Pythagoras Theorem — Exercise 11

Lengths of sides of triangle are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse:

(i) 3 cm, 8 cm, 6 cm

(ii) 13 cm, 12 cm, 5 cm

(iii) 1.4 cm, 4.8 cm, 5 cm

Answer

Choose the greatest length. Check whether the square of greatest length is equal to the sum of squares of other two lengths.

(i) Here greatest length is 8 cm and other lengths are 3 cm, 6 cm.

Note that 8² = 64 and 3² + 6² = 9 + 36 = 45.

Note that, 64 ≠ 45.

Hence, the triangle with given lengths of sides is not a right triangle.

(ii) Here greatest length is 13 cm and other lengths are 5 cm, 12 cm.

Note that 13² = 169 and 5² + 12² = 25 + 144 = 169.

Thus, 13² = 5² + 12².

Hence, the triangle with given lengths of sides is a right triangle and length of hypotenuse is 13 cm.

(iii) Here greatest length is 5 cm and other lengths are 1.4 cm, 4.8 cm.

Note that 5² = 25 and (4.8)² + (1.4)² = 23.04 + 1.96 = 25.

Thus, 5² = (4.8)² + (1.4)².

Hence, the triangle with given lengths of sides is a right triangle and length of hypotenuse is 5 cm.

Foot of a 10 m long ladder leaning against a vertical well is 6 m away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches.

Answer

Let AB be the ladder and BC be the vertical well.

So, △ABC is right triangle.

By pythagoras theorem,

⇒ AB² = BC² + AC²

⇒ 10² = BC² + (6)²

⇒ 100 = BC² + 36

⇒ BC² = 100 - 36 = 64

⇒ BC = $\sqrt{64}$ = 8 m.

Hence, the top of the ladder reaches 8 m above the wall of well.

A guy attached a wire 24 m long to a vertical pole of height 18 m and has a stake attached to other end. How far from the base of the pole should the stake be driven so that the wire will be taught ?

Answer

Let BC be the pole and AB be the wire and let the distance of stake from base of pole be x.

Since, pole is vertical so △ABC is right triangle.

By pythagoras theorem,

⇒ AB² = BC² + AC²

⇒ 24² = 18² + x²

⇒ 576 = x² + 324

⇒ x² = 576 - 324 = 252

⇒ x = $\sqrt{252} = 6\sqrt{7}$ m.

Hence, the stake should be at a distance of $6\sqrt{7}$ m from the base of the pole.

Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.

Answer

Let AB be the smaller pole and CD the bigger pole.

From figure,

⇒ CE = AB = 6 m and AE = BC = 12 m

⇒ CD = CE + ED

⇒ 11 = 6 + ED

⇒ ED = 5 m

From figure,

△ADE is right triangle.

By pythagoras theorem,

⇒ AD² = AE² + ED²

⇒ AD² = (12)² + (5)²

⇒ AD² = 144 + 25

⇒ AD² = 169

⇒ AD = $\sqrt{169}$ = 13 m.

Hence, the distance between tops of poles is 13 m.

In a right-angled triangle, if hypotenuse is 20 cm and the ratio of the other sides is 4 : 3, find the sides.

Answer

Ratio of other two sides = 4 : 3

Let other sides be 4x cm and 3x cm.

By pythagoras theorem,

⇒ 20² = (4x)² + (3x)²

⇒ 400 = 16x² + 9x²

⇒ 400 = 25x²

⇒ x² = $\dfrac{400}{25}$

⇒ x² = 16

⇒ x = 4

Sides ⇒ 4x = 16 cm and 3x = 12 cm.

Hence, the sides are 12 cm and 16 cm.

If the sides of the triangle are in the ratio 3 : 4 : 5, prove that it is right-angled triangle.

Answer

Ratio of sides = 3 : 4 : 5

Let sides be 3x, 4x and 5x cm.

Here, greatest length is 5x cm and other lengths are 3x cm, 4x cm.

Note that (5x)² = 25x² and (3x)² + (4x)² = 9x² + 16x² = 25x².

Thus, (5x)² = (3x)² + (4x)².

Hence, proved that triangle having sides in the ratio 3 : 4 : 5 is a right-angle triangle.

For going to a city B from the city A, there is route via city C such that AC ⊥ CB. AC = 2x km and CB = 2(x + 7) km. It is proposed to construct a 26 km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of highway.

Answer

From figure,

△ABC is right triangle.

