RBSE Solutions for Class 9 Maths Chapter 10 Area of Triangles and Quadrilaterals Additional Questions is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 10 Area of Triangles and Quadrilaterals Additional Questions.

Board |
RBSE |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 10 |

Chapter Name |
Area of Triangles and Quadrilaterals |

Exercise |
Additional Questions |

Number of Questions Solved |
28 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 10 Area of Triangles and Quadrilaterals Additional Questions

**Multiple Choice Questions**

Question 1.

The area of a parallelogram and a triangle are equal. If their base (RBSESolutions.com) is common and altitude of parallelogram is 2 cm, then altitude of the triangle is:

(A) 4 cm

(B) 2 cm

(C) 1 cm

(D) 3 cm

Solution

(A) 4 cm

Question 2.

Let ABCD be a parallelogram and ABEF be a rectangle with EF lying along the line CD. If AB = 7 cm and BE = 6.5 cm, then area of the parallelogram is:

(A) 22.75 cm²

(B) 11.375 cm²

(C) 45.5 cm²

(D) 45 cm²

Solution

(C) 45.5 cm²

Question 3.

Two parallelograms are on the same (RBSESolutions.com) base and between the same parallels. Then ratio between their areas is:

(A) 2 : 1

(B) 1 : 1

(C) 1 : 2

(D) 3 : 1

Solution

(B) 1 : 1

Question 4.

In the adjoining figure, ABCD is a rectangle with AE = EF = FB then the ratio of the area of the triangle CEF and that of the rectangle ABCD is:

(A) 1 : 4

(B) 1 : 6

(C) 2 : 5

(D) 2 : 3

Solution

(B) 1 : 6

Question 5.

In the given figure, if area (RBSESolutions.com) of the parallelogram ABCD is 30 cm², then ar (∆ADE) + ar (∆BCE) is equal to:

(A) 20 cm²

(B) 30 cm²

(C) 15 cm²

(D) 25 cm²

Solution

(C) 15 cm²

Question 6.

In figure, ABCD is a parallelogram if (RBSESolutions.com) area of ∆AEB is 16 cm², then area of ∆BFC is:

(A) 32 cm²

(B) 24 cm²

(C) 8 cm²

(D) 16 cm²

Solution

(D) 16 cm²

Question 7.

In the figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, then AD is equal to:

(A) 1.28 cm

(B) 12.8 cm

(C) 6.4 cm

(D) 14.4 cm

Solution

(B) 12.8 cm

Question 8.

In the given figure, ABCD is a parallelogram (RBSESolutions.com) whose area is 20 cm² then area of ∆AOD is:

(A) 10 cm²

(B) 15 cm²

(C) 5 cm²

(D) 12 cm²

Solution

(C) 5 cm²

Question 9.

AD is a median of ∆ABC. If area (RBSESolutions.com) of ∆ABD is 25 cm² then area of ∆ABC would be:

(A) 100 cm²

(B) 25 cm²

(C) 75 cm²

(D) 50 cm²

Solution

(D) 50 cm²

Question 10.

In an ∆ABC, E is the mid-point of median AD, then ar (∆BED) is:

Solution

(C) \(\frac { 1 }{ 4 }\) ar(∆ABC)

**Very Short Answer Type Questions**

Question 1.

In parallelogram ABCD, AB = 12 cm. The altitudes (RBSESolutions.com) corresponding to the two sides AB and AD are 8 cm and 10 cm respectively. Compute BC.

Solution.

Area of parallelogram ABCD = base x corresponding altitude = AB x CM = 12 cm x 8 cm = 96 cm^{2}

Again area of parallelogram ABCD = BC x CN

⇒ 96 = BC x 10

⇒ BC = 9.6 cm.

Question 2.

If the ratio of the altitude and the area of the (RBSESolutions.com) parallelogram is 2 : 11, then find the length of the base of the parallelogram.

Solution.

