## 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.**

**In the adjoining figure, show that ABCD is a parallelogram. Calculate the area of ||gm ABCD.**

**Solution:**

Given : In the figure, ABCD is a quadrilateral and

AB = CD = 5cm

**Question 2.**

**In a parallelogram ABCD, it is being given that AB = 10 cm and the altitude corresponding to the sides AB and AD are DL = 6 cm and BM = 8 cm, respectively. Find AD.**

**Solution:**

In ||gm ABCD,

AB = 10cm, altitude DL = 6cm

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

**Question 3.**

**Find the area of a rhombus, the lengths of whose diagonals are 16cm and 24cm respectively.**

**Solution:**

Diagonals of rhombus are 16cm and 24 cm.

Area = x product of diagonals

= x 1st diagonal x 2nd diagonal

= x 16 x 24

= 192 cm² Ans.

**Question 4.**

**Find the area of a trapezium whose parallel sides are 9cm and 6cm respectively and the distance between these sides is 8cm.**

**Solution:**

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

**Question 5.**

**(i) Calculate the area of quad. ABCD, given in fig. (i).**

**(ii) Calculate the area of trap. PQRS, given in fig. (ii).**

**Solution:**

from the figure

(i) In ∆ BCD, ∠ DBC = 90°

**Question 6.**

**In the adjoining figure, ABCD is a trapezium in which AB || DC ; AB = 7 cm ; AD = BC = 5 cm and the distance between AB and DC is 4 cm. Find the length of DC and hence, find the area of trap.ABCD.**

**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.**

**BD is one of the diagonals of a quad. ABCD. If AL⊥BD and CM⊥BD, show that:**

**Solution:**

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

To prove : ar(quad. ABCD)

**Question 8.**

**In the adjoining figure, ABCD is a quadrilateral in which diag. BD = Hem. If AL ⊥ BD and and CM ⊥ BD such that AL = 8cm and CM 6cm, find the area of quad. ABCD.**

**Solution:**

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

**Question 9.**

**In the adjoining figure, ABCD is a trapezium in which AB || DC and its diagonals AC and BD intersect at O. Prove that ar( ∆ AOD) = ar( ∆ BOC).**

**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.**

**In the adjoining figure, DE || BC. Prove that**

**(i) ar( ∆ ACD) = ar( ∆ ABE),**

**(ii) ar( ∆ OCE) = ar( ∆ OBD).**

**Solution:**

Given : In the figure,

DE || BC.

**Question 11.**

**In the adjoining figure, D and E are points on the sides AB and AC of ∆ ABC such that:**

**ar( ∆ BCE) = ar( ∆ BCD).**

**Show that DE || BC.**

**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.**

**In the adjoining figure, O is any point inside a parallelogram ABCD.Prove that**

**(i) ar( ∆ OAB) + ar( ∆ OCD) = ar(||gm ABCD),**

**(ii) ar( ∆ OAD) + ar( ∆ OBC) = ar(||gm ABCD).**

**Solution:**

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

**Question 13.**

**In the adjoining figure, ABCD is a quadrilateral. A line through D, parallel to AC, meets BC produced in P.**

**Prove that ar( ∆ ABP) = ar(quad. ABCD).**

**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.**

**In the adjoining figure, ∆ ABC and ∆ DBC are on the same base BC with A and D on opposite sides of BC such that ar( ∆ ABC) = ar( ∆ DBC). Show that BC bisects AD.**

**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.**

**In the adjoining figure, AD is the medians of a ∆ ABC and P is a point on AD.Prove that (i) ar( ∆ BDP) = ar( ∆ CDP), (ii) ar( ∆ ABP) = ar( ∆ ACP).**

**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.**

**In-the adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O.If BO = OD,**

**prove that : ar( ∆ ABC) = ar( ∆ ADC).**

**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.**

**ABC is a triangle in which D is the midpoint of BC and E is the midpoint of AD.**

**Prove that ar(∆ BED) = ar(∆ ABC)**

**Solution:**

In ∆ ABC,D is mid point of BC

and E is midpoint of AD and BE is joined.

**Question 18.**

**The vertex A of ∆ ABC is joined to a f point D on the side BC. The midpoint of AD is E,**

**Prove that ar( ∆ BEC) = ar( ∆ ABC).**

**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) = ar( ∆ ABC)

Proof : In ∆ ABD,

E is midpoint of AD

BE is its median

ar(∆ EBD) = ar(∆ ABE)

**Question 19.**

**D is die midpoint of side BC of ∆ ABC and E is the midpoint of BD. If O is the midpoint of AB, prove that ar( ∆ BOE) = ar(∆ ABC).**

**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) = ar(∆ ABC).

**Question 20.**

**In the adjoining figure, ABCD is a parallelogram and O is any point on the diagonal AC. Show that ar(∆ AOB) = ar(∆ AOD).**

**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.**

**P, Q, R, S are respectively the midpoints of the sides AB, BC, CD and DA of ||gm ABCD. Show that PQRS is a parallelogram and also show that:**

**ar(||gm PQRS) = x ar(||gm ABCD).**

**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.**

**The given figure shows a pentagon ABCDE. EG, drawn parallel to DA, meets BA produced at G, and CF, drawn parallel to DB, meets AB produced at F. Show that ar(pentagon ABCDE) = ar(∆ DGF).**

**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.**

**Prove that a median divides a triangle into two triangles of equal area.**

**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.**

**Show that a diagonal divides a parallelogram into two triangles of equal area.**

**Solution:**

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

**Question 25.**

**The base BC of ∆ ABC is divided at D such that BD = DC. Prove that ar( ∆ ABD) = x ar(∆ ABC).**

**Solution:**

Given : In ∆ ABC,

D is a point on BC such that

BD = DC

AD is joined.

**Question 26.**

**In the adjoining figure, the point D divides the side BC of ∆ ABC in the ratio m : n.**

**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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Divam says

August 3, 2018 at 1:50 pmUpdate rs pls