RD Chapter 15- Areas of Parallelograms and Triangles Ex-15.1 |
RD Chapter 15- Areas of Parallelograms and Triangles Ex-15.2 |
RD Chapter 15- Areas of Parallelograms and Triangles Ex-15.3 |

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

**Answer
1** :

**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 theinstruction 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 midpointof BC.*

*------(1)*

*------(2)*

*Now, ar(ÄABC) : ar(ÄBDE) = (from 1 and 2)*

*=*

*Hence we get the result ar(ÄABC) : ar(ÄBDE) = *

**Answer
2** :

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 =

**Answer
3** :

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

**Answer
4** :

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 ÄGEF =

**Answer
5** :

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 ÄRAS =

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

**Answer
6** :

Given: Here from thegiven question we get

(1) ABCD is a square,

(2) P is the midpoint ofAB

(3) Q is the midpoint ofCD

(4) PQ and BD intersectat O.

(5) AB = 8cm

To find : Area of ΔOPB

Calculation: Since P is themidpoint of AB,

BP = 4cm ……(1)

Hence we get the Areaof ÄOBP =

**Answer
7** :

Given: Here from thegiven question we get

(1) ABC is a triangle

(2) D is the midpoint ofBC

(3) E is the midpoint ofCD

(4) F is the midpoint ofA

Area of ÄABC = 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 themedian

For ΔADC , AE is themedian .

Similarly, For ΔAED , DFis the median .

Hence we get Areaof ÄDEF =

**Answer
8** :

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 BPQ = 17 cm^{2}

To find : Area of ΔASR.

Calculation: We know that ‘If a triangle and a parallelogram are on thesame base and the same parallels, the area of the triangle is equal to half thearea 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 thetriangle ΔASR is 17 cm^{2}.

**Answer
9** :

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 ofthe parallelogram ABCD.

From the figure,

Area (PBQ) =

And,

Area (BQC) =

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

Area (PCB) = …… (1)

Thus the area of theparallelogram ABCD is,

Area (ABCD) = AB × H

⇒ Area (ABCD) = 2BP × H

From equation (1), weget

Area (ABCD) = 4 × 30 =120 cm^{2}

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

**Answer
10** :

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

Since, AP||ED, and weknow that the area of triangles between the same parallel and on the same baseare equal. So,

Area (APE) = Area (APD)

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

⇒ Area (AME) = Area (PMD) …… (1)

Since, median dividetriangles 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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