Here we are given that

altitude : area = 2 : 11

altitude = 2x and area = 11x (say)

Then, area = base x altitude

11x = base x 2x

⇒ base = \(\frac { 11x }{ 2x }\) = 5.5 units

Question 3.

In figure, D divides the side BC of ∆ABC in (RBSESolutions.com) the ratio 3 : 5. Show that ar (∆ABD) = \(\frac { 3 }{ 8 }\) ar (∆ABC)

Solution.

We are given that D divides BC in the ratio 3 : 5

i.e. BD : CD = 3 : 5

⇒ BD = \(\frac { 3 }{ 8 }\) DC …(i)

Draw AE ⊥ BC

Question 4.

In ∆ABC (see figure), E is the mid-point of the (RBSESolutions.com) median AD. If area of the triangle ABC is 184 sq. units, find area of ∆DEC.

Solution.

Since AD is the median of ∆ABC

ar (∆ADC) = \(\frac { 1 }{ 2 }\) ar (∆ABC) = \(\frac { 1 }{ 2 }\) x 184 = 92 sq. unit.

Again E is the mid-point of side AD.

CE divides AADC into two parts equal in area.

area ∆DEC = \(\frac { 1 }{ 2 }\) (ar ∆ADC) = \(\frac { 1 }{ 2 }\) x 92 = 46 sq. unit.

Question 5.

In figure, area (∆BCE) = 75.2 m2. Find the area (RBSESolutions.com) of parallelogram ABCD.

Solution.

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

i.e Area of ∆BCE = \(\frac { 1 }{ 2 }\) ar (||gm ABCD)

⇒ 75.2 m2 = \(\frac { 1 }{ 2 }\) ar(||gm ABCD)

⇒ ar (||gm ABCD) = 75.2 x 2 m^{2} = 150.4 m^{2}.

Question 6.

Prove that a median of a triangle divides it (RBSESolutions.com) into two triangles of equal area.

Solution.

Given: A triangle PQR in which RS is a median

To prove: ar (∆RPS) = ar (∆RSQ)

Construction: Draw RL ⊥ PQ

Proof: RS is median

⇒ PS = SQ

⇒ \(\frac { 1 }{ 2 }\) x PS x RL = \(\frac { 1 }{ 2 }\) x SQ x RL

⇒ ar (∆RPS) = ar (∆RSQ)

Question 7.

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect (RBSESolutions.com) each other at O. Prove that ar (∆AOD) = ar (∆BOC).

Solution.

Since ∆ACD and ∆BDC are on the same base CD and between the same parallels AB and CD.

Therefore, ar (∆ACD) = ar (∆BDC)

Subtracting ar (∆ACD) from both side, we get

ar (∆ACD) – ar (∆COD) = ar (∆BDC) – ar (∆COD)

⇒ ar (∆AOD) = ar (∆BOC)

Question 8.

Two parallelograms PQRS and PQMN have (RBSESolutions.com) common base PQ as shown in figure. PQ = 9 cm, SM = 3 cm, and ST = 5 cm. Find the area of PQRN.

Solution.

Since parallelograms PQRS and PQMN are on the same base PQ and between the same parallels PQ and NR.

ar (PQRS) = ar (PQMN) = 9 x 5 = 45 cm2

Also, area of trapezium PQMS

= \(\frac { 1 }{ 2 }\) x (PQ + MS) x ST

= \(\frac { 1 }{ 2 }\) x (9 + 3) x 5

= \(\frac { 1 }{ 2 }\) x 12 x 5 =30 cm^{2}

ar (PQRN) = ar (PQRS) + ar (PQMN) – ar (PQMS) = (45 + 45 – 30) cm^{2} = 60 cm^{2}

Question 9.

In figure, ABCD is a quadrilateral in which (RBSESolutions.com) diagonal BD =14 cm. If AL ⊥ BD and CM ⊥ BD such that AL = 8 cm and CM = 6 cm, find the area of quadrilateral ABCD.

Solution.

