ML Aggarwal Solution Class 9 Chapter 12 Pythagoras Theorem Test

 Test

Question 1

(a) In fig. (i) given below, AD ⊥ BC, AB = 25 cm, AC = 17 cm and AD = 15 cm. Find the length of BC.

(b) In figure (ii) given below, ∠BAC = 90°, ∠ADC = 90°, AD = 6 cm, CD = 8 cm and BC = 26 cm. Find :

(i) AC (ii) AB (iii) area of the shaded region.

(c) In figure (iii) given below, triangle ABC is right angled at B. Given that AB = 9 cm, AC = 15 cm and D, E are mid-points of the sides AB and AC respectively, calculate

(i) the length of BC (ii) the area of ∆ADE.

Sol :
















(a) Given : In ΔABC, AD⟂BC, AB=25 cm, AC=17 cm and AD=15 cm

Required : The length of BC

In right angled ΔABD, 

⇒AB2=AD2-AD2

(By Pythagoras theorem)
∴BD2=AB2-AD2

⇒(25)2-(15)2

⇒625-225=400

⇒BD=√400=20 cm


Now , in right angled ΔADC

⇒AC2=AD2+DC2 (By Pythagoras theorem)

∴DC2=AC2-AD2

⇒(17)2-(15)2

⇒289-225=64
⇒DC=√64=8 cm
Hence, BC=BD+DC=20+8=28 cm

(b) Given : In ΔABC,
∠BAC=90° , ∠ADC=90° , AD=6 cm, CD=8 cm and BC=26 cm

Required : (i) AC

(ii) AB

(iii) Area of the shaded region

Sol :

In right angled ΔADC

⇒AC2=AD2+DC2 (By Pythagoras theorem)

⇒(6)2+(8)2 

⇒36+64=100

∴=√100=10 cm


In right angled ΔABC

⇒BC2=AB2+(10)2 (By Pythagoras theorem)

⇒(26)2-(10)2 

⇒676-100=576

AB2=576

∴=√576=24 cm


Now , Area of ΔABC$=\frac{1}{2}\times AB\times AC$

$=\frac{1}{2} \times 24 \times 10$

=12×10=120 cm2 


Area of ΔADC$=\frac{1}{2}\times AD\times DE$

$=\frac{1}{2} \times 6 \times 8$

=3×8 cm2 

=24 cm2 


Now , Area of ΔABC$=\frac{1}{2} \times \mathrm{AB} \times \mathrm{AC}$

$=\frac{1}{2} \times 24 \times 10$

=12×10=120 cm2 


Area of ΔADC$=\frac{1}{2} \times \mathrm{AD} \times \mathrm{DC}$

$=\frac{1}{2} \times 6 \times 8$

=3×8=24 cm2 

Hence , area of shaded region=Area of ΔABC-Area of ΔADC

=120 cm2 -24 cm2 

=96 cm2 


(c) Given : In right angled ΔABC, AB=9 cm, AC=15 cm , and D, E are mid points of the sides AB and AC respectively

Required :  (i) Length of BC

(ii) The area of ΔADE

In right angled ΔADE (by Pythagoras theorem)

⇒AE2=AD2+DE2 

⇒$\left(\frac{A C}{2}\right)^{2}=\left(\frac{A B}{2}\right)^{2}+D E^{2}$

(∵D and E are mid points of AB and AC respectively)

⇒$\left(\frac{15}{2}\right)^{2}=\left(\frac{9}{2}\right)^{2}+\mathrm{DE}^{2}$

⇒$\mathrm{DE}^{2}=\frac{225}{4}-\frac{81}{4}$

⇒$\mathrm{DE}^{2}=\frac{144}{4}=36$

⇒DE=√36=6 cm

Since D and E are mid points of AB and AC respectively 

DE||BC and DE$=\frac{1}{2}$BC

⇒BC=2DE=2×6 cm=12 cm


(ii) Area of ΔADE $=\frac{1}{2} \times \mathrm{AD} \times \mathrm{DE}$

$=\frac{1}{2} \times\left(\frac{\mathrm{AB}}{2}\right) \times \mathrm{DE}$

$=\frac{1}{2} \times \frac{9}{2} \times 6 \mathrm{~cm}^{2}$

$=\frac{9}{2} \times 3 \mathrm{~cm}^{2}$

$=\frac{27}{2} \mathrm{~cm}^{2}$

=13.5 cm2 


Question 2

If in ∆ABC, AB > AC and ADI BC, prove that AB² – AC² = BD² – CD².

