WBBSE Solutions For Class 10 Maths Chapter 22 Pythagoras Theorem Exercise 22.1

WBBSE Class 10 Maths Solutions Chapter 22 Pythagoras Theorem Exercise 22.1

 

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Application 1. Taking different proper values of m and n, let us write the sides of two right-angled triangles. 

Solution: In the case of a right-angled triangle, if two sides & the hypotenuse of the triangle are (m² – n²) units, 2mn units & (m²+n²) units

respectively then (m²- n²)² & (2mn)² = (m2+N2)²

Now let m = 10, n = 6

∴ LHS = (10262)² + (2.10.6)²

= (64)² + (120)²=4096 +14400 = 18496

RHS = (102 +62)² = (136)² = 18496

∴ LHS = RHS

Application 2. In our garden, a ladder of 25 m in length is inclined to a guard wall at a height of 24 m above the ground. Let us write by calculating, the distance of the foot of the ladder from the guard wall.

Solution:

Given

In our garden, a ladder of 25 m in length is inclined to a guard wall at a height of 24 m above the ground.

Let the length of the ladder AC = 25 m, AB = 24 m,

Pythagoras’ theorem in right-angled triangle ABC, we get,

AB²+BC² = AC²

or, (24m)² + (BC)² = (25m)²

∴ BC = 7

or, BC² (25m)²- (24m)² = 49

The foot of the ladder is in 7m

distance from the guard wall.

Application 3. If the lengths of two diagonals of a rhombus are 12 cm and 16 cm respectively, then let us write by calculating, the length of one side of this rhombus. 

Hints:

Given

If the lengths of two diagonals of a rhombus are 12 cm and 16 cm respectively

The two diagonals of a rhombus bisect each other, perpendicularly.

∴ OA = 12/2  cm = cm,       OB =   cm.

AOB is a right-angled triangle.      

∴ From Pythagoras’ theorem, we get, AB²+ OB²

We know, the diagonals of a rhombus bisect each other perpendicularly.

∴ Let BD 16 cm.    ∴ BO = OD = 8cm

& AC =12 cm.      ∴ AO OC = 6 cm.

∴ ∠AOB = ∠BOC = ∠COD = ∠AOD = 90° 

∴ BOC is a right-angled triangle.

BC = √BO² + CO²

= √82 +62

= √100 

= 10 cm.

Application 4. If in ΔABC, ADL BC, let us prove that, AB²+ CD² = AC²+ BD²

Solution: To prove AB²+ CD² = AC² + BD²

In ΔABC, perpendicular drawn from

A, on BC, is AD.

∴ AD ⊥ BC

∴ ∠ADB = ∠ADC = 90°

From two right-angled triangles ABD & ADC,

AB² = AD² + BD²——-(1)

& AC² = AD²+ DC²———(2)

Subtracting (1) from (2), AB² – AC² = BD² – DC²

or, AB²+ CD² = AC²+ BD² Proved.

Application 5. Let us prove that the area of a square drawn on the diagonal of a square is twice of the area of that square. 

Solution: Area of ABCD square = AD²

Area of ACEF square = AC²

In ΔACD, ∠ADC = 90°

AD = CD [Side of a square]

AC = √AD²+DC² = √AD²+DC² = AD √2

AC² = (AD √2)²= 2AD²

Question 1. If the followings are the lengths of the three sides of a triangle, then let us write by calculating, the cases where the triangles are right-angled triangles

1. 8 cm, 15 cm. and 17 cm

2. 9 cm, 11 cm, and 6 cm.

Solution: In a right angle triangle, (Perpendicular)² + (Base)² = (Hypotenuse)² Here, 

(8)² + (15)² = (17)²

or, 64+ 225

= 289

∴ It is a right-angled triangle.

Question 2. In the road of our locality there is a ladder of 15 m in length kept in such a way that it has touched Millis’ window at a height of 9 m above the ground. Now keeping the foot of the ladder at the same point of that road, the ladder is rotated in such a way that it touched our window situated on the other side of the road. If our window is 12 m above the ground, then let us determine the breadth of that road in our locality. 

