ar (∥ gm AEFB) = AB × h₂

ar (∥ gm ABCD) + ar(∥ gm AEFB) = AB × h₁ + AB × h₂

ar (∥ gm ABCD) + ar(∥ gm AEFB) = AB × (h₁ + h₂) ......(1)

From figure,

In ∥ gm EFCD,

Height = Height of ||gm ABCD + Height of ||gm AEFB = h₁ + h₂

ar (∥ gm EFCD) = DC × (h₁ + h₂)

Since, AB = DC [Opposite sides of parallelogram are equal]

ar (∥ gm EFCD) = AB × (h₁ + h₂).....(2)

From equation (1) and (2),

ar(∥ gm ABCD) + ar(∥ gm AEFB) = ar (∥ gm EFCD)

Hence, proved that ar(∥ gm ABCD) + ar(∥ gm AEFB) = ar (∥ gm EFCD).

In ∆ABD, AP is the median (As P is mid-point of BD because diagonals of ||gm bisect each other).

Since, median of triangle divides it into two triangles of equal area.

∴ Area of ∆ABP = Area of ∆ADP ......(1)

Similarly,

PO is median of ∆BOD,

∴ Area of ∆BOP = Area of ∆POD ......(2)

Now, adding equations (1) and (2), we get :

⇒ Area of ∆ABP + Area of ∆BOP = Area of ∆ADP + Area of ∆POD

⇒ Area of ∆AOB = Area of ∆AOD.

Hence, proved that area of ∆AOB = area of ∆AOD.

Parallelograms BCYE and BCFX stand on the same base BC and lie between the same parallels BC and XY.

ar (∥ gm BCYE) = ar (∥ gm BCFX) = x (let) .....(1)

Δ ABE shares the base BE with parallelogram BCYE and lies between the same parallels BE and AC.

ar (Δ ABE) = $\frac{1}{2}$ ar (∥ gm BCYE) = $\frac{1}{2}x$ .....(2)

Δ ACF shares the base CF with parallelogram BCFX and lies between the same parallels CF and AB.

ar (Δ ACF) = $\frac{1}{2}$ ar (∥ gm BCFX) = $\frac{1}{2}x$ .....(3)

From equation (2) and (3), we get :

ar (Δ ABE) = ar (Δ ACF)

Hence, proved that ar (Δ ABE) = ar (Δ ACF).

Δ AQC and Δ AQP shares the base AQ lies between the same parallels AQ and CP.

ar (ΔAQC) = ar (ΔAQP) ......(1)

Both triangles share a common part, which is Δ AQB. Subtract this area from both sides of Equation (1):

ar (ΔAQC) - ar (ΔAQB) = ar (ΔAQP) - ar (ΔAQB)

ar (ΔABC) = ar (ΔQPB) .......(2)

We know that a diagonal of a parallelogram divides it into two triangles of equal area.

In parallelogram ABCD, AC is the diagonal,

ar(Δ ABC) = $\frac{1}{2}$ ar(∥ gm ABCD)

In parallelogram BPRQ, QP is the diagonal,

ar(Δ QPB) = $\frac{1}{2}$ ar(∥ gm BPRQ)

Substituting the values in equation (2), we get :

$\frac{1}{2}$ ar(∥ gm ABCD) = $\frac{1}{2}$ ar(∥ gm BPRQ)

ar(∥ gm ABCD) = ar(∥ gm BPRQ).

Hence, proved that ar(∥ gm ABCD) = ar(∥ gm BPRQ).

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.

From figure,

AB // DC thus, AP // DC since AP is a straight line.

In parallelogram ABCD, the diagonal BD divides it into two triangles of equal area.

ar (ΔABD) = ar (ΔDBC) .....(1)

△ BPD and △ BPC lie on the same base BP and along the same parallel lines AP and DC.

ar (ΔBPD) = ar (ΔBPC) ....(2)

From figure,

ar (ΔAPD) = ar (ΔABD) + ar (ΔBPD)

Substituting values from equations (1) and (2) in above equation, we get :

ar (ΔAPD) = ar (ΔDBC) + ar (ΔBPC) ....(3)

From figure,

ar(quad. BPCD) = ar(△DBC) + ar(△BPC) ....(4)

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

∴ ar (ΔAPD) = ar(quad. BPCD)

Hence, proved that ar (ΔAPD) = ar(quad. BPCD).