By pythagoras theorem,

⇒ AB² = AC² + BC²

⇒ 26² = (2x)² + [2(x + 7)]²

⇒ 676 = 4x² + (2x + 14)²

⇒ 676 = 4x² + 4x² + 196 + 56x

⇒ 676 - 196 = 8x² + 56x

⇒ 8x² + 56x = 480

⇒ 8(x² + 7x) = 480

⇒ x² + 7x = 60

⇒ x² + 7x - 60 = 0

⇒ x² + 12x - 5x - 60 = 0

⇒ x(x + 12) - 5(x + 12) = 0

⇒ (x - 5)(x + 12) = 0

⇒ x = 5 or x = -12.

Since, distance cannot be negative so, x ≠ -12.

Distance taken to reach from B to A without highway = BC + AC

= 2(x + 7) + 2x

= 2x + 14 + 2x

= 4x + 14

= 4(5) + 14

= 20 + 14 = 34 km

Distance taken to reach from B to A through highway = 26 km

Distance saved = 34 - 26 = 8 km.

Hence, 8 km will be saved in reaching city A from B after construction of highway.

The hypotenuse of a right triangle is 6 m more than twice the shortest side. If the third side is 2m less than the hypotenuse, find the sides of the triangle.

Answer

Let the shortest side be x meters.

Hypotenuse = 2x + 6 meters

Third side = 2x + 6 - 2 = 2x + 4 meters.

Since, the sides are of a right triangle.

By pythagoras theorem,

⇒ (Hypotenuse)² = (First Side)² + (Second side)²

⇒ (2x + 6)² = (x)² + (2x + 4)²

⇒ 4x² + 36 + 24x = x² + 4x² + 16 + 16x

⇒ 4x² + 36 + 24x = 5x² + 16 + 16x

⇒ 5x² - 4x² + 16x - 24x + 16 - 36 = 0

⇒ x² - 8x - 20 = 0

⇒ x² - 10x + 2x - 20 = 0

⇒ x(x - 10) + 2(x - 10) = 0

⇒ (x + 2)(x - 10) = 0

⇒ x = -2 or x = 10.

Since, side cannot be negative,

x ≠ -2.

Shortest side = 10 m

Hypotenuse = 2x + 6 = 2(10) + 6 = 26 m

Third side = 2x + 4 = 2(10) + 6 = 24 m.

Hence, sides of triangle = 10 m, 24 m and 26 m.

ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².

Answer

Since, the triangle is right triangle.

Hence, the side opposite to right angle will be hypotenuse and will be the greatest side.

From figure,

Hypotenuse = AB.

The other two sides will be equal as the triangle is also isosceles i.e. (AC = BC)

By pythagoras theorem,

⇒ (AB)² = (AC)² + (BC)²

⇒ (AB)² = (AC)² + (AC)²

⇒ (AB)² = 2AC².

Hence, proved that (AB)² = 2AC².

In a triangle ABC, AD is perpendicular to BC. Prove that AB² + CD² = AC² + BD².

Answer

From figure,

Considering right triangle ABD,

By pythagoras theorem,

AB² = AD² + BD² .......(1)

Considering right triangle ACD,

By pythagoras theorem,

AC² = AD² + CD² .......(2)

Subtracting eqn. 2 from 1 we get,

⇒ AB² - AC² = AD² + BD² - AD² - CD²

⇒ AB² - AC² = BD² - CD²

⇒ AB² + CD² = BD² + AC².

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

In △PQR, PD ⊥ QR such that D lies on QR. If PQ = a, PR = b, QD = c and DR = d, prove that (a + b)(a - b) = (c + d)(c - d).

Answer

In right angle △PDQ,

By pythagoras theorem we get,

⇒ PQ² = PD² + QD²

⇒ a² = PD² + c²

⇒ PD² = a² - c² ........(i)

In right angle △PDR,

By pythagoras theorem we get,

⇒ PR² = PD² + DR²

⇒ b² = PD² + d²

PD² = b² - d² ........(ii)

From (i) and (ii) we get,

⇒ a² - c² = b² - d²

⇒ a² - b² = c² - d²

⇒ (a - b)(a + b) = (c - d)(c + d).

Hence, proved that (a - b)(a + b) = (c - d)(c + d).

ABC is an isosceles triangle with AB = AC = 12 cm and BC = 8 cm. Find the altitude on BC and hence calculate its area.

Answer

Let AD be altitude on BC and BD = x cm and CD = (8 - x) cm.

