## ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 14 Theorems on Area Chapter Test

Question 1.

(a) In the figure (1) given below, ABCD is a rectangle (not drawn to scale ) with side AB = 4 cm and AD = 6 cm. Find :

(i) the area of parallelogram DEFC

(ii)area of ∆EFG

(b) In the figure (2) given below, PQRS is a parallelogram formed by drawing lines parallel to the diagonals of a quadrilateral ABCD through its corners. Prove that area of || gm PQRS = 2 × area of quad. ABCD.

Answer:

(a)Given. ABCD is a rectangle AB = 4 cm and AD = 6 cm.

D C cm and AD = 6 cm.

Required. (i) The area of || gm DEFC.

(ii) area of ∆ EFG

Answer:

(i) Since AB = 4cm and AD = 6 cm (given)

∴ Area of rectangle ABCD = AB × AD

= 4 cm × 6 cm = 24 cm^{2}

Now, area of rectangle ABCD = area of || gm DEFC (∵ Both are on the same Base and between the same parallel lines)

⇒ Area of || gm DEFC = 24 cm^{2}

(ii) Area of ∆EFG = \(\frac{1}{2}\) (area of || gm DEFC)

(∴ Both are on the same base and between the same parallel lines)

∴ Area of AEFG = \(\frac{1}{2}\)7 × 24 cm^{2} =12 cm^{2}

(b) Given. PQRS is a || gm formed by drawing lines parallel to the diagonals of quadrilateral ABCD through its comers.

To prove. Area of II gm PQRS = 2. area of quad. ABCD

Proof. ar(∆ACD) = \(\frac{1}{2}\)ar(|| gm ACRS)

[∴ both are on same base AC and between the same || AC and SR ]

⇒ ar (|| gmACRS) = 2ar(∆ACD) …….(1)

Similarly,

ar ( ∆ ABC) = \(\frac{1}{2}\)ar (|| gm ∆APQC)

⇒ ar (|| gm APQC) = 2ar ( ∆ ABC) …..(2)

Adding (1) from (2),

ar (|| gm ACRS) + ar (|| gm APQC) = 2ar(∆ACD) + 2ar (∆ABC)

⇒ (|| gm PQRS) = 2[ar (∆ ACD) + ar ( ∆ ABC)]

⇒ ar (|| gm PQRS) = 2ar (quad. ABCD)

Hence, area of || gm PQRS = 2.area of quad. ABCD. (Q.E.D.)

Question P.Q.

In the adjoining figure, ABCD and ABEF are parallelogram and P is any point on DC. If area of || gm ABCD = 90 cm^{2}, find:

(i) area of || gm ABEF

(ii) area of ∆ABP.

(iii) area of ∆BEF.

Answer:

In the given figure,

ABCD and ABEF are parallelogram P is an point on DC

Area of ||gm ABCD = 90 cm^{2}

||gm ABCD and ABEF are on the same base AB are between the same parallels

(i) ∴ Area of ||gm ABEF = area of ||gm ABCD = 90 cm^{2}

(ii) ∵ ∆ABP and ||gm ABCD are on the same base AB and between the same parallels

∴ Area ∆ABP = \(\frac{1}{2}\) area ||gm ABCD

= \(\frac{1}{2}\) × 90 cm^{2} = 45 cm^{2}

(iii) ∵∆BEF and ||gm ABEF are on the same base EF and between the same parallels

∵ Area ∆BEF = \(\frac{1}{2}\) area ||gm ABEF

= \(\frac{1}{2}\) × 90 = 45 cm^{2}

Question 2.

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

(i) area of ∆CPD = area of ∆AQD

(ii)area of ∆ADQ = area of ∆APD + area of ∆CPB.

Answer:

Given. || gm ABCD in which P is a point on AB and Q is a point on BC.

To prove. (i) ar (∆ CPD) = ar ( ∆ AQD)

(ii) ar (∆ ADQ) = ar ( ∆ APD) + ar (∆ CPB)

Proof. ∆ CPD and II gm ABCD are on the same base CD and between the same parallels lines AB and CD.

∴ ar (∆ CPD) = \(\frac{1}{2}\) ar (|| gm ABCD) …..(1)

∆ ADQ and || gm ABCD are on the same base AD and between the same II lines AD and BC,

ar ( ∆ ADQ) = \(\frac{1}{2}\) (|| gm ABCD) ………(2)

From (1) and (2),

ar ( ∆ CPD) = ar( ∆ ADQ)

or ar ( ∆ CPD) = ar (∆ ADQ) (Q.E.D.)

