PYTHAGORAS THEOREM - Questions & Answers
EXERCISE 12(A)
1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) If the lengths of the sides of a triangle are in the ratio 5 : 12 : 13; then the triangle is:
(i) acute-angled triangle
(ii) scalene triangle
(iii) scalene right-angled triangle
(iv) obtuse-angled triangle
Step 2: Calculate the sum of the squares of the two smaller sides: (5x)² + (12x)² = 25x² + 144x² = 169x².
Step 3: Calculate the square of the largest side: (13x)² = 169x².
Step 4: Since the sum of the squares of the two smaller sides is equal to the square of the largest side (169x² = 169x²), the triangle is right-angled.
Step 5: Because all three sides have different lengths, it is a scalene right-angled triangle.
Answer: (iii) scalene right-angled triangle
(b) In a right-angled triangle, hypotenuse is 10 cm and the ratio of the other two sides is 3 : 4, the sides are :
(i) 6 cm and 4 cm
(ii) 8 cm and 6 cm
(iii) 3 cm and 4 cm
(iv) 8 cm and 4 cm
Step 2: According to the Pythagoras theorem, the sum of the squares of the two sides equals the square of the hypotenuse.
Step 3: (3x)² + (4x)² = 10².
Step 4: 9x² + 16x² = 100.
Step 5: 25x² = 100, which means x² = 4.
Step 6: Taking the square root, x = 2.
Step 7: The sides are 3(2) = 6 cm and 4(2) = 8 cm.
Answer: (ii) 8 cm and 6 cm
(c) ABC is an isosceles triangle with AB = AC = 12 cm and BC = 8 cm. The area of the triangle is :
(i) 32√2 cm²
(ii) 16√2 cm²
(iii) 8√2 cm²
(iv) 12√2 cm²
Step 2: In an isosceles triangle, the altitude bisects the base. So, BD = DC = 8 / 2 = 4 cm.
Step 3: In right-angled triangle ABD, apply Pythagoras theorem: AB² = AD² + BD².
Step 4: 12² = AD² + 4², which gives 144 = AD² + 16.
Step 5: AD² = 144 - 16 = 128.
Step 6: AD = √128 = √(64 × 2) = 8√2 cm. This is the height of the triangle.
Step 7: Area of the triangle = 1/2 × base × height = 1/2 × 8 × 8√2.
Step 8: Area = 4 × 8√2 = 32√2 cm².
Answer: (i) 32√2 cm²
(d) In a rhombus, its diagonals are 30 cm and 40 cm, its perimeter is :
(i) 20 cm
(ii) 10 cm
(iii) 60 cm
(iv) 100 cm
Step 2: Let the diagonals intersect at point O. The halves of the diagonals are 30 / 2 = 15 cm and 40 / 2 = 20 cm.
Step 3: These half-diagonals form the perpendicular legs of a right-angled triangle, where the hypotenuse is the side of the rhombus.
Step 4: Side² = 15² + 20² = 225 + 400 = 625.
Step 5: Side = √625 = 25 cm.
Step 6: The perimeter of a rhombus is 4 × side.
Step 7: Perimeter = 4 × 25 = 100 cm.
Answer: (iv) 100 cm
(e) In the given figure, AD = 13 cm, DC = 12 cm and BC = 3 cm, then AB is equal to :
(i) 4 cm
(ii) 3 cm
(iii) 5 cm
(iv) 6 cm
Step 2: By Pythagoras theorem in triangle ADC: AD² = AC² + DC².
Step 3: 13² = AC² + 12², so 169 = AC² + 144.
Step 4: AC² = 169 - 144 = 25, which gives AC = √25 = 5 cm.
Step 5: From the figure, triangle ABC is right-angled at B.
Step 6: By Pythagoras theorem in triangle ABC: AC² = AB² + BC².
Step 7: 5² = AB² + 3², so 25 = AB² + 9.
Step 8: AB² = 25 - 9 = 16.
Step 9: AB = √16 = 4 cm.
Answer: (i) 4 cm
2. A ladder 13 m long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the wall, find the distance of the other end of the ladder from the ground.
Step 1: Let the ladder be the hypotenuse of a right-angled triangle formed by the wall, the ground, and the ladder.Step 2: Length of the ladder (hypotenuse) = 13 m.
Step 3: Distance of the foot of the ladder from the wall (base) = 5 m.
Step 4: Let the height of the ladder on the wall be 'h'.
Step 5: By Pythagoras theorem, Hypotenuse² = Base² + Height².
Step 6: 13² = 5² + h².
Step 7: 169 = 25 + h².
Step 8: h² = 169 - 25 = 144.
Step 9: h = √144 = 12 m.
Answer: 12 m
3. A man goes 40 m due north and then 50 m due west. Find his distance from the starting point.
Step 1: The man's path forms a right-angled triangle with the starting point.Step 2: The distance travelled north is one perpendicular side (40 m).
Step 3: The distance travelled west is the second perpendicular side (50 m).
Step 4: The shortest distance from the starting point is the hypotenuse of this triangle.
Step 5: By Pythagoras theorem, Distance² = 40² + 50².
Step 6: Distance² = 1600 + 2500 = 4100.
Step 7: Distance = √4100 = √(100 × 41) = 10√41 m.
Answer: 10√41 m
4. In the figure : ∠PSQ = 90°, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.
Step 1: In the right-angled triangle PSQ, angle S is 90°.Step 2: By Pythagoras theorem, PQ² = PS² + QS².
Step 3: 10² = PS² + 6², which gives 100 = PS² + 36.
Step 4: PS² = 100 - 36 = 64, so PS = √64 = 8 cm.
Step 5: Since R, Q, and S lie on a straight line, the total base length RS = RQ + QS.
Step 6: RS = 9 cm + 6 cm = 15 cm.
Step 7: Now, look at the larger right-angled triangle PSR, where angle S is 90°.
Step 8: By Pythagoras theorem, PR² = PS² + RS².
Step 9: PR² = 8² + 15² = 64 + 225 = 289.
Step 10: PR = √289 = 17 cm.
Answer: 17 cm
5. In a quadrilateral PQRS, ∠Q = ∠S = 90° then prove that 2PR² - QR² = PQ² + PS² + SR².
Step 1: In the right-angled triangle PQR (since ∠Q = 90°), apply Pythagoras theorem.Step 2: PR² = PQ² + QR².
Step 3: In the right-angled triangle PSR (since ∠S = 90°), apply Pythagoras theorem.
Step 4: PR² = PS² + SR².
Step 5: Add the equations from Step 2 and Step 4 together.
Step 6: PR² + PR² = PQ² + QR² + PS² + SR².