From figure,

In right angle △ABD,

By pythagoras theorem,

AB² = AD² + BD²

12² = AD² + x²

AD² = 12² - x² ........(i)

In right angle △ADC,

By pythagoras theorem,

AC² = AD² + DC²

12² = AD² + (8 - x)²

AD² = 12² - (8 - x)² ........(ii)

From (i) and (ii) we get,

12² - x² = 12² - (8 - x)²

144 - x² = 144 - (64 + x² - 16x)

144 - x² = 144 - 64 - x² + 16x

144 - 144 - x² + x² + 64 = 16x

16x = 64

x = 4.

Substituting value of x in (i) we get,

AD² = 12² - x² = 144 - (4)² = 144 - 16 = 128

AD² = 128

AD = $\sqrt{128} = 8\sqrt{2}$ cm.

Area = $\dfrac{1}{2} \times AD \times BC = \dfrac{1}{2} \times 8\sqrt{2} \times 8 = 32\sqrt{2}$ cm².

Hence, AD = $8\sqrt{2}$ cm and area = $32\sqrt{2}$ cm².

Find the area and the perimeter of a square whose diagonal is 10 cm long.

Answer

Let each side of square be a cm.

Since, sides of square are perpendicular to each other,

In right angle △ABC,

By pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ 10² = a² + a²

⇒ 100 = 2a²

⇒ a² = 50

⇒ a = $\sqrt{50} = 5\sqrt{2}$ cm.

Perimeter = 4a = $20\sqrt{2}$ cm.

Area = a² = 50 cm².

Hence, perimeter = $20\sqrt{2}$ cm and area = 50 cm².

In figure given below, ABCD is a quadrilateral in which AD = 13 cm, DC = 12 cm, BC = 3 cm, ∠ABD = ∠BCD = 90°. Calculate the length of AB.

Answer

In right angle △DBC,

By pythagoras theorem,

⇒ DB² = DC² + BC²

⇒ DB² = 12² + 3²

⇒ DB² = 144 + 9

⇒ DB² = 153.

⇒ DB = $\sqrt{153}$.

In right angle △ABD,

By pythagoras theorem,

⇒ AD² = AB² + DB²

⇒ 13² = AB² + 153

⇒ 169 - 153 = AB²

⇒ AB² = 16

⇒ AB = $\sqrt{16}$ = 4 cm.

Hence, AB = 4 cm.

In figure given below, ABCD is a quadrilateral in which AB = AD, ∠A = 90° = ∠C, BC = 8 cm and CD = 6 cm. Find AB and calculate the area of △ABD.

Answer

Let AB = AD = x cm.

In right angle △BCD,

By pythagoras theorem,

⇒ BD² = BC² + CD²

⇒ BD² = 8² + 6²

⇒ BD² = 64 + 36

⇒ BD² = 100

⇒ BD = $\sqrt{100}$ = 10 cm.

In right angle △ABD,

By pythagoras theorem,

⇒ BD² = AB² + AD²

⇒ 10² = x² + x²

⇒ 100 = 2x²

⇒ x² = 50

⇒ x = $\sqrt{50} = 5\sqrt{2}$ cm.

Area of right angle △ABD

$= \dfrac{1}{2} \times \text{base} \times \text{ height} \\[1em] = \dfrac{1}{2} \times AB \times AD \\[1em] = \dfrac{1}{2} \times 5\sqrt{2} \times 5\sqrt{2} \\[1em] = 25 \text{ cm}^2.$

Hence, AB = $5\sqrt{2}$ cm and area of right angle △ABD = 25 cm².

In figure given below, AB = 12 cm, AC = 13 cm, CE = 10 cm and DE = 6 cm. Calculate the length of BD.

Answer

In right angle △ABC,

By pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ 13² = 12² + BC²

⇒ BC² = 13² - 12²

⇒ BC² = 169 - 144 = 25

⇒ BC = $\sqrt{25}$ = 5 cm.

In right angle △CDE,

By pythagoras theorem,

⇒ CE² = CD² + DE²

⇒ 10² = CD² + 6²

⇒ CD² = 10² - 6²

⇒ CD² = 100 - 36 = 64

⇒ CD = $\sqrt{64}$ = 8 cm.

From figure,

BD = BC + CD = 5 + 8 = 13 cm.

Hence, BD = 13 cm.

In figure given below, ∠PSR = 90°, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.

Answer

In right angle △PQS,

By pythagoras theorem,

⇒ PQ² = PS² + QS²

⇒ 10² = PS² + 6²

⇒ PS² = 10² - 6²

⇒ PS² = 100 - 36 = 64

⇒ PS = $\sqrt{64}$ = 8 cm.

