Updated on: 2026-03-31 | Author: Aarti Kulkarni

Geometry (Math 2) Chapter 2 Pythagoras Theorem Solutions

Q1. A congruent side of an isosceles right angled triangle is 7 cm, Find its perimeter

Diagram: Refer textbook

Q2. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of wall. Complete the given activity. Activity: As shown in figure suppose PR is the length of ladder = 10 m At P – window, At Q – base of wall, At R – foot of ladder ∴ PQ = 8 m ∴ QR = ? In ∆PQR, m∠PQR = 90° By Pythagoras Theorem, ∴ PQ 2 + `square` = PR 2 .....(I) Here, PR = 10, PQ = `square` From equation (I) 8 2 + QR 2 = 10 2 QR 2 = 10 2 – 8 2 QR 2 = 100 – 64 QR 2 = `square` QR = 6 ∴ The distance of foot of the ladder from the base of wall is 6 m.

As shown in figure suppose PR is the length of ladder = 10 m,

At P – window, At Q – base of wall, At R – foot of ladder

∴ PQ = 8 m
∴ QR = ?
In ∆PQR, m∠PQR = 90°

By Pythagoras Theorem,

∴ PQ2 + QR2 = PR2 .....(I)
Here, PR = 10, PQ = 8

From equation (I)

82 + QR2 = 102
QR2 = 102 – 82
QR2 = 100 – 64
QR2 = 36
QR = 6
∴ The distance of foot of the ladder from the base of wall is 6 m.

Q3. A rectangle having dimensions 35 m × 12 m, then what is the length of its diagonal?

Diagram: Refer textbook

Q4. As shown figure, ∠DFE = 90°, FG ⊥ ED, if GD = 8, FG = 12, then EG = ?

  1. In ∆DEF,
∠DFE = 90° and seg FG ⊥ hypotenuse ED ...[Given]
∴ FG2 = EG × GD ......[By theorem of geometric mean]
∴ (12)2 = EG × 8 ......[Given]
∴ 144 = EG × 8
∴ EG = `144/8`
∴ EG = 18 units
  1. In ∆DGF,
∠DGF = 90° .....[ ⸪ FG ⊥ ED]
∴ FD2 = FG2 + GD2 ......[Pythagoras theorem]
∴ FD2 = (12)2 + (8)2 ......[Given]
∴ FD2 = 144 + 64
∴ FD2 = 208
∴ FD = `sqrt(16 xx 13)` ......[Taking square root of both sides]
∴ FD = `4sqrt(13)` units
  1. In EGF,
∠EGF = 90° ......[⸪ FG ⊥ ED]
∴ EF2 = EG2 + FG2 ......[Pythagoras theorem]
∴ EF2 = (18)2 + (12)2 ......[From (i) and given]
∴ EF2 = 324 + 144
∴ EF2 = 468
∴ EF = `sqrt(36 xx 13)` .......[Taking square root of both sides]
∴ EF = `6sqrt(13)` units

Q5. As shown in figure, LK = `6sqrt(2)` then MK = ? ML = ? MN = ?

  1. In ∆MLK,
∠MLK = 90° and ∠LKM = 30° ......[Given]
∴ ∠LMK = 60° ......[Remaining angle of a triangle]
∴ ∆MLK is a 30° – 60° – 90° triangle.
∴ LK = `sqrt(3)/2` MK ......[Side opposite to 60°]
∴ `6sqrt(2) = sqrt(3)/2` MK ......[Given]
∴ MK = `6sqrt(2) xx 2/sqrt(3)`

= `(12sqrt(2))/sqrt(3)`

= `(12 xx sqrt(2) xx sqrt(3))/(sqrt(3) xx sqrt(3))` ......[Multiply numerator and denominator by `sqrt(3)`]

= `(12 xx sqrt(2 xx 3))/3`

MK = `4sqrt(6)` units

  1. In ∆MLK,
∠MLK = 90° ......[Given]
∴ MK2 = ML2 + LK2 ......[Pythagoras theorem]
∴ `(4sqrt(6))^2 = "ML"^2 + (6sqrt(2))^2` ......[From (i) and given]
∴ (16 × 6) = ML2 + (36 × 2)
∴ 96 = ML2 + 72
∴ ML2 = 24
∴ ML = `sqrt(24)` .......[Taking square root of both sides]
∴ ML = `sqrt(4 xx 6)`
∴ ML = `2sqrt(6)` units
  1. In ∆NKM,
∠NKM = 90° and ∠MNK = 45° ......[Given]
∴ ∠KMN = 45° ......[Remaining angle of a triangle]
∴ ∆NKM is 45° – 45° – 90° triangle.
∴ MK = `1/sqrt(2)` MN ......[Theorem of 45° – 45° – 90° triangle]
∴ `4sqrt(6) = 1/sqrt(2)` MN ......[From (i)]
∴ MN = `4sqrt(6) xx sqrt(2)`