Step 7: 2PR² = PQ² + QR² + PS² + SR².
Step 8: Subtract QR² from both sides to rearrange the formula.
Step 9: 2PR² - QR² = PQ² + PS² + SR².
Step 10: Hence Proved.
6. AD is drawn perpendicular to base BC of an equilateral triangle ABC. Given BC = 10 cm, find the length of AD, correct to 1 place of decimal.
Step 1: In equilateral triangle ABC, all sides are equal. Therefore, AB = BC = AC = 10 cm.Step 2: The perpendicular AD drawn to the base BC bisects the base.
Step 3: So, BD = DC = BC / 2 = 10 / 2 = 5 cm.
Step 4: In right-angled triangle ABD, apply Pythagoras theorem: AB² = AD² + BD².
Step 5: 10² = AD² + 5².
Step 6: 100 = AD² + 25.
Step 7: AD² = 100 - 25 = 75.
Step 8: AD = √75 = √(25 × 3) = 5√3 cm.
Step 9: Using the value of √3 ≈ 1.732, AD = 5 × 1.732 = 8.66 cm.
Step 10: Correcting to 1 place of decimal, AD = 8.7 cm.
Answer: 8.7 cm
7. In triangle ABC, given below, AB = 8 cm, BC = 6 cm and AC = 3 cm. Calculate the length of OC.
Step 1: From the figure, triangle ABO is right-angled at B. Point C lies on the line segment BO.Step 2: Therefore, triangle ABC is a right-angled triangle at B.
Step 3: By Pythagoras theorem in triangle ABC, AC² should equal AB² + BC².
Step 4: Calculating AB² + BC² gives 8² + 6² = 64 + 36 = 100.
Step 5: This means AC should be √100 = 10 cm.
Step 6: However, the given problem states AC = 3 cm, which contradicts the geometric properties shown in the diagram.
Step 7: Assuming the question intended to give AO = 17 cm (a common right triangle setup) to find OC, we would find BO = √17² - 8² = 15 cm, making OC = 15 - 6 = 9 cm.
Step 8: Because of the apparent misprint in the textbook data (stating AC = 3 cm instead of proper values), a direct numerical calculation using strictly the written numbers leads to a contradiction.
Answer: Data provided contains a contradiction (AC must be 10 cm for the right angle to hold).
8. In triangle ABC, AB = AC = x; BC = 10 cm and the area of the triangle is 60 cm². Find x.
Step 1: Draw an altitude AD from vertex A to the base BC. Let the height be h.Step 2: The area of the triangle is given by 1/2 × base × height.
Step 3: 60 = 1/2 × 10 × h.
Step 4: 60 = 5 × h, which gives h = 12 cm. So, AD = 12 cm.
Step 5: Since ABC is an isosceles triangle (AB = AC), the altitude AD bisects the base BC.
Step 6: Therefore, BD = DC = 10 / 2 = 5 cm.
Step 7: In the right-angled triangle ABD, apply Pythagoras theorem: AB² = AD² + BD².
Step 8: x² = 12² + 5².
Step 9: x² = 144 + 25 = 169.
Step 10: x = √169 = 13 cm.
Answer: 13 cm
9. If the sides of a triangle are in the ratio 1 : √2 : 1, show that it is a right-angled triangle.
Step 1: Let the sides of the triangle be x, x√2, and x.Step 2: The squares of the lengths of these sides are x², (x√2)², and x².
Step 3: Simplify the squares: x², 2x², and x².
Step 4: Add the squares of the two smaller sides: x² + x² = 2x².
Step 5: The sum of the squares of the two smaller sides exactly equals the square of the largest side (2x² = 2x²).
Step 6: According to the converse of the Pythagoras theorem, if the square of one side equals the sum of the squares of the other two sides, the triangle is right-angled.
Step 7: Hence, it is a right-angled triangle. Proved.
10. Two poles of heights 6 m and 11 m stand vertically on a plane ground. If the distance between their feet is 12 m; find the distance between their tips.
Step 1: Let the two vertical poles be AB = 6 m and CD = 11 m.Step 2: The distance between their feet on the ground is BC = 12 m.
Step 3: Draw a line AE from the top of the shorter pole AB parallel to the ground, meeting the taller pole CD at E.
Step 4: This forms a rectangle ABCE, where AE = BC = 12 m and CE = AB = 6 m.
Step 5: The remaining height of the taller pole is ED = CD - CE = 11 m - 6 m = 5 m.
Step 6: Now consider the right-angled triangle AED, where we need to find the hypotenuse AD (distance between tips).
Step 7: By Pythagoras theorem, AD² = AE² + ED².
Step 8: AD² = 12² + 5² = 144 + 25 = 169.
Step 9: AD = √169 = 13 m.
Answer: 13 m
EXERCISE 12(B)
1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) In ΔABC, ∠C = 90° and AC = BC, then AB² is equal to :
(i) AC²
(ii) 2AC²
(iii) BC²
(iv) 2BC² - AC²
Step 2: It is given that AC = BC.
Step 3: Substitute BC with AC in the equation: AB² = AC² + AC².
Step 4: Therefore, AB² = 2AC².
Answer: (ii) 2AC²
(b) In the given diagram, AE² + BD² is equal to :
(i) AB² - DE²
(ii) DE² - AB²
(iii) AB² + DE²
(iv) DE × AB
Step 2: In the right-angled triangle BCD, BD² = BC² + CD².
Step 3: Add the two equations: AE² + BD² = AC² + CE² + BC² + CD².
Step 4: Rearrange the terms: AE² + BD² = (AC² + BC²) + (CE² + CD²).
Step 5: In right-angled triangle ACB, AB² = AC² + BC².
Step 6: In right-angled triangle DCE, DE² = CE² + CD².
Step 7: Substitute these into the rearranged equation: AE² + BD² = AB² + DE².
Answer: (iii) AB² + DE²
(c) In the given figure, the value of AB² is :
(i) AC × BC
(ii) AC × CD
(iii) AC × AD
(iv) AC² + BC²
Step 2: This makes triangle ADB similar to the large triangle ABC.
Step 3: By the property of similar triangles, the ratio of their corresponding sides is equal.
Step 4: AB / AC = AD / AB.
Step 5: Cross-multiplying gives AB² = AC × AD.
Answer: (iii) AC × AD
(d) ABC is an isosceles triangle right-angled at C. Then 2AC² is equal to :
(i) BC²
(ii) AC²
(iii) AC² - BC²
(iv) AB²
Step 2: Since it is an isosceles triangle, the two legs are equal, so AC = BC.