From figure,

RS = RQ + QS = 9 + 6 = 15 cm.

In right angle △PRS,

By pythagoras theorem,

⇒ PR² = RS² + PS²

⇒ PR² = 15² + 8²

⇒ PR² = 225 + 64

⇒ PR² = 289

⇒ PR = $\sqrt{289}$ = 17 cm.

Hence, PR = 17 cm.

In figure given below, ∠D = 90°, AB = 16 cm, BC = 12 cm and CA = 6 cm. Find CD.

Answer

Let CD be x cm.

In right angle △ACD,

By pythagoras theorem,

⇒ AC² = AD² + CD²

⇒ 6² = AD² + x²

⇒ AD² = 36 - x² .......(i)

In right angle △ABD,

By pythagoras theorem,

⇒ AB² = AD² + BD²

⇒ 16² = (36 - x²) + (12 + x)²

⇒ 256 = 36 - x² + 144 + x² + 24x

⇒ 256 = 180 + 24x

⇒ 24x = 76

⇒ x = $\dfrac{76}{24} = 3\dfrac{1}{6}$ cm.

Hence, CD = $3\dfrac{1}{6}$ cm.

In figure given below, BC = 5 cm, ∠B = 90°, AB = 5AE, CD = 2AE and AC = ED. Calculate the lengths of EA, CD, AB and AC.

Answer

In right angle △ABC,

By pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = (5AE)² + 5²

⇒ AC² = 25AE² + 25

⇒ AC² = 25(AE² + 1) .......(i)

From figure,

EB = AB - AE = 5AE - AE = 4AE.

In right angle △BED,

⇒ ED² = EB² + BD²

⇒ ED² = (4AE)² + (5 + 2AE)²

⇒ ED² = 16AE² + 25 + 4AE² + 20AE

⇒ ED² = 20AE² + 20AE + 25 .......(ii)

Given, AC = ED.

∴ From (i) and (ii) we get,

⇒ 25(AE² + 1) = 20AE² + 20AE + 25

⇒ 25AE² + 25 = 20AE² + 20AE + 25

⇒ 25AE² - 20AE² - 20AE + 25 - 25 = 0

⇒ 5AE² - 20AE = 0

⇒ 5AE(AE - 4) = 0

⇒ 5AE = 0 or AE - 4 = 0

⇒ AE = 0 or AE = 4 cm.

Since, side cannot be 0 so AE ≠ 0.

AE = 4 cm,

CD = 2AE = 8 cm,

AB = 5AE = 20 cm,

Substituting value of AE in (i) we get,

⇒ AC² = 25(AE² + 1)

⇒ AC² = 25(4² + 1)

⇒ AC² = 25(16 + 1) = 25 × 17 = 425

⇒ AC = $\sqrt{425} = 5\sqrt{17}$ cm.

Hence, EA = 4 cm, CD = 8 cm, AB = 20 cm and AC = $5\sqrt{17}$ cm.

In figure given below, ABC is a right triangle right angled at C. If D is mid-point of BC, prove that AB² = 4AD² - 3AC².

Answer

Since, D is mid-point of BC so we get,

DC = $\dfrac{BC}{2}$.

ABC is a right triangle.

By pythagoras theorem,

⇒ AB² = AC² + BC² .......(i)

ADC is a right triangle.

By pythagoras theorem,

⇒ AD² = AC² + DC² .......(ii)

⇒ AC² = AD² - DC²

⇒ AC² = AD² - $\Big(\dfrac{1}{2}\text{BC}\Big)^2$

⇒ AC² = AD² - $\dfrac{1}{4}\text{BC}^2$

⇒ AC² = $\dfrac{4\text{AD}^2 - \text{BC}^2}{4}$

⇒ 4AC² = 4AD² - BC²

⇒ AC² + 3AC² = 4AD² - BC²

⇒ AC² + BC² = 4AD² - 3AC²

⇒ AB² = 4AD² - 3AC² [....From (i)]

Hence, proved that AB² = 4AD² - 3AC².

In △ABC, AB = AC = x, BC = 10 cm and the area of △ABC is 60 cm². Find x.

Answer

Let AD be the altitude. In isosceles triangle, the altitude to base bisects the base.

∴ BD = CD = 5 cm.

From figure,

In right triangle ABD,

AB² = AD² + BD²

x² = AD² + 5²

AD² = x² - 25.