Here we are given that BD = 14 cm

AL ⊥ BD and CM ⊥ BD and AL = 8 cm, CM = 6 cm

Area of quadrilateral ABCD = ar (∆ABD) + ar (∆BDC)

= \(\frac { 1 }{ 2 }\) x BD x AL + \(\frac { 1 }{ 2 }\) x BD x CM

= \(\frac { 1 }{ 2 }\) x BD x (AL + CM)

= \(\frac { 1 }{ 2 }\) x 14 x (8 + 6)

= \(\frac { 1 }{ 2 }\) x 14 x 14 = 98 cm^{2}

Question 10.

In figure, ABCD is a quadrilateral. BP is drawn (RBSESolutions.com) parallel to AC and BP meets DC produced at P. Prove that ar (∆ADP) = ar (quad. ABCD).

Solution.

Here ∆ABC and ∆ACP are on the same base AC and between the same parallels AC and BP.

Therefore,

ar (∆ABC) = ar (∆ACP)

⇒ ar (∆ADC) + ar (∆ABC) = ar (∆ADC) + ar (∆ACP)

⇒ ar (quad. ABCD) = ar (∆ADP)

Hence, ar (∆ADP) = ar (quad. ABCD)

**Short Answer Type Questions**

Question 1.

XY is a line parallel to side BC of (RBSESolutions.com) a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively. Show that ar (∆ABE) = ar (∆ACF).

Solution.

Since ∆ABE and parallelogram BCYE are on the same base BE and between the same parallels EB and AC. Therefore

ar (∆ABE) = \(\frac { 1 }{ 2 }\) ar (||gm BCYE) …(i)

Similarly, ∆ACF and parallelogram BCFX are on the same (RBSESolutions.com) base CF and between the same parallels FC and AB. Therefore

ar (∆ACF) = \(\frac { 1 }{ 2 }\) ar (||gm BCFX) …(ii)

But parallelogram BCYE and BCFX are on the same base BC and between the same parallels BC and EF.

ar (||gm BCYE) = ar (||gm BCFX) …….. (iii)

Using relations (i), (ii) and (iii), we have

ar (∆ABE) = ar (∆ACF).

Question 2.

Diagonals AC and BD of trapezium ABCD with AB || DC intersect (RBSESolutions.com) each other at O. Prove that ar (∆AOD) = ar (∆BOC).

Solution.

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

To prove: ar (∆AOD) = ar (∆BOC)

Proof: Since ∆’s ABC and ABD lie on the same base AB and between the same parallels AB and DC.

Therefore ar (∆ABC) = ar (∆ABD) …(i)

Subtracting ar (∆AOB) from both sides of (i), we get

ar (∆ABC) – ar (∆AOB) = ar (∆ABD) – ar (∆AOB)

⇒ ar (∆BOC) = ar (∆AOD)

i.e. ar (∆AOD) = ar (∆BOC)

Question 3.

Diagonals AC and BD of a quadrilateral ABCD intersect at O in (RBSESolutions.com) such a way that ar (∆AOD) = ar (∆BOC). Prove that ABCD is a trapezium.

Solution.

Given: ABCD is a quadrilateral in which AC and BD are diagonals which intersect at O, such that ar (∆AOD) = ar (∆BOC)

To prove: ABCD is a trapezium.

Proof: We are given that

ar (∆AOD) = ar (∆BOC) …(i)

Adding ar (∆AOB), we get on both sides

ar (∆AOD) + ar (∆AOB) = ar (∆BOC)+ ar (∆AOB)

⇒ ar (∆ABD) = ar (∆ABC)

But triangles ABD and ABC are on the same base AB and have equal in areas.

Therefore they must lie in between the same parallels, i.e. AB || DC

Hence, ABCD is a trapezium.

Question 4.

Prove that “Parallelograms on the (RBSESolutions.com) same base and between the same parallels are equal in area”.

Solution.

Given: Two parallelograms PQRS and PQMN, which have the same base PQ and which are between the same parallels lines PQ and NR.

