Area Theorems — ICSE Class 9 Maths Important Questions

Answer: (b) Equal in area.

Parallelograms on the same base and between the same parallels have the same base and the same height, so $\text{area}=b\times h$area=b× h is equal, though they need not be congruent.

Answer: (c) $24\text{ cm}^2$24 cm^2.

The triangle is half the parallelogram, so area $=\dfrac12\times48=24\text{ cm}^2$=12×48=24 cm^2.

Answer: (c) $30\text{ cm}^2$30 cm^2.

$\triangle APB$APB and parallelogram $ABCD$ABCD are on the same base $AB$AB and between the same parallels $AB$AB and $DC$DC, so $\text{area}(\triangle APB)=\dfrac12\times60=30\text{ cm}^2$area( APB)=12×60=30 cm^2.

Answer: (b) $10\text{ cm}^2$10 cm^2.

Median $AD$AD gives $\text{area}(\triangle ABD)=\dfrac12\times40=20\text{ cm}^2$area( ABD)=12×40=20 cm^2. In $\triangle ABD$ABD, $BE$BE is a median (as $E$E is mid-point of $AD$AD), so $\text{area}(\triangle BED)=\dfrac12\times20=10\text{ cm}^2$area( BED)=12×20=10 cm^2.

Answer: (a) A diagonal splits a parallelogram into two triangles that are congruent by SSS (opposite sides equal, diagonal common); congruent triangles have equal area, so R correctly explains A.

Area of parallelogram $=b\times h=8\times5=40\text{ cm}^2$=b× h=8×5=40 cm^2.

Diagonal $AC$AC divides it into two triangles of equal area, so $\text{area}(\triangle ABC)=\dfrac12\times40=20\text{ cm}^2$area( ABC)=12×40=20 cm^2.

The three triangles $ABX$ABX, $AXY$AXY and $AYC$AYC have equal bases $BX=XY=YC$BX=XY=YC and the same height (the perpendicular distance from $A$A to line $BC$BC).

Since area $=\dfrac12\times\text{base}\times\text{height}$=12×base×height and both base and height are equal,
$\text{area}(\triangle ABX)=\text{area}(\triangle AXY)=\text{area}(\triangle AYC).$area( ABX)=area( AXY)=area( AYC).

Let parallelogram $ABCD$ABCD and $\triangle ABP$ABP be on the same base $AB$AB and between the same parallels $AB$AB and $DC$DC (so $P$P lies on line $DC$DC).

Complete parallelogram $ABPQ$ABPQ on base $AB$AB between the same parallels. Parallelograms $ABCD$ABCD and $ABPQ$ABPQ are on the same base $AB$AB and between the same parallels, so
$\text{area}(ABCD)=\text{area}(ABPQ).$area(ABCD)=area(ABPQ).

Diagonal $BP$BP... more directly: diagonal $AP$AP of parallelogram $ABPQ$ABPQ divides it into two congruent triangles, so
$\text{area}(\triangle ABP)=\dfrac12\,\text{area}(ABPQ)=\dfrac12\,\text{area}(ABCD).$area( ABP)=12\,area(ABPQ)=12\,area(ABCD).

Hence the area of the triangle is half the area of the parallelogram on the same base and between the same parallels. Hence proved.

Draw diagonal-like reasoning using the same parallels $AB$AB and $DC$DC.

$\triangle DEC$DEC and parallelogram $ABCD$ABCD are on the same base $DC$DC and between the same parallels $DC$DC and $AB$AB, so
$\text{area}(\triangle DEC)=\dfrac12\,\text{area}(ABCD).$area( DEC)=12\,area(ABCD).

Also $\text{area}(\triangle ADE)+\text{area}(\triangle BCE)+\text{area}(\triangle DEC)=\text{area}(ABCD)$area( ADE)+area( BCE)+area( DEC)=area(ABCD).

Let $\text{area}(ABCD)=S$area(ABCD)=S. Then $\text{area}(\triangle DEC)=\dfrac{S}{2}$area( DEC)=S/2, so
$12+18+\dfrac{S}{2}=S\ \Rightarrow\ 30=\dfrac{S}{2}\ \Rightarrow\ S=60\text{ cm}^2.$12+18+S/2=S 30=S/2 S=60 cm^2.

Area of parallelogram $ABCD=60\text{ cm}^2$ABCD=60 cm^2.

In parallelogram $ABCD$ABCD the diagonals $AC$AC and $BD$BD bisect each other at $O$O, so $O$O is the mid-point of both diagonals: $OA=OC$OA=OC and $OB=OD$OB=OD.

In $\triangle ABD$ABD, $AO$AO is a median (as $O$O is mid-point of $BD$BD), so
$\text{area}(\triangle AOB)=\text{area}(\triangle AOD). \quad(1)$area( AOB)=area( AOD). (1)

In $\triangle ABC$ABC, $BO$BO is a median (as $O$O is mid-point of $AC$AC), so
$\text{area}(\triangle AOB)=\text{area}(\triangle BOC). \quad(2)$area( AOB)=area( BOC). (2)

In $\triangle BCD$BCD, $CO$CO is a median, so
$\text{area}(\triangle BOC)=\text{area}(\triangle COD). \quad(3)$area( BOC)=area( COD). (3)

From (1), (2) and (3),
$\text{area}(\triangle AOB)=\text{area}(\triangle BOC)=\text{area}(\triangle COD)=\text{area}(\triangle AOD).$area( AOB)=area( BOC)=area( COD)=area( AOD).