= `4sqrt(6 xx 2)`

= `4sqrt(4 xx 3)`

= `4 xx2 xx sqrt(3)`

∴ MN = `8sqrt(3)` units

Q6. Choose the correct alternative: A rectangle having length of a side is 12 and length of diagonal is 20, then what is length of other side?

Diagram: Refer textbook

Q7. Choose the correct alternative: If length of both diagonals of rhombus are 60 and 80, then what is the length of side?

Diagram: Refer textbook

Q8. Choose the correct alternative: If length of sides of a triangle are a, b, c and a 2 + b 2 = c 2 , then which type of triangle it is?

Right angled triangle

Q9. Choose the correct alternative: If the length of diagonal of square is √2, then what is the length of each side?

Diagram: Refer textbook

Q10. Choose the correct alternative: In ∆ABC, AB = `6sqrt(3)` cm, AC = 12 cm, and BC = 6 cm, then m∠A = ?

Diagram: Refer textbook

Q11. Choose the correct alternative: In right angled triangle, if sum of the squares of the sides of right angle is 169, then what is the length of the hypotenuse?

Diagram: Refer textbook

Q12. Choose the correct alternative: Out of all numbers from given dates, which is a Pythagoras triplet?

15/8/17

Here,

152 + 82 = 225 + 64 = 289, and 172 = 289
∴ 152 + 82 = 172.

Q13. Choose the correct alternative : Out of given triplets, which is a Pythagoras triplet?

(3, 4, 5)

Here, 52 = 25
32 + 42 = 9 + 16 = 25
∴ 52 = 32 + 42

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ (3, 4, 5) is a Pythagoras triplet.

Q14. Choose the correct alternative: Out of given triplets, which is not a Pythagoras triplet?

(7, 8, 15)

Here, 152 = 225
72 + 82 = 49 + 64 = 113
∴ 152 ≠ 72 + 82

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴ (7, 8, 15) is not a Pythagoras triplet.

Q15. Choose the correct alternative: Out of given triplets, which is not a Pythagoras triplet?

(6, 14, 15)

Here, 152 = 225
62 + 142 = 36 + 196 = 232
∴ 152 ≠ 62 + 142

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴(6, 14, 15) is not a Pythagoras triplet.

Q16. Choose the correct alternative: The diagonal of a square is `10sqrt(2)` cm then its perimeter is ______

Diagram: Refer textbook

Q17. Complete the following activity to find the length of hypotenuse of right angled triangle, if sides of right angle are 9 cm and 12 cm. Activity: In ∆PQR, m∠PQR = 90° By Pythagoras Theorem, PQ 2 + `square` = PR 2 ......(I) ∴ PR 2 = 9 2 + 12 2 ∴ PR 2 = `square` + 144 ∴ PR 2 = `square` ∴ PR = 15 ∴ Length of hypotenuse of triangle PQR is `square` cm.

In ∆PQR, m∠PQR = 90°

By Pythagoras Theorem,

PQ2 + QR2 = PR2 ......(I)
∴ PR2 = 92 + 122
∴ PR2 = 81 + 144
∴ PR2 = 225
∴ PR = 15 ......[Taking square roots of both sides]
∴ Length of hypotenuse of triangle PQR is 15 cm.

Q18. Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq.cm. Complete the following activity. Activity: As shown in figure LMNT is a reactangle. ∴ Area of rectangle = length × breadth ∴ Area of rectangle = `square` × breadth ∴ 192 = `square` × breadth ∴ Breadth = 12 cm Also, ∠TLM = 90° ......[Each angle of reactangle is right angle] In ∆TLM, By Pythagoras theorem ∴ TM 2 = TL 2 + `square` ∴ TM 2 = 12 2 + `square` ∴ TM 2 = 144 + `square` ∴ TM 2 = 400 ∴ TM = 20

As shown in figure LMNT is reactangle.

