Pythagoras Theorem — Chapter Test

In the figure given below, AD ⊥ BC, AB = 25 cm, AC = 17 cm and AD = 15 cm. Find the length of BC.

Answer

By pythagoras theorem,

In right angle triangle ADB,

⇒ AB² = AD² + BD²

⇒ 25² = 15² + BD²

⇒ 625 = 225 + BD²

⇒ BD² = 625 - 225 = 400

⇒ BD = $\sqrt{400}$ = 20 cm

In right angle triangle ADC,

⇒ AC² = AD² + DC²

⇒ 17² = 15² + DC²

⇒ 289 = 225 + DC²

⇒ DC² = 289 - 225 = 64

⇒ DC =$\sqrt{64}$ = 8 cm

From figure,

⇒ BC = BD + DC = 20 + 8 = 28 cm.

Hence, BC = 28 cm.

In the figure given below, ∠BAC = 90°, ADC = 90°, AD = 6 cm, CD = 8 cm and BC = 26 cm. Find

(i) AC

(ii) AB

(iii) area of the shaded region.

Answer

(i) By pythagoras theorem,

In right angle triangle ADC,

⇒ AC² = AD² + DC²

⇒ AC² = 6² + 8²

⇒ AC² = 36 + 64

⇒ AC² = 100

⇒ AC = $\sqrt{100}$ = 10 cm.

Hence, AC = 10 cm.

(ii) By pythagoras theorem,

In right angle triangle ABC,

⇒ BC² = AB² + AC²

⇒ 26² = AB² + 10²

⇒ AB² = 676 - 100

⇒ AB² = 576

⇒ AB = $\sqrt{576}$ = 24 cm.

Hence, AB = 24 cm.

(iii) Area of shaded region = Area of △ABC - Area of △ADC

$= \dfrac{1}{2} \times AB \times AC - \dfrac{1}{2} \times AD \times DC \\[1em] = \dfrac{1}{2} \times 24 \times 10 - \dfrac{1}{2} \times 6 \times 8 \\[1em] = 120 - 24 \\[1em] = 96 \text{ cm}^2.$

Hence, area of shaded region = 96 cm².

In figure given below, triangle ABC is right angled at B. Given that AB = 9 cm, AC = 15 cm and D, E are mid-points of the sides AB and AC respectively, calculate (i) the length of BC (ii) the area of △ADE.

Answer

(i) In right angle triangle ABC,

By pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ 15² = 9² + BC²

⇒ BC² = 225 - 81

⇒ BC² = 144

⇒ BC = $\sqrt{144}$ = 12 cm.

Hence, BC = 12 cm.

(ii) Given, D and E are midpoints of AB and AC.

∴ By mid-point theorem,

DE = $\dfrac{1}{2}BC$ = 6 cm.

Area of △ADE = $\dfrac{1}{2} \times AD \times DE = \dfrac{1}{2} \times 4.5 \times 6$ = 13.5 cm²

Hence, area of △ADE = 13.5 cm².

If in △ABC, AB > AC and AD ⊥ BC, prove that AB² - AC² = BD² - CD².

Answer

From figure,

In right angle △ADB,

By pythagoras theorem we get,

AB² = AD² + BD² ........(i)

In right angle △ADC,

By pythagoras theorem we get,

AC² = AD² + CD² ........(ii)

Subtracting (ii) from (i),

AB² - AC² = AD² + BD² - (AD² + CD²)

AB² - AC² = BD² - CD².

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

In a right angled triangle ABC, right angled at C, P and Q are the points on the sides CA and CB respectively which divide these sides in the ratio 2 : 1. Prove that

(i) 9AQ² = 9AC² + 4BC²

(ii) 9BP² = 9BC² + 4AC²

(iii) 9(AQ² + BP²) = 13AB²

Answer

(i) In right angle triangle ACQ,

By pythagoras theorem we get,

⇒ AQ² = AC² + CQ²

Multiplying both sides by 9 we get,

⇒ 9AQ² = 9AC² + 9CQ²

⇒ 9AQ² = 9AC² + (3CQ)² .......(i)

Given, BQ : CQ = 1 : 2

$\Rightarrow \dfrac{CQ}{BC} = \dfrac{CQ}{BQ + CQ} \\[1em] = \dfrac{2}{1 + 2} = \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{CQ}{BC} = \dfrac{2}{3} \\[1em] \Rightarrow 3CQ = 2BC$

Substituting above value in (i) we get,

⇒ 9AQ² = 9AC² + (2BC)²

⇒ 9AQ² = 9AC² + 4BC².

Hence, proved that 9AQ² = 9AC² + 4BC².

(ii) In right angle triangle BPC,

By pythagoras theorem we get,

⇒ BP² = BC² + CP²

Multiplying both sides by 9 we get,

⇒ 9BP² = 9BC² + 9CP²

⇒ 9BP² = 9BC² + (3CP)² .......(ii)

Given, AP : PC = 1 : 2

$\Rightarrow \dfrac{CP}{AC} = \dfrac{CP}{AP + PC} \\[1em] = \dfrac{2}{1 + 2} = \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{CP}{AC} = \dfrac{2}{3} \\[1em] \Rightarrow 3CP = 2AC$

Substituting above value in (ii) we get,

⇒ 9BP² = 9BC² + (2AC)²

⇒ 9BP² = 9BC² + 4AC².

Hence, proved that 9BP² = 9BC² + 4AC².

(iii) In right angle triangle ABC,

By pythagoras theorem we get,

⇒ AB² = AC² + BC² ......(iii)

Adding resultants from part (i) and (ii) we get,

9AQ² + 9BP² = 9AC² + 4BC² + 9BC² + 4AC²

9(AQ² + BP²) = 13AC² + 13BC²

9(AQ² + BP²) = 13(AC² + BC²)

9(AQ² + BP²) = 13AB² [From (iii)].