Thus the four triangles are equal in area. Hence proved.

Given: Parallelograms $ABCD$ABCD and $ABEF$ABEF on the same base $AB$AB and between the same parallels $AB$AB and $DF$DF (so $D,C,E,F$D,C,E,F lie on the line parallel to $AB$AB).

To prove: $\text{area}(ABCD)=\text{area}(ABEF)$area(ABCD)=area(ABEF).

Proof: Consider $\triangle ADF$ADF and $\triangle BCE$BCE.

$AD=BC$AD=BC (opposite sides of parallelogram $ABCD$ABCD).
$AF=BE$AF=BE (opposite sides of parallelogram $ABEF$ABEF).
$\angle DAF=\angle CBE$DAF= CBE (each equal to the angle between $AB$AB-parallel directions; corresponding angles, as $AD\parallel BC$AD BC and $AF\parallel BE$AF BE).

By SAS, $\triangle ADF\cong\triangle BCE$ADF BCE, so $\text{area}(\triangle ADF)=\text{area}(\triangle BCE)$area( ADF)=area( BCE).

Now,
$\text{area}(ABCD)=\text{area}(ABED)-\text{area}(\triangle BCE)+\text{area}(\triangle ADF)$area(ABCD)=area(ABED)-area( BCE)+area( ADF)
using the common region. More precisely, from the whole trapezium $ABED$ABED:
$\text{area}(ABCD)=\text{area}(ABED)-\text{area}(\triangle BCE),$area(ABCD)=area(ABED)-area( BCE),
$\text{area}(ABEF)=\text{area}(ABED)-\text{area}(\triangle ADF).$area(ABEF)=area(ABED)-area( ADF).

Since $\text{area}(\triangle ADF)=\text{area}(\triangle BCE)$area( ADF)=area( BCE), we get
$\text{area}(ABCD)=\text{area}(ABEF).$area(ABCD)=area(ABEF).

Hence the two parallelograms are equal in area. Hence proved.

$D$D is the mid-point of $BC$BC, so $AD$AD is a median of $\triangle ABC$ABC. A median divides a triangle into two triangles of equal area:
$\text{area}(\triangle ABD)=\text{area}(\triangle ACD)=\dfrac12\,\text{area}(\triangle ABC). \quad(1)$area( ABD)=area( ACD)=12\,area( ABC). (1)

Also, in $\triangle ABD$ABD, $BE$BE is a median (since $E$E is mid-point of $AD$AD), so
$\text{area}(\triangle AEB)=\text{area}(\triangle BED)=\dfrac12\,\text{area}(\triangle ABD). \quad(2)$area( AEB)=area( BED)=12\,area( ABD). (2)

From (1) and (2),
$\text{area}(\triangle AEB)=\dfrac12\times\dfrac12\,\text{area}(\triangle ABC)=\dfrac14\,\text{area}(\triangle ABC).$area( AEB)=12×12\,area( ABC)=14\,area( ABC).

For $\triangle BEC$BEC: since $E$E is mid-point of $AD$AD, $CE$CE is a median of $\triangle ADC$ADC, giving $\text{area}(\triangle CED)=\dfrac12\,\text{area}(\triangle ACD)=\dfrac14\,\text{area}(\triangle ABC)$area( CED)=12\,area( ACD)=14\,area( ABC).

Then
$\text{area}(\triangle BEC)=\text{area}(\triangle BED)+\text{area}(\triangle CED)=\dfrac14+\dfrac14=\dfrac12\,\text{area}(\triangle ABC).$area( BEC)=area( BED)+area( CED)=14+14=12\,area( ABC).

Hence proved.

(i) $AD$AD is a median of $\triangle ABC$ABC, so it bisects the area:
$\text{area}(\triangle ABD)=\dfrac12\times2400=1200\text{ m}^2.$area( ABD)=12×2400=1200 m^2.

(ii) In $\triangle ABD$ABD, $BE$BE is a median ($E$E is mid-point of $AD$AD), so
$\text{area}(\triangle ABE)=\dfrac12\times1200=600\text{ m}^2.$area( ABE)=12×1200=600 m^2.

(iii) $\text{area}(\triangle BED)=600\text{ m}^2$area( BED)=600 m^2 and $\text{area}(\triangle CED)=\dfrac12\,\text{area}(\triangle ACD)=\dfrac12\times1200=600\text{ m}^2$area( CED)=12\,area( ACD)=12×1200=600 m^2, so
$\text{area}(\triangle BEC)=600+600=1200\text{ m}^2.$area( BEC)=600+600=1200 m^2.

(iv) $\dfrac{\text{area}(\triangle ABE)}{\text{area}(\triangle ABC)}=\dfrac{600}{2400}=\dfrac14$area( ABE)/area( ABC)=600/2400=14. So $\triangle ABE$ABE is one-fourth of the park.