Rd Sharma 2021 Solutions for Class 9 Maths Chapter 14 Areas Of Parallelograms And Triangles are provided here with simple step-by-step explanations. These solutions for Areas Of Parallelograms And Triangles are extremely popular among Class 9 students for Maths Areas Of Parallelograms And Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2021 Book of Class 9 Maths Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2021 Solutions. All Rd Sharma 2021 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 14.15:

#### Question 1:

If the given figure, *ABCD* is a parallelogram, *AE* ⊥ *DC* and *CF* ⊥ *AD*. If *AB* = 16 cm. *AE* = 8 cm and *CF* = 10 cm, find *AD*.

#### Answer:

**Given:** Here in the question it is given

(1) ABCD is a parallelogram,

(2) and

(3) , AB = 16 cm

(4) AE = 8cm

(5) CF = 10cm

**To Find :** AD =?

**Calculation:** We know that formula for calculating the

Therefore,

Area of paralleogram ABCD = DC × AE (Taking base as DC and Height as AE )

Area of paralleogram ABCD = AB × AE (AB = DC as opposite side of the parallelogram are equal)

Therefore,

Area of paralleogram ABCD = 16 × 8 ……(1)

Taking the base of Parallelogram ABCD as AD we get

Area of paralleogram ABCD = AD × CF (taking base as AD and height as CF)

Area of paralleogram ABCD = AD × 10 ……(2)

Since equation 1 and 2 both represent the Area of the same Parallelogram ABCD , both should be equal.

Hence fro equation (1) and (2),

This means that,

Hence we get the result as

#### Page No 14.15:

#### Question 2:

In Q.No. 1, if *AD *= 6 cm, *CF* = 10 cm, and *AE *= 8 cm, find *AB*.

#### Answer:

**Given:** Here in the question it is given that

(1) ABCD is a parallelogram,

(2) and

(3)

(4) AD = 6 cm

(5) AE = 8cm

(6) CF = 10cm

**To Find :** AB =?

**Calculation:** We know that formula for calculating the

_{Area of paralleogram = base × height}

Therefore,

Area of paralleogram ABCD = DC × AE (Taking base as DC and Height as AE )

Area of paralleogram ABCD = AB × AE (AB = DC as opposite side of the parallelogram are equal)

Therefore, Area of paralleogram ABCd = 16 × 8

Area of Parallelogram ABCD = AB× 8 ** **……(1)

Taking the base of Parallelogram ABCD as AD we get

Area of paralleogram ABCD = AD × CF (taking base as AD and height as CF)

Area of paralleogram ABCD = 6 × 10 ……(2)

Since equation 1and 2 both represent the Area of the same Parallelogram ABCD , both should be equal.

Hence equation 1 is equal to equation 2

Which means that,

Hence we got the measure of AB equal to

#### Page No 14.15:

#### Question 3:

Let *ABCD* be a parallelogram of area 124 cm^{2}. If *E* and *F* are the mid-points of sides *AB* and *CD* respectively, then find the area of parallelogram *AEFD*.

#### Answer:

**Given:** Here in the question it is given that

(1) Area of paralleogram ABCD = 124 cm^{2}

(2) E is the midpoint of AB, which means

(3) F is the midpoint of CD, which means

**To Find :** Area of Parallelogram AEFD

**Calculation:** We know that formula for calculating the

**Area of Parallelogram = base × height**

Therefore,

Area of paralleogram ABCD = AB × AD (Taking base as AB and Height as AD ) ……(1)

Therefore,

Area of paralleogram AEFD = AE × AD (Taking base as AB and Height as AD ) ……(2)

()

= Area of Parallelogram ABCD (from equation1)

Hence we got the result **Area of Parallelogram AEFD **

#### Page No 14.15:

#### Question 4:

If *ABCD* is a parallelogram, then prove that

ar (Δ*ABD*) = ar (Δ*BCD*) = ar (Δ*ABC*) = ar (Δ*ACD*) = $\frac{1}{2}$ ar (||^{gm} *ABCD*)

#### Answer:

**Given:** Here in the question it is given that

(1) ABCD is a Parallelogram

**To Prove :**

(1)

(2)

(3)

(4)

**Construction:** Draw

**Calculation:** We know that formula for calculating the

**Area of Parallelogram = base × height**

Area of paralleogram ABCD = BC × AE (Taking base as BC and Height as AE ……(1)

We know that formula for calculating the

Area of ΔADC = Base × Height

(AD is the base of ΔADC and AE is the height of ΔADC)

= Area of Parallelogram ABCD (from equation1)

Hence we get the result

Similarly we can show that** **

(2)

(3)

(4)

#### Page No 14.3:

#### Question 1:

Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and two parallels:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

#### Answer:

GIVEN: Here in the question figure 1 to 6 are shown.

To Find :

(1) The figures which lie on the same base and between the same parallels.

(2) Write the common base and parallels.

**As we know that ***‘Two geometric figures are said to be on the same base and between the same parallels, if they have a common side(base) and the vertices (or vertex) opposite to the common base of each figure lie on a line parallel to the base.’*

(1) ΔAPB and trapezium ABCD are on the same base CD and between the same parallels AB and CD.

(2) Parallelograms ABCD and APQD are on the same base AD and between the same parallels AD and BQ.

(3) Parallelogram ABCD and ΔPQR are between the same parallels AD and BC,. but they are not on the same base AD

(4) Parallelogram ABCD and ΔPQR are between the same parallels AD and BC,. but they are not on the same base AD .

(5) ΔQRT and Parallelogram PQRS are on the same base QR and between the same parallels PS and QR.

(6) Parallelogram PQRS, AQRD, and BCQR are between the same parallels. Also Parallelogram PQRS, BPSC and APSD are between the same parallels.

#### Page No 14.44:

#### Question 1:

In the given figure, compute the area of quadrilateral *ABCD*.

#### Answer:

**Given: **Here from the given figure we get

(1) ABCD is a quadrilateral with base AB,

(2) ΔABD is a right angled triangle

(3) ΔBCD is a right angled triangle with base BC right angled at B

To Find**:** Area of quadrilateral ABCD

**Calculation:**

In right triangle ΔBCD, by using Pythagoreans theorem

.So

In right triangle ABD

Hence we get Area of quadrilateral ABCD =

#### Page No 14.44:

#### Question 2:

In the given figure, PQRS is a square and *T* and *U* are, respectively, the mid-points of *PS* and *QR*. Find the area of Δ *OTS* if *PQ* = 8 cm.

#### Answer:

**Given: **Here from the given figure we get

(1) PQRS is a square,

(2) T is the midpoint of PS which means

(3) U is the midpoint of PS which means

(4) QU = 8 cm

**To find:** Area of ΔOTS

**Calculation:**

Since it is given that PQ = 8 cm. So

Since T and U are the mid points of PS and QR respectively. So

Therefore area of triangle OTS is equals to

Hence we get the result that Area of triangle OTS is

#### Page No 14.44:

#### Question 3:

Compute the area of trapezium* PQRS *in the given figure.

#### Answer:

**Given:**

(1) PQRS is a trapezium in which SR||PQ..

(2) PT = 5 cm.

(3) QT = 8 cm.

(4) RQ = 17 cm.

To Calculate: Area of trapezium PQRS.

**Calculation:**

In triangle

.So

No area of rectangle PTRS

Therefore area of trapezium PQRS is

Hence the answer is

#### Page No 14.44:

#### Question 4:

In the given figure, ∠*AOB* = 90°, *AC* = *BC*, *OA* = 12 cm and *OC* = 6.5 cm. Find the area of Δ *AOB*.

#### Answer:

**Given: **In figure:

(1) ∠AOB = 90°

(2) AC = BC,

(3) OA = 012 cm,

(4) OC = 6.5 cm.

**To find:** Area of ΔAOB

**Calculation:**

It is given that AC = BC where C is the mid point of AB

We know that the mid point of hypotenuse of right triangle is equidistant from the vertices

Therefore

CA = BC = OC

⇒ CA = BC = 6.5

⇒ AB = 2 × 6.5 = 13 cm

Now inn triangle OAB use Pythagoras Theorem

So area of triangle OAB

Hence area of triangle is

#### Page No 14.44:

#### Question 5:

In the given figure, *ABCD* is a trapezium in which *AB* = 7 cm, *AD* = *BC* = 5 cm, *DC* = x cm, and distance between *AB* and *DC* is 4 cm. Find the value of *x* and area of trapezium *ABCD*.

#### Answer:

**Given: **Here from the given figure we get

(1) ABCD is a trapezium

(2) AB = 7 cm,

(3) AD = BC = 5 cm,

(4) DC = *x* cm

(5) Distance between AB and DC is 4 cm

**To find:**

(a) The value of x

(b) Area of trapezium

**Construction:** Draw AL⊥ CD, and BM ⊥ CD

**Calculation:**

Since AL ⊥ CD, and BM ⊥ CD

Since distance between AB and CD is 4 cm. So

AL = BM = 4 cm, and LM = 7 cm

In triangle ADL use Pythagoras Theorem

Similarly in right triangle BMC use Pythagoras Theorem

Now

We know that,

We get the result as

Area of trapezium is

#### Page No 14.45:

#### Question 6:

In the given figure, *OCDE* is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If *OE* = 2$\sqrt{5}$, find the area of the rectangle.

#### Answer:

**Given: **Here from the given figure we get

(1) OCDE is a rectangle inscribed in a quadrant of a circle with radius 10cm,

(2) OE = 2√5cm

**To find:** Area of rectangle OCDE.

**Calculation:**

In right triangle ΔODE use Pythagoras Theorem

We know that,

Hence we get the result as area of Rectangle OCDE =

#### Page No 14.45:

#### Question 7:

In the given figure, *ABCD* is a trapezium in which *AB* || *DC*. Prove that

ar(Δ *AOD*) = ar(Δ *BOC**)*.