(ii) ar (∆ ADQ) = \(\frac{1}{2}\)ar (|| gm ABCD) ( Proved in part (i) above )

⇒ 2ar (∆ ADQ) = ar (|| gm ABCD)

⇒ ar (∆ ADQ) + ar( ∆ ADQ) = ar (|| gm ABCD) ……..(3)

But ar ( ∆ ADQ) = ar (∆ CPD) …………(4)

(Proved in part (i) above)

From (3) and (4),

ar ( ∆ ADQ) + ar ( ∆ CPD) = ar (|| gm ABCD)

⇒ ar ( ∆ ADQ) + ar ( ∆ CPD)

⇒ ar (∆ APD) + ar (∆ CPD) + ar (∆ CPB)

⇒ or (∆ ADQ) = ar( ∆ APD) + ar (∆ CPB) (Q.E.D.)

Question 3.

In the adjoining figure, X and Y are points on the side LN of triangle LMN. Through X, a line is drawn parallel to LM to meet MN at Z. Prove that area of ∆LZY = area of quad. MZYX.

Answer:

Given : In the figure,

X and Y are points on side LN of ALMN. Through X, a line XZ || LM is drawn which meets MN at Z.

To prove : area of ∆LZY = area of quad. MZYX

Construction : Join MX, ZY and LZ

Proof: ∆ LM || XZ

and ∆LZX and ∆MZX are on the same base XZ and between the same parallels

∴ area ∆LZX = area ∆MZX

Adding area ∆XZY to both sides area ∆LZX + area ∆XZY

= area ∆MZX + area ∆XZY

⇒ area ∆LZY = area quadrilateral MZYX

Question P.Q.

If D is a point on the base BC of a triangle ABC such that 2BD = DC, prove that area of ∆ABD= \(\frac{1}{2}\) area of ∆ ABC.

Answer:

Given. ∆ ABC in which base BC. D is a point on BC such that 2BD = DC.

To prove, ar ( ∆ ABD) = \(\frac{1}{3}\) ar (ABC)

Construction. Let P is the mid-point of DC join AD = DC

⇒ BD = \(\frac{1}{2}\)DC ^{2}

i.e. BD = DP (P is mid-point of DC)

∴ D is mid-point of BP.

In ∆ ABP, AD is median of BP (D is mid-point of BP)

∴ ar (∆ABD) = ar ( ∆ ADP) …..(1)

Again in ∆ ADC, AP is the median of DC. (P is mid-point of DC)

ar (∆ ADP) = ar (∆ APC) ……………(2)

From (1) and (2),

∴ ar (∆ ABD) = ar (∆ ADP) = ar ( ∆ APC)

∴ ∆ ABC is divided into three equal triangles and each A will be of \(\frac{1}{3}\) ∆ ABC.

∴ ar (∆ ABD) = \(\frac{1}{3}\)ar (∆ ABC) (Q.E.D.)

Question 4.

Perpendiculars are drawn from a point within an equilateral triangle to the three sides. Prove that the sum of the three perpendiculars is equal to the altitude of the triangle.

Answer:

ABC is an equilateral triangle, i.e. AB = BC = CA. P is any point within an equilateral triangle to the three sides.

PN, PM, and PL are perpendicular on side AB, AC and BC respectively. AD is any altitude from point A on side BC.

To prove. AD = NP + LP + MP

Construction. Join PA, PB and PC.

Proof. Area of ∆ ABC = \(\frac{1}{2}\) × Base × Altitude

or(∆ABC)= \(\frac{1}{2}\) × BC × AD ………(1)

Now, area of ∆ APB = \(\frac{1}{2}\) × AB × NP ………….(2)

area of ∆ APC = \(\frac{1}{2}\) × AC × MP

area of ∆ BPC = \(\frac{1}{2}\) × BC × LP …….(4)

Adding (2), (3) and (4)

ar ( ∆ APB) + ar (∆ APC) + ar (∆ BPC)

= \(\frac{1}{2}\) × AB × NP + \(\frac{1}{2}\) × AC × MP + \(\frac{1}{2}\) × BC × LP

ar(∆ ABC) = \(\frac{1}{2}\) [AB × NP + AC × MP × BC × LP]

= \(\frac{1}{2}\) [BC × NP + BC MP × BC × LP] (∵ Ab = AC = CA)

ar(∆ABC) = \(\frac{1}{2}\) × BC[NP + MP + LP ] ..(5)

From (4) and (5),

\(\frac{1}{2}\) × BC × AD = \(\frac{1}{2}\) × BC × (NP + LP + MP)

⇒ AD = NP + LP + MP

⇒ NP + LP + MP = AD

i.e.sum of three perpendiculars is equal to the altitude of the triangle.

