Pythagoras Theorem — Exercise 12(B)

In △ ABC, ∠C = 90° and AC = BC, then AB² is equal to :

1. AC²

2. 2AC²

3. BC²

4. 2BC² - AC²

Answer

ABC is a right-angled triangle at C.

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ AB² = AC² + BC²

⇒ AB² = AC² + AC² (Since, AC = BC)

⇒ AB² = 2AC².

Hence, Option 2 is the correct option.

In the given diagram, AE² + BD² is equal to :

1. AB² - DE²

2. DE² - AB²

3. AB² + DE²

4. DE × AB

Answer

By Formula,

By pythagoras theorem,

(Hypotenuse)² = (Perpendicular)² + (Base)².

In right-angled triangle ABC,

⇒ AB² = AC² + BC² ........(1)

In right-angled triangle DEC,

⇒ DE² = CD² + EC² ........(2)

In right-angled triangle AEC,

⇒ AE² = AC² + EC² ........(3)

In right-angled triangle BDC,

⇒ BD² = CD² + BC² ........(4)

Adding equations (3) and (4), we get :

⇒ AE² + BD² = AC² + EC² + CD² + BC²

= (AC² + BC²) + (CD² + EC²)

= AB² + DE² [From equations (1) and (2)].

Hence, Option 3 is the correct option.

In the given figure, the value of AB × CD is :

1. AC × BC

2. AC × CD

3. AC × AB

4. AC² + BC²

Answer

By formula,

⇒ Area of triangle = $\dfrac{1}{2}$ × base × height

From figure,

⇒ Area of △ ABC = $\dfrac{1}{2}$ × AB × CD .........(1)

⇒ Area of △ ABC = $\dfrac{1}{2}$ × BC × AC .........(2)

From equations (1) and (2), we get :

⇒ $\dfrac{1}{2}$ × AB × CD = $\dfrac{1}{2}$ × BC × AC

⇒ AB × CD = BC × AC.

Hence, Option 1 is the correct option.

ABC is an isosceles triangle right-angled at C. Then 2AC² is equal to :

1. BC²

2. AC²

3. AC² - BC²

4. AB²

Answer

Since, ABC is an isosceles triangle right-angled at C.

∴ ∠A = ∠B

∴ BC = AC [Sides opposite to equal angles are equal] ........(1)

In right-angled △ ABC,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ AB² = AC² + BC²

⇒ AB² = AC² + AC² [From equation (1)]

⇒ AB² = 2AC².

Hence, Option 4 is the correct option.

In the figure, given below, AD ⊥ BC. Prove that :

c² = a² + b² - 2ax.

Answer

By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

In right-angled △ ABD,

⇒ AB² = AD² + BD²

⇒ c² = h² + (a - x)² .............(1)

In right-angled △ ACD,

⇒ AC² = AD² + CD²

⇒ b² = h² + x²

⇒ h² = b² - x² ..........(2)

Substituting value of h² from equation (2) in (1), we get :

⇒ c² = b² - x² + (a - x)²

⇒ c² = b² - x² + a² + x² - 2ax

⇒ c² = a² + b² - 2ax.

Hence, proved that c² = a² + b² - 2ax.

In equilateral △ ABC, AD ⊥ BC and BC = x cm. Find, in terms of x, the length of AD.

Answer

In △ ABD and △ ACD,

⇒ ∠ADB = ∠ADC (Both equal to 90°)

⇒ AD = AD (Common side)

⇒ AB = AC (Since, ABC is an equilateral triangle)

∴ △ ABD ≅ △ ACD (By S.A.S. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

∴ BD = CD = $\dfrac{BC}{2} = \dfrac{x}{2}$ cm.

In right-angled triangle ABD,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ AB² = AD² + BD²

⇒ x² = AD² + $\Big(\dfrac{x}{2}\Big)^2$

⇒ AD² = x² - $\Big(\dfrac{x}{2}\Big)^2$

⇒ AD² = $x^2 - \dfrac{x^2}{4}$

⇒ AD² = $\dfrac{4x^2 - x^2}{4}$

⇒ AD² = $\dfrac{3x^2}{4}$

⇒ AD = $\sqrt{\dfrac{3x^2}{4}} = \dfrac{\sqrt{3}x}{2}$ cm.

Hence, AD = $\dfrac{\sqrt{3}x}{2}$ cm.

ABC is a triangle, right-angled at B. M is a point on BC. Prove that :

AM² + BC² = AC² + BM².

Answer

By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

In right-angled △ ABM,

⇒ AM² = AB² + BM²

⇒ AB² = AM² - BM² ...........(1)

In right-angled △ ABC,

⇒ AC² = AB² + BC²

⇒ AB² = AC² - BC² ...........(2)

From equations (1) and (2), we get :

⇒ AM² - BM² = AC² - BC²

⇒ AM² + BC² = AC² + BM².