#### Answer:

**Given:**

ABCD is a trapezium with AB||DC

**To prove:** Area of ΔAOD = Area of ΔBOC

**Proof:**

We know that ‘triangles between the same base and between the same parallels have equal area’

Here ΔABC and ΔABD are between the same base and between the same parallels AB and DC.

Therefore

Hence it is proved that

#### Page No 14.45:

#### Question 8:

In the given figure, *ABCD*, *ABFE* and *CDEF* are parallelograms. Prove that

ar(Δ *ADE*) = ar(Δ *BCF*)

#### Answer:

**Given:**

(1) ABCD is a parallelogram,

(2) ABFE is a parallelogram

(3) CDEF is a parallelogram

**To prove:** Area of ΔADE = Area of ΔBCF

**Proof:**

We know that,” opposite sides of a parallelogram are equal”

Therefore for

Parallelogram ABCD, AD = BC

Parallelogram ABFE, AE = BF

Parallelogram CDEF, DE = CF.

Thus, in ΔADE and ΔBCF, we have

So be SSS criterion we have

This means that

Hence it is proved that

#### Page No 14.45:

#### Question 9:

In the given figure, *ABC* and *ABD *are two triangles on the base *AB*. If line segment *CD* is bisected by *AB* at* O*, show that ar (Δ *ABC*) = ar (Δ *ABD*)

#### Answer:

**Given:**

(1) ABC and ABD are two triangles on the same base AB,

(2) CD bisect AB at O which means AO = OB

**To Prove:** Area of ΔABC = Area of ΔABD

**Proof:**

Here it is given that CD bisected by AB at O which means O is the midpoint of CD.

Therefore AO is the median of triangle ACD.

Since the median divides a triangle in two triangles of equal area

Therefore Area of ΔCAO = Area of ΔAOD ...... (1)

Similarly for Δ CBD, O is the midpoint of CD

Therefore BO is the median of triangle BCD.

Therefore Area of ΔCOB = Area of ΔBOD ...... (2)

Adding equation (1) and (2) we get

Area of ΔCAO + Area of ΔCOB = Area of ΔAOD + Area of ΔBOD

⇒ Area of ΔABC = Area of ΔABD

Hence it is proved that

#### Page No 14.45:

#### Question 10:

If *AD* is a median of a triangle *ABC*, then prove that triangles *ADB* and *ADC* are equal in area. If *G* is the mid-point of median *AD*, prove that ar(Δ *BGC*)= 2 ar (Δ *AGC*).

#### Answer:

**Given:**

(1) ABC is a triangle

(2) AD is the median of ΔABC

(3) G is the midpoint of the median AD

**To prove:**

(a) Area of Δ ADB = Area of Δ ADC

(b) Area of Δ BGC = 2 Area of Δ AGC

**Construction: **Draw a line AM perpendicular to AC

**Proof: **Since AD is the median of ΔABC.

Therefore BD = DC

So multiplying by AM on both sides we get

In ΔBGC, GD is the median

Since the median divides a triangle in to two triangles of equal area. So

Area of ΔBDG = Area of ΔGCD

⇒ Area of ΔBGC = 2(Area of ΔBGD)

Similarly In ΔACD, CG is the median

⇒ Area of ΔAGC = Area of ΔGCD

From the above calculation we have

Area of ΔBGD = Area of ΔAGC

But Area of ΔBGC = 2(Area of ΔBGD)

So we have

Area of ΔBGC = 2(Area of ΔAGC)

Hence it is proved that

(1)

(2)

#### Page No 14.45:

#### Question 11:

A point *D* is taken on the side *BC* of a Δ*ABC* such that *BD* = 2*DC*. Prove that ar (Δ *ABD*) = 2 ar (Δ *ADC*).

#### Answer:

**Given:**

(1) ABC is a triangle

(2) D is a point on BC such that BD = 2DC

**To prove: **Area of ΔABD = 2 Area of ΔAGC

**Proof:**

In ΔABC, BD = 2DC

Let E is the midpoint of BD. Then,

BE = ED = DC

Since AE and AD are the medians of ΔABD and ΔAEC respectively

and

The median divides a triangle in to two triangles of equal area. So

Hence it is proved that

#### Page No 14.46:

#### Question 12:

*ABCD* is a parallelogram whose diagonals intersect at *O*. If *P* is any point on *BO*, prove that:

(i) ar (Δ *ADO*) = ar(Δ *CDO*)

(ii) ar (Δ *ABP*) = ar (Δ *CBP*).

#### Answer:

**Given:** Here from the given figure we get

(1) ABCD is a parallelogram

(2) BD and CA are the diagonals intersecting at O.

(3) P is any point on BO

**To prove:**

(a) Area of ΔADO = Area ofΔ CDO

(b) Area of ΔAPB = Area ofΔ CBP

**Proof: **We know that diagonals of a parallelogram bisect each other.

O is the midpoint of AC and BD.

Since medians divide the triangle into two equal areas

In ΔACD, DO is the median

Area of ΔADO = Area ofΔ CDO

Again O is the midpoint of AC.

In ΔAPC, OP is the median

⇒ Area of ΔAOP = Area of ΔCOP …… (1)

Similarly O is the midpoint of AC.

In ΔABC, OB is the median

⇒ Area of ΔAOB = Area of ΔCOB …… (2)

Subtracting (1) from (2) we get,

Area of ΔAOB − Area of ΔAOP = Area of ΔCOB − Area of ΔCOP

⇒ Area of ΔABP = Area of ΔCBP

Hence it is proved that

(a)

(b)

#### Page No 14.46:

#### Question 13:

*ABCD* is a parallelogram in which *BC* is produced to *E* such that *CE* = *BC*. *AE* intersects *CD* at *F.*

(i) Prove that ar (Δ *ADF*) = ar (Δ *ECF*)

(ii) If the area of Δ *DFB* = 3 cm^{2}, find the area of ||^{gm} *ABCD*.

#### Answer:

**Given: **Here from the given figure we get

(1) ABCD is a parallelogram with base AB,

(2) BC is produced to E such that CE = BC

(3) AE intersects CD at F

(4) Area of ΔDFB = 3 cm

**To find:**

(a) Area of ΔADF = Area of ΔECF

(b) Area of parallelogram ABCD

**Proof: **Δ ADF and ΔECF, we can see that

∠ADF = ∠ECF (Alternate angles formed by parallel sides AD and CE)

AD = EC

∠DFA = ∠CFA (Vertically opposite angles)

(ASA condition of congruence)

As

DF = CF

Since DF = CF. So BF is a median in ΔBCD

Since median divides the triangle in to two equal triangles. So

Since .So

Hence Area of parallelogram ABCD

Hence we get the result

(a)

(b)

#### Page No 14.46:

#### Question 14:

*ABCD* is a parallelogram whose diagonals *AC* and *BD* intersect at *O*. *A* line through *O* intersects *AB* at *P* and *DC* at *Q*. Prove that ar (Δ *POA*) = ar (Δ *QOC*).

#### Answer:

**Given:**

(1) Diagonals AC and BD of a parallelogram ABCD intersect at point O.

(2) A line through O intersects AB at P point.

(3) A line through O intersects DC at Q point.

**To find:** Area of (ΔPOA) = Area of (ΔQOC)

**Proof:**

From ΔPOA and ΔQOC* *we get that

=

OA = OC

=

So, by ASA congruence criterion, we have

So

Area (ΔPOA) = Area (ΔQOC)

Hence it is proved that

#### Page No 14.46:

#### Question 15:

In the given figure, *D* and *E* are two points on *BC* such that *BD* = *DE* = *EC*. Show that a (Δ *ABD*) = ar (Δ *ADE*) = ar (Δ *AEC*).

#### Answer:

**Given: **In ΔABCD, D and E are two points on BC such that BD = DE = EC

**To prove:**

**Proof:** The ΔABD, ΔADE, and ΔAEC, are on the equal bases and their heights are equal

Therefore their areas are equal

So

Hence we get the result as

#### Page No 14.46:

#### Question 16:

Diagonals *AC* and *BD* of a quadrilateral *ABCD* intersect each other at *P*. Show that:

ar(Δ *APB*) ✕ ar (Δ *CPD*) = ar (Δ *APD*) ✕ ar (Δ *BPC*)

#### Answer:

**Given:**

(1) ABCD is a quadrilateral,

(2) Diagonals AC and BD of quadrilateral ABCD intersect at P.

**To prove:** Area ofΔ APB ×Area of ΔCPD = Area of ΔAPD × Area of ΔBPC

**Con****struction: **Draw AL perpendicular to BD and CM perpendicular to BD

**Proof:**

We know that

Area of triangle = × base× height

Area of ΔAPD = × DP × AL …… (1)

Area of ΔBPC = × CM × BP …… (2)

Area of ΔAPB = × BP × AL …… (3)

Area of ΔCPD = × CM × DP …… (4)

Therefore

Hence it is proved that

#### Page No 14.46:

#### Question 17:

If *P *is any point in the interior of a parallelogram *ABCD*, then prove that area of the triangle *APB* is less than half the area of parallelogram.

#### Answer:

**Given: **Here from the question we get

(1) ABCD is a parallelogram

(2) P is any point in the interior of parallelogram ABCD

**To prove:**

**Construction:** Draw DN perpendicular to AB and PM perpendicular AB

**Proof: **Area of triangle = × base× height

Area of ΔAPB = × AB × PM …… (1)

Also we know that: Area of parallelogram = base× height

Area of parallelogram ABCD = AB × DN …… (2)

Now PM < DN (Since P is a point inside the parallelogram ABCD)

Hence it is proved that

#### Page No 14.46:

#### Question 18:

*ABCD* is a parallelogram. *E* is a point on *BA* such that *BE* = 2 *EA* and *F* is a point on *DC* such that *DF* = 2 *FC*. Prove that *AE* *CF* is a parallelogram whose area is one third of the area of parallelogram *AB* *CD*.

#### Answer:

**Given:**

(1) ABCD is a parallelogram.

(2) E is a point on BA such that BE = 2EA

(3) F is a point on DC such that DF = 2FC.