To prove: ar (||gm PQRS) = ar (||gm PQMN)

Proof: ar (||gm PQRS) = ar (quad. PQMS) + ar (AQRM) ….(i)

Also ar (||gm PQMN) = ar (quad. PQMS) + ar (APSN) …(ii)

Now, in triangles RQM and SPN

RQ = SP (opposite sides of a ||gm)

QM = NP (opposite sides of a ||gm)

∠RQM = ∠SPN (∵ QR || SP and QM || PN)

∆RQM = ∆SPN (by SAS congruency property)

⇒ ar (∆RQM) = ar (∆SPN) …(iii)

Using relations (i), (ii) and (iii), we get

ar (||gm PQRS) = ar (||gm PQMN).

Question 5.

The side AB of a parallelogram ABCD is produced to (RBSESolutions.com) any point P. A line through A, parallel to CP meets CB produced at Q and the parallelogram PBQR is completed. Prove that ar (||gm ABCD) = ar (||gm PBQR).

Solution.

Given: The side AB of a parallelogram ABCD is produced to P. AQ || CP where Q is on CB produced. PBQR is a parallelogram.

To prove: ar(||gm ABCD) = ar(||gm PBQR)

Construction: Join AC and QP

Proof: In parallelogram ABCD, AC is a diagonal

ar (||gm ABCD) = 2 ar (∆ABC) …(i)

Similarly, in (RBSESolutions.com) parallelogram PBQR, QP is a diagonal

ar (||gm PBQR) = 2 ar (∆BPQ) …(ii)

Now ∆AQC and ∆AQP are on the same base AQ and between the same parallels AQ and CP.

ar (∆AQC) = ar (∆AQP)

ar (∆AQC) – ar (AAQB) = ar (∆AQP) – ar (∆AQB)

⇒ ar (∆ABC) = ar (∆BPQ) …(iii)

From (i), (ii) and (iii), we get

ar (||gm ABCD) = ar (||gm PBQR).

**Long Answer Type Questions**

Question 1.

D, E and F are respectively the mid-points (RBSESolutions.com) of the sides BC, CA and AB of a ∆ABC. Show that:

(i) BDEF is a parallelogram

(ii) ar (∆DEF) = \(\frac { 1 }{ 4 }\) ar (∆ABC)

(iii) ar (BDEF) = \(\frac { 1 }{ 2 }\) ar (∆ABC)

Solution.

Given: In ∆ABC, D, E and F are mid-points

of sides BC, CA and AB respectively.

To prove:

(i) BDEF is a parallelogram

(ii) ar (∆DEF) = \(\frac { 1 }{ 4 }\) ar (∆ABC)

(iii) ar (BDEF) = \(\frac { 1 }{ 2 }\) ar (∆ABC)

Proof: (i) In ∆ABC, F is the (RBSESolutions.com) mid-point of AB and E is the mid-point of AC.

Therefore from mid-point theorem

EF || BC and EF = \(\frac { 1 }{ 2 }\) AB = BD …(i)

Similarly, DE || AB and DE = \(\frac { 1 }{ 2 }\) AB = BF … (ii)

EF || BD and EF = BD

⇒ BDEF is a parallelogram.

(ii) Here DF is diagonal of ||gm BDEF

ar (∆BDF) = ar (∆DEF) …(iii)

Similarly, ar (ADEF) = ar (∆AEF) …(iv)

and ar (∆CDE) = ar (∆DEF) …(v)

From (iii), (iv) and (v), we have

ar (∆BDF) = ar (∆AFE)

ar (∆DEF) = ar (∆CDE)

But ar (∆BDF) + ar (∆AFE) + ar (∆DEF) + ar (∆CDE) = ar (∆ABC)

⇒ 4 ar (∆DEF) = ar (∆ABC)

⇒ ar (∆DEF) = \(\frac { 1 }{ 4 }\) ar (∆ABC) …(vi)

(iii) Now, area (RBSESolutions.com) of parallelogram BDEF = ar (∆BDF) + ar (∆DEF) [∵ ar (∆BDF) ar (∆DEF)]

⇒ ar (||gm BDEF) = 2 ar(∆DEF) = 2x \(\frac { 1 }{ 4 }\) ar (∆ABC) [using relation (vi)]

= \(\frac { 1 }{ 2 }\) ar (∆ABC)

Hence, ar (||gm BDEF) = \(\frac { 1 }{ 2 }\) ar (∆ABC)

Question 2.

In figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:

(i) ar (∆DOC) = ar (∆AOB)

(ii) ar (∆DCB) = ar (∆ACB)

(iii) DA || CB or ABCD is a parallelogram.

[Hint: From D and B, draw perpendiculars to AC]

Solution.

Given: ABCD is (RBSESolutions.com) a quadrilateral in which AC and BD are diagonals and they intersect each other at O such that OB = OD.

To prove:

(i) ar (∆DOC) = ar (∆AOB)

(ii) ar (∆DCB) = ar (∆ACB)

(iii) DA || CB or ABCD is a parallelogram.

Construction: Draw DN ⊥ AC and BM ⊥ AC

Proof: (i) O is the mid-point of diagonal BD.

Therefore AO and CO will (RBSESolutions.com) be the medians of ∆ABD and ∆BCD respectively.

ar (∆AOD) = ar (∆AOB) …(i)

and ar (∆COD) = ar (∆BOC) …(ii)

Reason: Median of a triangle divides it into two triangles equal in areas.

ar (∆AOD) + ar (∆COD) = ar (∆AOB) + ar (∆BOC)

⇒ ar (∆ACD) = ar (∆ABC)

⇒ \(\frac { 1 }{ 2 }\) AC x DN = \(\frac { 1 }{ 2 }\) AC x BM

⇒ DN = BM

Now in ∆CND and ∆ABM

AB = CD (given)

DN = BM

and ∠DNC = ∠BMA = 90°

⇒ ∆CND = ∆ABM (by R.H.S. congruence rule)

⇒ ar (∆CND) = ar (∆ABM) …(iii)

Similarly, in ∆DON and ∆BOM

DN = BM

DO = OB (given)

and ∠DNO = ∠BOM = 90° (each)

∆DON = ∆BOM (by R.H.S. congruence rule)

⇒ ar (∆DON) = ar (∆BOM) …(iv)

On adding (iii) and (iv), we get

ar (∆CND) + ar (∆DON) = ar (∆ABM) + ar (∆BOM)

⇒ ar (∆DOC) = ar (∆AOB)

(ii) ar (∆DOC) = ar (∆AOB)

On Adding ar (∆BOC) to both side, we get

ar (∆DOC) + ar (∆BOC) = ar (∆AOB) + ar (∆BOC)

⇒ ar (∆DCB) = ar (∆ACB)

(iii) Since A’s DCB and ACB are on the (RBSESolutions.com) same base BC and equal areas, therefore they must lie between the same parallels

i.e. DA || CB …(v)

∆CDN = ∆ABM and ∆DNO = ∆BMO [by using relations (iii) and (iv)]

∠CDN = ∠ABM and ∠NDO = ∠MBO

On adding, we get

∠CDN + ∠NDO = ∠ABM + ∠MBO

⇒ ∠CDO = ∠ABO

But these are alternate angles. Therefore CD || BA and DA || CB.

Hence, ABCD is a parallelogram.

Question 3.

ABCD is a parallelogram, E and F are (RBSESolutions.com) the mid-points of BC and CD respectively.

Prove that: ar (∆AEF) = \(\frac { 3 }{ 8 }\) ar (||gm ABCD)

Solution.

Given: ABCD is a parallelogram in which E and F are the mid-points of sides BC and CD respectively.

To prove: ar (∆AEF) = \(\frac { 3 }{ 8 }\) ar (||gm ABCD)

Construction: Join BD and EF.

Proof: In ∆BCD, E is the mid-point of BC and F is the mid-point of CD

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