AD = $\sqrt{x^2 - 25}$

$\text{Area of △ABC } = \dfrac{1}{2} \times \text{AD} \times \text{BC} \\[1em] \Rightarrow 60 = \dfrac{1}{2} \times \sqrt{x^2 - 25} \times 10 \\[1em] \Rightarrow 120 = 10\sqrt{x^2 - 25} \\[1em] \Rightarrow \sqrt{x^2 - 25} = 12 \\[1em] \Rightarrow x^2 - 25 = 144 \\[1em] \Rightarrow x^2 = 169 \\[1em] \Rightarrow x = \sqrt{169} = 13.$

Hence, x = 13 cm.

In rhombus if diagonals are 30 cm and 40 cm, find its perimeter.

Answer

Let AC = 30 cm and BD = 40 cm.

We know that,

Diagonals of rhombus are perpendicular and bisect each other,

OB = $\dfrac{1}{2}BD$ = 20 cm and AO = $\dfrac{1}{2}$AC = 15 cm.

In right triangle AOB,

By pythagoras theorem we get,

⇒ AB² = AO² + OB²

⇒ AB² = 15² + 20²

⇒ AB² = 225 + 400

⇒ AB² = 625

⇒ AB = 25 cm.

Hence, each side of rhombus is 25 cm as all sides of rhombus are equal.

Perimeter = 4(Side) = 100 cm.

Hence, perimeter of rhombus = 100 cm.

In figure given below, AB || DC, BC = AD = 13 cm, AB = 22 cm and DC = 12 cm. Calculate the height of the trapezium ABCD.

Answer

From figure,

EF = DC = 12 cm

Since,

AD = BC and DE = CF = h (height)

∴ AE = FB = x.

⇒ AB = AE + EF + FB

⇒ 22 = x + 12 + x

⇒ 2x = 22 - 12

⇒ 2x = 10

⇒ x = 5 cm.

In right triangle ADE,

By pythagoras theorem,

⇒ AD² = DE² + AE²

⇒ 13² = h² + 5²

⇒ 169 - 25 = h²

⇒ h² = 144

⇒ h = 12 cm.

Hence, height = 12 cm.

In figure given below, AB || DC, ∠A = 90°, DC = 7 cm, AB = 17 cm and AC = 25 cm. Calculate BC.

Answer

Since, AB || DC

∠D = ∠A = 90°

In right angle triangle ACD,

By pythagoras theorem,

⇒ AC² = AD² + CD²

⇒ 25² = AD² + 7²

⇒ 625 = AD² + 49

⇒ AD² = 625 - 49

⇒ AD² = 576

⇒ AD = $\sqrt{576}$ = 24 cm.

From figure,

CE = AD = 24 cm

In right angle triangle ACE,

By pythagoras theorem,

⇒ AC² = AE² + CE²

⇒ 25² = AE² + 24²

⇒ 625 = AE² + 576

⇒ AE² = 625 - 576

⇒ AE² = 49

⇒ AE = $\sqrt{49}$ = 7 cm.

So, EB = AB - AE = 17 - 7 = 10 cm.

In right angle triangle BCE,

By pythagoras theorem,

⇒ BC² = CE² + EB²

⇒ BC² = 24² + 10²

⇒ BC² = 576 + 100

⇒ BC² = 676

⇒ BC = $\sqrt{676}$ = 26 cm.

Hence, BC = 26 cm.

In the figure given below, ABCD is a square of side 7 cm. If

AE = FC = CG = HA = 3 cm,

(i) prove that EFGH is a rectangle.

(ii) find the area and perimeter of EFGH.

Answer

(i) Given, ABCD is a square of side 7 cm and AE = FC = CG = HA = 3 cm.

From figure,

DH = DA - HA = 7 - 3 = 4 cm,

GD = DC - CG = 7 - 3 = 4 cm,

EB = AB - AE = 7 - 3 = 4 cm,

FB = BC - FC = 7 - 3 = 4 cm.

Since in square, sides are perpendicular to each other,

By pythagoras theorem,

In right angle triangle HAE,

⇒ HE² = HA² + AE²

⇒ HE² = 3² + 3²

⇒ HE² = 9 + 9

⇒ HE² = 18

⇒ HE = $\sqrt{18} = 3\sqrt{2}$ cm.

In right angle triangle FCG,

⇒ GF² = GC² + FC²

⇒ GF² = 3² + 3²

⇒ GF² = 9 + 9

⇒ GF² = 18

⇒ GF = $\sqrt{18} = 3\sqrt{2}$ cm.

In right angle triangle GDH,

⇒ GH² = GD² + DH²

⇒ GH² = 4² + 4²

⇒ GH² = 16 + 16

⇒ GH² = 32

⇒ GH = $\sqrt{32} = 4\sqrt{2}$ cm.