**To find:**

Area of parallelogram

**Proof: **We have,

BE = 2EA and DF = 2FC

AB − AE = 2AE and DC − FC = 2FC

AB = 3AE and DC = 3FC

AE = AB and FC = DC

AE = FC [since AB = DC]

Thus, AE || FC such that AE = FC

Therefore AECF is a parallelogram.

Clearly, parallelograms ABCD and AECF have the same altitude and

AE = AB.

Therefore

Hence proved that

#### Page No 14.46:

#### Question 19:

In a Δ *ABC*, *P* and *Q *are respectively the mid-points of *AB* and *BC* and *R* is the mid-point of *AP*. Prove that:

(i) ar (Δ *PBQ*) = ar (Δ *ARC*)

(ii) ar (Δ *PRQ*) = $\frac{1}{2}$ ar (Δ *ARC*)

(iii) ar (Δ *RQC*) = $\frac{3}{8}$ ar (Δ *ABC*).

#### Answer:

Given:

(1) In a triangle ABC, P is the mid-point of AB.

(2) Q is mid-point of BC.

(3) R is mid-point of AP.

To prove:

(a) Area of ΔPBQ = Area of ΔARC

(b) Area of ΔPRQ = $\frac{1}{2}$ Area of ΔARC

(c) Area of ΔRQC = $\frac{3}{8}$ Area of ΔABC

**Proof: **We know that each median of a triangle divides it into two triangles of equal area.

(a) Since CR is a median of ΔCAP

Therefore …… (1)

Also, CP is a median of ΔCAB.

Therefore …… (2)

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

Therefore …… (3)

PQ is a median of ΔABQ

Therefore

Since

Put this value in the above equation we get

…… (4)

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

Therefore …… (5)

(b)

…… (6)

…… (7)

From equation (6) and (7)

…… (8)

From equation (7) and (8)

(c)

= …… (9)

#### Page No 14.46:

#### Question 20:

*ABCD* is a parallelogram, *G* is the point on *AB* such that *AG* = 2 *GB*, *E* is a point of *DC* such that *CE* = 2*DE* and *F* is the point of *BC* such that *BF* = 2*FC*. Prove that:

(i) ar (A*DEG*) = ar (*GBCE*)

(ii) ar (Δ *EGB*) = $\frac{1}{6}$ ar (*ABCD*)

(iii) ar (Δ *EFC*) = $\frac{1}{2}$ ar (Δ *EBF*)

(iv) ar (Δ *EBG*) = ar (Δ *EFC*)

(v) Find what portion of the area of parallelogram is the area of Δ *EFG*

#### Answer:

**Given:**

ABCD is a parallelogram

G is a point such that AG = 2GB

E is a point such that CE = 2DE

F is a point such that BF = 2FC

**To prove:**

(i)

(ii)

(iii)

(iv)

What portion of the area of parallelogram ABCD is the area of ΔEFG

**Construction:** draw a parallel line to AB through point F and a perpendicular line to AB through

PROOF:

(i) Since ABCD is a parallelogram,

So AB = CD and AD = BC

Consider the two trapeziums ADEG and GBCE:

Since AB = DC, EC = 2DE, AG = 2GB

, and

, and

So, and

Since the two trapeziums ADEG and GBCE have same height and their sum of two parallel sides are equal

Since

So

Hence

(ii) Since we know from above that

. So

Hence

(iii) Since height of triangle EFC and triangle EBF are equal. So

Hence

(iv) Consider the trapezium in which

(From (*iii*))

Now from (ii) part we have

(v) In the figure it is given that FB = 2CF. Let CF = *x* and FB = 2*x*

Now consider the tow triangles CFI and CBH which are similar triangles

So by the property of similar triangle CI = *k* and IH = 2*k*

Now consider the triangle EGF in which

Now

(Multiply both sides by 2)

…… (2)

From (1) and (2) we have

#### Page No 14.47:

#### Question 21:

In the given figure, *CD* || *AE* and *CY* || *BA*.

(i) Name a triangle equal in area of Δ*CBX*

(ii) Prove that ar (Δ *ZDE*) = ar (Δ *CZA*)

(iii) Prove that ar (*BCZY*) = ar (Δ *EDZ*).

#### Answer:

**Given:**

(1) CD||AE.

(2) CY||BA.

**To find:**

(i) Name a triangle equal in area of ΔCBX.

(ii) .

(iii) .

**Proof:**

(i) Since triangle BCY and triangle YCA are on the same base and between same parallel, so their area should be equal. Therefore

Therefore area of triangle CBX is equal to area of triangle AXY

(ii) Triangle ADE and triangle ACE are on the same base AE and between the same parallels AE and CD.

(iii) Triangle ACY and BCY are on the same base CY and between same parallels CY and BA. So we have

Now we know that

#### Page No 14.47:

#### Question 22:

In the given figure, *PSDA* is a parallelogram in which *PQ* = *QR* = *RS* and *AP* || *BQ* || *CR*. Prove that ar (*PQE*) = ar (Δ *CFD*).

#### Answer:

**Given:**

(i) PSDA is a parallelogram.

(ii) .

(iii)

**To find:**

**Proof:**

Since AP||BQ||CR||DS and AD||PS

So PQ = CD …… (1)

In ΔBED, C is the mid point of BD and CF||BE

This implies that F is the mid point of ED. So

EF = FD …… (2)

In ΔPQE and ΔCFD, we have

PE = FD

, and [Alternate angles]

PQ = CD.

So, by SAS congruence criterion, we have

ΔPQE = ΔDCF

Hence proved that

#### Page No 14.47:

#### Question 23:

In the given figure, *ABCD* is a trapezium in which *AB* || *DC* and *DC* = 40 cm and *AB* = 60 cm. If *X* and *Y *are, respectively, the mid-points of *AD* and *BC*, prove that:

(i) *XY* = 50 cm

(ii) *DCYX* is a trapezium

(iii) ar (trap. *DCYX*) = $\frac{9}{11}$ ar (trap. (*XYBA*)

#### Answer:

**Given: **ABCD IS A trapezium in which

(a) AB||DC

(b) DC = 40 cm

(c) AB = 60 cm

(d) X is the midpoint of AD

(e) Y is the midpoint of BC

**To prove:**

(i) XY = 50 cm

(ii) DCYX is a trapezium

(iii)

**Construction:** Join DY and produce it to meet AB produced at P.

**Proof:**

(i) In ΔBYP and ΔCYD

Y is the midpoint of BC also X is the midpoint of AD

Therefore XY||AP and

(ii) We have proved above that XY||AP

XY|| AP and AB||DC (Given in question)

XY|| DC

(iii) Since X and Y are the midpoints of AD and BC respectively.

Therefore DCYX and ABYX are of the same height say *h* cm.

#### Page No 14.47:

#### Question 24:

*D* is the mid-point of side *BC* of Δ *ABC* and *E* is the mid-point of *BD*. If O is the mid-point of *AE*, prove that ar (Δ *BOE*) = $\frac{1}{8}$ ar (Δ *ABC*).

#### Answer:

**Given: **In ΔABC

(1) D is the midpoint of the side BC

(2) E is the midpoint of the side BD

(3) O is the midpoint of the side AE

**To prove:**

**Proof:** We know that the median of a triangle divides the triangle into two triangles of equal area.

Since AD and AE are the medians of ΔABC and ΔABD respectively. And OB is the median of ΔABE

…… (1)

…… (2)

…… (3)

Therefore

Hence we have proved that

#### Page No 14.47:

#### Question 25:

In the given figure, *X* and *Y* are the mid-points of AC and AB respectively, *QP* || *BC* and *CYQ* and *BXP* are straight lines. Prove that ar (Δ *ABP*) = ar (Δ *ACQ*).

#### Answer:

**Given:**

(1) X and Y are the, midpoints of AC and AB respectively.

(2) QP|| BC

(3) CYQ and BXP are straight lines.

**To prove:**

**Proof:** Since X and Y are the, midpoints of AC and AB respectively.

So XY||BC

ΔBYC and ΔBXC are on the same base BC and between the same parallels XY and BC.

Therefore

…… (1)

Similarly the quadrilaterals XYAP and XYQA are on the same base XY and between the same parallels XY and PQ. Therefore

…… (2)

Adding equation 1 and 2 we get

Hence we had proved that

#### Page No 14.47:

#### Question 26:

In the given figure, *ABCD* and *AEFD* are two parallelograms. Prove that

(i) *PE* = *FQ*

(ii) ar (Δ *APE*): ar (Δ*PFA*) = ar Δ (*QFD*) : ar (Δ *PFD*)

(iii) ar (Δ *PEA*) = ar (Δ *QFD*).

#### Answer:

**Given:** ABCD and AEFD are two parallelograms

**To prove:**

(i) PE = FQ

(ii)

(iii)

Proof: (i) and (iii)

In ΔAPE and ΔDQF

Therefore

, and

(ii) ΔPFA and ΔPFD are on the same base PF and between the same parallels PQ and AD.

From (1) and (2) we get

#### Page No 14.48:

#### Question 27:

In the given figure, *ABCD* is a ||^{gm}.* O* is any point on *AC*. *PQ* || *AB* and *LM* || *AD*. Prove that ar (||^{gm} *DLOP*) = ar (||^{gm} *BMOQ*)

#### Answer:

**Given:**

(1) ABCD is a parallelogram

(2) O is any point of AC.

