## ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 12 Pythagoras Theorem Chapter 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.

Answer: (a) Given. In ∆ ABC , AD ⊥ BC, AB = 25 cm, AC = 17 cm and AD = 15 cm

Required. The length of BC.

Solution:

In right angled ∆ABD,

AB^{2} = AD^{2} + BD^{2} (By Pythagoras theorem)

∴ BD^{2} = AB^{2} – AD^{2}

= (25)^{2} – (15)^{2}

= 625 – 225 = 400

⇒ BD = \(\sqrt{400}\) = 20 cm.

Now, in right angled ∆ ADC

AC^{2} = AD^{2} + DC^{2} (By Pythagoras theorem)

∴ DC^{2}= AC^{2} – AD^{2}

⇒ DC^{2} = (17)^{2} – (15)^{2}

⇒ DC^{2} = 289 – 225 = 64

DC = \(\sqrt{64}\) = 8cm

Hence, BC = BD + DC = 20 cm + 8 cm = 28 cm.

(b) Given. In ∆ ABC,

∠BAC = 90°, ∠ ADC = 90° AD = 6cm, CD = 8 cm and BC = 26 cm.

Required. (i) AC (ii) AB

(iii) area of the shaded region

Solution:

In right angled ∆ ADC

AC^{2} = AD^{2} + DC^{2} (By Pythagoras theorem)

= (6)^{2} + (8)^{2}

= 36 + 64 = 100

∴ = \(\sqrt{10}\) = 10 cm Answer:

In right angled ∆ ABC

BC^{2} = AB^{2} + AC^{2} (By Pythagoras theorem)

⇒ (26)^{2} = AB^{2} + (10)^{2}

⇒ AB^{2} = (26)^{2} – (10)^{2}

⇒ AB^{2} = 676 – 100 = 576

⇒ AB^{2} = 576

⇒ AB = \(\sqrt{576}\) = 24 cm

Now, Area of A ABC = \(\frac{1}{2}\) × AB × AC

= \(\frac{1}{2}\) × 24 × 10 cm^{2} = 12 × 10 cm^{2} = 120 cm^{2}

Area of ∆ ADC = \(\frac{1}{2}\) × AD × DE

= \(\frac{1}{2}\) × 6 × 8 cm^{2} =3 × 8 cm^{2} = 24 cm^{2}

Now, Area of ∆ ABC = \(\frac{1}{2}\) × AB × AC

= \(\frac{1}{2}\) × 24 × 10 cm^{2} = 12 × 10 cm^{2} = 120 cm^{2}

Area of ∆ ADC = \(\frac{1}{2}\) × AD × DC

= \(\frac{1}{2}\) × 6 × 8 cm^{2} = 3 × 8 cm^{2} = 24 cm^{2}

Hence, area of shaded region = Area of ∆ ABC – Area of ∆ ADC = 120 cm^{2} – 24 cm^{2} = 96 cm^{2}

(c) Given. In right angled ∆ ABC, 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 ∆ ADE

Solution:

In right angled ∆ ADE, (By Pythagoras theorem)

AE^{2} = AD^{2} + DE^{2}

Question 2.

If in ∆ ABC, AB > AC and AD ⊥ BC, prove that AB^{2} – AC^{2} = BD^{2} – CD^{2}.

Answer:

Given. In ∆ ABC, AB > AC and AD ⊥ BC

To prove. AB^{2} – AC^{2} = BD^{2} – CD^{2}

Proof. In right angled ∆ ABD

AB^{2} = AD^{2} + BD^{2} …….(1) (By Pythagoras theorem

In right angled ∆ACD

AC^{2} = AD^{2} + CD^{2} ……(2)

Subtracting (2) from (1), we get

AB^{2} – AC^{2} = (AD^{2} + BD^{2}) – (AD^{2} + CD^{2})

= AD^{2} + BD^{2} – AD^{2} – CD^{2} = BD^{2} – CD^{2}

∴ AB^{2} – AC^{2} = BD^{2} – CD^{2} 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) 9AQ^{2} = 9AC^{2} + 4BC^{2}

(ii) 9BP^{2} = 9BC^{2} + 4AC^{2}

(iii) 9(AQ^{2} + BP^{2}) = 13AB^{2}.

Solution:

Given. 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) 9AQ^{2} = 9AC^{2} + 4BC^{2}

(ii) 9BP^{2} = 9BC^{2} + 4AC^{2}

(iii) 9 (AQ^{2} + BP^{2}) = 13 AB^{2}

Construction. Join AQ and BP.

Proof. (i) In right angled ∆ ACQ

AQ^{2} = AC^{2} + QC^{2} (By Pythagoras theorem)

9AQ^{2} = 9AC^{2} + 9QC^{2} ( Multiplying both sides by 9)

= 9AC^{2} + (3QC)^{2} = 9AC^{2} + (2BC)^{2}

= 9AC^{2} + 4BC^{2}

∴ 9AQ^{2} = 9AC^{2} + 4BC^{2} …..(1)

(ii) In right angled ∆ BPC

BP^{2} = BC^{2} + CP^{2} (By Pythagoras theorem)

9BP^{2} = 9BC^{2} + 9CP^{2} (∵ Multiplying both side by 9)

= 9BC^{2} + (3 CP)^{2} = 9BC^{2} + (2AC)^{2}

= 9BC^{2} + 4AC^{2}

∴ 9BP^{2} = 9BC^{2} + 4AC^{2} ….(2)

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

9AQ^{2} + 9BP^{2} = 9AC^{2} + 4BC^{2} + 9BC^{2} + 4AC^{2}

= 13AC^{2} + 13BC^{2} = 13 (AC^{2} + BC^{2}) = 13 AB^{2} [In right angled ∆ ABC = AB^{2} = AC^{2} + BC^{2}]

∴ 9AQ + 9BP^{2} = 13 AB^{2}

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 8PT^{2} = 3PR^{2} + 5PS^{2}.