Step 3: Substitute BC with AC in the Pythagoras equation: AB² = AC² + AC².
Step 4: Therefore, AB² = 2AC².
Answer: (iv) AB²
2. In the figure, given below, AD ⊥ BC. Prove that : c² = a² + b² - 2ax.
Step 1: In the right-angled triangle ADC, apply Pythagoras theorem: AC² = AD² + DC².Step 2: This gives b² = h² + x², which can be rewritten as h² = b² - x².
Step 3: In the right-angled triangle ADB, apply Pythagoras theorem: AB² = AD² + BD².
Step 4: This gives c² = h² + (a - x)².
Step 5: Expand the square term: c² = h² + a² + x² - 2ax.
Step 6: Substitute the value of h² from Step 2 into this equation.
Step 7: c² = (b² - x²) + a² + x² - 2ax.
Step 8: The -x² and +x² cancel each other out.
Step 9: c² = b² + a² - 2ax, which can be rearranged to c² = a² + b² - 2ax.
Step 10: Hence Proved.
3. In equilateral ΔABC, AD ⊥ BC and BC = x cm. Find, in terms of x, the length of AD.
Step 1: Since ABC is an equilateral triangle, AB = BC = AC = x cm.Step 2: The perpendicular AD bisects the base BC. Therefore, BD = DC = x / 2 cm.
Step 3: In right-angled triangle ABD, apply Pythagoras theorem: AB² = AD² + BD².
Step 4: x² = AD² + (x / 2)².
Step 5: x² = AD² + x² / 4.
Step 6: AD² = x² - x² / 4 = (4x² - x²) / 4 = 3x² / 4.
Step 7: Taking the square root of both sides, AD = √(3x² / 4) = (x√3) / 2 cm.
Answer: (x√3) / 2 cm
4. ABC is a triangle, right-angled at B. M is a point on BC. Prove that : AM² + BC² = AC² + BM².
Step 1: In right-angled triangle ABM, apply Pythagoras theorem: AM² = AB² + BM².Step 2: From this, we can write AB² = AM² - BM².
Step 3: In right-angled triangle ABC, apply Pythagoras theorem: AC² = AB² + BC².
Step 4: Substitute the expression for AB² from Step 2 into the equation from Step 3.
Step 5: AC² = (AM² - BM²) + BC².
Step 6: Add BM² to both sides of the equation.
Step 7: AC² + BM² = AM² + BC².
Step 8: Rearrange to match the required form: AM² + BC² = AC² + BM².
Step 9: Hence Proved.
5. M and N are the mid-points of the sides QR and PQ respectively of a ΔPQR, right-angled at Q. Prove that :
(i) PM² + RN² = 5 MN²
(ii) 4 PM² = 4 PQ² + QR²
(iii) 4 RN² = PQ² + 4 QR²
(iv) 4 (PM² + RN²) = 5 PR²
Step 2: In right triangle PQM, PM² = PQ² + QM² = PQ² + (QR/2)² = PQ² + QR²/4.
Step 3: In right triangle RQN, RN² = QR² + QN² = QR² + (PQ/2)² = QR² + PQ²/4.
Step 4: Proof for (ii): Multiply the equation from Step 2 by 4: 4PM² = 4(PQ² + QR²/4) = 4PQ² + QR². Proved (ii).
Step 5: Proof for (iii): Multiply the equation from Step 3 by 4: 4RN² = 4(QR² + PQ²/4) = 4QR² + PQ². Proved (iii).
Step 6: Proof for (i): Add the equations from Step 2 and Step 3: PM² + RN² = (PQ² + QR²/4) + (QR² + PQ²/4).
Step 7: Combine like terms: PM² + RN² = 5/4 PQ² + 5/4 QR² = (5/4)(PQ² + QR²).
Step 8: In large right triangle PQR, PR² = PQ² + QR². So, PM² + RN² = (5/4)PR².
Step 9: In right triangle MQN, MN² = QM² + QN² = (QR/2)² + (PQ/2)² = (QR² + PQ²) / 4 = PR² / 4.
Step 10: Substitute PR²/4 with MN²: PM² + RN² = 5(PR² / 4) = 5MN². Proved (i).
Step 11: Proof for (iv): From Step 8, we have PM² + RN² = (5/4)PR².
Step 12: Multiply the entire equation by 4: 4(PM² + RN²) = 5PR². Proved (iv).
6. In triangle ABC, ∠B = 90° and D is the mid-point of BC. Prove that : AC² = AD² + 3CD².
Step 1: In the right-angled triangle ABC, apply Pythagoras theorem: AC² = AB² + BC².Step 2: Since D is the mid-point of BC, BC = 2CD. Therefore, BC² = (2CD)² = 4CD².
Step 3: Substitute BC² into the first equation: AC² = AB² + 4CD².
Step 4: In the smaller right-angled triangle ABD, apply Pythagoras theorem: AD² = AB² + BD².
Step 5: This means AB² = AD² - BD².
Step 6: Since D is the mid-point, BD = CD, so BD² = CD².
Step 7: Substitute BD² with CD²: AB² = AD² - CD².
Step 8: Substitute this expression for AB² into the equation from Step 3: AC² = (AD² - CD²) + 4CD².
Step 9: Simplify the terms: AC² = AD² + 3CD².
Step 10: Hence Proved.
7. In a rectangle ABCD, prove that : AC² + BD² = AB² + BC² + CD² + DA².
Step 1: In a rectangle, all interior angles are 90°, and opposite sides are equal (AB = CD and BC = DA).Step 2: Draw the diagonal AC. In the right-angled triangle ABC, AC² = AB² + BC².
Step 3: Draw the diagonal BD. In the right-angled triangle BCD, BD² = BC² + CD².
Step 4: We also know that the diagonals of a rectangle are equal in length, so AC = BD, and AC² = BD².
Step 5: Add AC² and BD² by doubling the expression for AC²: AC² + BD² = (AB² + BC²) + (AB² + BC²).
Step 6: Replace one AB² with CD² and one BC² with DA², because opposite sides are equal.
Step 7: AC² + BD² = AB² + BC² + CD² + DA².
Step 8: Hence Proved.
8. In a quadrilateral ABCD, ∠B = 90° and ∠D = 90°. Prove that : 2AC² - AB² = BC² + CD² + DA².
Step 1: Draw the diagonal AC. This divides the quadrilateral into two right-angled triangles, ABC and ADC.Step 2: In right-angled triangle ABC (since ∠B = 90°), apply Pythagoras theorem: AC² = AB² + BC².
Step 3: In right-angled triangle ADC (since ∠D = 90°), apply Pythagoras theorem: AC² = AD² + CD².