(3) PQ||AB and LM||AD

**To prove:**

**Calculation:**

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

Therefore we have

Since OC and AO are diagonals of parallelogram OQCL and AMOP respectively. Therefore

Subtracting (2) and (3) from (1) we get

Hence we get the result

#### Page No 14.48:

#### Question 28:

In a Δ *ABC*, if *L* and *M* are points on *AB* and *AC* respectively such that *LM* || *BC*. Prove that:

(i) ar (Δ *LCM*) = ar (Δ *LBM*)

(ii) ar (Δ *LBC*) = (Δ *MBC*)

(iii) ar (Δ *ABM*) = ar (Δ *ACL*)

(iv) ar (Δ *LOB*) = ar (Δ *MOC*)

#### Answer:

**Given:**

In ΔABC, if L and M are points on AB and AC such that LM||BC

**To prove:**

(i)

(ii)

(iii)

(iv)

**Proof: **We know that triangles between the same base and between the same parallels are equal in area.

(i) Here we can see that ΔLMB and ΔLMC are on the same base BC and between the same parallels LM and BC

Therefore

…… (1)

(ii) Here we can see that ΔLBC and ΔLMC are on the same base BC and between the same parallels LM and BC

Therefore

…… (2)

(iii) From equation (1) we have,

(iv) From (2) we have,

#### Page No 14.48:

#### Question 29:

In the given figure, *ABC* and *BDE* are two equilateral triangles such that *D* is the mid-point of *BC*. *AE* intersects *BC* in *F.* Prove that

(i) ar (Δ *BDE*) = $\frac{1}{4}$ ar (Δ *ABC*)

(ii) ar (Δ *BDE*) = $\frac{1}{2}$ ar (Δ *BAE*)

(iii) ar (Δ *BFE*) =ar (Δ *AFD*)

(iv) ar (Δ*ABC*) = 2 ar (Δ *BEC*)

(v) ar (Δ *FED*) = $\frac{1}{8}$ ar (Δ *AFC*)

(vi) ar (Δ *BFE*) = 2 ar (*EFD*)

#### Answer:

**Given:**

(a) ΔABC and Δ BDE are two equilateral triangles

(b) D is the midpoint of BC

(c) AE intersect BC in F.

**To prove:**

(i)

(ii)

(iii)

(iv)

(v)

(vi)

**Proof:** Let AB = BC = CA = *x* cm.

Then BD = = DE = BE

(i) We have

(ii) We Know that ΔABC and ΔBED are equilateral triangles

BE||AC

(iii) We Know that ΔABC and ΔBED are equilateral triangles

AB || DE

(iv) Since ED is a median of Δ BEC

(v) We basically want to find out FD. Let FD = *y*

Since triangle BED and triangle DEA are on the same base and between same parallels ED and BE respectively. So

Since altitude of altitude of any equilateral triangle having side *x *is

So

…… (1)

Now

…… (2)

From (1) and (2) we get

(vi) Now we know *y *in terms of *x*. So

……. (3)

…… (4)

From (3) and (4) we get

#### Page No 14.48:

#### Question 30:

If the given figure, *ABC *is a right triangle right angled at A, *BCED*, *ACFG*and *ABMN* are squares on the sides *BC*, *CA* and *AB* respectively. Line segment *AX* ⊥ *DE* meets *BC* at* Y*.

Show that

(i) Δ *MBC* $\cong $ Δ *ABD*

(ii) ar (*BYXD*) = 2 ar (Δ *MBC*)

(iii) ar (*BYXD*) = ar (*ABMN*)

(iv) Δ *FCB* $\cong $ Δ *ACE*

(v) ar (*CYXE*) = 2 ar (Δ*FCB*)

(vi) ar (*CYXE*) = ar (*ACFG*)

(vii) ar (*BCED*) = ar (*ABMN*) + ar (*ACFG*)

#### Answer:

**Given:**

(1) ABCD is a right angled triangle at A

(2) BCED, ACFG and ABMN are the squares on the sides of BC, CA and AB respectively.

(3) , meets BC at Y.

**To prove:**

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

**Proof:**

(i)

…… (1)

(ii) Triangle ABD and rectangle BYXD are on the same base BD and between the same parallels AX and BD.

Therefore

(Using (1)) …… (2)

(iii) Since ΔMBC and square MBAN are on the same base MB and between the same parallels MB and NC.

…… (3)

From (2) and (3) we get

(iv) In triangle FCB and ACE

…… (4)

(v) Since ΔACE and rectangle CYXE are on the same base CE and between the same parallels CE and AX.

…… (5)

(vi) Since ΔFCB and rectangle FCAG are on the same base FC and between the same parallels FC and BG

…… (6)

From (5) and (6) we get

(vii) Applying Pythagoras Theorem in ΔACB, WE get

#### Page No 14.59:

#### Question 1:

Two parallelograms are on the same base and between the same parallels. The ratio of their areas is

(a) 1 : 2

(b) 2 : 1

(c) 1 : 1

(d) 3 : 1

#### Answer:

**Given: **Two parallelogram with the same base and between the same parallels.

**To find:** Ratio of their area of two parallelogram with the same base and between the same parallels.

**Calculation:** We know that “Two parallelogram with the same base and between the same parallels, are equal in area”

Hence their ratio is ,

So the correct answer is,i.e. option (c).

#### Page No 14.59:

#### Question 2:

A triangle and a parallelogram are on the same base and between the same parallels. The ratio of the areas of triangle and parallelogram is

(a) 1 : 1

(b) 1 : 2

(c) 2 : 1

(d) 1 : 3

#### Answer:

**Given:** A triangle and a parallelogram with the same base and between the same parallels.

**To find:** The ratio of the area of a triangle and a parallelogram with the same base and between the same parallels.

**Calculation: **We know that,” the area of a triangle is half the area of a parallelogram with the same base and between the same parallels.”

Hence the ratio of the area of a triangle and a parallelogram with the same base and between the same parallels is

Therefore the correct answer is option is (b).

#### Page No 14.59:

#### Question 3:

Let *ABC* be a triangle of area 24 sq. units and *PQR* be the triangle formed by the mid-points of the sides of Δ *ABC*. Then the area of Δ*PQR* is

(a) 12 sq. units

(b) 6 sq. units

(c) 4 sq. units

(d) 3 sq. units

#### Answer:

**Given: **(1) The Area of ΔABC = 24 sq units.

(2) ΔPQR is formed by joining the midpoints of ΔABC

**To find:** The area of ΔPQR

**Calculation:** In ΔABC, we have

Since Q and R are the midpoints of BC and AC respectively.

PQ || BA PQ || BP

Similarly, RQ || BP. So BQRP is a parallelogram.

Similarly APRQ and PQCR are parallelograms.

We know that diagonal** of a parallelogram bisect the parallelogram into two triangles of equal area.**

Now, PR is a diagonal of APQR.

∴ Area of ΔAPR = Area of ΔPQR ……(1)

Similarly,

PQ is a diagonal of PBQR

∴ Area of ΔPQR = Area of ΔPBQ ……(2)

QR is the diagonal of PQCR

∴ Area of ΔPQR = Area of ΔRCQ ……(3)

From (1), (2), (3) we have

Area of ΔAPR = Area of ΔPQR = Area of ΔPBQ = Area of ΔRCQ

But

Area of ΔAPR + Area of ΔPQR + Area of ΔPBQ + Area of ΔRCQ = Area of ΔABC

4(Area of ΔPBQ) = Area of ΔABC

∴ Area of ΔPBQ

Hence the correct answer is option (b).

#### Page No 14.59:

#### Question 4:

The median of a triangle divides it into two

(a) congruent triangle

(b) isosceles triangles

(c) right triangles

(d) triangles of equal areas

#### Answer:

**Given:** A triangle with a median.

**Calculation: **We know that a ,”**median of a triangle divides it into two triangles of equal area.”**

Hence the correct answer is option (d).

#### Page No 14.59:

#### Question 5:

In a Δ*ABC*, *D*, *E*, *F* are the mid-points of sides *BC*, *CA* and *AB* respectively. If ar (Δ*ABC*) = 16cm^{2}, then ar (trapezium *FBCE*) =

(a) 4 cm^{2}

(b) 8 cm^{2}

(c) 12 cm^{2}

(d) 10 cm^{2}

#### Answer:

**Given:** In ΔABC

(1) D is the midpoint of BC

(2) E is the midpoint of CA

(3) F is the midpoint of AB

(4) Area of ΔABC = 16 cm^{2}

**To find:** The area of Trapezium FBCE

**Calculation: **Here we can see that in the given figure,

Area of trapezium FBCE = Area of ||^{gm} FBDE + Area of ΔCDE

Since D and E are the midpoints of BC and AC respectively.

∴ DE || BA DE || BF

Similarly, FE || BD. So BDEF is a parallelogram.

Now, DF is a diagonal of ||^{gm} BDEF.

∴ Area of ΔBDF = Area of ΔDEF ……(1)

Similarly,

DE is a diagonal of ||^{gm} DCEF

∴ Area of ΔDCE = Area of ΔDEF ……(2)

FE is the diagonal of ||^{gm} AFDE

∴ Area of ΔAFE = Area of ΔDEF ……(3)

From (1), (2), (3) we have

Area of ΔBDF = Area of ΔDCF = Area of ΔAFE = Area of ΔDEF

But

Area of ΔBDF + Area of ΔDCE + Area of ΔAFE + Area of ΔDEF = Area of ΔABC

∴ 4 Area of ΔBDF = Area of ΔABC

Area of ΔBDF = Area of ΔDCE = Area of ΔAFE = Area of ΔDEF = 4 cm^{2} …….(4)

Now

Hence we get

Area of trapezium FBCE

There fore the correct answer is option (c).

#### Page No 14.59:

#### Question 6:

*ABCD* is a parallelogram. *P* is any point on *CD*. If ar (Δ*DPA*) = 15 cm2 and ar (Δ*APC*) = 20 cm^{2}, then ar (Δ*APB*) =

(a) 15 cm^{2}

(b) 20 cm^{2}

(c) 35 cm^{2}

(d) 30 cm^{2}

#### Answer:

**Given: **(1) ABCD is a parallelogram

(2) P is any point on CD

(3) Area of ΔDPA = 15 cm^{2}

(4) Area of ΔAPC = 20 cm^{2}

**To find:** Area of ΔAPB

**Calculation:** We know that , “If a parallelogram and a a triangle are on the base between the same parallels, the area of triangle is equal to half the area of the parallelogram.”

Here , ΔAPB and ΔACB are on the same base and between the same parallels.

(since AC is the diagonal of parallelogram ABCD, diagonal of a parallelogram divides the parallelogram in two triangles of equal area)

Hence the correct answer is option (c).