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 = $\frac{1}{2}$ 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 = $\frac{1}{2}$ Area of ||gm ABCD .....(2)

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

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

(i) Draw QN perpendicular CB as shown in the figure below :

$$\begin{aligned}\Rightarrow \frac{\text{Area of △BPQ ​}}{\text{Area of △CPQ}} = \frac{\frac{1}{2}BP \times QN}{\frac{1}{2} PC \times QN} \\[1em] \Rightarrow \frac{\text{Area of △BPQ ​}}{\text{Area of △CPQ}} = \frac{BP}{PC} \\[1em] \Rightarrow \frac{\text{Area of △BPQ ​}}{\text{Area of △CPQ}} = \frac{1}{2} \\[1em] \therefore \text{Area of △BPQ​} = \frac{1}{2} \text{Area of △CPQ} = \frac{1}{2} \times 20 = 10 \text{ cm}^2\end{aligned}$$

Considering △CDP and △BQP,

∠CPD = ∠QPB (Vertically opposite angles are equal)

∠PDC = ∠PQB (Alternate angles are equal)

Hence, by AA axiom △CDP ~ △BQP.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △CDP}}{\text{Area of △BQP}} = \frac{PC^2}{BP^2} \\[1em] \therefore \frac{\text{Area of △CDP}}{\text{Area of △BQP}} = \frac{2^2}{1^2} \\[1em] \therefore \frac{\text{Area of △CDP}}{\text{Area of △BQP}} = \frac{4}{1} .\end{aligned}$$

∴ Area of △CDP = 4 × Area of △BQP = 4 × 10 = 40 cm².

Hence, the area of △CDP = 40 cm².

(ii) Area of ||gm ABCD = 2 Area of △DCQ (As △DCQ and ||gm ABCD have same base DC and are between same parallels AQ and DC)

= 2(Area of △CDP + Area of △CPQ)

= 2(40 + 20)

= 2 × 60

= 120 cm².

Hence, the area of ||gm ABCD = 120 cm².

(i) Since,

∆APB and ||gm ABCD have same base AB and are between same parallel lines AB and DC. So,

area of ∆APB = $\frac{1}{2}$ area of ||gm ABCD

From figure,

⇒ $\frac{1}{2}$ area of ||gm ABCD = area of (∆DAP + ∆BCP)

⇒ $\frac{1}{2}$ area of ||gm ABCD = 25 + 15

⇒ $\frac{1}{2}$ area of ||gm ABCD = 40

⇒ area of ||gm ABCD = 2 × 40 = 80 cm².

Hence, area of ||gm ABCD = 80 cm².

(ii) From figure,

∆DAP and ∆BCP are on the same base CD and between the same parallel lines CD and AB.

$$\begin{aligned}\Rightarrow \frac{\text{Area of ∆DAP}}{\text{Area of ∆BCP}} = \frac{DP}{PC} \\[1em] \Rightarrow \frac{25}{15} = \frac{DP}{PC} \\[1em] \Rightarrow \frac{DP}{PC} = \frac{5}{3}.\end{aligned}$$

Hence, DP : PC = 5 : 3.

We know that,

AD || BE ⇒ AD || EG

ED || FA ⇒ ED || GA

Since, opposite sides are parallel.

Hence, ADEG is a parallelogram.

Since ||gm ABCD and ||gm ADEG lie on same base AD and between same parallel lines AD and EB,

area of ||gm ABCD = area of ||gm ADEG .......(1)

We know that,

ED || FA ⇒ DE || FH

DC || EF ⇒ DH || EF

Since, opposite sides are parallel.

Hence, DEFH is a parallelogram.

Since ||gm DEFH and ||gm ADEG lie on same base DE and between same parallel lines DE and FA,

area of ||gm DEFH = area of ||gm ADEG ........(2)

From (1) and (2) we get,

⇒ area of ||gm ABCD = area of ||gm DEFH

Hence, proved that area of ||gm ABCD = area of ||gm DEFH.

(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

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.