Step 4: Add the equations from Step 2 and Step 3 together: AC² + AC² = AB² + BC² + AD² + CD².
Step 5: 2AC² = AB² + BC² + CD² + DA² (since AD = DA).
Step 6: Subtract AB² from both sides of the equation.
Step 7: 2AC² - AB² = BC² + CD² + DA².
Step 8: Hence Proved.
9. O is any point inside a rectangle ABCD. Prove that : OB² + OD² = OC² + OA².
Step 1: Draw a line through point O parallel to AB and CD, meeting AD at P and BC at Q.Step 2: Because ABCD is a rectangle and PQ is parallel to the top and bottom sides, $\angle P$ and $\angle Q$ are 90°, making APOQ and DPCQ rectangles? No, ABQP and CDQP are rectangles.
Step 3: Thus, AP = BQ and PD = QC. Also, $\triangle OPA$, $\triangle OPD$, $\triangle OQB$, and $\triangle OQC$ are right-angled triangles.
Step 4: In right $\triangle OPA$, OA² = OP² + AP².
Step 5: In right $\triangle OPD$, OD² = OP² + PD².
Step 6: In right $\triangle OQB$, OB² = OQ² + BQ².
Step 7: In right $\triangle OQC$, OC² = OQ² + QC².
Step 8: Add OB² and OD²: OB² + OD² = (OQ² + BQ²) + (OP² + PD²).
Step 9: Add OA² and OC²: OA² + OC² = (OP² + AP²) + (OQ² + QC²).
Step 10: Since AP = BQ and PD = QC, substitute them into the OA²+OC² sum: OA² + OC² = OP² + BQ² + OQ² + PD².
Step 11: Comparing the two sums, they contain the exact same terms.
Step 12: Therefore, OB² + OD² = OC² + OA². Hence Proved.
10. In the following figure, OP, OQ and OR are drawn perpendiculars to the sides BC, CA and AB respectively of triangle ABC. Prove that : AR² + BP² + CQ² = AQ² + CP² + BR².
Join OA, OB and OC. Now find the values of AR² + BP² + CQ² and AQ² + CP² + BR²
Step 2: In right $\triangle ARO$ and $\triangle AQO$, AR² = OA² - OR² and AQ² = OA² - OQ².
Step 3: In right $\triangle BPO$ and $\triangle BRO$, BP² = OB² - OP² and BR² = OB² - OR².
Step 4: In right $\triangle CQO$ and $\triangle CPO$, CQ² = OC² - OQ² and CP² = OC² - OP².
Step 5: Add AR², BP², and CQ²: AR² + BP² + CQ² = (OA² - OR²) + (OB² - OP²) + (OC² - OQ²).
Step 6: Add AQ², CP², and BR²: AQ² + CP² + BR² = (OA² - OQ²) + (OC² - OP²) + (OB² - OR²).
Step 7: Rearrange terms in both sums: both equal OA² + OB² + OC² - OP² - OQ² - OR².
Step 8: Since both expressions evaluate to the same sum, AR² + BP² + CQ² = AQ² + CP² + BR². Hence Proved.
Step 9: To find the values, observe Step 7.
Answer: The value of both expressions is OA² + OB² + OC² - OP² - OQ² - OR².
TEST YOURSELF
1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) Angle AOB is :
(i) 60°
(ii) 90°
(iii) 45°
(iv) none of these
Step 2: Calculate the sum of squares of the two smaller sides: 5² + 12² = 25 + 144 = 169.
Step 3: Calculate the square of the longest side: 13² = 169.
Step 4: Since 5² + 12² = 13², the triangle satisfies the converse of Pythagoras theorem.
Step 5: The angle opposite to the longest side (13) is a right angle. Angle AOB is 90°.
Answer: (ii) 90°
(b) Ranbeer runs 10 km due North and 24 km due West. The distance between his two positions is :
(i) 34 km
(ii) 17 km
(iii) 26 km
(iv) none of these
Step 2: The lengths of the perpendicular sides are 10 km and 24 km.
Step 3: By Pythagoras theorem, Distance² = 10² + 24².
Step 4: Distance² = 100 + 576 = 676.
Step 5: Distance = √676 = 26 km.
Answer: (iii) 26 km
(c) Angle AOB is :
(i) 60°
(ii) 90°
(iii) 45°
(iv) none of these
Step 2: The two shorter sides are equal to x.
Step 3: Sum of squares of the two smaller sides = x² + x² = 2x².
Step 4: The square of the longest side is (x√2)² = 2x².
Step 5: Since x² + x² = (x√2)², the triangle satisfies the Pythagoras theorem condition.
Step 6: The angle opposite the longest side is 90°. Therefore, Angle AOB is 90°.
Answer: (ii) 90°
(d) The sides of a rectangle are 12 cm and 16 cm. The length of its diagonal is :
(i) 28 cm
(ii) 4 cm
(iii) √(12² + 16²) cm
(iv) √(16² - 12²) cm
Step 2: The two sides and the diagonal form a right-angled triangle.
Step 3: By Pythagoras theorem, Diagonal² = Side1² + Side2².
Step 4: Diagonal² = 12² + 16².
Step 5: Diagonal = √(12² + 16²) cm. Evaluating it gives √400 = 20 cm, but option (iii) matches the exact mathematical form.
Answer: (iii) √(12² + 16²) cm
(e) Statement (1) : ABCD is a rhombus, its diagonal AC = 16 cm and diagonal BD = 12 cm, perimeter of rhombus = 64 cm.
Statement (2) : OA = 8 cm, OB = 6 cm. Then, AB = 10 cm. And, perimeter of rhombus = 40 cm
(i) Both the statements are true.
(ii) Both the statements are false.
(iii) Statement 1 is true, and statement 2 is false.
(iv) Statement 1 is false, and statement 2 is true.
Step 2: In right triangle AOB, Side AB = √(8² + 6²) = √(64 + 36) = √100 = 10 cm.
Step 3: Perimeter = 4 × side = 4 × 10 = 40 cm.
Step 4: Statement (1) claims the perimeter is 64 cm, which is incorrect.
Step 5: Statement (2) states OA=8, OB=6, AB=10, perimeter=40, which matches our calculation exactly.
Step 6: Therefore, Statement (1) is false, and Statement (2) is true.
Answer: (iv) Statement 1 is false, and statement 2 is true.
(f) Statement (1) : Area of given triangle ABC = 6 × 5 cm².
Statement (2) : Area of given triangle ABC = 1/2 × 6 × 4 cm².
(i) Both the statements are true.
(ii) Both the statements are false.
(iii) Statement 1 is true, and statement 2 is false.