Solution:

Given

In the road of our locality there is a ladder of 15 m in length kept in such a way that it has touched Millis’ window at a height of 9 m above the ground. Now keeping the foot of the ladder at the same point of that road, the ladder is rotated in such a way that it touched our window situated on the other side of the road. If our window is 12 m above the ground

In ΔABE, BE2 (AE)²- (AB)²

= (15)² – (9)²

= 225-81

= 144

∴ BE = √144 = 12

In Δ EDC = EC² = ED² – CD²

=152 – 122

= 225-144

= 81

∴ EC = 9

∴ BC =  BE + EC = 12+9 = 21 m.

∴ The breadth of the road = 21 m.

 

Question 3. If the length of one diagonal of a rhombus having a side of 10 cm length is 12 cm, then let us write, by calculating the length of the other diagonal.

Solution:

Given

If the length of one diagonal of a rhombus having a side of 10 cm length is 12 cm

The length of one diagonal of a rhombus of side 10 cm is 12 cm.

In △AOB, \(O B^2+O A^2=A B^2\)

Here, OB = \(\frac{12}{2}\) = 6 cm

& OA = 10cm.

∴ \(\mathrm{AO}^2=\mathrm{AB}^2-\mathrm{OB}^2\)

= \((10)^2-(6)^2\)

= 100 – 36 = 64

AO = √64 = 8

 

Question 4. I have drawn a triangle PQR whose ∠Q is a right angle. If S is any point on QR, let us prove that PS²+ QR² = PR²+ QS².

Solution: In a right-angled triangle PQR, ∠Q = 90°

S is any point in QR

To prove \(P S^2+Q R^2=P R^2+Q S^2\)

Proof:  △PQS &  △PQR are two right angled triangles.

∴ \(\mathrm{PS}^2=\mathrm{PQ}^2+\mathrm{QS}^2\)

\(P R^2=P Q^2+Q^2\)

or, \(\mathrm{QR}^2=P R^2-\mathrm{PQ}^2\)

∴ \(P S^2+Q R^2=P Q^2+Q S^2+P R^2-P Q^2\)

= \(Q S^2+P R^2\)

∴ \(\mathrm{PS}^2+\mathrm{QR}^2=\mathrm{PR}^2+\mathrm{QS}^2\) Proved.

 

Question 5. Let us prove that the sum of squares drawn on the sides of a rhombus is equal to the sum of squares drawn on two diagonals.

Solution: AC & BD two diagonals of the rhombus ABCD, intersect each other at O.

To prove, \(A B^2+B C^2+C D^2+D A^2=A C^2+B D^2\)

We know the diagonals of a rhombus bisect each other perpendicularly.

i.e., AC ⊥ BD, & AO = OC & BO = OD

∴ △AOB, △BOC, △COD & △DOA

are four right angled triangles.

\(\mathrm{AB}^2=\mathrm{OA}^2+\mathrm{OB}^2 ; \mathrm{BC}^2=\mathrm{OB}^2+\mathrm{OC}^2\) \(\mathrm{CD}^2=\mathrm{OC}^2+\mathrm{OD}^2 ; A D^2=\mathrm{OD}^2+\mathrm{OA}^2\) \(\mathrm{AB}^2+\mathrm{BC}^2+\mathrm{CD}^2+\mathrm{AD}^2=\mathrm{OA}^2+\mathrm{OB}^2+\mathrm{OB}^2+\mathrm{OC}^2+\mathrm{OC}^2+\mathrm{OD}^2+\mathrm{OD}^2+\mathrm{OA}^2\)

= \(2\left(\mathrm{OA}^2+\mathrm{OB}^2+\mathrm{OC}^2+\mathrm{OD}^2\right)\)

= \(2\left(\mathrm{OA}^2+\mathrm{OC}^2+\mathrm{OB}^2+\mathrm{OB}^2\right)\)

 

Question 6. ABC is an equilateral triangle. AD is perpendicular on the side BC; let us prove that, AB²+ BC²+ CA² = 4AD².

Solution: In the equilateral triangle ABC, AD is perpendicular to BC.

To prove \(A B^2+B C^2+C A^2=4 A D^2\)

Proof: In  △ABC, AD ⊥ BC ∴ BD = DC

△ABD & △ACD are two right angled triangles

\(\mathrm{AB}^2=\mathrm{AD}^2+\mathrm{BD}^2 \& \mathrm{CA}^2=\mathrm{AD}^2+\mathrm{DC}^2\)

\(A B^2=A D^2+B D^2=A D^2+D C^2\)    [∵ BD = DC]

 

Question 7. I have drawn a right-angled triangle ABC whose A is a right angle. I took two points P and Q on the sides AB and AC respectively. By joining P, Q; B, Q and C, P, let us prove that BQ²+ PC² = BC² + PQ².