Sol :
Given : In ∆ABC , AB>AC and AD⟂BC
To prove : AB2-AC2=BD2-CD2 





Proof : In right angled ΔABD

AB2=AD2+BD2 ...(1)

(by Pythagoras theorem)

In right angled ΔACD

AC2=AD2+CD2 ...(2)

Subtracting (2) from (1), we get

AB2-AC2=(AD2+BD2)-(AD2+BD2)

=AD2+BD2-AD2-CD2

=BD2-CD2

∴AB2-AC2=BD2-CD2

Hence, the result


Question 3

In a right angled triangle ABC, right angled at C, P and Q are the points on the sides CA and CB respectively which divide these sides in the ratio 2:1. Prove that

(i) 9AQ² = 9AC² + 4BC²

(ii) 9BP² = 9BC² + 4AC²

(iii) 9(AQ² + BP²) = 13AB².

Sol :

A right angled ∆ ABC in which ∠C 90° , P and Q are points on the side CA and CB respectively such that CP : AP=2 : 1 and CQ : BQ=2 : 1

To prove : 

(i) 9AQ2=9AC2+4BC2

(ii) 9BP2=9BC2+4AC2

(iii) 9(AQ2+BP2)=13AB2











Construction : Join AQ and BP

Proof : (i) In right angled ΔACQ

AQ2=AC2+QC2

(by Pythagoras theorem)

9AQ2=9AC2+9QC2

Multiplying both sides by 9)

=9AC2+(3QC)2=9AC2+(2BC)2

[∵BQ : CQ : 1 : 2

⇒$\frac{QC}{BC}=\frac{QC}{BQ+CQ}=\frac{2}{3}$

⇒3QC=2BC

=9AC2+4BC2

∴9AQ2=9AC2+4BC2...(1)


(ii) In right angled ΔBPC

BP2=BC2+CP2 (by Pythagoras theorem)

9BP2=9BC2+9CP2

(∵Multiplying both sides by 9)

=9BC2+(3CP)2=9BC2+(2AC)2

[∵AP : CP= 1 : 2 

$\frac{C P}{A C}=\frac{C P}{A P+C P}=\frac{2}{3}$ ,

3CP=2AC]

=9BC2+4AC2

∴9BP2=9BC2+4AC2...(2)


(iii) Adding (1) and (2) ,

9AQ2+9BP2=9AC2+4BC2+9BC2+4AC2

=13AC2+13BC2=13(AC2+BC2)=13AB2

[In right angled ΔABC=AB2=AC2+BC2]

∴9AQ+9BP2=13AB2

Hence , the result


Question 4

In the given figure, ∆PQR is right angled at Q and points S and T trisect side QR. Prove that 8PT² – 3PR² + 5PS².

Sol :












In the ∆PQR ,∠Q=90°

T and S are points on RQ such that there trisect it

i.e. RT=TS=SQ

To prove : 8PT2=3PR2+5PS2

Proof : Let RT=TS=SQ=x

In right ∆PRQ 

PR2=RQ2+PQ2

=(3x)2+PQ2

=9x2+PQ2

Similarly in right PTS

PT2=TQ2+PQ2

=(2x)2+PQ2

=4x2+PQ2

and in PSQ,


PS2=SQ2+PQ2=x2+PQ2

8PT2=8(4x2+PQ2)=32x2+8PQ2

3PR2=3(9x2+PQ2)=27x2+3PQ2

5PS2=5(x2+PQ2)=5x2+5PQ2

LHS=8PT2=32x2+8PQ2

RHS=3PR2+5PS2=27x2+3PQ2+5x2+5PQ2

=32x2+8PQ2

∴L.H.S=R.H.S

Hence proved


Question 5

In a quadrilateral ABCD, ∠B = 90°. If AD² = AB² + BC² + CD², prove that ∠ACD = 90°.