(iv) Statement 1 is false, and statement 2 is true.
Step 2: Drop an altitude from the top vertex to the base. It bisects the base into 3 cm and 3 cm.
Step 3: By Pythagoras theorem, Height = √(5² - 3²) = √(25 - 9) = √16 = 4 cm.
Step 4: Area of triangle = 1/2 × base × height = 1/2 × 6 × 4 cm² = 12 cm².
Step 5: Statement (1) calculates it as 6 × 5 = 30, which is incorrect.
Step 6: Statement (2) calculates it correctly as 1/2 × 6 × 4.
Step 7: Therefore, Statement (1) is false, and Statement (2) is true.
Answer: (iv) Statement 1 is false, and statement 2 is true.
(g) Assertion (A) : Angle BOC = 90°.
Reason (R) : OC² = 3² + 4² = 25, OB² = 6² + 8² = 100, OC² + OB² = 125 = BC²
(i) A is true, R is false.
(ii) A is false, R is true.
(iii) Both A and R are true and R is the correct reason for A.
(iv) Both A and R are true and R is the incorrect reason for A.
Step 2: The length of BC is given as 5√5 cm, which means BC² = (5√5)² = 125.
Step 3: The Reason states OC² + OB² = 25 + 100 = 125.
Step 4: Since OC² + OB² = 125 and BC² = 125, the equation OC² + OB² = BC² holds true.
Step 5: By the converse of Pythagoras theorem, this confirms that triangle BOC is right-angled at O, so Angle BOC = 90°.
Step 6: The Assertion is true, and the Reason mathematically proves it.
Answer: (iii) Both A and R are true and R is the correct reason for A.
(h) Assertion (A) : x = 5√2.
Reason (R) : AC² = 8² + 6² = x² + x²
(i) A is true, R is false.
(ii) A is false, R is true.
(iii) Both A and R are true and R is the correct reason for A.
(iv) Both A and R are true and R is the incorrect reason for A.
Step 2: In the upper right-angled triangle ADC (right-angled at D), the legs are equal (AD = DC = x). By Pythagoras theorem, AC² = x² + x².
Step 3: Equating the two expressions for AC²: 8² + 6² = x² + x². This is exactly what Reason (R) states.
Step 4: 64 + 36 = 2x², which means 100 = 2x².
Step 5: x² = 50, so x = √50 = √(25 × 2) = 5√2.
Step 6: This perfectly matches Assertion (A).
Step 7: Both are true, and R is the exact correct calculation to prove A.
Answer: (iii) Both A and R are true and R is the correct reason for A.
2. In the given figure, AB//CD, AB = 7 cm, BD = 25 cm and CD = 17 cm; find the length of side BC.
Step 1: From the figure, assume AD is vertical and perpendicular to parallel lines AB and CD, making $\angle A = 90^\circ$ and $\angle D = 90^\circ$.Step 2: In right-angled triangle ABD, AB = 7 cm and the hypotenuse BD = 25 cm.
Step 3: By Pythagoras theorem, AD² + AB² = BD².
Step 4: AD² + 7² = 25², so AD² + 49 = 625.
Step 5: AD² = 625 - 49 = 576, which gives AD = √576 = 24 cm.
Step 6: Draw a perpendicular BE from B to CD. Since AB // CD and AD $\perp$ CD, ABED is a rectangle.
Step 7: Therefore, BE = AD = 24 cm and ED = AB = 7 cm.
Step 8: The length of CE = CD - ED = 17 cm - 7 cm = 10 cm.
Step 9: In right-angled triangle BEC, BC² = BE² + CE².
Step 10: BC² = 24² + 10² = 576 + 100 = 676.
Step 11: BC = √676 = 26 cm.
Answer: 26 cm
3. In the given figure, ∠B = 90°, XY//BC, AB = 12 cm, AY = 8 cm and AX : XB = 1 : 2 = AY : YC. Find the lengths of AC and BC.
Step 1: We are given the ratio AY : YC = 1 : 2. Since AY = 8 cm, YC must be 8 × 2 = 16 cm.Step 2: The total length of AC = AY + YC = 8 cm + 16 cm = 24 cm.
Step 3: Now we know AC = 24 cm and AB = 12 cm in the right-angled triangle ABC (∠B = 90°).
Step 4: By Pythagoras theorem, AC² = AB² + BC².
Step 5: 24² = 12² + BC².
Step 6: 576 = 144 + BC².
Step 7: BC² = 576 - 144 = 432.
Step 8: BC = √432 = √(144 × 3) = 12√3 cm.
Answer: AC = 24 cm, BC = 12√3 cm
4. In ΔABC, ∠B = 90°. Find the sides of the triangle, if :
(i) AB = (x - 3) cm, BC = (x + 4) cm and AC = (x + 6) cm
Step 2: Substitute the expressions: (x - 3)² + (x + 4)² = (x + 6)².
Step 3: Expand all squares: (x² - 6x + 9) + (x² + 8x + 16) = x² + 12x + 36.
Step 4: Combine terms on the left: 2x² + 2x + 25 = x² + 12x + 36.
Step 5: Move all terms to one side: x² - 10x - 11 = 0.
Step 6: Factorize the quadratic equation: x² - 11x + 1x - 11 = 0, so (x - 11)(x + 1) = 0.
Step 7: The possible values are x = 11 or x = -1. Since side lengths must be positive, x = 11.
Step 8: Calculate the sides: AB = 11 - 3 = 8 cm, BC = 11 + 4 = 15 cm, AC = 11 + 6 = 17 cm.
Answer: AB = 8 cm, BC = 15 cm, AC = 17 cm
(ii) AB = x cm, BC = (4x + 4) cm and AC = (4x + 5) cm
Step 1: By Pythagoras theorem: AB² + BC² = AC².Step 2: Substitute the expressions: x² + (4x + 4)² = (4x + 5)².
Step 3: Expand the squares: x² + (16x² + 32x + 16) = 16x² + 40x + 25.
Step 4: Subtract 16x² from both sides: x² + 32x + 16 = 40x + 25.
Step 5: Move all terms to the left: x² - 8x - 9 = 0.
Step 6: Factorize the quadratic equation: x² - 9x + 1x - 9 = 0, so (x - 9)(x + 1) = 0.
Step 7: The positive root is x = 9.
Step 8: Calculate the sides: AB = 9 cm, BC = 4(9) + 4 = 36 + 4 = 40 cm, AC = 4(9) + 5 = 36 + 5 = 41 cm.