Solution:

Given

I have drawn a right-angled triangle ABC whose A is a right angle. I took two points P and Q on the sides AB and AC respectively.

To prove BQ²+ PC² = BC² + PQ²

△ABC, △ABQ, △APQ are all right angled triangles

where ∠A = 90°

\(\mathrm{BC}^2=\mathrm{AB}^2+\mathrm{AC}^2\) \(B Q^2=A Q^2+A B^2 ; P C^2=A P^2+A C^2 ; P Q^2=A P^2+A Q^2\)

Now, \(B Q^2+P C^2=A Q^2+A B^2+A P^2+A C^2\)

& \(\mathrm{BQ}^2+\mathrm{PQ}^2=A B^2+A C^2+A P^2+A Q^2=A Q^2+A B^2+\mathrm{AP}^2+\mathrm{AC}^2\)

 

BQ²+ PC² = BC² + PQ². Proved

Question 8. If two diagonals of a quadrilateral ABCD intersect each other perpendicularly, then let us prove that AB²+ CD² = BC² + DA².

Solution: To prove, AB²+ CD² = BC² + DA²

AC & BD, the two diagonals of the quadrilateral ABCD, intersect each other perpendicularly at O.

∴  △AOB,  △BOC,  △COD &  △DOA are all right angled triangles.

∴ \(\mathrm{AB}^2=\mathrm{OA}^2+\mathrm{OB}^2\)

\(\mathrm{BC}^2=\mathrm{OB}^2+\mathrm{OC}^2\) \(\mathrm{CD}^2=\mathrm{OC}^2+\mathrm{OD}^2\) \(\mathrm{DA}^2=\mathrm{OD}^2+\mathrm{OA}^2\)

∴ \(A B^2+C D^2=O A^2+O B^2+O C^2+O D^2\)

& \(\mathrm{BC}^2+\mathrm{DA}=\mathrm{OB}^2+\mathrm{OC}^2+\mathrm{OD}^2+\mathrm{OA}^2\)

 

∴ AB²+ CD² = BC²+ DA². Proved.

 

Question 9. I have drawn a triangle ABC whose height is AD. If AB > AC, let us prove that, AB² – AC² = BD² – CD².

Solution: In ΔABC, AD is the height of the AABC & AB > AC.

To prove, \(A B^2-A C^2=B D^2-C D^2\)

Proof: AD is the height of ABC, AD ⊥ BC.

△ABD &  △ACD are two right angled triangle

\(A B^2=A D^2+B D^2\)

& \(A C^2=A D^2+C D^2\)

\(A B^2-A C^2=A D^2+B D^2-A D^2-C D^2\)

= \(B D^2-C D^2\)

\(A B^2-A C^2=B D^2-C D^2\)   Proved.

 

Question 10. ABC is an isosceles triangle ‘whose C is a right angle. If D is any point on AB, then let us prove that AD²+ DB² = 2CD².

Solution:

Given

ABC is an isosceles triangle ‘whose C is a right angle. If D is any point on AB

To prove AD²+ BD² = 2CD²

From point D, DE & DF are two perpendicular on BC

& CA respectively,

& ∠DFC = 1 right angle

& ∠DEC = 1 right angle

In CEDF quadrilateral,

∠FCE = ∠DFC = ∠DEC = 1r. angle

In CEDF rectangle, CF || ED

∴ CA || ED & ∠ADB is the intercept

∴ ∠CAD = ∠EDB (corresponding)

∴ ∠EDB = ∠DBE ∴ DE = BE

as CAB = ABC,  △ABC is an isosceles triangle.

 

∴CF = DE = BF: Now, ΔADF, ΔBDE & ΔCDE are all right angles.

∴ AD² = AF² + FD²;

DB² = BE² + DE²;

CD² = CE² + DE²

Now, AD² + DB² = AF² + FD²+ BE² + DE²

= CE²+ CE² + DE² + DE² = 2CE² + 2DE²

=2(CE² + DE²) = 2CD² [AC – CF BC – BE, AF = CE & BE = DE]

= AD²+ DB²

2CD² Proved.

 

Question 11. In the triangle ABC, ZA is the right angle; if CD is the median, let us prove that BC² = CD² + 3AD².

