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Geometry · Textbook solutions

Pythagoras Theorem

Every solved example, exercise, and miscellaneous question, in the order the textbook teaches them. Plus every board question asked from this chapter. · 56 questions

Solved Examples

7 q
  1. Solved Ex.1
    See fig 2.11. In △ABC\triangle ABC, ∠B=90∘\angle B = 90^\circ, ∠A=30∘\angle A = 30^\circ, AC = 14, then find AB and BC.
  2. Solved Ex.2
    See fig 2.12, In △ABC\triangle ABC, seg AD ⊥\perp seg BC, ∠C=45∘\angle C = 45^\circ, BD = 5 and AC = 828\sqrt{2} then find AD and BC.
  3. Solved Ex.3
    In fig 2.13, ∠PQR=90∘\angle PQR = 90^\circ, seg QN ⊥\perp seg PR, PN = 9, NR = 16. Find QN.
  4. Solved Ex.4
    See figure 2.14. In △PQR\triangle PQR, ∠PQR=90∘\angle PQR = 90^\circ, seg QS ⊥\perp seg PR then find xx, yy, zz. (From fig. 2.14: S lies on seg PR, PS = 10, SR = 8, QS = xx, QR = yy, PQ = zz.)
  5. Solved Ex.5
    In the right angled triangle, sides making right angle are 9 cm and 12 cm. Find the length of the hypotenuse.
  6. Solved Ex.6
    In △LMN\triangle LMN, l=5l = 5, m=13m = 13, n=12n = 12. State whether △LMN\triangle LMN is a right angled triangle or not.
  7. Solved Ex.7
    See fig 2.16. In △ABC\triangle ABC, seg AD ⊥\perp seg BC. Prove that: AB2+CD2=BD2+AC2AB^2 + CD^2 = BD^2 + AC^2

Practice set 2.1

10 q
  1. Ex 2.1 Q.1
    Identify, with reason, which of the following are Pythagorean triplets. (i) (3, 5, 4) (ii) (4, 9, 12) (iii) (5, 12, 13) (iv) (24, 70, 74) (v) (10, 24, 27) (vi) (11, 60, 61)
  2. Ex 2.1 Q.2
    In figure 2.17, ∠MNP=90∘\angle MNP = 90^\circ, seg NQ ⊥\perp seg MP, MQ = 9, QP = 4, find NQ.
  3. Ex 2.1 Q.3
    In figure 2.18, ∠QPR=90∘\angle QPR = 90^\circ, seg PM ⊥\perp seg QR and Q-M-R, PM = 10, QM = 8, find QR.
  4. Ex 2.1 Q.4
    See figure 2.19. Find RP and PS using the information given in △PSR\triangle PSR. (From fig. 2.19: ∠PSR=90∘\angle PSR = 90^\circ, ∠SPR=30∘\angle SPR = 30^\circ, SR = 6.)
  5. Ex 2.1 Q.5
    For finding AB and BC with the help of information given in figure 2.20, complete following activity. (From fig. 2.20: in △ABC\triangle ABC, ∠ABC=90∘\angle ABC = 90^\circ, seg AB and seg BC carry equal marks, AC = 8\sqrt{8}.) AB=BCAB = BC .......... □\square ∴∠BAC=□\therefore \angle BAC = \square ∴AB=BC=□×AC\therefore AB = BC = \square \times AC                 =□×8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \square \times \sqrt{8}                 =□×22\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \square \times 2\sqrt{2}                 =□\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \square
  6. Ex 2.1 Q.6
    Find the side and perimeter of a square whose diagonal is 10 cm.
  7. Ex 2.1 Q.7
    In figure 2.21, ∠DFE=90∘\angle DFE = 90^\circ, FG ⊥\perp ED, If GD = 8, FG = 12, find (1) EG (2) FD and (3) EF (From fig. 2.21: G lies on seg ED, between E and D.)
  8. Ex 2.1 Q.8
    Find the diagonal of a rectangle whose length is 35 cm and breadth is 12 cm.
  9. Ex 2.1 Q.9
    In the figure 2.22, M is the midpoint of QR. ∠PRQ=90∘\angle PRQ = 90^\circ. Prove that, PQ2=4PM2−3PR2PQ^2 = 4PM^2 - 3PR^2 [Note: marked ★ (challenging) in the textbook.]
  10. Ex 2.1 Q.10
    Walls of two buildings on either side of a street are parallel to each other. A ladder 5.8 m long is placed on the street such that its top just reaches the window of a building at the height of 4 m. On turning the ladder over to the other side of the street, its top touches the window of the other building at a height 4.2 m. Find the width of the street. [Note: marked ★ (challenging) in the textbook.]