Answer: AB = 9 cm, BC = 40 cm, AC = 41 cm
5. If a side of a rhombus is 10 cm and one of the diagonals is 16 cm, find the other diagonal.
Step 1: Let the rhombus be ABCD, with side = 10 cm, and diagonal AC = 16 cm.Step 2: The diagonals of a rhombus bisect each other at right angles (90°) at point O.
Step 3: The half-diagonal OA = 16 / 2 = 8 cm.
Step 4: In right-angled triangle AOB, the side AB is the hypotenuse (10 cm).
Step 5: By Pythagoras theorem, OA² + OB² = AB².
Step 6: 8² + OB² = 10².
Step 7: 64 + OB² = 100, which gives OB² = 36.
Step 8: OB = √36 = 6 cm.
Step 9: The full length of the other diagonal BD is 2 × OB = 2 × 6 = 12 cm.
Answer: 12 cm
6. In the given figure, diagonals AC and BD intersect at right angle. Show that : AB² + CD² = AD² + BC²
Step 1: Let the diagonals intersect at point O. We have four right-angled triangles: AOB, BOC, COD, and AOD.Step 2: In right triangle AOB, AB² = OA² + OB².
Step 3: In right triangle COD, CD² = OC² + OD².
Step 4: Add the two equations: AB² + CD² = OA² + OB² + OC² + OD².
Step 5: In right triangle AOD, AD² = OA² + OD².
Step 6: In right triangle BOC, BC² = OB² + OC².
Step 7: Add these two equations: AD² + BC² = OA² + OD² + OB² + OC².
Step 8: The sum of terms on the right side of equations in Step 4 and Step 7 are identical.
Step 9: Therefore, AB² + CD² = AD² + BC². Hence Proved.
7. Diagonals of rhombus ABCD intersect each other at point O. Prove that : OA² + OC² = 2AD² - BD²/2
Step 1: In rhombus ABCD, diagonals bisect each other at 90°, so triangle AOD is right-angled at O.Step 2: By Pythagoras theorem, AD² = OA² + OD².
Step 3: Since diagonals bisect each other, OD = BD / 2. Also, OA = OC.
Step 4: Substitute OD into the equation: AD² = OA² + (BD / 2)² = OA² + BD² / 4.
Step 5: Multiply the entire equation by 2: 2AD² = 2OA² + BD² / 2.
Step 6: Rearrange the terms to isolate 2OA²: 2OA² = 2AD² - BD² / 2.
Step 7: Since OA = OC, we can write 2OA² as OA² + OA² = OA² + OC².
Step 8: Substitute OA² + OC² into the equation: OA² + OC² = 2AD² - BD² / 2.
Step 9: Hence Proved.
8. In the figure AB = BC and AD is perpendicular to CD. Prove that : AC² = 2.BC.DC.
Step 1: From the figure, D, B, and C lie on a straight line. AD is perpendicular to CD, so angle ADC is 90°.Step 2: In right-angled triangle ADC, apply Pythagoras theorem: AC² = AD² + DC².
Step 3: In right-angled triangle ADB, apply Pythagoras theorem: AD² = AB² - DB².
Step 4: Substitute AD² from Step 3 into the equation from Step 2: AC² = (AB² - DB²) + DC².
Step 5: From the line segment, DC = DB + BC. Substitute DC with this sum: AC² = AB² - DB² + (DB + BC)².
Step 6: Expand the square: AC² = AB² - DB² + DB² + BC² + 2 × DB × BC.
Step 7: Simplify the terms: AC² = AB² + BC² + 2 × DB × BC.
Step 8: We are given that AB = BC. Substitute AB² with BC²: AC² = BC² + BC² + 2 × DB × BC.
Step 9: AC² = 2BC² + 2 × DB × BC.
Step 10: Factor out 2BC: AC² = 2BC(BC + DB).
Step 11: Since BC + DB = DC, substitute DC back into the equation: AC² = 2 × BC × DC.
Step 12: Hence Proved.
9. In an isosceles triangle ABC; AB = AC and D is a point on BC produced. Prove that : AD² = AC² + BD.CD.
Step 1: Draw an altitude AE perpendicular to BC. Since ABC is isosceles (AB = AC), E is the midpoint of BC, so BE = CE.Step 2: In right-angled triangle AED, apply Pythagoras theorem: AD² = AE² + ED².
Step 3: In right-angled triangle AEC, apply Pythagoras theorem: AC² = AE² + CE², which means AE² = AC² - CE².
Step 4: Substitute AE² in the equation from Step 2: AD² = (AC² - CE²) + ED².
Step 5: Since D is on BC produced, the distance ED = CE + CD.
Step 6: Expand ED²: ED² = (CE + CD)² = CE² + CD² + 2 × CE × CD.
Step 7: Substitute ED² back: AD² = AC² - CE² + CE² + CD² + 2 × CE × CD.
Step 8: Simplify: AD² = AC² + CD² + 2 × CE × CD.
Step 9: Factor out CD: AD² = AC² + CD(CD + 2CE).
Step 10: Since BE = CE, 2CE = BE + CE = BC.
Step 11: Substitute 2CE with BC: AD² = AC² + CD(CD + BC).
Step 12: From the figure, CD + BC = BD. Thus, AD² = AC² + BD × CD.
Step 13: Hence Proved.
10. In triangle ABC, angle A = 90°, CA = AB and D is a point on AB produced. Prove that : DC² - BD² = 2AB.AD.
Step 1: Since D is on AB produced, the length BD = AD - AB.Step 2: Square both sides: BD² = (AD - AB)² = AD² + AB² - 2 × AB × AD.
Step 3: In right-angled triangle CAD, apply Pythagoras theorem: DC² = CA² + AD².
Step 4: Since we are given CA = AB, substitute CA with AB: DC² = AB² + AD².
Step 5: We need to evaluate DC² - BD². Substitute the expressions from Step 4 and Step 2.
Step 6: DC² - BD² = (AB² + AD²) - (AD² + AB² - 2 × AB × AD).
Step 7: Simplify the expression: DC² - BD² = AB² + AD² - AD² - AB² + 2 × AB × AD.
Step 8: DC² - BD² = 2 × AB × AD.
Step 9: Hence Proved.
11. In triangle ABC, AB = AC and BD is perpendicular to AC. Prove that : BD² - CD² = 2CD × AD
Step 1: BD is perpendicular to AC, making triangles ADB and CDB right-angled at D.Step 2: In right triangle ADB, AB² = AD² + BD², which gives BD² = AB² - AD².
Step 3: We need to evaluate BD² - CD². Substitute BD²: BD² - CD² = AB² - AD² - CD².
Step 4: We are given that AB = AC. Since D is on AC, AC = AD + CD.
