## RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A.

**Question 1.**

**Solution:**

Given : In the figure, ABCD is a quadrilateral and

AB = CD = 5cm

**Question 2.**

**Solution:**

In ||gm ABCD,

AB = 10cm, altitude DL = 6cm

and BM is altitude on AD, and BM = 8 cm.

**Question 3.**

**Solution:**

Diagonals of rhombus are 16cm and 24 cm.

Area = \(\frac { 1 }{ 2 } \) x product of diagonals

= \(\frac { 1 }{ 2 } \) x 1st diagonal x 2nd diagonal

= \(\frac { 1 }{ 2 } \) x 16 x 24

= 192 cm² Ans.

**Question 4.**

**Solution:**

Parallel sides of a trapezium are 9cm and 6cm and distance between them is 8cm

**Question 5.**

**Solution:**

from the figure

(i) In ∆ BCD, ∠ DBC = 90°

**Question 6.**

**Solution:**

In the fig, ABCD is a trapezium. AB || DC

AB = 7cm, AD = BC = 5cm.

Distance between AB and DC = 4 cm.

i.e. ⊥AL = ⊥BM = 4cm.

**Question 7.**

**Solution:**

Given : In quad. ABCD. AL⊥BD and CM⊥BD.

To prove : ar(quad. ABCD)

**Question 8.**

**Solution:**

In quad. ABCD, BD is its diagonal and AL⊥BD, CM⊥BD

**Question 9.**

**Solution:**

Given : ABCD is a trapezium in which AB || DC and its diagonals AC and BD intersect each other at O.

To prove : ar(∆ AOD) = ar(∆ BOC)

**Question 10.**

**Solution:**

Given : In the figure,

DE || BC.

**Question 11.**

**Solution:**

Given : In ∆ ABC, D and E are the points on AB and AC such that

ar( ∆ BCE) = ar( ∆ BCD)

To prove : DE || BC.

Proof : (∆ BCE) = ar(∆ BCD)

But these are on the same base BC.

Their altitudes are equal.

Hence DE || BC

Hence proved.

**Question 12.**

**Solution:**

Given : In ||gm ABCD, O is any. point inside the ||gm. OA, OB, OC and OD are joined.

**Question 13.**

**Solution:**

Given : In quad. ABCD.

A line through D, parallel to AC, meets ‘BC produced in P. AP in joined which intersects CD at E.

To prove : ar( ∆ ABP) = ar(quad. ABCD).

Const. Join AC

**Question 14.**

**Solution:**

Given : ∆ ABC and ∆ DBC are on the same base BC with points A and D on , opposite sides of BC and

ar( ∆ ABC) = ar( ∆ DBC).

**Question 15.**

**Solution:**

Given : In ∆ ABC, AD is the median and P is a point on AD

BP and CP are joined

To prove : (i) ar(∆BDP) = ar(∆CDP)

(ii) ar( ∆ ABP) = ar( ∆ ACP)

**Question 16.**

**Solution:**

Given : In quad. ABCD, diagonals AC and BD intersect each other at O and BO = OD

To prove : ar(∆ ABC) = ar(∆ ADC)

Proof : In ∆ ABD,

**Question 17.**

**Solution:**

In ∆ ABC,D is mid point of BC

and E is midpoint of AD and BE is joined.

**Question 18.**

**Solution:**

Given : In ∆ ABC. D is a point on AB and AD is joined. E is mid point of AD EB and EC are joined.

To prove : ar( ∆ BEC) = \(\frac { 1 }{ 2 } \) ar( ∆ ABC)

Proof : In ∆ ABD,

E is midpoint of AD

BE is its median

ar(∆ EBD) = ar(∆ ABE)

**Question 19.**

**Solution:**

Given : In ∆ ABC, D is midpoint of BC and E is die midpoint of BO is the midpoint of AE.

To prove that ar( ∆ BOE) = \(\frac { 1 }{ 8 } \) ar(∆ ABC).

**Question 20.**

**Solution:**

Given : In ||gm ABCD, O is any point on diagonal AC.

To prove : ar( ∆ AOB) = ar( ∆ AOD)

Const. Join BD which intersects AC at P

Proof : In ∆ OBD,

P is midpoint of BD

(Diagonals of ||gm bisect each other)

**Question 21.**

**Solution:**

Given : ABCD is a ||gm.

P, Q, R and S are the midpoints of sides AB, BC, CD, DA respectively.

PQ, QR, RS and SP are joined.

**Question 22.**

**Solution:**

Given : In pentagon ABCDE,

EG || DA meets BA produced and

CF || DB, meets AB produced.

To prove : ar(pentagon ABCDE) = ar(∆ DGF)

**Question 23.**

**Solution:**

Given ; A ∆ ABC in which AD is the median.

To prove ; ar( ∆ ABD) = ar( ∆ ACD)

Const : Draw AE⊥BC.

Proof : Area of ∆ ABD

**Question 24.**

**Solution:**

Given : A ||gm ABCD in which AC is its diagonal which divides ||gm ABCD in two ∆ ABC and ∆ ADC.

**Question 25.**

**Solution:**

Given : In ∆ ABC,

D is a point on BC such that

BD = \(\frac { 1 }{ 2 } \) DC

AD is joined.

**Question 26.**

**Solution:**

Given : In ∆ ABC, D is a point on BC such that

BD : DC = m : n

AD is joined.

Hope given RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A are helpful to complete your math homework.

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