Hence, proved AM² + BC² = AC² + BM².

M and N are the mid-points of the sides QR and PQ respectively of a △ PQR, 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

Answer

Since, M and N are the mid-points of the sides QR and PQ respectively.

PN = NQ and QM = RM

(i) In △ MNQ,

By pythagoras theorem,

⇒ MN² = NQ² + QM² ..........(1)

In △ PQM,

By pythagoras theorem,

⇒ PM² = PQ² + QM²

⇒ PM² = (PN + NQ)² + QM²

⇒ PM² = PN² + NQ² + 2.PN.NQ + QM²

⇒ PM² = MN² + PN² + 2.PN.NQ [From equation (1)] ...........(2)

In △ RNQ,

By pythagoras theorem,

⇒ RN² = NQ² + RQ²

⇒ RN² = NQ² + (QM + RM)²

⇒ RN² = NQ² + QM² + RM² + 2.QM.RM

⇒ RN² = MN² + RM² + 2.QM.RM [From equation (1)] ...........(3)

Adding equations (2) and (3), we get :

⇒ PM² + RN² = MN² + PN² + 2.PN.NQ + MN² + RM² + 2.QM.RM

⇒ PM² + RN² = 2MN² + PN² + RM² + 2.PN.NQ + 2.QM.RM

Substituting PN = QN and RM = QM in above equation, we get :

⇒ PM² + RN² = 2MN² + QN² + QM² + 2.NQ.NQ + 2.QM.QM

⇒ PM² + RN² = 2MN² + QN² + QM² + 2 QN² + 2 QM²

⇒ PM² + RN² = 2MN² + (QN² + QM²) + 2 (QN² + QM²)

⇒ PM² + RN² = 2MN² + MN² + 2MN² [From equation (1)]

⇒ PM² + RN² = 5MN².

Hence, proved that PM² + RN² = 5MN².

(ii) In △ PQM,

By pythagoras theorem,

⇒ PM² = PQ² + QM²

Multiplying both sides of the above equation by 4, we get :

⇒ 4PM² = 4PQ² + 4QM²

⇒ 4PM² = 4PQ² + $4 \times \Big(\dfrac{1}{2}QR\Big)^2$

⇒ 4PM² = 4PQ² + $4 \times \dfrac{1}{4} \times QR^2$

⇒ 4PM² = 4PQ² + QR².

Hence, proved that 4PM² = 4PQ² + QR².

(iii) In △ RQN,

By pythagoras theorem,

⇒ RN² = NQ² + QR²

Multiplying both sides of the above equation by 4, we get :

⇒ 4RN² = 4NQ² + 4QR²

⇒ 4RN² = 4QR² + $4 \times \Big(\dfrac{1}{2}PQ\Big)^2$

⇒ 4RN² = 4QR² + $4 \times \dfrac{1}{4} \times PQ^2$

⇒ 4RN² = 4QR² + PQ².

Hence, proved that 4RN² = PQ² + 4QR².

(iv) Proved in part (i), we get :

⇒ PM² + RN² = 5MN²

Multiplying both side of the above equation by 4, we get :

⇒ 4(PM² + RN²) = 4 × 5 MN²

⇒ 4(PM² + RN²) = 4 × 5 (NQ² + MQ²)

⇒ 4(PM² + RN²) = 4 × 5 $\times \Big[\Big(\dfrac{1}{2}PQ\Big)^2 + \Big(\dfrac{1}{2}RQ\Big)^2\Big]$

⇒ 4(PM² + RN²) = 4 × 5 $\times \Big[\dfrac{1}{4}PQ^2 + \dfrac{1}{4}RQ^2\Big]$

⇒ 4(PM² + RN²) = 4 × 5 $\times \dfrac{1}{4}$ (PQ² + RQ²)

⇒ 4(PM² + RN²) = 5 (PQ² + RQ²) .....(4)

In right angled triangle PQR,

By pythagoras theorem,

⇒ PR² = PQ² + RQ²

Substituting above value of PR² in equation (4), we get :

⇒ 4(PM² + RN²) = 5 PR².

Hence, proved that 4(PM² + RN²) = 5 PR².

In triangle ABC, ∠B = 90° and D is the mid-point of BC. Prove that : AC² = AD² + 3CD².

Answer

Given,

D is the mid-point of BC.