Step 5: Substitute AB with (AD + CD): AB² = (AD + CD)² = AD² + CD² + 2 × AD × CD.
Step 6: Substitute this back into the expression from Step 3: BD² - CD² = (AD² + CD² + 2 × AD × CD) - AD² - CD².
Step 7: Cancel out AD² and CD²: BD² - CD² = 2 × AD × CD.
Step 8: Hence Proved.
12. In the following figure, AD is perpendicular to BC and D divides BC in the ratio 1 : 3. Prove that : 2AC² = 2AB² + BC²
Step 1: The ratio BD : DC = 1 : 3 means BD is 1 part and DC is 3 parts of the whole base BC.Step 2: Therefore, BD = (1/4)BC and CD = (3/4)BC.
Step 3: In right-angled triangle ADC, AC² = AD² + CD².
Step 4: In right-angled triangle ADB, AB² = AD² + BD².
Step 5: Subtract the equation in Step 4 from Step 3: AC² - AB² = CD² - BD².
Step 6: Substitute the expressions from Step 2: AC² - AB² = ((3/4)BC)² - ((1/4)BC)².
Step 7: Square the fractions: AC² - AB² = (9/16)BC² - (1/16)BC².
Step 8: Subtract them: AC² - AB² = (8/16)BC² = (1/2)BC².
Step 9: Multiply the entire equation by 2: 2(AC² - AB²) = BC².
Step 10: 2AC² - 2AB² = BC², which can be rewritten as 2AC² = 2AB² + BC².
Step 11: Hence Proved.
13. In the given figure, AB = 16 cm, BC = 12 cm and CA = 6 cm; find the length of CD.
Step 1: From the figure, triangle ABD is right-angled at D. The hypotenuse is AB.Step 2: Point C lies on the line segment BD, meaning BD = BC + CD = 12 + CD.
Step 3: Triangle ACD is also right-angled at D. The hypotenuse is AC = 6 cm.
Step 4: By Pythagoras theorem in triangle ACD: AD² + CD² = AC² = 6² = 36.
Step 5: By Pythagoras theorem in triangle ABD: AD² + BD² = AB² = 16² = 256.
Step 6: Substitute BD = (12 + CD) into the second equation: AD² + (12 + CD)² = 256.
Step 7: Expand the square: AD² + 144 + CD² + 24 × CD = 256.
Step 8: Group AD² and CD² together: (AD² + CD²) + 144 + 24 × CD = 256.
Step 9: Substitute the value from Step 4 (AD² + CD² = 36): 36 + 144 + 24 × CD = 256.
Step 10: 180 + 24 × CD = 256.
Step 11: 24 × CD = 256 - 180 = 76.
Step 12: CD = 76 / 24 = 19 / 6 = 3.167 cm (approximately).
Answer: 19 / 6 cm (or 3.167 cm)
14. In a quadrilateral ABCD, ∠A + ∠D = 90°, prove that : AC² + BD² = AD² + BC².
Step 1: Extend lines AB and DC to meet at a point E outside the quadrilateral.Step 2: In triangle ADE, the sum of internal angles is 180°. So, ∠E = 180° - (∠A + ∠D).
Step 3: We are given that ∠A + ∠D = 90°. Therefore, ∠E = 180° - 90° = 90°.
Step 4: This makes triangle ADE a right-angled triangle at E. By Pythagoras theorem, AD² = AE² + DE².
Step 5: Similarly, triangle BCE is right-angled at E. By Pythagoras theorem, BC² = BE² + CE².
Step 6: Triangles ACE and BDE are also right-angled at E.
Step 7: Therefore, AC² = AE² + CE² and BD² = BE² + DE².
Step 8: Add the two equations from Step 7: AC² + BD² = AE² + CE² + BE² + DE².
Step 9: Group the terms differently: AC² + BD² = (AE² + DE²) + (BE² + CE²).
Step 10: Substitute AD² and BC² from Steps 4 and 5: AC² + BD² = AD² + BC².
Step 11: Hence Proved.
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Quick Review Flashcards - Click to flip and test your knowledge!
Question
Which Indian mathematician developed a relationship between the squares of the sides of a right-angled triangle around 600 B.C.?
Answer
The Indian mathematician Buddhayan developed this relationship.
Question
According to the Pythagoras Theorem, what is the relationship between the hypotenuse and the remaining two sides of a right-angled triangle?
Answer
The square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
Question
In the area-based proof of Pythagoras Theorem, what is the geometric relationship between $\triangle GAC$ and square $ABFG$?
Answer
The area of $\triangle GAC$ is half the area of square $ABFG$.
Question
In the area-based proof, why are $\triangle GAC$ and $\triangle BAE$ considered congruent?
Answer
They are congruent by the Side-Angle-Side (SAS) postulate.
Question
In the area-based proof, the area of square $ABFG$ is proved to be equal to the area of which rectangle?
Answer
It is equal to the area of rectangle $AMNE$.
Question
What is the Converse of Pythagoras Theorem?
Answer
If the square on the longest side of a triangle equals the sum of the squares on the other two sides, the angle opposite the longest side is a right-angle.
Question
What is the similarity-based construction used to prove $AC^2 = AB^2 + BC^2$ in $\triangle ABC$ where $\angle ABC = 90^{\circ}$?
Answer
The construction involves drawing a perpendicular $BD$ from the right-angle vertex to the hypotenuse $AC$.
Question
Using similar triangles in the proof of Pythagoras Theorem, what does $BC^2$ equal in terms of the hypotenuse $AC$ and segment $DC$?
Answer
$BC^2$ equals $AC \times DC$.
Question
Using similar triangles in the proof of Pythagoras Theorem, what does $AB^2$ equal in terms of the hypotenuse $AC$ and segment $AD$?
Answer
$AB^2$ equals $AC \times AD$.
Question
Which postulate is used to prove that $\triangle ABC \sim \triangle BDC$ in the similarity-based proof of Pythagoras Theorem?
Answer
The Angle-Angle (A.A.) postulate is used.
Question
In any right-angled triangle, which side is always the largest?
Answer
The hypotenuse is always the largest side.
Question
If $AB$ is the largest side of $\triangle ABC$ and $AB^2 > AC^2 + BC^2$, what type of triangle is it?
Answer
It is an obtuse-angled triangle.
Question
If $AB$ is the largest side of $\triangle ABC$ and $AB^2 < AC^2 + BC^2$, what type of triangle is it?
Answer
It is an acute-angled triangle.
Question
Define 'Pythagorean triplets' using positive numbers $a$, $b$, and $c$ (where $c$ is the largest).
Answer
They are three positive numbers such that $a^2 + b^2 = c^2$.
