Area Theorems Class 9 ICSE Important Questions with Answers

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 cm².

The triangle is half the parallelogram, so area $=\dfrac12\times48=24\text{ cm}^2$=1/2×48=24 cm².

Answer: (c) $30\text{ cm}^2$30 cm².

$\triangle APB$△ 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)=1/2×60=30 cm².

Answer: (b) $10\text{ cm}^2$10 cm².

Median $AD$AD gives $\text{area}(\triangle ABD)=\dfrac12\times40=20\text{ cm}^2$area(△ ABD)=1/2×40=20 cm². 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)=1/2×20=10 cm².

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 cm².

Diagonal $AC$AC divides it into two triangles of equal area, so $\text{area}(\triangle ABC)=\dfrac12\times40=20\text{ cm}^2$area(△ ABC)=1/2×40=20 cm².

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 $=\dfrac12\times\text{base}\times\text{height}$=1/2× 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).

Given: Parallelogram $ABCD$ABCD and $\triangle ABP$△ ABP on the same base $AB$AB and between the same parallels $AB$AB and $DC$DC ($P$P lies on line $DC$DC).

To prove: $\text{area}(\triangle ABP)=\dfrac12\,\text{area}(ABCD)$area(△ ABP)=1/2 area(ABCD).

Construction: Through $B$B draw $BQ\parallel AP$BQ∥ AP to meet line $DC$DC at $Q$Q.

Proof: $AB\parallel PQ$AB∥ PQ and $AP\parallel BQ$AP∥ BQ, so $ABQP$ABQP is a parallelogram.

Parallelograms $ABQP$ABQP and $ABCD$ABCD are on the same base $AB$AB and between the same parallels, so
$\text{area}(ABQP)=\text{area}(ABCD).\quad(1)$area(ABQP)=area(ABCD). (1)

The diagonal $PB$PB divides parallelogram $ABQP$ABQP into two congruent triangles, $\triangle ABP\cong\triangle QPB$△ ABP≅ △ QPB, so
$\text{area}(\triangle ABP)=\dfrac12\,\text{area}(ABQP)=\dfrac12\,\text{area}(ABCD)\quad\text{[by (1)]}.$area(△ ABP)=1/2 area(ABQP)=1/2 area(ABCD) [by (1)].

Hence proved.

$\triangle DEC$△ 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)=1/2 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 cm².

Area of parallelogram $ABCD=60\text{ cm}^2$ABCD=60 cm².

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.

(a) Triangles on the same base and between the same parallels are equal in area. $\triangle ADC$△ ADC and $\triangle BDC$△ BDC stand on the same base $DC$DC and lie between the same parallels $DC$DC and $AB$AB, so
$\text{area}(\triangle ADC)=\text{area}(\triangle BDC).\quad(1)$area(△ ADC)=area(△ BDC). (1)

(b) $\text{area}(\triangle ADC)=\text{area}(\triangle AOD)+\text{area}(\triangle DOC)$area(△ ADC)=area(△ AOD)+area(△ DOC) and $\text{area}(\triangle BDC)=\text{area}(\triangle BOC)+\text{area}(\triangle DOC)$area(△ BDC)=area(△ BOC)+area(△ DOC).

Subtracting the common $\text{area}(\triangle DOC)$area(△ DOC) from both sides of (1):
$\text{area}(\triangle AOD)=\text{area}(\triangle BOC).$area(△ AOD)=area(△ BOC). Hence proved.

(c) By (b), $\text{area}(\triangle AOD)=12\text{ cm}^2$area(△ AOD)=12 cm². So
$\text{area}(\triangle ADC)=\text{area}(\triangle AOD)+\text{area}(\triangle DOC)=12+8=20\text{ cm}^2.$area(△ ADC)=area(△ AOD)+area(△ DOC)=12+8=20 cm².

(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)=1/2×2400=1200 m².

(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)=1/2×1200=600 m².

(iii) $\text{area}(\triangle BED)=600\text{ m}^2$area(△ BED)=600 m² and $\text{area}(\triangle CED)=\dfrac12\,\text{area}(\triangle ACD)=\dfrac12\times1200=600\text{ m}^2$area(△ CED)=1/2 area(△ ACD)=1/2×1200=600 m², so
$\text{area}(\triangle BEC)=600+600=1200\text{ m}^2.$area(△ BEC)=600+600=1200 m².

(iv) $\dfrac{\text{area}(\triangle ABE)}{\text{area}(\triangle ABC)}=\dfrac{600}{2400}=\dfrac14$(area(△ ABE))/(area(△ ABC))=600/2400=1/4. So $\triangle ABE$△ ABE is one-fourth of the park.