∴ Area of rectangle = length × breadth
∴ Area of rectangle = 16 × breadth
∴ 192 = 16 × breadth
∴ Breadth = `192/16`
∴ Breadth = 12 cm
i.e., TL = 12 cm and LM = 16 cm ......(i)
Also, ∠TLM = 90° ......[Each angle of reactangle is right angle]

In ∆TLM, By Pythagoras theorem

∴ TM2 = TL2 + LM2
∴ TM2 = 122 + 162 ......[From (i)]
∴ TM2 = 144 + 256
∴ TM2 = 400
∴ TM = 20

Q19. Find the height of an equilateral triangle having side 4 cm?

Diagram: Refer textbook

Q20. From given figure, In ∆ABC, AB ⊥ BC, AB = BC, AC = `2sqrt(2)` then l (AB) = ?

AB = BC ......[Given]
∴ ∠A = ∠C ......[Isosceles triangle theorem]
Let ∠A = ∠C = x ......(i)
In ∆ABC, ∠A + ∠B + ∠C = 180° ......[Sum of the measures of the angles of a triangle is 180°]
∴ x + 90° + x = 180° ......[From (i)]
∴ 2x = 90°
∴ x = `90^circ/2` ......[From (i)]
∴ x = 45°
∴ ∠BAC = ∠BCA = 45°
∴ ∆ABC is a 45° – 45° – 90° triangle.
∴ AB = BC = `1/sqrt(2) xx "AC"` ......[Side opposite to 45°]

= `1/sqrt(2) xx 2sqrt(2)`

∴ `l("AB")` = 2 units

Q21. From given figure, In ∆ABC, AB ⊥ BC, AB = BC, AC = `5sqrt(2)` , then what is the height of ∆ABC?

AB = BC ......[Given]
∴ ∠A = ∠C ......[Isosceles triangle theorem]
Let ∠A = ∠C = x ......(i)
In ∆ABC, ∠A + ∠B + ∠C = 180° ......[Sum of the measures of the angles of a triangle is 180°]
∴ x + 90° + x = 180° .......[From (i)]
∴ 2x = 90°
∴ x = `90^circ/2` .......[From (i)]
∴ x = 45°
∴ ∠A = ∠C = 45°
∴ ∆ABC is a 45° – 45° – 90° triangle.
∴ AB = BC = `1/sqrt(2) xx "AC"` ......[Side opposite to 45°]

= `1/sqrt(2) xx 5sqrt(2)`

∴ AB = BC = 5 units
∴ The height of ∆ABC is 5 units.

Q22. From given figure, In ∆ABC, AB ⊥ BC, AB = BC then m∠A = ?

AB = BC ......[Given]
∴ ∠A = ∠C ......[Isosceles triangle theorem]
Let ∠A = ∠C = x

In ∆ABC,

∠A + ∠B + ∠C = 180° ...[Sum of the measures of the angles of a triangle is 180°]
∴ x + 90° + x = 180°
∴ 2x = 90°
∴ x = `(90°)/2`
∴ x = 45°
∴ m∠A = 45°

Q23. From given figure, In ∆ABC, AD ⊥ BC, then prove that AB 2 + CD 2 = BD 2 + AC 2 by completing activity. Activity: From given figure, In ∆ACD, By pythagoras theorem AC 2 = AD 2 + `square` ∴ AD 2 = AC 2 – CD 2 ......(I) Also, In ∆ABD, by pythagoras theorem, AB 2 = `square` + BD 2 ∴ AD 2 = AB 2 – BD 2 ......(II) ∴ `square` − BD 2 = AC 2 − `square` ∴ AB 2 + CD 2 = AC 2 + BD 2

From given figure, in ∆ACD, By pythagoras theorem

AC2 = AD2 + CD2
∴ AD2 = AC2 – CD2 ......(I)

Also, In ∆ABD, by pythagoras theorem,

AB2 = AD2 + BD2
∴ AD2 = AB2 – BD2 ......(II)
∴ AB2 − BD2 = AC2 − CD2 ......[From (i) and (ii)]
∴ AB2 + CD2 = AC2+ BD2

Q24. From given figure, In ∆ABC, If ∠ABC = 90° ∠CAB=30°, AC = 14 then for finding value of AB and BC, complete the following activity. Activity: In ∆ABC, If ∠ABC = 90°, ∠CAB=30° ∴ ∠BCA = `square` By theorem of 30° – 60° – 90° triangle, ∴ `square = 1/2` AC and `square = sqrt(3)/2` AC ∴ BC = `1/2 xx square` and AB = `sqrt(3)/2 xx 14` ∴ BC = 7 and AB = `7sqrt(3)`