In right angle triangle EBF,

⇒ EF² = EB² + FB²

⇒ EF² = 4² + 4²

⇒ EF² = 16 + 16

⇒ EF² = 32

⇒ EF = $\sqrt{32} = 4\sqrt{2}$ cm.

In isosceles triangle AEH,

∠E = ∠H = x (let) (As angles opposite to equal side in isosceles triangle are equal).

So, ∠A + ∠E + ∠H = 180°

90° + x + x = 180°

2x = 90°

x = 45°

Similar is the case for triangles EBF, GDH, FCG.

From figure,

∠AEH + ∠HEF + ∠FEB = 180° (Linear pairs)

45° + ∠HEF + 45° = 180°

∠HEF = 90°

Since, angles of EFGH = 90° and EH = GF and HG = EF.

Hence, proved that EFGH is a rectangle.

(ii) Area of EFGH = EF × GF = $4\sqrt{2} \times 3\sqrt{2}$ = 24 cm².

Perimeter of EFGH = 2(EF + GF) = $2(4\sqrt{2} + 3\sqrt{2}) = 14\sqrt{2}$ cm.

Hence, area = 24 cm² and perimeter = $14\sqrt{2}$ cm.

AD is perpendicular to the side BC of an equilateral △ABC. Prove that 4AD² = 3AB².

Answer

Given, AD ⊥ BC and AB = BC = CA (Equilateral triangle).

The perpendicular to base in equilateral triangle bisects the base.

∴ BD = $\dfrac{BC}{2}$.

From figure,

In right triangle ABD,

By pythagoras theorem,

⇒ AB² = AD² + BD²

⇒ AB² = AD² + $\Big(\dfrac{BC}{2}\Big)^2$

⇒ AB² = AD² + $\dfrac{BC^2}{4}$

⇒ AB² = $\dfrac{4\text{AD}^2 + \text{BC}^2}{4}$

⇒ 4AB² = 4AD² + BC²

⇒ 4AB² = 4AD² + AB² (∵ BC = AB)

⇒ 4AD² = 3AB².

Hence, proved that 4AD² = 3AB².

In the adjoining figure, D and E are mid-points of the sides BC and CA respectively of a △ABC, right angled at C. Prove that :

(i) 4AD² = 4AC² + BC²

(ii) 4BE² = 4BC² + AC²

(iii) 4(AD² + BE²) = 5AB²

Answer

(i) Since, D is mid-point of BC,

∴ CD = $\dfrac{BC}{2}$

In right triangle ACD,

By pythagoras theorem,

⇒ AD² = AC² + CD²

⇒ AD² = AC² + $\Big(\dfrac{BC}{2}\Big)^2$

⇒ AD² = AC² + $\dfrac{BC^2}{4}$

⇒ AD² = $\dfrac{4\text{AC}^2 + \text{BC}^2}{4}$

⇒ 4AD² = 4AC² + BC² .....(1)

Hence, proved that 4AD² = 4AC² + BC².

(ii) As E is mid-point of AC,

∴ CE = $\dfrac{AC}{2}$

⇒ AC = 2CE

BCE is right triangle,

By pythagoras theorem,

⇒ BE² = BC² + CE²

Multiplying both sides by 4 we get,

⇒ 4BE² = 4BC² + 4CE²

⇒ 4BE² = 4BC² + (2CE)²

⇒ 4BE² = 4BC² + AC² ......(2)

Hence, proved that 4BE² = 4BC² + AC².

(iii) As, ABC is a right triangle,

By pythagoras theorem we get,

⇒ AB² = AC² + BC²

Adding 1 and 2 from above parts we get,

⇒ 4AD² + 4BE² = 4AC² + BC² + 4BC² + AC²

⇒ 4(AD² + BE²) = 5AC² + 5BC²

⇒ 4(AD² + BE²) = 5(AC² + BC²)

⇒ 4(AD² + BE²) = 5AB².

Hence, proved that 4(AD² + BE²) = 5AB².

If AD, BE and CF are medians of △ABC, prove that

3(AB² + BC² + CA²) = 4(AD² + BE² + CF²).

Answer

Draw AP perpendicular to BC.

In right triangle APD,

By pythagoras theorem,

AD² = AP² + PD² .......(i)

In right triangle APB,

By pythagoras theorem,

⇒ AB² = AP² + BP²

⇒ AB² = AP² + (BD - PD)²

⇒ AB² = AP² + BD² + PD² - 2BD.PD

Since, BD = $\dfrac{\text{BC}}{2}$ as AD is median of BC.