Hence, proved that 9(AQ² + BP²) = 13AB².

In the adjoining figure, △PQR is right angled at Q and points S and T trisect side QR. Prove that

8PT² = 3PR² + 5PS².

Answer

Let RT = TS = SQ = x

In right angle triangle PQR,

By pythagoras theorem we get,

⇒ PR² = QR² + PQ²

⇒ PR² = (3x)² + PQ²

⇒ PR² = 9x² + PQ²

Multiplying above equation by 3 we get,

⇒ 3PR² = 27x² + 3PQ² .......(i)

In right angle triangle PTQ,

By pythagoras theorem we get,

⇒ PT² = QT² + PQ²

⇒ PT² = (2x)² + PQ²

⇒ PT² = 4x² + PQ²

Multiplying above equation by 8 we get,

⇒ 8PT² = 32x² + 8PQ² .......(ii)

In right angle triangle PSQ,

By pythagoras theorem we get,

⇒ PS² = SQ² + PQ²

⇒ PS² = (x)² + PQ²

⇒ PS² = x² + PQ²

Multiplying above equation by 5 we get,

⇒ 5PS² = 5x² + 5PQ² .......(iii)

Adding (i) and (iii) we get,

⇒ 3PR² + 5PS² = 27x² + 5x² + 3PQ² + 5PQ²

⇒ 3PR² + 5PS² = 32x² + 8PQ²

From (ii) we get,

⇒ 3PR² + 5PS² = 8PT².

Hence, proved that 8PT² = 3PR² + 5PS².

In a quadrilateral ABCD, ∠B = 90°. If AD² = AB² + BC² + CD², Prove that ∠ACD = 90°.

Answer

In right angle triangle ABC,

By pythagoras theorem,

AC² = AB² + BC² .......(i)

Given,

AD² = AB² + BC² + CD²

Putting value of AB² + BC² from eqn (i) we get,

AD² = AC² + CD²

∴ △ACD is a right angled triangle.

In △ACD,

∠ACD = 90° i.e. angle opposite to hypotenuse = 90° (By converse of pythagoras theorem.)

Hence, proved that ∠ACD = 90°.

In the adjoining figure, find the length of AD in terms of b and c.

Answer

In right angle triangle ABC,

By pythagoras theorem,

BC² = AB² + AC²

BC² = c² + b²

BC = $\sqrt{b^2 + c^2}$.

From figure,

Area of △ABC = Area of △ABD + Area of △ADC

⇒ $\dfrac{1}{2}$ x AB x AC = $\dfrac{1}{2}$ x AD x BD + $\dfrac{1}{2}$ x AD x CD

⇒ $\dfrac{1}{2}$(AB.AC) = $\dfrac{1}{2}$(AD.BD + AD.CD)

⇒ AB.AC = AD.BD + AD.CD

⇒ AB.AC = AD(BD + CD)

⇒ AB.AC = AD.BC [∵ BD + CD = BC]

⇒ AD = $\dfrac{\text{AB.AC}}{\text{BC}}$

Putting values of AB, AC and BC in above equation we get,

AD = $\dfrac{c.b}{\sqrt{b^2 + c^2}}$

Hence, AD = $\dfrac{bc}{\sqrt{b^2 + c^2}}$.

ABCD is a square, F is mid-point of AB and BE is one third of BC. If area of △FBE is 108 cm², find the length of AC.

Answer

Let x cm be the side of each square.

Since, F is midpoint of AB so, FB = $\dfrac{x}{2}$ cm

BE is one third of BC, BE = $\dfrac{x}{3}$.

Given, area of △FBE is 108 cm²

$\therefore \dfrac{1}{2} \times FB \times BE = 108 \\[1em] \Rightarrow \dfrac{1}{2} \times \dfrac{x}{2} \times \dfrac{x}{3} = 108 \\[1em] \Rightarrow \dfrac{x^2}{12} = 108 \\[1em] \Rightarrow x^2 = 1296 \\[1em] \Rightarrow x = \sqrt{1296} = 36.$

Length of diagonal of square = $\sqrt{2}$ Side = $36\sqrt{2}$ cm.

Hence, length of diagonal of square = $36\sqrt{2}$ cm.

In a triangle ABC, AB = AC and D is a point on side AC such that BC² = AC × CD.

Prove that BD = BC.

Answer

Draw BE ⊥ AC.

In right triangle BEC,

By pythagoras theorem,

⇒ BC² = BE² + EC²

⇒ BC² = BE² + (AC - AE)²

⇒ BC² = BE² + AC² + AE² - 2AC.AE .........(i)

In right triangle ABE,

By pythagoras theorem,

⇒ AB² = BE² + AE².........(ii)

Substituting value of BE² + AE² from (ii) in (i) we get,

⇒ BC² = AB² + AC² - 2AC.CE

⇒ BC² = AC² + AC² - 2AC.AE [∵ AB = AC]

⇒ BC² = 2AC² - 2AC.AE

⇒ BC² = 2AC(AC - AE)

⇒ BC² = 2AC × EC .........(iii)

Given, BC² = AC × CD ..........(iv)

Comparing (iii) and (iv) we get,

⇒ 2AC × EC = AC × CD

⇒ 2EC = CD or EC = $\dfrac{\text{CD}}{2}$

So, we can say that ED = EC as E is mid-point of CD.

In △BDE and △BCE,

BE = BE (Common)

ED = EC

∠BED = ∠BEC = 90°

△BDE ≅ △BCE by SAS axiom of congruency.

We know that corresponding parts of congruent triangle are equal.

∴ BD = BC.

Hence, proved that BD = BC.