#### Page No 14.60:

#### Question 7:

The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 16 cm and 12 cm is

(a) 28 cm^{2}

(b) 48 cm^{2}

(c) 96 cm^{2}

(d) 24 cm^{2}

#### Answer:

**Given:** Rhombus with diagonals measuring 16cm and 12 cm.

**To find:** Area of the figure formed by lines joining the midpoints of the adjacent sides.

**Calculation: **We know that, ‘**Area of a rhombus is half the product of their diagonals’.**

H and F are the midpoints of AD and BC respectively.

Now ABCD is a parallelogram which means

……..(1)

……(2)

From 1 and 2 we get that ABFH is a parallelogram.

Since Parallelogram FHAB and ΔFHE are on the base FH and between the same parallels HF and AB.

……(3)

Similarly ,

……(4)

Adding 3 and 4 we get,

Hence the correct answer is option (b).

#### Page No 14.60:

#### Question 8:

A, B, C, D are mid-points of sides of parallelogram *PQRS*. If ar (*PQRS*) = 36 cm^{2}, then ar (*ABCD*) =

(a) 24 cm^{2}

(b) 18cm^{2}

(c) 30 cm^{2}

(d) 36 cm^{2}

#### Answer:

**Given:**

(1) PQRS is a parallelogram.

(2) A, B, C, D are the midpoints of the adjacent sides of Parallelogram PQRS.

(3)

**To find:**

**Calculation:**

A and C are the midpoints of PS and QR respectively.

Now PQRS is a parallelogram which means

AP = CQ ……..(1)

Also, PS || QR

AP || CQ ……(2)

From 1 and 2 we get that APCQ is a parallelogram.

Since Parallelogram APCQ and ΔABC are on the base AC and between the same parallels AC and PQ.

……(3)

Similarly ,

……(4)

Adding 3 and 4 we get,

Hence the correct answer is option (b).

#### Page No 14.60:

#### Question 9:

The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is

(a) a rhombus of area 24 cm^{2}

(b) a rectangle of area 24 cm^{2}

(c) a square of area 26 cm^{2}

(d) a trapezium of area 14 cm^{2}

#### Answer:

**Given:** Rectangle with sides 8 cm and 6cm

**To find: **Area of the figure which is formed by joining the midpoints of the adjacent sides of rectangle.

**Calculation: **Since we know that

For rhombus EFGH, EG is the one diagonal which is equal to DA

FH is the other diagonal which is equal to AB

Hence the result is option (a).

#### Page No 14.60:

#### Question 10:

If *AD* is median of Δ*ABC* and *P *is a point on *AC* such that

ar (Δ*ADP*) : ar (Δ*ABD*) = 2 : 3, then ar (Δ *PDC*) : ar (Δ *ABC*)

(a) 1 : 5

(b) 1 : 5

(c) 1 : 6

(d) 3 : 5

#### Answer:

**Given:** (1) AD is the Median of ΔABC

(2) P is a point on AC such that ar (ΔADP) : ar (ΔABD) =

**To find:** ar (ΔPDC) : ar (ΔABC)

We know that” **the medians of the triangle divides the triangle in two two triangles of equal area.”**

Since AD is the median of ΔABC,

ar (ΔABD) = ar (ΔADC) ……(1)

Also it is given that

ar (ΔADP) : ar (ΔABD) = ……(2)

Now,

Therefore,

Hence the correct answer is option (c).

#### Page No 14.60:

#### Question 11:

Medians of Δ*ABC* intersect at G. If ar (Δ*ABC*) = 27 cm^{2}, then ar (Δ*BGC*) =

(a) 6 cm^{2}

(b) 9 cm^{2}

(c) 12 cm^{2}

(d) 18 cm^{2}

#### Answer:

**Given: **(1) Median of ΔABC meet at G.

(2) Area of ΔABC = 27 cm^{2}

**To find:** Area of ΔBCG.

We know that the medians of the triangle divides each other in the ratio of 2:1

Hence,

Hence the correct answer is option (b).

#### Page No 14.60:

#### Question 12:

In a Δ*ABC* if *D* and *E* are mid-points of *BC* and *AD* respectively such that ar (Δ*AEC*) = 4cm^{2}, then ar (Δ*BEC*) =

(a) 4 cm^{2}

(b) 6 cm^{2}

(c) 8 cm^{2}

(d) 12 cm^{2}

#### Answer:

**Given: **In ΔABC

(1) D is the midpoint of BC

(2) E is the midpoint of AD

(3) ar (ΔAEC) = 4 cm^{2}

**To find:** ar (ΔBEC)

**Calculation:** We know that”**the median of the triangle divides the triangle into two triangle of equal area”**

Since AD is the median of ΔABC,

ar (ΔABD) = ar (ΔADC) …… (1)

EC is the median of ΔADC,

ar (ΔAEC) = ar (ΔDEC) …… (2)

⇒ ar (ΔDEC) = 4 cm^{2}

EC is the median of ΔBED

ar (ΔBED) = ar (ΔDEC) …… (3)

From 2 and 3 we get,

ar (ΔBED) = ar (ΔAEC) …… (4)

⇒ ar (ΔBED) = 4 cm^{2}

Now,

Hence the correct answer is option (c).

#### Page No 14.60:

#### Question 13:

In the given figure, *ABCD* is a parallelogram. If *AB* = 12 cm, *AE* = 7.5 cm, *CF* = 15 cm, then *AD* =

(a) 3 cm

(b) 6 cm

(c) 8 cm

(d)10.5 cm

**Graphic**

#### Answer:

**Given:** (1) ABCD is a parallelogram.

(2) AB = 12 cm

(3) AE = 7.5 cm

(4) CF = 15cm

**To find:** AD

**Calculation:** We know that,

**Area of a parallelogram = base × height**

Area of a parallelogram ABCD = DC ×AE (with DC as base and AE as height) ……(1)

Area of a parallelogram ABCD = AD ×CF (with DC as base and AE as height) ……(2)

Since equation 1 and 2 both are Area of a parallelogram ABCD

Hence the correct answer is option (b).

#### Page No 14.60:

#### Question 14:

In the given figure, *PQRS* is a parallelogram. If *X* and *Y* are mid-points of *PQ* and *SR* respectively and diagonal *Q* is joined. The ratio ar (||^{gm} *XQRY*) : ar (Δ*QSR*) =

(i) 1 : 4

(ii) 2 : 1

(iii) 1 : 2

(iv) 1 : 1

#### Answer:

**Given:** (1) PQRS is a parallelogram.

(2) X is the midpoint of PQ.

(3) Y is the midpoint of SR.

(4) SQ is the diagonal.

**To find:** Ratio of area of ||^{gm} XQRY : area of ΔQRS.

**Calculation: **We know that **the triangle and parallelogram on the same base and between the same parallels are equal in area.**

∴ Ar (||^{gm} PQRS) = Ar (ΔQRS)

(since X is the mid point of PQ and Y is the midpoint of SR)

Hence the correct answer is option (d).

#### Page No 14.60:

#### Question 15:

Diagonal *AC* and *BD* of trapezium *ABCD*, in which *AB* || *DC*, intersect each other at *O*. The triangle which is equal in area of Δ*AOD* is

(a) Δ*AOB*

(b) Δ*BOC*

(c) Δ*DOC*

(d) Δ*ADC*

#### Answer:

**Given:** (1) ABCD is a trapezium, with parallel sides AB and DC

(2) Diagonals AC and BD intersect at O

**To find:** Area of ΔAOD equals to ?

**Calculation:** We know that ,” **two triangles with the same base and between the same parallels are equal in area.”**

Therefore,

Hence the correct answer is option (b).

#### Page No 14.60:

#### Question 16:

*ABCD* is a trapezium in which *AB* || *DC*. If ar (Δ*ABD*) = 24 cm^{2} and *AB* = 8 cm, then height of Δ*ABC* is

(a) 3 cm

(b) 4 cm

(c) 6 cm

(d) 8 cm

#### Answer:

**Given: **(1) ABCD is a trapezium, with parallel sides AB and DC

(2) Area of ΔADB = 24 cm^{2}

(3) AB = 8 cm

**To find:** Height of ΔABC.

**Calculation:** We know that ,” **two triangles with the same base and between the same parallels are equal in area.”**

Here we can see that, ΔADB and ΔACB are on the same base AB.

Hence,

Area of ΔACB = Area of ΔADB

Area of ΔACB = 24

Hence the correct answer is option (c).

#### Page No 14.61:

#### Question 17:

*ABCD* is a trapezium with parallel sides *AB* =*a* and *DC* = *b*. If *E* and *F* are mid-points of non-parallel sides *AD* and *BC* respectively, then the ratio of areas of quadrilaterals *ABFE* and *EFCD* is

(a) *a* : *b*

(b) (*a* + 3*b*): (3*a* + *b*)

(c) (3*a* + *b*) : (*a* + 3*b*)

(d) (2*a* + *b*) : (3*a* + *b*)

#### Answer:

**Given: **(1) ABCD is a trapezium, with parallel sides AB and DC

(2) AB = a cm

(3) DC = b cm

(4) E is the midpoint of non parallel sides AD.

(5) G is the midpoint of non parallel sides BC.

**To find:** Ratio of the area of the Quadrilaterals ABFE and EFCD.

**Calculation: **We know that, ‘**Area of a trapezium is half the product of its height and the sum of the parallel sides.’**

Since, E and F are mid points of AD and BC respectively, so h_{1} = h_{2}

Area of trapezium ABFE

Now, Area (trap ABCD) = area (trap EFCD) + Area (ABFE)

Therefore,

Thus,

Hence, the correct option is (c)

#### Page No 14.61:

#### Question 18:

*ABCD* is a rectangle with *O* as any point in its interior. If ar (Δ*AOD*) = 3 cm^{2} and ar (Δ*ABOC*) = 6 cm^{2}, then area of rectangle *ABCD* is

(a) 9 cm^{2}

(b) 12 cm^{2}

(c) 15 cm^{2}

(d) 18 cm^{2}

#### Answer:

**Given:** A rectangle ABCD , O is a point in the interior of the rectangles such that

(1) ar (ΔAOB) = 3 cm^{2}

(2) ar (ΔBOC) = 6 cm^{2}

**To find:** ar (rect.ABCD)

**Construction:** Draw a line LM passing through O and parallel to AD and BC.

**Calculation:** We know that **,” If a triangle and a parallelogram are on the same base and between the same parallels the area of the triangle is equal to half the area of the parallelogram”**

Here we can see that ΔAOD and rectangle AMLD are on the same base AD and between the same parallels AD and LM.