Solution :

Given

In the triangle ABC, ZA is the right angle; if CD is the median

To prove BC² = CD² + 3AD²

In △ABC, CD is median AD = DB.

△CAB & △CAD are right angled triangles.

\(B C^2=A B^2+A C^2 \& C D^2=A D^2+A C^2\) \(A C^2=C D^2-A D^2\) \(B C^2=(2 A D)^2+C D^2-A D^2\)

= \(4 A D^2+C D^2-A D^2\)

= \(3 A D^2+C D^2\)

∴ \(\mathrm{BC}^2=\mathrm{CD}^2+3 A D^2\) Proved.

Question 12. From point O within a triangle ABC, I have drawn the perpendiculars OX, OY, and OZ on BC, CA, and AB respectively. Let us prove that AZ + BX2+ CY² = AY² + CX² + BZ².

Solution:

Given

From point O within a triangle ABC, I have drawn the perpendiculars OX, OY, and OZ on BC, CA, and AB respectively.

To prove, AZ²+ BX²+ CY² = AY² + CX²

AAZO; ABZO; ABXO; ACXO; ACYO; AAYO are all right angle triangles.

∴ AO² = AZ² – ZO²    or, AZ²2 = AO² – ZO²

CX² = CO²-OX²       or, CY² = CO² – YO²

or, AY² = AO² – OY²

∴ AZ² + BX² + CY² – (AY²+ CX² + BZ²)

= (AO²-OZ + BO²-OX²+ CO²-OY²)-(AO²-OY² + CO² – OX² + BO² – OZ²)

=(AO² – AO²)+(OZ²-OZ²) + (BO² – BO²)+ (OX² – OX²) + (CO²- CO²)+ (OY² – OY²) = 0 

∴ AZ²+ BX²+ CY² = AY²+ BZ²+ CX² Proved.

Question 13. In ARST, ZS is a right angle. The midpoints of the two sides RS and ST are X and Y respectively; let us prove that, RY² + XT² = 5XY².

Solution: To prove, RY² + XT² = 5XY²

Proof: From △RSY, \(R Y^2=R^2+S Y^2\)

From △XST, \(\mathrm{XTT}^2=\mathrm{ST}^2+\mathrm{SX}^2\)

∴ \(R Y^2+X T^2=\left(R S^2+S T^2\right)+\left(S Y^2+S X^2\right)\)

= \(R T^2+\left(\frac{1}{2} S T\right)^2+\left(\frac{1}{2} R S\right)^2\)

as \(S Y=\frac{1}{2} S T \& S \dot{x}^2=\frac{1}{2} R S\)

= \(R \mathrm{~T}^2+\frac{1}{4} R \mathrm{~T}^2\)

as \(X Y=\frac{1}{2} R T\)

 

∴ From (1) RY²+ XT² = 5XY² Proved.

WBBSE Solutions Guide Class 10 Maths Chapter 22 Pythagoras Theorem Exercise 22.1 Multiple Choice Questions

Question 1. A person goes 24 m west from a place and then he goes 10 m north. The distance of the person from starting point is

1. 34 m.
2. 17 m.
3. 26 m.
4. 25 m.

Solution: The distance of the person from the starting point

= √242+102 = √576+100 = √676 = 26 m.

Answer.  3. 26 m.

Question 2. If ABC is an equilateral triangle and AD BC, then AD2 =

1. 3/2DC²
2. 2DC²
3. 3DC²
4. 4DC²

Solution: AD (Height of equilateral triangle)

= (√3/2)Side

= 3/4 Side²

= 3/4 (BC)²

= 3/4 × (2DC)²

=3/4 x 4DC²

= 3DC²

Answer. 3. 3DC²

Question 3. In an isosceles triangle ABC, If AC BC and AB² = 2AC², then the measure of ZC is

1. 30°
2. 90°
3. 45°
4. 60°

Solution: Here, AC BC & AB² = 2AC²

∴ AB² = AC² + BC²    ∴ ZC = 90°

Answer. 2. 90°

Question 4. Two rods of 13 m in length and 7 m in length are situated perpendicularly on the ground and the distance between their foots is 8 m. The distance between their top parts is

1. 9 m.
2. 10 m.
3. 11 m.
4. 12 m.

Solution: Here AC² = AB²+ BC²

= 62+82 = 100

∴ AC = 10 m.

Ans. 2. 10 m.