Sol :
In quadrilateral ABCD , ∠B = 90° and AD² = AB² + BC² + CD²

AB2+BC2+CD2

To prove :  ∠ACD=90°

Proof : In ∠ABC, ∠B=90°

∴AC2=AB2+BC2...(i)

(Pythagoras theorem)

But AD2=AB2+BC2+CD2 (given)










⇒AD2=AC2+CD2  [From (i)]

∴In ΔACD,

∠ACD=90°

(converse of Pythagoras theorem)


Question 6

In the given figure, find the length of AD in terms of b and c.

Sol :










In the given figure,

ABC is a triangle , ∠A=90°

AB=c, AC=b

To find : AD in terms of b and c

Area of ΔABC$=\frac{1}{2}\times AB\times AC$

$=\frac{1}{2}bc$...(i)

and ΔABC$=\frac{1}{2}\times BC\times AD$...(ii)

But BC$=\sqrt{\mathrm{AB}^{2}+\mathrm{AC}^{2}}=\sqrt{c^{2}+b^{2}}$

$=\sqrt{b^{2}+c^{2}}$...(iii)

From (1) and (ii)

$=\frac{1}{2} \mathrm{BC} \times \mathrm{AD}=\frac{1}{2} b c$

⇒BC×AD=bc

⇒$\sqrt{b^{2}+c^{2}} \times A D$=bc [from (iii)]

Hence AD$=\frac{b c}{\sqrt{b^{2}+c^{2}}}$


Question 7

ABCD is a square, F is mid-point of AB and BE is one-third of BC. If area of ∆FBE is 108 cm², find the length of AC.

Sol :
Given : In square ABCD . F is mid point of AB and BE$=\frac{1}{3} \mathrm{BC}$

Area of ΔFBE=108 cm2

AC and EF are joined











To find : AC

Let each side of a square is =a

$\mathrm{FB}=\frac{1}{2} \mathrm{AB}$  [F is mid point of AB]

$=\frac{1}{2} a$

and $\mathrm{BE}=\frac{1}{3} \mathrm{BC}=\frac{1}{3} a$

Now in square ABCD

$\mathrm{AC}=\sqrt{2} \times$ Side $=\sqrt{2} a$

and area of ΔFBE$=\frac{1}{2} \mathrm{FB} \times \mathrm{BE}$

$=\frac{1}{2} \mathrm{FB} \times \mathrm{BE}$

∴$\frac{1}{12} a^{2}=108$

⇒a2=12×108=1296

⇒a=√1296=36

∴$A C=\sqrt{2} a=\sqrt{2} \times 36=36 \sqrt{2} \mathrm{~cm}$


Question 8

In a triangle ABC, AB = AC and D is a point on side AC such that BC² = AC x CD, Prove that BD = BC.

Sol :
Given : In a triangle ABC, AB=AC and D is point on side AC such that BC2=AC×CA
To prove : BD=BC







Construction : Draw BE⊥AC

Proof : In right angled ΔBCE

⇒BC2=BE2+EC2 (by Pythagoras theorem)

⇒BE2+(AC-AE)2 

⇒BE2+AC2-AE2-2AC.AE 

⇒(BE2+AE2)+AC2-2AC.AE

⇒AB2+AC2-2AC.AE

(In right angled ΔABC ,AB2=BE2+AE2 )

⇒AC2+AC2-2AC.AE (given AB=AC)

⇒2AC2-2AC.AE=2AC(AC-AE)

=2AC.EC

But BC2=AC×CD (given)

⇒AC×CD=2AC.EC
⇒CD=2EC
∴E is mid point of CD
⇒EC=DE

Now , in ΔBED and ΔBEC

EC=DE (above proved)

BE=BE (common)

∠BED=∠BEC  (each 90°)

∴ΔBED≅ΔBEC (by S.A.S axiom of congruency)

∴BD=BC (c.p.c.t)

Hence , the result

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