⇒ AB² = AP² + PD² + $\Big(\dfrac{\text{BC}}{2}\Big)^2$ - 2 x $\dfrac{\text{BC}}{2}$ x PD

⇒ AB² = AD² + $\dfrac{\text{BC}^2}{4}$ - BC.PD ....(ii) (From i)

In right triangle APC,

By pythagoras theorem,

⇒ AC² = AP² + PC²

⇒ AC² = AP² + (PD + DC)²

⇒ AC² = AP² + PD² + DC² + 2PD.DC

Since, DC = $\dfrac{BC}{2}$

⇒ AC² = AD² + DC² + 2PD.DC [....From (i)]

⇒ AC² = AD² + $\Big(\dfrac{\text{BC}}{2}\Big)^2$ + 2 x PD x $\dfrac{\text{BC}}{2}$

⇒ AC² = AD² + $\dfrac{\text{BC}^2}{4}$ + PD.BC ....(iii)

Adding (ii) and (iii) we get,

⇒ AB² + AC² = AD² + $\dfrac{\text{BC}^2}{4}$ - BC.PD + AD² + $\dfrac{\text{BC}^2}{4}$ + PD.BC

⇒ AB² + AC² = 2AD² + $\dfrac{\text{BC}^2}{2}$ ....(iv)

Draw BQ perpendicular to AC,

In right triangle BQE,

By pythagoras theorem,

BE² = BQ² + QE² .......(v)

In right triangle ABQ,

By pythagoras theorem,

⇒ AB² = BQ² + AQ²

⇒ AB² = BQ² + (AE - QE)²

⇒ AB² = BQ² + QE² + AE² - 2AE.QE

Since, AE = $\dfrac{\text{AC}}{2}$ as BE is median of AC.

⇒ AB² = BE² + $\Big(\dfrac{\text{AC}}{2}\Big)^2$ - 2 x $\dfrac{\text{AC}}{2}$ x QE (From v)

⇒ AB² = BE² + $\dfrac{\text{AC}^2}{4}$ - AC.QE ....(vi)

In right triangle BQC,

By pythagoras theorem,

⇒ BC² = BQ² + QC²

⇒ BC² = BQ² + (QE + EC)²

⇒ BC² = BQ² + QE² + EC² + 2QE.EC

Since, EC = $\dfrac{AC}{2}$ as BE is median of AC.

⇒ BC² = BE² + EC² + 2QE.EC [....From (v)]

⇒ BC² = BE² + $\Big(\dfrac{\text{AC}}{2}\Big)^2$ + 2 x QE x $\dfrac{\text{AC}}{2}$

⇒ BC² = BE² + $\dfrac{\text{AC}^2}{4}$ + QE.AC ....(vii)

Adding (vi) and (vii) we get,

⇒ AB² + BC² = BE² + $\dfrac{\text{AC}^2}{4}$ - AC.QE + BE² + $\dfrac{\text{AC}^2}{4}$ + QE.AC

⇒ AB² + BC² = 2BE² + $\dfrac{\text{AC}^2}{2}$ ....(viii)

Draw CR perpendicular to AB,

In right triangle CFR,

By pythagoras theorem,

CF² = CR² + FR² .......(ix)

In right triangle CBR,

By pythagoras theorem,

⇒ CB² = CR² + RB²

⇒ CB² = CR² + (BF - FR)²

⇒ CB² = CR² + FR² + BF² - 2BF.FR

Since, BF = $\dfrac{\text{AB}}{2}$ as CF is median of AB.

⇒ CB² = CF² + $\Big(\dfrac{\text{AB}}{2}\Big)^2$ - 2 x $\dfrac{\text{AB}}{2}$ x FR (From ix)

⇒ CB² = CF² + $\dfrac{\text{AB}^2}{4}$ - AB.FR ....(x)

In right triangle ACR,

By pythagoras theorem,

⇒ AC² = RC² + AR²

⇒ AC² = RC² + (AF + FR)²

⇒ AC² = RC² + FR² + AF² + 2AF.FR

Since, AF = $\dfrac{\text{AB}}{2}$ as CF is median of AB.