Question 5. If the lengths of two diagonals of a rhombus are 24 cm and 10 cm, the perimeter of the rhombus is

1. 13 cm.
2. 26 cm.
3. 52 cm.
4. 25 cm.

Solution: ∴ BD = 24 cm.          ∴ DO = 12 cm.

AC = 12 cm.       CO = 5 cm.

DC² = AO² + OC² = 122 + 52

= 144 +25

= 169

∴ DC = √169

= 13 cm.

:. Perimeter = 4 x 13 cm.

= 52 cm.

Answer. 3. 52 cm.

WBBSE Class 10 Maths Solutions Pdf In English Chapter 22 Pythagoras Theorem Exercise 22.1 True Or False

1. If the ratio of the lengths of three sides of a triangle is 3: 45, then the triangle will always be a right-angled triangle.

Answer: True

2. If in a circle of radius 10 cm in length, a chord subtends a right angle at the center, then the length of the chord will be 5 cm.

Answer: False

Fill In The Blanks

1. In a right-angled triangle, the area of a square drawn on the hypotenuse is equal to the area of the sum of squares drawn on the other two sides.

2. In an isosceles right-angled triangle if the length of each of two equal sides is 4 cm, then the length of the hypotenuse will be 8 cm.

3. In a rectangular figure ABCD, the two diagonals AC and BD intersect each other at point O, if AB = 12 cm, AP = 6.5, then the length of BC is 15 cm.

WBBSE Class 10 Maths Solutions Chapter 22 Pythagoras Theorem Exercise 22.1 Short Answer

Question 1. In AABC, if AB = (2a-1) cm AC = 2√2a  cm. and BC= (2a + 1) cm, then let us write the value of <BAC.

Solution.

Given

In AABC, if AB = (2a-1) cm AC = 2√2a  cm. and BC= (2a + 1) cm

AB = (2a – 1) cm.

AC = 2√2a cm.

BC (2a + 1) cm.

∴ AB²+ AC² = (2a-1)+(2√2a)²

= 4a² – 4a+1 + 8a

= 4a² + 4a + 1

= (2a + 1)²

O = BC²

∴ ∠BAC 90°

Question 2. In the adjoining figure, point O is situated within the triangle PQR in such a way that, POQ = 90°, OP = 6 cm, and OQ = 8 cm. If PR = 24 cm and ∠QPR = 90°, then let us write the length of QR.

Solution.

Given

In the adjoining figure, point O is situated within the triangle PQR in such a way that, POQ = 90°, OP = 6 cm, and OQ = 8 cm. If PR = 24 cm and ∠QPR = 90°,

As ∠POQ = 90°

∴ PQ² = OP² + OQ² = 62+ 82 = 100

∴ PQ = 10 cm.

Again, ∠QPR = 90°

∴ QR² = PQ²+ PR²

= 100+ 576

= 676

QR = √676

= 26 cm.

Question 3. point O is situated within the rectangular figure ABCD in such a way that OB = 6 cm, OD = 8cm, and OA = 5 cm. Let us determine the length of OC.

Solution. 5√3

Question 4. In the triangle, ABC the perpendicular AD from point A on side BC meets side BC at point D. If BD = 8 cm, DC = 2 cm, and AD = 4 cm, then let us write the measure of <BAC.

Solution.

Given

In the triangle, ABC the perpendicular AD from point A on side BC meets side BC at point D. If BD = 8 cm, DC = 2 cm, and AD = 4 cm,

∠ADB 90°

∴ AB² = AD²+ DB²

= 42+ 82

= 16+6480

AC² = CD²+ AD²

= 4+16

=20

∴ Here, BC² = (2+8)² (10)²

= 100

∴ AB²+ AC²=80+20

=100

∴ <BAC = 90°.

Question 5. In a right-angled triangle, ABC, ABC = 90°, AB = 3 cm, BC = 4 cm, and the perpendicular BD on the side AC from point B which meets the side AC at point D. Let us determine the length of BD.

Solution.

Given

In a right-angled triangle, ABC, ABC = 90°, AB = 3 cm, BC = 4 cm, and the perpendicular BD on the side AC from point B which meets the side AC at point D.

Here ABC ∼ BDC

AB/AC = BD/BC     AC² = AB² + BC²

3/5 = BD/4        

= 9+16

=25

∴ AC = √125

=5

∴ BD= 12/5 

= 2.4 cm.

The length of BD = 2.4 cm.