Solution:

In the ∆PQR, ∠Q = 90°

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

i.e., RT = TS = SQ

To prove : 8PT^{2} = 3PR^{2} + 5PS^{2}

Proof: Let RT = TS = SQ = x

In right ∆PRQ

PR^{2} = RQ^{2} + PQ^{2} = (3x)^{2} + PQ^{2} = 9x^{2} + PQ^{2}

Similarly in right PTS,

PT^{2} = TQ^{2} + PQ^{2} = (2x)^{2} + PQ^{2} = 4x^{2} + PQ^{2}

and in PSQ,

PS^{2} = SQ^{2} + PQ^{2} = x^{2} + PQ^{2}

8PT^{2} = 8(4x^{2} + PQ^{2}) = 32x^{2} + 8PQ^{2}

3PR^{2} = 3(9x^{2} + PQ^{2}) = 27x^{2} + 3PQ^{2}

5PS^{2} = 5(x^{2} + PQ^{2}) = 5x^{2} + 5PQ^{2}

LHS = 8PT^{2} = 32x^{2} + 8PQ^{2}

RHS = 3PR^{2} + 5PS^{2} = 27x^{2} + 3PQ^{2} + 5x^{2} + 5PQ^{2}.

= 32x^{2} + 8PQ^{2}.

LHS = RHS

Hence proved.

Question 5.

In a quadrilateral ABCD, ∠B = 90°. If AD^{2} = AB^{2} + BC^{2} + CD^{2}, prove that ∠ACD = 90°.

Solution:

In quadrilateral ABCD, ∠B = 90° and AD^{2} = AB^{2} + BC^{2} + CD^{2}

To prove : ∠ACD = 90°

Proof: In ∠ABC, ∠B = 90°

∴ AC^{2} = AB^{2} + BC^{2} …(i) (Pythagoras Theorem)

But AD^{2} = AB^{2} + BC^{2} + CD^{2} (Given)

⇒ AD^{2} = AC^{2} + CD^{2} [From (i)]

∴ In ∆ACD,

∠ACD = 90° (Converse of Pythagoras Theorem)

Question 6.

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

Solution:

In the given figure,

ABC is a triangle, ∠A = 90°

AB = c, AC = b

AD ⊥ BC

To find : AD in terms of b and c

Solution:

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 cm^{2}, find the length of AC.

Solution:

Given : In square ABCD. F is mid piont of

AB and BE = \(\frac{1}{3}\) BC

Area of AFBE = 108 cm^{2}

AC and EF are joined

To find: AC

Solution:

Let each side of square is = a

FB = \(\frac{1}{2}\) AB (F is mid point of AB)

= \(\frac{1}{2}\) a

and BE = \(\frac{1}{3}\)BC = \(\frac{1}{3}\)a

Now in square ABCD

AC = \( \sqrt{{2}} \) × Side = \( \sqrt{{2}} \)a

and area ∆FBE = \(\frac{1}{2}\) FB × BE

= \(\frac{1}{2}\) × \(\frac{1}{2}\)a × \(\frac{1}{2}\)a = \(\frac{1}{12}\)a^{2}

∴ \(\frac{1}{12}\) a^{2} = 108 ⇒ a^{2} = 12 × 108 = 1296

⇒ a = \( \sqrt{{1296}} \) = 36

∴ AC = \( \sqrt{{2}} \) a = \( \sqrt{{2}} \) × 36 = 36 \( \sqrt{{2}} \) cm

Question 8.

In a triangle ABC, AB = AC and D is a point on side AC such that BC^{2} = AC × CD. Prove that BD = BC.

Solution:

Given. In a triangle ABC, AB = AC and D is point on side AC such that BC^{2} = AC × CD

To prove. BD = BC

Construction. Draw BE ⊥ AC

Proof. In right angled ∆ BCE

BC^{2} = BE^{2} + EC^{2} (By Pythagoras theorem)

= BE^{2} + (AC – AE)^{2}

= BE^{2} + AC^{2} + AE^{2} -2AC.AE

= (BE^{2} + AE^{2}) + AC^{2} – 2AC.AE

= AB^{2} + AC^{2} – 2AC.AE (In rt. ∠ ed ∆ ABC, AB^{2} = BE^{2} + AE^{2})

= AC^{2} + AC^{2} – 2AC.AE (given AB = AC)

= 2AC^{2} – 2AC. AE = 2AC (AC – AE)

= 2AC.EC

But BC^{2} = AC × CD (given)

⇒ AC × CD = 2AC.EC ⇒ CD = 2EC

∴ E is mid-points of CD ⇒ EC = DE

Now, in ∆BED and ∆ BEC

EC = DE (above proved)

BE = BE (common)

∠ BED = ∠ BEC (each 90°)

∴ ∆BED ≅ ∆ BEC (By S. A. S. axiom of congruency)

∴ BD = BC (c.p.c.t.)

Hence, the result.