Hence ,

Similarly, we can see that ΔBOC and rectangle BCLM are on the same base BC and between the same parallels BC and LM

Hence,

We known that

Hence the correct answer is option (d).

#### Page No 14.61:

#### Question 19:

In the given figure, a parallelogram *ABCD *and a rectangle *ABEF *are of equal area. Then,

(a) perimeter of *ABCD *= perimeter of *ABEF*

(b) perimeter of *ABCD *< perimeter of *ABEF*

(c) perimeter of *ABCD *> perimeter of *ABEF*

(d) perimeter of *ABCD *= $\frac{1}{2}$ perimeter of *ABEF*

#### Answer:

We know, opposite sides of a rectangle are equal.

Therefore, in rectangle *ABEF*,

*AB = EF* ...(1)

Also, opposite sides of a parallelogram are equal.

Therefore, in parallelogram *ABCD*,

*AB = DC* ...(3)

From (1) and (3),

*DC = EF*

⇒ *AB + EF = **AB + DC* ...(5)

Now, we know that, of all the line segments, perpendicular segment is the shortest.

∴ *AF < AD*

*BE < BC*

⇒ AF + BE < AD + BC ...(6)

Adding (5) and (6), we get

*AB + EF + AF + BE < AB + DC + AD + BC*

⇒ Perimeter of rectangle < perimeter of parallelogram

Hence, the correct option is (c).

#### Page No 14.61:

#### Question 20:

In the given figure, the area of parallelogram *ABCD *is

(a) *AB × BM*

(b) *BC × BN*

(c) *DC × DL*

(d) *AD × DL*

#### Answer:

We know, area of the parallelogram = Base × Altitude

Therefore,

The area of parallelogram *ABCD = AB *× *DL*

* = DC *× *DL*

Hence, the correct option is (c).

#### Page No 14.61:

#### Question 21:

*ABCD *is quadrilateral whose diagonal *AC *divided it into two parts, equal in area, then *ABCD*

(a) is a rectangle

(b) is always a rhombus

(c) is a parallelogram

(d) need not be any of (a), (b) or (c)

#### Answer:

The diagonal of a parallelogram, rectangle, rhombus, or square divides them into two parts, equal in area.

Therefore, if *ABCD *is a quadrilateral whose diagonal *AC *divides it into two parts, equal in area, then *ABCD *need not be a rectangle, parallelogram, or rhombus as it can be a square as well.

Hence, the correct answer is option (d).

#### Page No 14.61:

#### Question 22:

The mid-points of the sides of a triangle ABC along with any of the vertices as the fourth point make a parallelogram of area equal to

(a) ar (Δ*ABC*)

(b) $\frac{1}{2}$ar (Δ*ABC*)

(c) $\frac{1}{3}$ar (Δ*ABC*)

(d) $\frac{1}{4}$ar (Δ*ABC*)

#### Answer:

**Given:** (1) ABCD is a triangle.

(2) mid points of the sides of ΔABC with any of the vertices forms a parallelogram.

**To find:** Area of the parallelogram

**Calculation: **We know that:** ****Area of a parallelogram = base × height**

Hence area of ||^{gm} DECF = EC × EG

area of ||^{gm} DECF = EC × EG

area of ||^{gm} DECF = (E is the midpoint of BC)

area of ||^{gm} DECF =

area of ||^{gm} DECF =

Hence the result is option (b).

#### Page No 14.61:

#### Question 23:

In the given figure, *ABCD* and *FECG* are parallelograms equal in area. If ar (Δ*AQE*) = 12 cm^{2}, then ar (||^{gm} *FGBQ*) =

(a) 12 cm^{2}

(b) 20 cm^{2}

(c) 24 cm^{2}

(d) 36 cm^{2}

#### Answer:

**Given:** (1) Area of parallelogram ABCD is equal to Area of parallelogram FECG.

(2) If Area of ΔAQE is 12cm.

**To find:** Area of parallelogram FGBQ

**Calculation: **We know that **diagonal of a parallelogram divides the parallelogram into two triangles of equal area.**

It is given that,

Hence the correct answer is option (c).

#### Page No 14.62:

#### Question 1:

The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm is _________.

#### Answer:

Given:

A rhombus *ABCD* with diagonals 12 cm and 16 cm

i.e., *AC* = 16 cm and *BD* = 12 cm

And a quadrilateral *PQRS* formed by joining the mid-points of the adjacent sides of *ABCD.*

Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.

In ∆*ABC*,

*PQ || AC*

*PQ* = $\frac{1}{2}$*AC
⇒ PQ = *$\frac{1}{2}$(16)

*⇒ PQ =*8 cm

In ∆

*ADC*,

*RS || AC*

*RS*= $\frac{1}{2}$

*AC*

⇒ RS =$\frac{1}{2}$(16)

⇒ RS =

*⇒ RS =*8 cm

In ∆

*BCD*,

*RQ || BD*

*RQ*= $\frac{1}{2}$

*BD*

⇒ RQ =$\frac{1}{2}$(12)

⇒ RQ =

*⇒ RQ =*6 cm

In ∆

*BAD*,

*SP || BD*

*SP*= $\frac{1}{2}$

*BD*

⇒ SP =$\frac{1}{2}$(12)

⇒ SP =

*⇒ SP =*6 cm

Since,

*PQ =*8 cm =

*RS*and

*RQ =*6 cm =

*SP*

and Diagonals of a rhombus intersect at right angle.

*⇒*angle between

*AC*and

*BD*

__is 90°__

*⇒*angle between

*PQ*and

*QR*is 90°

Therefore,

*PQRS*is a rectangle

Thus, Area of rectangle =

*PQ × QR*

= 8 × 6

= 48 cm

^{2}

Hence, the area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm is

__48 cm__.

^{2}#### Page No 14.62:

#### Question 2:

The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is a ________ of area _________.

#### Answer:

Given:

A rectangle *ABCD* of sides 8 cm and 6 cm

i.e., *AB* = 8 cm and *AD* = 6 cm

And a quadrilateral *PQRS* formed by joining the mid-points of the adjacent sides of *ABCD.*

We can see that *QS || AB* and *PR || AD*

Also,* QS = AB* = 8 cm and *PR = AD *= 6 cm

Thus, angle between *QS* and *PR* is 90°

Therefore, *PQRS* is a rhombus.

Area of rhombus = $\frac{1}{2}$*× **SQ × PR
= *$\frac{1}{2}$

*×*8 × 6

*=*4 × 6

*=*24 cm

^{2}

Hence, the figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is a

__rhombus__of area

__24 cm__

^{2}.#### Page No 14.62:

#### Question 3:

If *P* is any point on the median *AD* of a Δ*ABC*, then $\frac{\mathrm{ar}\left(\u2206ABP\right)}{\mathrm{ar}\left(\u2206ACP\right)}$= _________.

#### Answer:

Given: *P* is any point on the median *AD* of a Δ*ABC*

We know, median of a triangle divides it into two triangles of equal area.

Therefore, ar (Δ*ADB*) = ar (Δ*ADC*) ...(1)

Also, ar (Δ*PDB*) = ar (Δ*PDC*) ...(2)

Subtracting (2) from (1), we get

ar (Δ*ADB*) − ar (Δ*PDB*) = ar (Δ*ADC*) − ar (Δ*PDC*)

⇒ ar (Δ*ABP*) = ar (Δ*ACP*)

⇒$\frac{\mathrm{ar}\left(\u2206ABP\right)}{\mathrm{ar}\left(\u2206ACP\right)}$= 1

Hence, $\frac{\mathrm{ar}\left(\u2206ABP\right)}{\mathrm{ar}\left(\u2206ACP\right)}$= __1__.

#### Page No 14.62:

#### Question 4:

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

#### Answer:

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.

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

$\Rightarrow \frac{\mathrm{Area}\mathrm{of}\mathrm{triangle}}{\mathrm{Area}\mathrm{of}\mathrm{parallelogram}}=\frac{1}{2}$

Hence, if a triangle and a parallelogram are on the same base and between the same parallels, then the ratio of the area of the triangle to the area of the parallelogram is __1 : 2__.

#### Page No 14.62:

#### Question 5:

Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is _________.

#### Answer:

Parallelograms on equal bases and between the same parallels are equal in area.

Area of first parallelogram = Area of the second parallelogram

$\Rightarrow \frac{\mathrm{Area}\mathrm{of}\mathrm{first}\mathrm{parallelogram}}{\mathrm{Area}\mathrm{of}\mathrm{second}\mathrm{parallelogram}}=\frac{1}{1}$

Hence, the ratio of their areas is __1 : 1__.

#### Page No 14.62:

#### Question 6:

*ABCD *is a parallelogram and *X* is the mid-point of *AB*. If ar(*AXCD*) = 24 cm^{2} , then ar(Δ*ABC*) = ________.

#### Answer:

Given:

*ABCD *is a parallelogram

*X* is the mid-point of *AB*

ar(*AXCD*) = 24 cm^{2}

We know, 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.

