Area Theorem Class-9th Concise ICSE Maths Selina Solutions Chapter-16. We provide step by step Solutions of Exercise / lesson-16 Area Theorem  for  ICSE  Class-9th Concise Selina Mathematics by R K Bansal.

Our Solutions contain all type Questions with Exe-16 B, Exe-16 B,  Exe-16 C, to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-9th Mathematics .

## Area Theorem Class-9th Concise ICSE Maths Selina Solutions Chapter-16

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Exercise 16 A,

Exercise-16 B,

Exercise-16 C,

### Exercise – 16 A, Area Theorem Class-9th Concise ICSE Maths Selina Solutions

#### Question 1

In the given figure, if area of triangle ADE is 60 cm2, state, given reason, the area of :

(i) Parallelogram ABED;

(ii) Rectangle ABCF;

(iii) Triangle ABE.

………………..

(i)ΔADE and parallelogram ABED are on the same base AB and between the same parallels DE//AB, so area of the triangle ΔADE is half the area of parallelogram ABED.

Area of ABED = 2 (Area of ADE) = 120 cm2

(ii)Area of parallelogram is equal to the area of rectangle on the same base and of the same altitude i.e, between the same parallels

Area of ABCF = Area of ABED = 120 cm2

(iii)We know that area of triangles on the same base and between same parallel lines are equal

Area of ABE=Area of ADE =60 cm2

#### Question 2

The given figure shows a rectangle ABDC and a parallelogram ABEF; drawn on opposite sides of AB. Prove that:

(i) Quadrilateral CDEF is a parallelogram;

= Area of rect. ABDC

+ Area of // gm. ABEF.

……………..

After drawing the opposite sides of AB, we get Since from the figure, we get CD//FE therefore FC must parallel to DE. Therefore it is proved that the quadrilateral CDEF is a parallelogram.

Area of parallelogram on same base and between same parallel lines is always equal and area of parallelogram is equal to the area of rectangle on the same base and of the same altitude i.e, between same parallel lines.

So Area of CDEF= Area of ABDC + Area of ABEF

Hence Proved

#### Question 3

In the given figure, diagonals PR and QS of the parallelogram PQRS intersect at point O and LM is parallel to PS. Show that:

…………………………

(i)

Since POS and parallelogram PMLS are on the same base PS and between the same parallels i.e. SP//LM.

As O is the center of LM and Ratio of area of triangles with same vertex and bases along the same line is equal to ratio of their respective bases.

The area of the parallelogram is  twice the area of the triangle if they lie on the same base and in between the same parallels.

So 2(Area of PSO)=Area of PMLS

Hence Proved.

(ii) #### Question 4

In parallelogram ABCD, P is a point on side AB and Q is a point on side BC.

Prove that:

……………… Given ABCD is a parallelogram. P and Q are any points on the sides AB and BC respectively, join diagonals AC and BD.

proof:

since triangles with same base and between same set of parallel lines have equal areas

area (CPD)=area(BCD)…… (1)

again, diagonals of the parallelogram bisects area in two equal parts

area (BCD)=(1/2) area of parallelogram ABCD…… (2)

from (1) and (2)

area(CPD)=1/2 area(ABCD)…… (3)

similarly area (AQD)=area(ABD)=1/2 area(ABCD)…… (4)

from (3) and (4)

area(CPD)=area(AQD),

hence proved.

(ii)

We know that area of triangles on the same base and between same parallel lines are equal

So Area of AQD= Area of ACD= Area of PDC = Area of BDC = Area of ABC=Area of APD + Area of BPC

Hence Proved

#### Question 5

In the given figure, M and N are the mid-points of the sides DC and AB respectively of the parallelogram ABCD.

If the area of parallelogram ABCD is 48 cm2;

(i) State the area of the triangle BEC.

(ii) Name the parallelogram which is equal in area to the triangle BEC.

(i)

Since triangle BEC and parallelogram ABCD are on the same base BC and between the same parallels i.e. BC//AD. Therefore, Parallelograms ANMD and NBCM have areas equal to triangle BEC

#### Question 6

In the following figure, CE is drawn parallel to diagonals DB of the quadrilateral ABCD which meets AB produced at point E.

Since ΔDCB and DEB are on the same base DB and between the same parallels i.e. DB//CE, therefore we get #### Question 7

ABCD is a parallelogram a line through A cuts DC at point P and BC produced at Q. Prove that triangle BCP is equal in area to triangle DPQ.

………….

ΔAPB and parallelogram ABCD are on the same base AB and between the same parallel lines AB and CD. Question 8

The given figure shows a pentagon ABCDE. EG drawn parallel to DA meets BA produced at G and CF draw parallel to DB meets AB produced at F.

Prove that the area of pentagon ABCDE is equal to the area of triangle GDF.

…………………..

#### Question 9

In the given figure, AP is parallel to BC, BP is parallel to CQ. Prove that the area of triangles ABC and BQP are equal.

…………………..

Joining PC we get Hence proved.

#### Question 10

In the figure given alongside, squares ABDE and AFGC are drawn on the side AB and the hypotenuse AC of the right triangle ABC.

…………..

(i) …………

(ii) Area of the square ABDE

= Area of the rectangle ARHF.

#### Question 11

In the following figure, DE is parallel to BC. Show that:

(ii) Area (ΔBOD) = Area (ΔCOE)..

…………. Hence Proved

(ii)

We know that area of triangles on the same base and between same parallel lines are equal

Area(triangle DBC)= Area(triangle BCE)

Area(triangle DOB) + Area(triangle BOC) = Area(triangle BOC) + Area(triangle COE)

So Area(triangle DOB) = Area(triangle COE)

Question 12

ABCD and BCFE are parallelograms. If area of triangle EBC = 480 cm2; AB = 30 cm and BC = 40 cm; Calculate;

(i) Area of parallelogram ABCD;

(ii) Area of the parallelogram BCFE;

(iii) Length of altitude from A on CD;

(iv) Area of triangle ECF.

………………..

#### Question 13

In the given figure, D is mid-point of side AB of ΔABC and BDEC is a parallelogram.

Prove that:

Area of ΔABC = Area of // gm BDEC. 