In ∆ABC, If ∠ABC = 90°, ∠CAB=30°
∴ ∠BCA = 60° ......[Remaining angle of a triangle]

By theorem of 30° – 60° – 90° triangle,

∴ BC = `1/2` AC .....[Side opposite to 30°]

and AB = `sqrt(3)/2` AC .....[Side opposite to 60°]

∴ BC = `1/2 xx 14` and AB = `sqrt(3)/2 xx 14`
∴ BC = 7 and AB = `7sqrt(3)`

Q25. From given figure, In ∆ABC, If AC = 12 cm. then AB =? Activity: From given figure, In ∆ABC, ∠ABC = 90°, ∠ACB = 30° ∴ ∠BAC = `square` ∴ ∆ABC is 30° – 60° – 90° triangle ∴ In ∆ABC by property of 30° – 60° – 90° triangle. ∴ AB = `1/2` AC and `square` = `sqrt(3)/2` AC ∴ `square` = `1/2 xx 12` and BC = `sqrt(3)/2 xx 12` ∴ `square` = 6 and BC = `6sqrt(3)`

From given figure, In ∆ABC, ∠ABC = 90°, ∠ACB = 30°
∴ ∠BAC = 60° ......[Remaining angle of a triangle]
∴ ∆ABC is 30° – 60° – 90° triangle.
∴ In ∆ABC,

by property of 30° – 60° – 90° triangle.

∴ AB = `1/2` AC .......[Side opposite to 30°]

and BC = `sqrt(3)/2` AC .......[Side opposite to 60°]

∴ AB = `1/2 xx 12` and BC = `sqrt(3)/2 xx 12`
∴ AB = 6 and BC = `6sqrt(3)`

Q26. From given figure, in ∆MNK, if ∠MNK = 90°, ∠M = 45°, MK = 6, then for finding value of MN and KN, complete the following activity. Activity: In ∆MNK, ∠MNK = 90°, ∠M = 45° …...[Given] ∴ ∠K = `square` .....[Remaining angle of ∆MNK] By theorem of 45° – 45° – 90° triangle, ∴ `square = 1/sqrt(2)` MK and `square = 1/sqrt(2)` MK ∴ MN = `1/sqrt(2) xx square` and KN = `1/sqrt(2) xx 6` ∴ MN = `3sqrt(2)` and KN = `3sqrt(2)`

In ∆MNK, ∠MNK = 90°, ∠M = 45° …...[Given]
∴ ∠K = 45° .....[Remaining angle of ∆MNK]

By theorem of 45° – 45° – 90° triangle,

∴ MN = `1/sqrt(2)` MK and KN = `1/sqrt(2)` MK
∴ MN = `1/sqrt(2) xx 6` and KN = `1/sqrt(2) xx 6`
∴ MN = `(1 xx 6 xx sqrt(2))/(sqrt(2) xx sqrt(2))` and KN = `(1 xx 6 xx sqrt(2))/(sqrt(2) xx sqrt(2))` .......[Multiply numerator and denominator by `sqrt(2)`]
∴ MN = `(6 xx sqrt(2))/2` and KN = `(6 xx sqrt(2))/2`
∴ MN = `3sqrt(2)` and KN = `3sqrt(2)`

Q27. From given figure, in ∆PQR, if ∠QPR = 90°, PM ⊥ QR, PM = 10, QM = 8, then for finding the value of QR, complete the following activity. Activity: In ∆PQR, if ∠QPR = 90°, PM ⊥ QR, ......[Given] In ∆PMQ, by Pythagoras Theorem, ∴ PM 2 + `square` = PQ 2 ......(I) ∴ PQ 2 = 10 2 + 8 2 ∴ PQ 2 = `square` + 64 ∴ PQ 2 = `square` ∴ PQ = `sqrt(164)` Here, ∆QPR ~ ∆QMP ~ ∆PMR ∴ ∆QMP ~ ∆PMR ∴ `"PM"/"RM" = "QM"/"PM"` ∴ PM 2 = RM × QM ∴ 10 2 = RM × 8 RM = `100/8 = square` And, QR = QM + MR QR = `square` + `25/2 = 41/2`

In ∆PQR, if ∠QPR = 90°, PM ⊥ QR, ......[Given]

In ∆PMQ,

by Pythagoras Theorem,

∴ PM2 + QM2 = PQ2 ......(I)
∴ PQ2 = 102 + 82
∴ PQ2 = 100 + 64
∴ PQ2 = 164
∴ PQ = `sqrt(164)`