Thus, ar(Δ*ABC*) *= $\frac{1}{2}$*ar(*ABCD*) ...(1)

Since, *X* is the mid-point of *AB*

Therefore, ar(Δ*XBC*) *= $\frac{1}{2}$*ar(*ABC*)

= *$\frac{1}{2}$×$\frac{1}{2}$*ar(*ABCD*) (From (1))

= *$\frac{1}{4}$*ar(*ABCD*) ...(2)

Thus, ar(*AXCD*) = ar(*ABCD*) − ar(Δ*XBC*)

⇒ 24 = ar(*ABCD*) − *$\frac{1}{4}$*ar(*ABCD*) (From (2))

⇒ 24 = $\frac{3}{4}$ar(*ABCD*)

⇒ ar(*ABCD*) = $24\times \frac{4}{3}$

⇒ ar(*ABCD*) = 8 × 4

⇒ ar(*ABCD*) = 32 cm^{2}

From (1)

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

= *$\frac{1}{2}$*× 32

= 16 cm^{2}

Hence, ar(Δ*ABC*) = __16 cm__^{2}.

#### Page No 14.62:

#### Question 7:

*PQRS *is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on *PQ*. If *PS *= 5 cm, then ar(Δ*PAS*) = _______.

#### Answer:

Given:

*PQRS *is a rectangle inscribed in a quadrant of a circle of radius 13 cm

*PS *= 5 cm

*A* is any point on *PQ*

*QS* = radius of the circle = 13 cm ...(1)

In Δ*PQS*

Using pythagoras theorem,

* QS*^{2}* = PS*^{2}* + PQ*^{2}

⇒ 13^{2} = 5^{2} + *PQ*^{2}

⇒ *PQ*^{2} = 169 − 25

⇒ *PQ*^{2} = 144

⇒ *PQ* = 12 cm = *SR* ...(2)

Thus,

ar(Δ*RAS*) *= $\frac{1}{2}$*× base × height

= *$\frac{1}{2}$*× *SR × PS*

= *$\frac{1}{2}$*× 12 × 5

= 30 cm^{2}

Hence, ar(Δ*RAS*) = __30 cm__^{2}.

Disclaimer: The question is to find the area of Δ*RAS* instead of the area of Δ*PAS*.

#### Page No 14.62:

#### Question 8:

If *ABC *and *BDE *are two equilateral triangles such that *D *is the mid-point of *BC *then ar(Δ*ABC*) : ar(*BDE*) = _________.

#### Answer:

Given:

*ABC *and *BDE *are two equilateral triangles

*D *is the mid-point of *BC*

$\mathrm{ar}\left(\u2206ABC\right)=\frac{\sqrt{3}}{4}\times {\left(\mathrm{side}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}\times {\left(BC\right)}^{2}...\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{ar}\left(\u2206BDE\right)=\frac{\sqrt{3}}{4}\times {\left(\mathrm{side}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}\times {\left(BD\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}\times {\left(\frac{1}{2}BC\right)}^{2}\left(\because D\mathrm{is}\mathrm{the}\mathrm{mid}-\mathrm{point}\mathrm{of}BC\right)\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}\times \frac{1}{4}\times {\left(BC\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{16}\times {\left(BC\right)}^{2}...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{ar}\left(\u2206ABC\right)}{\mathrm{ar}\left(\u2206BDE\right)}=\frac{{\displaystyle \frac{\sqrt{3}}{4}}{\left(BC\right)}^{2}}{{\displaystyle \frac{\sqrt{3}}{16}}{\left(BC\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{4}{1}\phantom{\rule{0ex}{0ex}}$

Hence, ar(Δ*ABC*) : ar(Δ*BDE*) = __4 : 1__.

#### Page No 14.62:

#### Question 9:

If *PQRS *is a parallelogram whose area is 180 cm^{2} and *A* is any point on the diagonal *QS*, then ar(Δ*ASR*) is _______ 90 cm^{2}?

#### Answer:

Given:

*PQRS *is a parallelogram with area 180 cm^{2}

*A* is any point on the diagonal *QS*

We know, the diagonal of the parallelogram bisects it into two triangles of equal area.

Thus, ar(Δ*PQS*) = ar(Δ*QRS*) = $\frac{1}{2}$ar(*PQRS*)

⇒ ar(Δ*PQS*) = ar(Δ*QRS*) = 90 cm^{2}

Therefore, ar(Δ*ASR*) is always less than 90 cm^{2} unless or until the point *A* coincides with *Q* or *S*.

Hence, ar(Δ*ASR*) is __less than__ 90 cm^{2}

#### Page No 14.63:

#### Question 10:

The mid points of the sides of a triangle *ABC *along with any of the vertices as the fourth point make a parallelogram of area equal to ________.

#### Answer:

Given:

*ABC *is a triangle

Let *D* is the mid-point of *AB*, *E* is the mid-point of *BC *and *F* is the mid-point of *AC*.

*ADEF* is a parallelogram having 2 triangles of equal area i.e., ∆*ADF* and ∆*DEF*.

But the ∆*ABC* is divided in 4 triangles of equal area i.e., ∆*ADF*, ∆*DEF*, ∆*BED* and ∆*CEF.*

Thus, area of ∆*ABC = *2 *× *area of the parallelogram *ADEF.*

Hence, the mid-points of the sides of a triangle *ABC *along with any of the vertices as the fourth point make a parallelogram of area equal to __half the area of the triangle ABC__.

#### Page No 14.63:

#### Question 11:

A median of a triangle divides it into two ___________.

#### Answer:

Let *ABC* be a triangle with a mid-point *D* on *BC*.

Therefore, *BD = DC*

Let *AE* be the altitude from *A* on *BC.*

Now, ar(∆*ABD*) = $\frac{1}{2}$*× *base × height

* =* $\frac{1}{2}$*× BD × AE*

Also, ar(∆*ACD*) = $\frac{1}{2}$*× *base × height

* =* $\frac{1}{2}$*× CD × AE
=* $\frac{1}{2}$

*× BD × AE*(∵

*BD = CD*)

*=*ar(∆

*ABD*)

Hence, a median of a triangle divides it into two

__triangles of equal area__.

#### Page No 14.63:

#### Question 12:

If *ABCD *is a rectangle, *E* and *F* are the mid-points of *BC *and *AD *respectively and *G *is any point on *EF*. Then ar(Δ*GAB*) : ar (rectangle *ABCD*) = _________.

#### Answer:

Given:

*ABCD *is a rectangle

*E* and *F* are the mid-points of *BC *and *AD *respectively

*G *is any point on *EF*

Since, *E* and *F* are the mid-points of *BC *and *AD *respectively.

Therefore, ar(*BEFA*) = ar(*ECDF*) = $\frac{1}{2}$*× *ar(*ABCD*) ...(1)

We know, if a triangle and a rectangle are on the same base and between the same parallels, then the area of the triangle is half the area of the rectangle.

Thus, ar(Δ*GAB*) *= $\frac{1}{2}$*ar(*BEFA*) ...(2)

From (1) and (2),

ar(Δ*GAB*) = *$\frac{1}{2}$×*$\frac{1}{2}$*× *ar(*ABCD*)

⇒ ar(Δ*GAB*) = $\frac{1}{4}$*× *ar(*ABCD*)

⇒ $\frac{\mathrm{ar}\left(\u2206GAB\right)}{\mathrm{ar}\left(\mathrm{rectangle}ABCD\right)}=\frac{1}{4}$

**

Hence, ar(Δ*GAB*) : ar(rectangle *ABCD*) = __1 : 4__.

#### Page No 14.63:

#### Question 13:

*PQRS *is a square of side 8 cm. *T *and *U *and respectively the mid points of *PS *and *QR*. If *TU *and *QS *intersect at 0, then ar(Δ*OTS*) = ________.

#### Answer:

Given:

*PQRS *is a square of side 8 cm.

*T *and *U *are respectively the mid-points of *PS *and *QR*

*TU *and *QS *intersect at *O*

In Δ*QOU *and Δ*OTS*,

∠*QOU* = ∠*TOS* (vertically opposite angles)

∠*OQU* = ∠*OST* (alternate angles)

*QU = TS* (mid-points of sides of a square)

By AAS property,

Δ*QOU* ≅ Δ*OTS*

Thus, *OU = OT* (by CPCT)

⇒ *OU + OT *= *PQ* = 8 cm

⇒ *OU = OT *= 4 cm ...(1)

Also, *TS* = 4 cm (*T* is the mid-point of *PS*) ...(2)

ar(Δ*OTS*) = *$\frac{1}{2}$*× base × height

* = $\frac{1}{2}$*× *TS × OT*

* = $\frac{1}{2}$*× 4 × 4 (From (1) and (2))

* = *8 cm^{2}

**

Hence, ar(Δ*OTS*) = __8 cm ^{2}__.

#### Page No 14.63:

#### Question 14:

In the given figure, *ABCD *and *EFGD *are two parallelograms and G is the mid point of CD. Then, ar(Δ*DPC*) : ar(*EFGD*) = ________.

#### Answer:

Given:

*ABCD *and *EFGD *are two parallelograms.

*G* is the mid point of *CD*

Since, *G* is the mid point of *CD*

Therefore*, DG = GC*

Since, Δ*DPG *and Δ*GPC *have equal base and common height,

Thus, ar(Δ*DPG*) = ar(Δ*GPC*)

⇒ ar(Δ*DPC*) = 2 ar(Δ*DPG*) ...(1)

Also, 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.

Thus, ar(Δ*DPG*) *= $\frac{1}{2}$*ar(*EFGD*) ...(2)

From (1) and (2),

ar(Δ*DPC*) = 2 × *$\frac{1}{2}$*ar(*EFGD*)

⇒ ar(Δ*DPC*) = ar(*EFGD*)

**

Hence, ar(Δ*DPC*) : ar(*EFGD*) = __1 : 1__.

#### Page No 14.63:

#### Question 15:

If the given figure, *PQRS *and *EFRS *are two parallelograms, then ar(||^{gm} *PQRS*) : ar(Δ*MFR*) = ________.

#### Answer:

Given:

*PQRS *and *EFRS *are two parallelograms

*PQRS *and *EFRS *are two parallelograms lying on the same base *SR* and between the same parallels *SR* and *PF*.

We know, if two parallelograms are on the same base and between the same parallels, then the area of the parallelograms are equal.