⇒ AC² = CF² + AF² + 2AF.FR [....From (ix)]

⇒ AC² = CF² + $\Big(\dfrac{\text{AB}}{2}\Big)^2$ + 2 x FR x $\dfrac{\text{AB}}{2}$

⇒ AC² = CF² + $\dfrac{\text{AB}^2}{4}$ + FR.AB ....(xi)

Adding (x) and (xi) we get,

⇒ CB² + AC² = CF² + $\dfrac{\text{AB}^2}{4}$ - AB.FR + CF² + $\dfrac{\text{AB}^2}{4}$ + FR.AB

⇒ CB² + AC² = 2CF² + $\dfrac{\text{AB}^2}{2}$ ....(xii)

On adding (iv), (viii) and (xii) we get,

AB² + AC² + AB² + BC² + CB² + AC² = 2AD² + $\dfrac{\text{BC}^2}{2}$ + 2BE² + $\dfrac{\text{AC}^2}{2}$ + 2CF² + $\dfrac{\text{AB}^2}{2}$

⇒ 2(AB² + BC² + CA²) = 2(AD² + BE² + CF²) + $\dfrac{1}{2}$(AB² + BC² + CA²)

⇒ 2(AB² + BC² + CA²) - $\dfrac{1}{2}$(AB² + BC² + CA²) = 2(AD² + BE² + CF²)

⇒ $\dfrac{3}{2}$(AB² + BC² + CA²) = 2(AD² + BE² + CF²)

⇒ $3$(AB² + BC² + CA²) = 4(AD² + BE² + CF²)

Hence, proved that 3(AB² + BC² + CA²) = 4(AD² + BE² + CF²).

In the adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O, at right angles. Prove that AB² + CD² = AD² + BC².

Answer

By pythagoras theorem,

In right angle triangle AOB,

AB² = OB² + OA² .......(i)

In right angle triangle COD,

CD² = OC² + OD² .......(ii)

In right angle triangle AOD,

AD² = AO² + OD² .......(iii)

In right angle triangle BOC,

BC² = OB² + OC² .......(iv)

Adding (i), (ii) we get,

AB² + CD² = OB² + OA² + OC² + OD²

AB² + CD² = (OA² + OD²) + (OC² + OB²)

Substituting value from (iii) and (iv) in above equation we get,

AB² + CD² = AD² + BC².

Hence, proved that AB² + CD² = AD² + BC².

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

2AC² - BC² = AB² + AD² + DC².

Answer

Given, ∠B = 90° = ∠D

By pythagoras theorem,

In right angle triangle ABC,

AC² = AB² + BC² ......(i)

By pythagoras theorem,

In right angle triangle ADC,

AC² = AD² + DC² ......(ii)

Adding (i) and (ii) we get,

2AC² = AB² + BC² + AD² + DC²

2AC² - BC² = AB² + AD² + DC².

Hence, proved that 2AC² - BC² = AB² + AD² + DC².

In a △ABC, ∠A = 90°, CA = AB and D is a point on AB produced. Prove that

DC² - BD² = 2AB × AD.

Answer

In right angle triangle ACD,

DC² = CA² + AD² (Pythagoras theorem)

DC² = CA² + (AB + BD)²

DC² = CA² + AB² + BD² + 2AB.BD

DC² - BD² = AB² + AB² + 2AB.BD [∵ CA = AB]

DC² - BD² = 2AB² + 2AB.BD

DC² - BD² = 2AB(AB + BD)

From figure, AB + BD = AD

DC² - BD² = 2AB.AD

Hence, proved that DC² - BD² = 2AB.AD.

In an isosceles triangle ABC, AB = AC and D is a point on BC produced. Prove that AD² = AC² + BD × CD.

Answer

Draw AP ⊥ BC.

APD is right triangle

By pythagoras theorem we get,

⇒ AD² = AP² + PD²

⇒ AD² = AP² + (PC + CD)²

⇒ AD² = AP² + PC² + CD² + 2PC.CD ......(i)

APC is right triangle,

By pythagoras theorem we get,

⇒ AC² = AP² + PC² ......(ii)

Substituting the value of AP² + PC² from (ii) in (i) we get,

⇒ AD² = AC² + CD² + 2PC.CD .......(iii)

Since, ABC is an isosceles triangle,

⇒ PC = $\dfrac{\text{BC}}{2}$ [∵ altitude to the base of an isosceles triangle bisects the base]

⇒ AD² = AC² + CD² + $2 \times \dfrac{1}{2} \times \text{BC} \times \text{CD}$

⇒ AD² = AC² + CD² + BC.CD

⇒ AD² = AC² + CD(CD + BC)

From figure, CD + BC = BD

⇒ AD² = AC² + CD.BD

Hence, proved that AD² = AC² + BD × CD.