Application of Pythagoras theorem

2 q

Solved Examples

Worked · 2
  1. Application of Pythagoras theorem SolvedEx.1
    In △ABC\triangle ABC, ∠C\angle C is an acute angle, seg AD ⊥\perp seg BC. Prove that: AB2=BC2+AC2−2BC×DCAB^2 = BC^2 + AC^2 - 2BC \times DC
  2. Application of Pythagoras theorem SolvedEx.2
    In △ABC\triangle ABC, ∠ACB\angle ACB is obtuse angle, seg AD ⊥\perp seg BC. Prove that: AB2=BC2+AC2+2BC×CDAB^2 = BC^2 + AC^2 + 2BC \times CD

Apollonius theorem

2 q

Solved Examples

Worked · 2
  1. Apollonius theorem SolvedEx.1
    In the figure 2.26, seg PM is a median of △PQR\triangle PQR. PM = 9 and PQ2+PR2=290PQ^2 + PR^2 = 290, then find QR.
  2. Apollonius theorem SolvedEx.2
    Prove that, the sum of the squares of the diagonals of a rhombus is equal to the sum of the squares of the sides.

Practice set 2.2

5 q
  1. Ex 2.2 Q.1
    In △PQR\triangle PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13, find QR.
  2. Ex 2.2 Q.2
    In △ABC\triangle ABC, AB = 10, AC = 7, BC = 9 then find the length of the median drawn from point C to side AB.
  3. Ex 2.2 Q.3
    In the figure 2.28 seg PS is the median of △PQR\triangle PQR and PT ⊥\perp QR. Prove that, (1) PR2=PS2+QR×ST+(QR2)2PR^2 = PS^2 + QR \times ST + \left(\frac{QR}{2}\right)^2 ii) PQ2=PS2−QR×ST+(QR2)2PQ^2 = PS^2 - QR \times ST + \left(\frac{QR}{2}\right)^2
  4. Ex 2.2 Q.4
    In △ABC\triangle ABC, point M is the midpoint of side BC. If AB2+AC2=290AB^2 + AC^2 = 290 cm2^2, AM = 8 cm, find BC.
  5. Ex 2.2 Q.5
    In figure 2.30, point T is in the interior of rectangle PQRS. Prove that, TS2+TQ2=TP2+TR2TS^2 + TQ^2 = TP^2 + TR^2 (As shown in the figure, draw seg AB ∥\parallel side SR and A-T-B) [Note: marked ★ (challenging) in the textbook.]

Problem set 2

30 q
  1. Some questions and their alternative answers are given. Select the correct alternative.
    PS2 Q.1 (1)
    Out of the following which is the Pythagorean triplet?

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  2. PS2 Q.1 (2)
    In a right angled triangle, if sum of the squares of the sides making right angle is 169 then what is the length of the hypotenuse?

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  3. PS2 Q.1 (3)
    Out of the dates given below which date constitutes a Pythagorean triplet?

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  4. PS2 Q.1 (4)
    If a, b, c are sides of a triangle and a2+b2=c2a^2 + b^2 = c^2, name the type of triangle.

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  5. PS2 Q.1 (5)
    Find perimeter of a square if its diagonal is 10210\sqrt{2} cm.

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  6. PS2 Q.1 (6)
    Altitude on the hypotenuse of a right angled triangle divides it in two parts of lengths 4 cm and 9 cm. Find the length of the altitude.

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  7. PS2 Q.1 (7)
    Height and base of a right angled triangle are 24 cm and 18 cm find the length of its hypotenuse

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  8. PS2 Q.1 (8)
    In △ABC\triangle ABC, AB = 636\sqrt{3} cm, AC = 12 cm, BC = 6 cm. Find measure of ∠A\angle A.