Thus, ar(*PQRS*) = ar(*EFRS*) ...(1)

Also, Δ*MFR *and parallelogram *EFRS *is lying on the same base *FR* and between the same parallels *SR* and *EF*.

We know, 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.

Thus, ar(Δ*MFR*) *= $\frac{1}{2}$*ar(*EFRS*) ...(2)

From (1) and (2),

ar(Δ*MFR*) = *$\frac{1}{2}$*ar(*PQRS*)

⇒ ar(||^{gm} *PQRS*) = 2 ar(Δ*MFR*)

**

Hence, ar(||^{gm} *PQRS*) : ar(Δ*MFR*) = __2 : 1__.

#### Page No 14.63:

#### Question 16:

In the given figure, if *ABCD *is a parallelogram of area 90 cm^{2}. Then, ar(||^{gm} *ABEF*) = ________ ar(Δ*ABD*) = _______ and ar(*ΔBEF*) = _________.

#### Answer:

Given:

*ABCD *is a parallelogram of area 90 cm^{2}

*ABCD *and *ABEF *are two parallelograms lying on the same base *AB* and between the same parallels *AB* and *CF*.

We know, if two parallelograms are on the same base and between the same parallels, then the area of the parallelograms are equal.

Thus, ar(*ABCD*) = ar(*ABEF*) = 90 cm^{2} ...(1)

Also, Δ*ABD *and parallelogram *ABEF *is lying on the same base *AB* and between the same parallels *AF* and *BE*.

We know, 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.

Thus, ar(Δ*ABD*) *= $\frac{1}{2}$*ar(*ABEF*)

⇒ ar(||^{gm} *ABEF*) = 2 ar(Δ*ABD*) ...(2)

From (1) and (2),

ar(||^{gm} *ABEF*) = 2 ar(Δ*ABD*) = 90 cm^{2}

**

Also, diagonal of a parallelogram divides it into two triangles of equal area.

Thus, ar(Δ*BEF*) *= $\frac{1}{2}$*ar(*ABEF*)

= *$\frac{1}{2}$*(90)

= 45 cm^{2}

Hence, ar(||^{gm} *ABEF*) = __2__ ar(Δ*ABD*) = __90 cm ^{2}__ and ar(

*ΔBEF*) =

__45 cm__.

^{2}#### Page No 14.63:

#### Question 1:

If *ABC* and *BDE* are two equilateral triangles such that *D* is the mid-point of *BC*, then find ar (Δ*ABC*) : ar (Δ*BDE*).

#### Answer:

**Given:** (1) ΔABC is equilateral triangle.

(2) ΔBDE is equilateral triangle.

(3) D is the midpoint of BC.

**To find:**

PROOF : Let us draw the figure as per the instruction given in the question.

We know that area of equilateral triangle = , where *a* is the side of the triangle.

Let us assume that length of BC is *a* cm.

This means that length of BD is cm, Since D is the midpoint of BC.

------(1)

------(2)

Now, ar(ΔABC) : ar(ΔBDE) = (from 1 and 2)

=

Hence we get the result ar(ΔABC) : ar(ΔBDE) =

#### Page No 14.64:

#### Question 2:

In the given figure, *ABCD* is a rectangle in which *CD* = 6 cm, *AD* = 8 cm. Find the area of parallelogram *CDEF*.

#### Answer:

**Given:** (1) ABCD is a rectangle.

(2) CD = 6 cm

(3) AD = 8cm

**To find:** Area of rectangle CDEF.

**Calculation:** We know that,

_{Area of parallelogram = base × height}

**The Area of parallelogram and a rectangle on the same base and between the same parallels are equal in area.**

Here we can see that rectangle ABCD and Parallelogram CDEF are between the same base and same parallels.

Hence,

Hence we get the result as Area of Rectangle CDEF =

#### Page No 14.64:

#### Question 3:

In the given figure, find the area of ΔGEF.

#### Answer:

**Given:** (1) ABCD is a rectangle.

(2) CD = 6 cm

(3) AD = 8cm

**To find:** Area of ΔGEF.

**Calculation:** We know that,

_{Area of Parallelogram = base × height}

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

Here we can see that Parallelogram ABCD and triangle GEF are between the same base and same parallels.

Hence,

Hence we get the result as

#### Page No 14.64:

#### Question 4:

In the given figure, *ABCD* is a rectangle with sides *AB* = 10 cm and *AD* = 5 cm. Find the area of ΔEFG.

#### Answer:

**Given:** (1) ABCD is a rectangle.

(2) AB = 10 cm

(3) AD = 5cm

**To find:** Area of ΔEGF.

**Calculation:** We know that,

Area of Rectangle = base × height

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

Here we can see that Rectangle ABCD and triangle GEF are between the same base and same parallels.

Hence,

Hence we get the result as Area of ΔGEF =

#### Page No 14.64:

#### Question 5:

*PQRS* is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on *PQ*. If *PS* = 5 cm, then find *ar (ΔRAS)*

#### Answer:

**Given:** Here from the given figure we get

(1) PQRS is a rectangle inscribed in a quadrant of a circle with radius 10cm,

(2) PS = 5cm

(3) PR = 13cm(radius of the quadrant)

**To find:** Area of ΔRAS.

**Calculation: **In right ΔPSR, (Using Pythagoras Theorem)

Hence we get the Area of ΔRAS =

#### Page No 14.64:

#### Question 6:

In square *ABCD*, *P* and *Q* are mid-point of *AB* and *CD* respectively. If *AB* = 8cm and *PQ* and *BD* intersect at *O*, then find area of Δ*OPB*.

#### Answer:

**Given:** Here from the given question we get

(1) ABCD is a square,

(2) P is the midpoint of AB

(3) Q is the midpoint of CD

(4) PQ and BD intersect at O.

(5) AB = 8cm

**To find :** Area of ΔOPB

**Calculation:** Since P is the midpoint of AB,

BP = 4cm ……(1)

Hence we get the Area of ΔOBP =

#### Page No 14.64:

#### Question 7:

*ABC* is a triangle in which *D* is the mid-point of *BC*. *E *and *F* are mid-points of *DC* and *AE* respectively. IF area of Δ*ABC* is 16 cm^{2}, find the area of Δ*DEF*.

#### Answer:

**Given:** Here from the given question we get

(1) ABC is a triangle

(2) D is the midpoint of BC

(3) E is the midpoint of CD

(4) F is the midpoint of A

Area of ΔABC = 16 cm^{2}

**To find :** Area of ΔDEF

**Calculation: **We know that ,

**The median divides a triangle in two triangles of equal area.**

For ΔABC, AD is the median

For ΔADC , AE is the median .

Similarly, For ΔAED , DF is the median .

Hence we get Area of ΔDEF =

#### Page No 14.64:

#### Question 8:

*PQRS* is a trapezium having *PS* and *QR* as parallel sides. *A* is any point on *PQ* and *B *is a point on *SR* such that *AB* || *QR*. If area of Δ*PBQ* is 17cm^{2}, find the area of Δ*ASR*.

#### Answer:

**Given:** Here from the given figure we get

(1) PQRS is a trapezium having PS||QR

(2) A is any point on PQ

(3) B is any point on SR

(4) AB||QR

(5) Area of ΔBPQ = 17 cm^{2}

**T**o find : Area of ΔASR.

**Calculation: **We know that **‘If a triangle and a parallelogram are on the same base and the same parallels, the area of the triangle is equal to half the area of the parallelogram’**

Here we can see that:

Area (ΔAPB) = Area (ΔABS) …… (1)

And, Area (ΔAQR) = Area (ΔABR) …… (2)

Therefore,

Area (ΔASR) = Area (ΔABS) + Area (ΔABR)

From equation (1) and (2), we have,

Area (ΔASR) = Area (ΔAPB) + Area (ΔAQR)

⇒ Area (ΔASR) = Area (ΔBPQ) = 17 cm^{2}

Hence, the area of the triangle ΔASR is 17 cm^{2}.

#### Page No 14.64:

#### Question 9:

*ABCD* is a parallelogram. *P* is the mid-point of *AB*. *BD* and *CP* intersect at *Q* such that *CQ* : *QP* = 3.1. If ar (Δ*PBQ*) = 10cm^{2}, find the area of parallelogram *ABCD*.

#### Answer:

It is given that CQ : QP = 3: 1 and Area (PBQ) = 10 cm^{2}

Let CQ = x and QP = 3x

We need to find area of the parallelogram ABCD.

From the figure,

Area (PBQ) =

And,

Area (BQC) =

Now, let H be the perpendicular distance between AP and CD. Therefore,

Area (PCB) = …… (1)

Thus the area of the parallelogram ABCD is,

Area (ABCD) = AB × H

⇒ Area (ABCD) = 2BP × H

From equation (1), we get

Area (ABCD) = 4 × 30 = 120 cm^{2}

Hence, the area of the parallelogram ABCD is 120 cm^{2}.

#### Page No 14.64:

#### Question 10:

*P* is any point on base *BC* of Δ*ABC** *and *D* is the mid-point of* BC*. *DE* is drawn parallel to* PA* to meet *AC* at *E*. If ar (Δ*ABC*) = 12 cm^{2}, then find area of Δ*EPC*.

#### Answer:

**Given:** Area (ABC) = 12 cm^{2}, D is midpoint of BC and AP is parallel to ED. We need to find area of the triangle EPC.

Since, AP||ED, and we know that the area of triangles between the same parallel and on the same base are equal. So,

Area (APE) = Area (APD)

⇒ Area (APM) + Area (AME) = Area (APM) + Area (PMD)

⇒ Area (AME) = Area (PMD) …… (1)

Since, median divide triangles into two equal parts. So,

Area (ADC) = Area (ABC) = = 6 cm^{2}

⇒ Area (ADC) = Area (MDCE) + Area (AME)

⇒Area (ADC) = Area (MDCE) + Area (PMD) (from equation (1))

⇒ Area (ADC) = Area (PEC)

Therefore,

Area (PEC) = 6 cm^{2}.

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