Steps of construction:

1. Draw AB = 3.2 cm.

2. With A as centre and radius 4.5 cm, draw an arc.

3. With B as centre and radius 5.3 cm draw another arc, cutting the previous arc at D.

4. Join AD.

5. With D as centre and radius 4 cm, draw an arc.

6. With B as centre and radius 2.8 cm, draw another arc cutting the previous arc at C.

7. Join BC and DC, to form quadrilateral ABCD.

8. Join BD and through C, construct a straight line parallel to DB to meet AB produced at E.

9. Join DE.

Since △DBC and △DBE have same base DB and are between the same parallels BD and EC, we have;

ar(△DBC) = ar(△DBE)

ar(quad ABCD) = ar(△ABD) + ar(△DBC)

ar(quad ABCD) = ar(△ABD) + ar(△DBE)

ar(quad ABCD) = ar(△AED)

Hence, triangle AED is the required triangle whose area is equal to the area of the quadrilateral ABCD.

We know that,

Area of the rhombus = $\frac{1}{2}$ × Product of diagonals

= $\frac{1}{2}$ × 16 × 24

= 8 × 24

= 192 cm²

Hence, option 4 is the correct option.

We know that,

Area of the trapezium = $\frac{1}{2}$ × sum of parallel sides × height

= $\frac{1}{2}$ × (9 + 6) × 8

= 15 × 4

= 60 cm²

Hence, option 2 is the correct option.

We know that,

Area of ∥gm = Base × Height

= AB × DB

= 5 × 7

= 35 cm².

Hence, option 3 is the correct option.

Median AD divides ΔABC into two Δs of equal area.

ar (ΔADC) = ar (ΔABD)

Hence, option 2 is the correct option.

In triangle RTQ,

RQ² = RT² + QT²

(17)² = RT² + (8)²

RT² = 289 - 64

RT² = 225

RT = $\sqrt{225}$ = 15 cm.

We know that,

Area of the trapezium = $\frac{1}{2}$ × sum of parallel sides × height

= $\frac{1}{2}$ × (8 + 16) × 15

= $\frac{1}{2}$ × (24) × 15

= 15 × 12

= 180 cm².

Hence, option 2 is the correct option.

We know that,

Area of ∥gm = Base × Height

Area of ∥gm ABCD using base AB = 3.6 × 5 = 18 cm²

Area of ∥gm ABCD using base AD = AD × 4

18 = AD × 4

AD = $\frac{18}{4}$

AD = 4.5 cm.

Hence, option 4 is the correct option.

Let the common base be b and the common height be h.

Area of triangle = $\frac{1}{2}$ × ​b × h

Area of parallelogram = ​b × h

Area of rectangle = ​b × h

Ratio of areas of triangle, parallelogram and rectangle:

= $\frac{1}{2}$ × ​b × h : ​b × h : ​b × h

= 1 : 2 : 2.

Hence, option 4 is the correct option.

Diagonal BD divides it into triangles ABD and BCD of equal area.

ar(ΔBCD) = $\frac{1}{2}$ ar(∥gm ABCD)

= $\frac{1}{2} \times 10$

= 5 cm².

Hence, option 2 is the correct option.

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

= $\frac{1}{2}$ × 5 × 4

= 10 cm²

A triangle and a parallelogram, both have same base and are between the same parallels.

Area of triangle = $\frac{1}{2}$ ar(∥gm ABCD)

10 = $\frac{1}{2}$ ar(∥gm ABCD)

ar(∥gm ABCD) = 10(2)

ar(∥gm ABCD) = 20 cm²

Hence, option 1 is the correct option.

ΔPQR and parallelogram PQRS, both have same base PQ and are between the same parallels PQ and SR.

Area of △PQR = $\frac{1}{2}$ Area of parallelogram PQRS

= $\frac{1}{2} \times 16$

= 8 sq.units.

M is the mid-point of PQ, QM = $\frac{1}{2}$ PQ

Thus, RM is the median of a triangle PQR, and divides it into two triangles of equal area.

Area of △QMR = $\frac{1}{2}$ Area of triangle PQR

= $\frac{1}{2} \times 8$

= 4 sq. units.

Hence, option 3 is the correct option.

In a rectangle, a diagonal divides it into two equal triangles.

So,

Area of rectangle ABCD = 2 × Area of △ABD

= 2 × 5

= 10 cm².

Rectangle ABCD and parallelogram ABQC are on the same base AB and between the same parallel lines AB and DQ.