Question 5.

If each diagonal of a quadrilateral’ divides it into two triangles of equal areas, then prove that the quadrilateral is a parallelogram.

Answer:

Given : In quadrilateral ABCD, diagonal AC bisects the quadrilateral ABCD in two triangle of equal area i.e.

ar (∆ ABC) = ar (∆ ADC)

To prove : ABCD is a parallelogram.

Proof : Join BD.

Proof: ∵ Diagonals of quad. ABCD divides the quad, into two triangles of equal area.

∴ ar( ∆ ABC) = ar( ∆ ABD)

= \(\frac{1}{2}\) ar (ABCD)

But, these are on the same base AB

∴ Their heights are equal

∴ DC || AB …(i)

Similarly, we can prove that :

ar (∆ABC) = ar (∆BDC)

∴ BC || AD …(ii)

From (i) and (ii)

ABCD is a parallelogram.

Hence proved.

Question 6.

In the given figure, ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. If area of ∆DFB = 3 cm^{2}, find the area of parallelogram ABCD.

Solution:

In the figure, ABCD is a parallelogram BC is produced to E such that CE = BC

Join BD and AE

which intersects DC at F

Join BF, AC and DE

∴ Area of ∆DFB = 3 cm^{2}

Find the area of ||gm ABCD

Solution:

∵ In ∆ABE, C is mid-point of BE and CD || AB

∴ F is mid-point of AE and CD

∴ ABED is a ||gm (∵ Diagonals AE and CD bisect each other at F)

∵ BD is the diagonal of ||gm ABCD

∆BCD = \(\frac{1}{2}\) ||gm ABCD

∵ F is mid-point of DC

∴ ADFB = \(\frac{1}{2}\) ∆BCD

⇒ ADFB = \(\frac{1}{2}\) × \(\frac{1}{2}\) (||gm ABCD)

⇒ ADFB = \(\frac{1}{2}\) (||gm ABCD)

∴ area ||gm ABCD = 4 area ∆DFB

= 4 × 3 = 12 cm^{2}

Question 7.

In the given figure, ABCD is a square. E and F are mid-points of sides BC and CD respectively. If R is mid-point of EF, prove that: area of ∆AER = area of ∆AFR.

Solution:

Given : In square ABCD, BD is diagonals E and F are mid-point of BC and CD respectively. R is mid-point of EF.

To prove : area (∆AER = area (∆AFR)

Proof: In ∆ABE and ∆ADF

AB = AD (Sides of a square)

∠B = ∠D (Each 90°)

BE = CE (E is mid-point of BC)

∴ ∆ABE ≅ ∆ADF (SAS axiom)

∴ AE = AF (c.p.c.t.)

Again in ∆AER and ∆AFR

AE = AF (Produced)

AR = AR (Common)

ER = FR (R is mid-point of EF)

∴ ∆AER = ∆AFR (SSS axiom)

∴ area(∆AER) = area (∆AFR)

Question 8.

In the given figure, X and Y are mid-points of the sides AC and AB respectively of ∆ABC. QP || BC and CYQ and BXP are straight lines. Prove that area of ∆ABP = area of ∆ACQ.

Solution:

Given : In the given figure,

X and Y are the mid-points of the sides AC and AB respectively of ∆ABC

QP || BC

CYQ and BXP are straight lines

To prove : area(∆ABP) = area(∆ACQ)

Proof: v X and Y are the mid-points of sides AC and AB respectively

∴ YX || BC

But QP || BC

∴ QP || BC || YX

In ∆BAP, Y is mid of AB and YX || QP

∴ X is mid-point of BP

∴YX = \(\frac{1}{2}\)AP …(i)

Similarly we can prove in ∆AQC

YX = \(\frac{1}{2}\) QA …(ii)

From (i) and (ii),

QA = AP

Now ∆ABP and ∆ACQ are on the equal base and between the same parallel lines

area(∆ABP) = area(∆ACQ)