Here, ∆QPR ~ ∆QMP ~ ∆PMR

∴ ∆QMP ~ ∆PMR
∴ `"PM"/"RM" = "QM"/"PM"` ......[Corresponding sides of similar triangles]
∴ PM2 = RM × QM
∴ 102 = RM × 8

RM = `100/8` = `25/2`

And,

QR = QM + MR
QR = 8 + `25/2 = 41/2`

Q28. From the given figure, in ∆ABC, if AD ⊥ BC, ∠C = 45°, AC = `8sqrt(2)` , BD = 5, then for finding value of AD and BC, complete the following activity. Activity: In ∆ADC, if ∠ADC = 90°, ∠C = 45° ......[Given] ∴ ∠DAC = `square` .....[Remaining angle of ∆ADC] By theorem of 45° – 45° – 90° triangle, ∴ `square = 1/sqrt(2)` AC and `square = 1/sqrt(2)` AC ∴ AD =`1/sqrt(2) xx square` and DC = `1/sqrt(2) xx 8sqrt(2)` ∴ AD = 8 and DC = 8 ∴ BC = BD +DC = 5 + 8 = 13

In ∆ADC, if ∠ADC = 90°, ∠C = 45° ......[Given]
∴ ∠DAC = 45° .....[Remaining angle of ∆ADC]

By theorem of 45° – 45° – 90° triangle,

∴ AD = `1/sqrt(2)` AC and DC = `1/sqrt(2)` AC
∴ AD =`1/sqrt(2) xx 8sqrt(2)` and DC = `1/sqrt(2) xx 8sqrt(2)`
∴ AD = 8 and DC = 8
∴ BC = BD +DC

= 5 + 8

= 13

Q29. From the given figure, in ∆ABQ, if AQ = 8 cm, then AB =?

In ∆ABQ,

∠B = 90°, ∠Q = 30° ...[Given]
∴ ∠A = 60° ...[Remaining angle of a triangle]
∴ ∆ABQ is a 30°–60°–90° triangle.
∴ AB = `1/2` AQ ....[Side opposite to 30°]
∴ AB = `1/2 xx 8`
∴ AB = 4 cm

Q30. Height and base of a right angled triangle are 24 cm and 18 cm. Find the length of its hypotenus?

Diagram: Refer textbook

Q31. If a triangle having sides 50 cm, 14 cm and 48 cm, then state whether given triangle is right angled triangle or not

The sides of the triangle are 50 cm, 14 cm and 48 cm.

The longest side of the triangle is 50 cm.

∴ (50)2 = 2500

Now, sum of the squares of the remaining sides is,

(14)2 + (48)2 = 196 + 2304 = 2500
∴ (50)2 = (14)2 + (48)2
∴ Square of the longest side is equal to the sum of the squares of the remaining two sides.
∴ The given sides will form a right angled triangle. .......[Converse of Pythagoras theorem]

Q32. If a triangle having sides 8 cm, 15 cm and 17 cm, then state whether given triangle is right angled triangle or not

The sides of the triangle are 8 cm, 15 cm, and 17 cm.

The longest side of the triangle is 17 cm.

∴ (17)2 = 289

Now, sum of the squares of the remaining sides is,

(8)2 + (15)2 = 64 + 225 = 289
∴ (17)2 = (8)2 + (15)2
∴ Square of the longest side is equal to the sum of the squares of the remaining two sides.
∴ The given sides will form a right-angled triangle. ......[Converse of Pythagoras theorem]

Q33. In a right angled triangle, if length of hypotenuse is 25 cm and height is 7 cm, then what is the length of its base?

Diagram: Refer textbook

Q34. In ∆LMN, l = 5, m = 13, n = 12 then complete the activity to show that whether the given triangle is right angled triangle or not. *(l, m, n are opposite sides of ∠L, ∠M, ∠N respectively) Activity: In ∆LMN, l = 5, m = 13, n = `square` ∴ l 2 = `square`, m 2 = 169, n 2 = 144. ∴ l 2 + n 2 = 25 + 144 = `square` ∴ `square` + l 2 = m 2 ∴By Converse of Pythagoras theorem, ∆LMN is right angled triangle.

In ∆LMN,

l = 5, m = 13, n = 12
∴ l2 = 25, m2 = 169, n2 = 144.
∴ l2 + n2 = 25 + 144 = 169
∴ n2 + l2 = m2
∴By Converse of Pythagoras theorem, ∆LMN is right angled triangle.

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