Area of the parallelogram ABQC = Area of rectangle ABCD = 10 cm².

Hence, option 2 is the correct option.

Triangles PAB and QAB have the same base AB and lie between the same parallels AB and DC.

ar(△QAB) = ar(△PAB) = 10 cm²

Hence, option 2 is the correct option.

The diagonals of a parallelogram divide it into four triangles of equal area.

So, Area(ΔAOB) = $\frac{1}{4}$ Area(∥gm ABCD)

= $\frac{1}{4} \times 20$

= 5 cm ².

Hence, option 4 is the correct option.

Draw diagonal BD.

A diagonal of a parallelogram divides it into two triangles of equal area :

Area (ΔABD) = $\frac{1}{2} \times$ Area(∥gm ABCD)

= $\frac{1}{2} \times$ (60)

= 30 cm².

Point E is the midpoint of AB, so:

BE = $\frac{1}{2}$ AB

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

Area(ΔBDE) = $\frac{1}{2}$ Area(ΔABD)

= $\frac{1}{2} \times$ (30)

= 15 cm².

Hence, option 3 is the correct option.

1. When point X is chosen on the side CD and joined to vertices A and B, the parallelogram is split into three distinct triangular regions: △ ADX, △ ABX, and △ BCX.

Hence, option (b) is the correct option.

2. Since X is a point on a line segment (CD) and it is connected to the endpoints of the opposite parallel side (A and B), all three resulting closed figures are three-sided polygons.

All three parts formed are triangles.

Hence, option (a) is the correct option.

3. Since ΔABX and parallelogram ABCD are on the same base AB and between the same parallels AB and CD,

Area of △XAB = $\frac{1}{2}$ Area of parallelogram ABCD

Since ΔXAB is half the total area, the sum of the remaining two triangles is also half the total area.

ar(△ADX) + ar(△BCX) = $\frac{1}{2}$ Area of parallelogram ABCD

Thus,

△XAB = △ADX + △BCX

Hence, option (d) is the correct option.

4. ΔABX and parallelogram ABCD stand on the same base AB and lie between the same parallels AB and CD.

Hence, option (c) is the correct option.

5. If Y is taken on BC and YA, YD are joined.

△ADY now shares the base AD with the parallelogram and lies between parallels AD and BC. Thus,

Area(△ ADY) = $\frac{1}{2}$ Area(ABCD) meaning it is equal to the sum of the remaining two triangles ΔABY and ΔDCY.

area of ΔADY = area of ΔABY + area of ΔDCY

Hence, option (a) is the correct option.

In a parallelogram ABCD, the diagonal BD divides it into two triangles ΔABD and ΔBCD of equal area.

Area of △ABD = $\frac{1}{2}$ × Area of parallelogram ABCD

Assertion (A) is true.

The statement is a standard area theorem:

If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.

Reason (R) is true.

Both A and R are true.

Hence, option 3 is the correct option.

In ΔABC, point D is the midpoint of AB.

Thus, CD is the median of trinagle ABC.

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

Thus, area of ΔBCD = area of ΔACD.

Assertion (A) is true.

We know that,

If a triangle and a parallelogram lie on the same base and between the same parallels then area of triangle is equal to half the area of parallelogram.

Area of the triangle = $\frac{1}{2}$ area of the parallelogram.

Reason (R) is false.

A is true, R is false.

Hence, option 1 is the correct option.

In third figure,

Join BD and AC.

Triangle ABC and triangle ABD. They both share the same base AB. Crucially, the vertices C and D both lie on the same line, and that line is parallel to the base AB.

Hence, option 3 is the correct option.

Join BD which intersects EF at M.

In ∆ABD,

E is the midpoint of AD and EM || AB

By midpoint theorem,

M is the midpoint of BD

EM = $\frac{1}{2} AB$ ....(1)

In ∆CBD,

F is mid-point of BC and M is mid-point of BD so by mid-point theorem,

MF = $\frac{1}{2} CD$ ....(2)

So EF ∥ AB and EF ∥ CD

That means:

AB ∥ EF ∥ CD

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

EM + MF = $\frac{1}{2} AB + \frac{1}{2} CD = \frac{x + y}{2}$

Since:

AB ∥ EF ∥ CD

Let total height between AB and CD = H

EF lies exactly halfway between them,

∴ Height of trapezium EFCD = Height of trapezium ABEF = $\frac{H}{2}$ = h (let)

Area of trapezium = $\frac{1}{2} \times (\text{sum of parallel sides}) \times h$

Area of trapezium ABFE

$$\begin{aligned}= \frac{1}{2} \times \Big(x + \frac{x + y}{2}\Big) \times h \\[1em] = \frac{1}{2} \times \Big(\frac{3x + y}{2}\Big) h \\[1em] = \frac{h}{4} \times (3x + y).\end{aligned}$$

Area of trapezium EFCD

$$\begin{aligned}= \frac{1}{2} \times \Big(y + \frac{x + y}{2}\Big) \times h \\[1em] = \frac{1}{2} \times \Big(\frac{x + 3y}{2}\Big) h \\[1em] = \frac{h}{4} \times (x + 3y).\end{aligned}$$

Required ratio = Area of trapezium ABFE / Area of trapezium EFCD

By substituting the values,

$$\begin{aligned}\text{Ratio} = \frac{\frac{h}{4}\times (3x + y)}{\frac{h}{4} \times (x + 3y)} \\[1em] \text{Ratio} = \frac{(3x + y)}{(x + 3y)} \\[1em] \text{Ratio} = (3x + y) : (x + 3y).\end{aligned}$$

The ratio of ar (ABFE) and ar (EFCD) is (3x + y) : (x + 3y).

Hence, option 2 is the correct option.

We know that,

Area of triangles on the same base and between the same parallel lines are equal.

△ ADF and △ DFB lie on same base DF and between same parallel lines AB and DC.

∴ Area of △ ADF = Area of △ DFB = 3 cm²

By converse of mid-point theorem,

If a line is drawn through the midpoint of one side of a triangle, and parallel to the other side, it bisects the third side.

In △ ABE,

C is the mid-point of BE and CF || AB.

∴ F is the mid-point of AE. (By converse of mid-point theorem)

∴ EF = AF.

In △ ADF and △ EFC,

⇒ ∠AFD = ∠EFC (Vertically opposite angles are equal)

⇒ EF = AF (Proved above)

⇒ ∠DAF = ∠CEF (Alternate interior angles are equal)

∴ △ ADF ≅ △ ECF (By A.S.A. axiom)

We know that,

Area of congruent triangles are equal.

∴ Area of △ EFC = Area of △ ADF = 3 cm².

In △ BFE,

Since, C is the mid-point of BE.

∴ CF is the median of triangle BFE.

We know that,

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

∴ Area of △ BFC = Area of △ EFC = 3 cm².

From figure,

⇒ Area of △ BDC = Area of △ BDF + Area of △ BFC

⇒ Area of △ BDC = 3 + 3 = 6 cm².

We know that,

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

From figure,

||gm ABCD and △ BDC lies on same base DC and between same parallel lines AB and DC.

∴ Area of △ BDC = $\frac{1}{2}$ Area of ||gm ABCD

⇒ Area of ||gm ABCD = 2 × Area of △ BDC

⇒ Area of ||gm ABCD = 2 × 6 = 12 cm².

Hence, option 3 is the correct option.

If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram.

△ ABP and parallelogram ABCD both share the base AB and lie between the same parallel lines, AB and CD.

Area (△ ABP) = $\frac{1}{2}$ Area(∥gm ABCD).

Son 1: Received the area of △ ABP.

Since the total area of the field is the sum of the three triangles, and one triangle takes up exactly half, the other two triangles must together make up the remaining half.

Son 2: Received the combined area of △ ADP and △ BCP

Hence, son 1 received the area of △ ABP and son 2 received the area of △ ADP and △ BCP.

Let distance between lines l and m be h units.

We know that,

Area of parallelogram ABCD = base × height = DC × h

$$\begin{aligned}\text{Area of trapezium} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{ height} \\[1em] \text{Area of trapezium EFCD} = \frac{1}{2} \times (DC + EF) \times h.\end{aligned}$$

The length EF is less than AB (which is equal to DC). Since the average of DC and EF $\Big(\frac{DC + EF}{2}\Big)$ is smaller than DC itself, thus the parallelogram has a larger area.

ar(ABCD) > ar(EFCD).

Hence, ar(ABCD) > ar(EFCD).