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  9. Solve the following examples.
    PS2 Q.2 (1)
    Find the height of an equilateral triangle having side 2a2a.
  10. PS2 Q.2 (2)
    Do sides 7 cm , 24 cm, 25 cm form a right angled triangle ? Give reason.
  11. PS2 Q.2 (3)
    Find the length of a diagonal of a rectangle having sides 11 cm and 60cm.
  12. PS2 Q.2 (4)
    Find the length of the hypotenuse of a right angled triangle if remaining sides are 9 cm and 12 cm.
  13. PS2 Q.2 (5)
    A side of an isosceles right angled triangle is xx. Find its hypotenuse.
  14. PS2 Q.2 (6)
    In △PQR\triangle PQR; PQ = 8\sqrt{8}, QR = 5\sqrt{5}, PR = 3\sqrt{3}. Is △PQR\triangle PQR a right angled triangle ? If yes, which angle is of 90∘90^\circ ?
  15. PS2 Q.3
    In △RST\triangle RST, ∠S=90∘\angle S = 90^\circ, ∠T=30∘\angle T = 30^\circ, RT = 12 cm then find RS and ST.
  16. PS2 Q.4
    Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq.cm.
  17. PS2 Q.5
    Find the length of the side and perimeter of an equilateral triangle whose height is 3\sqrt{3} cm. [Note: marked ★ (challenging) in the textbook.]
  18. PS2 Q.6
    In △ABC\triangle ABC seg AP is a median. If BC = 18, AB2+AC2=260AB^2 + AC^2 = 260 Find AP.
  19. PS2 Q.7
    △ABC\triangle ABC is an equilateral triangle. Point P is on base BC such that PC = 13\frac{1}{3} BC, if AB = 6 cm find AP. [Note: marked ★ (challenging) in the textbook.]
  20. PS2 Q.8
    From the information given in the figure 2.31, prove that PM = PN = 3×a\sqrt{3} \times a. In the figure, the points M, Q, S, R, N are collinear, seg PS ⊥\perp seg MN, PQ = PR = aa and MQ = QR = RN = aa.
  21. PS2 Q.9
    Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.
  22. PS2 Q.10
    Pranali and Prasad started walking to the East and to the North respectively, from the same point and at the same speed. After 2 hours distance between them was 15215\sqrt{2} km. Find their speed per hour.
  23. PS2 Q.11
    In △ABC\triangle ABC, ∠BAC=90∘\angle BAC = 90^\circ, seg BL and seg CM are medians of △ABC\triangle ABC. Then prove that: 4(BL2+CM2)=5BC24(BL^2 + CM^2) = 5 BC^2 [Note: marked ★ (challenging) in the textbook.]
  24. PS2 Q.12
    Sum of the squares of adjacent sides of a parallelogram is 130 sq.cm and length of one of its diagonals is 14 cm. Find the length of the other diagonal.
  25. PS2 Q.13
    In △ABC\triangle ABC, seg AD ⊥\perp seg BC, DB = 3CD. Prove that : 2AB2=2AC2+BC22AB^2 = 2AC^2 + BC^2
  26. PS2 Q.14
    In an isosceles triangle, length of the congruent sides is 13 cm and its base is 10 cm. Find the distance between the vertex opposite the base and the centroid. [Note: marked ★ (challenging) in the textbook.]
  27. PS2 Q.15
    In a trapezium ABCD, seg AB ∥\parallel seg DC, seg BD ⊥\perp seg AD, seg AC ⊥\perp seg BC, If AD = 15, BC = 15 and AB = 25. Find A(□\square ABCD)
  28. PS2 Q.16
    In the figure 2.35, △PQR\triangle PQR is an equilateral triangle. Point S is on seg QR such that QS = 13\frac{1}{3} QR. Prove that : 9PS2=7PQ29 PS^2 = 7 PQ^2 [Note: marked ★ (challenging) in the textbook.]
  29. PS2 Q.17
    Seg PM is a median of △PQR\triangle PQR. If PQ = 40, PR = 42 and PM = 29, find QR. [Note: marked ★ (challenging) in the textbook.]
  30. PS2 Q.18
    Seg AM is a median of △ABC\triangle ABC. If AB = 22, AC = 34, BC = 24, find AM