∴ CD = BD ........(1)

By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

In right-angled △ ABC,

⇒ AC² = AB² + BC²

⇒ AC² = AB² + (BD + CD)²

⇒ AC² = AB² + (CD + CD)² ..........[From equation (1)]

⇒ AC² = AB² + (2CD)²

⇒ AC² = AB² + 4 CD² ...........(2)

In right-angled △ ABD,

⇒ AD² = AB² + BD²

⇒ AD² = AB² + CD² [From equation (1)] .......(3)

Subtracting equation (3) from (2), we get :

⇒ AC² - AD² = AB² + 4 CD² - (AB² + CD²)

⇒ AC² - AD² = AB² - AB² + 4 CD² - CD²

⇒ AC² - AD² = 3 CD²

⇒ AC² = AD² + 3 CD².

Hence, proved that AC² = AD² + 3 CD².

In a rectangle ABCD, prove that :

AC² + BD² = AB² + BC² + CD² + DA².

Answer

In rectangle the interior angles equal to 90° and opposite sides are equal.

∴ ∠A = ∠B = ∠C = ∠D = 90°, AB = DC and BC = AD.

By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

In right-angled △ ACD,

⇒ AC² = AD² + CD² ........(1)

In right-angled △ BCD,

⇒ BD² = BC² + CD²

⇒ BD² = BC² + AB² (As CD = AB) .........(2)

Adding equation (1) and (2), we get :

⇒ AC² + BD² = AD² + CD² + BC² + AB².

Hence, proved that AC² + BD² = AB² + BC² + CD² + DA².

In a quadrilateral ABCD, ∠B = 90° and ∠D = 90°. Prove that :

2AC² - AB² = BC² + CD² + DA².

Answer

In right-angled triangle ABC,

By pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AB² = AC² - BC² .........(1)

In right-angled triangle ADC,

By pythagoras theorem,

⇒ AC² = AD² + DC² ........(2)

To prove :

2AC² - AB² = BC² + CD² + DA².

Solving L.H.S. of the equation :

⇒ 2AC² - AB² = 2AC² - (AC² - BC²) [From equation (1)]

= 2AC² - AC² + BC²

= AC² + BC²

= AD² + DC² + BC² [From equation (2)]

= R.H.S.

Hence, proved that 2AC² - AB² = BC² + CD² + DA².

O is any point inside a rectangle ABCD. Prove that :

OB² + OD² = OC² + OA².

Answer

Through point O, draw PQ || BC so that P lies on AB and Q lies on DC.

Since, PQ || BC and AB ⊥ BC,

∴ AB ⊥ PQ and DC ⊥ PQ.

∴ BPQC and APQD are both rectangles.

In right-angled triangle OBP,

By pythagoras theorem,

⇒ OB² = OP² + BP² ..........(1)

In right-angled triangle OQD,

By pythagoras theorem,

⇒ OD² = OQ² + DQ² ..........(2)

In right-angled triangle OQC,

By pythagoras theorem,

⇒ OC² = OQ² + QC² ..........(3)

In right-angled triangle OAP,

By pythagoras theorem,

⇒ OA² = OP² + AP² ..........(4)

Adding equations (1) and (2), we get :

⇒ OB² + OD² = OP² + BP² + OQ² + DQ²

= OP² + CQ² + OQ² + AP² (As, BP = CQ and DQ = AP)

= CQ² + OQ² + OP² + AP²

= OC² + OA². [From equation (3) and (4)]

Hence, proved that OB² + OD² = OC² + OA².

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.

Answer

Join OA, OB and OC.

By formula,

By pythagoras theorem,

(Hypotenuse)² = (Perpendicular)² + (Base)²

In right-angled triangle AOR,

⇒ AO² = AR² + OR²

⇒ AR² = AO² - OR² ..........(1)

In right-angled triangle BOP,

⇒ BO² = BP² + OP²

⇒ BP² = BO² - OP² ..........(2)

In right-angled triangle COQ,

⇒ CO² = CQ² + OQ²

⇒ CQ² = CO² - OQ² ..........(3)

In right-angled triangle AOQ,

⇒ AO² = AQ² + OQ²

⇒ AQ² = AO² - OQ² ..........(4)

In right-angled triangle COP,

⇒ CO² = CP² + OP²

⇒ CP² = CO² - OP² ..........(5)

In right-angled triangle BOR,

⇒ BO² = BR² + OR²

⇒ BR² = BO² - OR² ..........(6)

Adding equations (1), (2) and (3), we get :

⇒ AR² + BP² + CQ² = AO² - OR² + BO² - OP² + CO² - OQ² ..........(7)

Adding equations (4), (5) and (6), we get :

⇒ AQ² + CP² + BR² = AO² - OQ² + CO² - OP² + BO² - OR²

⇒ AQ² + CP² + BR² = AO² - OR² + BO² - OP² + CO² - OQ² ...........(8)

From equations (7) and (8), we get :

⇒ AR² + BP² + CQ² = AQ² + CP² + BR².

Hence, proved that AR² + BP² + CQ² = AQ² + CP² + BR².