Question
Provide a common example of a Pythagorean triplet mentioned in the text.
Answer
The numbers 3, 4, and 5 form a Pythagorean triplet.
Question
What is the formula for the width of a street $PQ$ when a ladder of length $L$ reaches heights $h_1$ and $h_2$ on opposite sides?
Answer
$PQ = \sqrt{L^2 - h_1^2} + \sqrt{L^2 - h_2^2}$.
Question
In a right-angled triangle with sides $a$ and $b$, hypotenuse $c$, and altitude $p$ to the hypotenuse, what is the relationship between $p$, $a$, and $b$?
Answer
The relationship is $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$.
Question
In a triangle where $AD \perp BC$ produced, what is the identity for the side $c$ opposite the obtuse angle?
Answer
The identity is $c^2 = a^2 + b^2 + 2ax$.
Question
In a triangle where $AD \perp BC$ and the angle at $C$ is acute, what is the identity for side $c$ (where $CD = x$)?
Answer
The identity is $c^2 = a^2 + b^2 - 2ax$.
Question
What is the geometric property regarding the sum of the squares of the diagonals of a parallelogram?
Answer
The sum of the squares on the diagonals is equal to the sum of the squares on its sides.
Question
Express the relationship between the side $s$ and diagonals $d_1, d_2$ of a rhombus using Pythagoras Theorem.
Answer
The relationship is $4s^2 = d_1^2 + d_2^2$.
Question
If the lengths of the sides of a triangle are in the ratio $5:12:13$, what specific type of triangle is it?
Answer
It is a scalene right-angled triangle.
Question
In a right-angled triangle, if the hypotenuse is 10 cm and the ratio of the other two sides is $3:4$, what are the lengths of those sides?
Answer
The sides are 6 cm and 8 cm.
Question
What is the area of an isosceles triangle where the equal sides $AB = AC = 12$ cm and the base $BC = 8$ cm?
Answer
The area is $16\sqrt{2}$ $\text{cm}^2$.
Question
In a rhombus with diagonals of 30 cm and 40 cm, what is its perimeter?
Answer
The perimeter is 100 cm.
Question
If a man goes 40 m due north and then 50 m due west, what is his distance from the starting point?
Answer
The distance is $\sqrt{4100}$ m (or $10\sqrt{41}$ m).
Question
In an equilateral triangle with side $x$, what is the length of the altitude $AD$ in terms of $x$?
Answer
The length is $\frac{\sqrt{3}}{2}x$.
Question
In $\triangle ABC$ right-angled at $B$, if $M$ is a point on $BC$, what is the relationship between $AM^2, BC^2, AC^2,$ and $BM^2$?
Answer
The relationship is $AM^2 + BC^2 = AC^2 + BM^2$.
Question
For any point $O$ inside a rectangle $ABCD$, what is the relationship between the distances to the vertices?
Answer
$OB^2 + OD^2 = OC^2 + OA^2$.
Question
The sides of a rectangle are 12 cm and 16 cm; what is the length of its diagonal?
Answer
The length of the diagonal is 20 cm.
Question
If a triangle has sides in the ratio $1 : \sqrt{2} : 1$, why is it a right-angled triangle?
Answer
Because $1^2 + 1^2 = (\sqrt{2})^2$, satisfying the Pythagoras Theorem.
Question
A ladder 13 m long rests against a vertical wall with its foot 5 m from the wall; how high does it reach?
Answer
It reaches a height of 12 m.
Question
If $AD$ is the altitude of an equilateral triangle $ABC$, what is the value of $3AB^2$ in terms of $AD^2$?
Answer
$3AB^2 = 4AD^2$.
Question
In a right-angled triangle $PQR$ at $Q$, with $M$ on $QR$ and $N$ on $PQ$, what does $PM^2 + RN^2$ equal?
Answer
It equals $PR^2 + MN^2$.
Question
In $\triangle ABC$, if $AB > AC$ and $E$ is the mid-point of $BC$ with $AD \perp BC$, what does $AB^2 - AC^2$ equal?
Answer
It equals $2BC \times ED$.
Question
What is the diagonal length of a square with side $x$?
Answer
The diagonal length is $x\sqrt{2}$.
Question
In a quadrilateral $ABCD$, if $\angle A + \angle D = 90^{\circ}$, what is the relationship between its diagonals and sides?
Answer
$AC^2 + BD^2 = AD^2 + BC^2$.
Question
If the sides containing the right angle of a triangle are 4 cm and $4\sqrt{3}$ cm, what is the length of the longest side?
Answer
The longest side is 8 cm.
Question
What is the perimeter of a right-angled triangle with sides 4 cm and $4\sqrt{3}$ cm?
Answer
The perimeter is $(12 + 4\sqrt{3})$ cm.
Question
In a triangle where $AB = AC$ (isosceles) and $AD \perp BC$, what is the property of point $D$?
Answer
Point $D$ is the mid-point of $BC$.
Question
If $BC^2 = AB^2 + AC^2$ in $\triangle ABC$, which angle is $90^{\circ}$?
Answer
$\angle A$ (or $\angle BAC$) is $90^{\circ}$.
Question
Two poles of heights 6 m and 11 m stand 12 m apart; what is the distance between their tips?
Answer
The distance between their tips is 13 m.
Question
In $\triangle ABC$ where $AB=8, BC=6, AC=3$, why is it not a right-angled triangle?
Answer
Because $3^2 + 6^2 \neq 8^2$ ($9 + 36 = 45 \neq 64$).
Question
How is the area of a triangle related to a rectangle if they share the same base and are between the same parallels?
Answer
The area of the triangle is half the area of the rectangle.
Question
In the identity $c^2 = a^2 + b^2 + 2ax$, what does the variable '$x$' represent?
Answer
It represents the length of the projection of one side onto the extension of the base.
Question
If the ratio of the sides of a triangle is $3:4:5$, what is the measure of the angle opposite the side with ratio 5?
Answer
The measure of the angle is $90^{\circ}$.
Question
Statement: In a right-angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides. Is this true?
Answer
Yes, this is the geometric interpretation of the Pythagoras Theorem.
Question
In $\triangle ABC$, if $AD$ is drawn perpendicular to $BC$ and $AD^2 = BD \times DC$, what is $\angle BAC$?
Answer
$\angle BAC$ is $90^{\circ}$.
Question
What is the length of the diagonal of a rectangle with sides '$l$' and '$b$'?
Answer
The length is $\sqrt{l^2 + b^2}$.
Question
In an isosceles right-angled triangle with hypotenuse $h$ and equal sides $s$, what is the formula for $h^2$?
Answer
$h^2 = 2s^2$.