Students can practice TS Class 10 Maths Solutions Chapter 8 Similar Triangles Ex 8.4 to get the best methods of solving problems.

## TS 10th Class Maths Solutions Chapter 8 Similar Triangles Exercise 8.4

Question 1.

Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. (AS_{2})

Solution:

ABCD is a rhombus. Let its diagonals AC and BD meet at ‘O’.

In ∆AOD, ∠AOD = 90°

∴ By Pythagoras theorem,

AD^{2} = OA^{2} + OD^{2} …………… (1)

In ∆COD, ∠COD = 90°

∴ CD^{2} = OC^{2} + OD^{2} …………… (2)

In ∆ AOB, ∠AOB = 90°

∴ AB^{2} = OA^{2} + OB^{2} ……………… (3)

In ∆BOC, ∠BOC = 90°

∴ BC^{2} = OB^{2} + OC^{2} ……………… (4)

Adding (1), (2), (3) & (4), we get

AD^{2} + CD^{2} + AB^{2} + BC^{2} = OA^{2} + OD^{2} + OC^{2} + OD^{2} + OA^{2} + OB^{2} + OB^{2} + OC^{2}

= OA^{2} + OA^{2} + OC^{2} + OC^{2} + OB^{2} + OB^{2} + OD^{2} + OD^{2}

= OA^{2} + OA^{2} + OA^{2} + OA^{2} + OB^{2} + OB^{2} + OB^{2} + OB^{2} (∵ OC = OA & OD = OB)

= 4OA^{2} + 4OB^{2}

= (2OA^{2} + 2OB^{2}) (∵ 2OA = AC & 2OB = BD)

Question 2.

ABC is a right triangle right angled at B. Let D and E be any points on AB and BC respectively.

Prove that AE^{2} + CD^{2} = AC^{2} + DE^{2}. (AS_{2})

Solution:

In ∆ABE, ∠B = 90°

By Pythagoras Theorem,

AE^{2} = AB^{2} + BE^{2} …………… (1)

In ∆CBD, ∠CBD = 90°

By Pythagoras Theorem,

CD^{2} = BD^{2} + BC^{2} ……………. (2)

Adding (1) & (2), we get AE^{2} + CD^{2} = AB^{2} + BE^{2} + BD^{2} + BC^{2}

= (AB^{2} + BC^{2}) + (BE^{2} + BD^{2})

= AC^{2} + DE^{2} (∵ In ∆ABC, ∠B = 90°)

(∴ AB^{2} + BC^{2} = AC^{2} and In ∆DBE, ∠B = 90° ∴ BD^{2} + BE^{2} = DE^{2})

Question 3.

Prove that three times the square of any side of an equilateral triangle is equal to four times the square of the altitude. (AS_{2})

Solution:

ABC is an equilateral triangle,

(∵ AD ⊥ BC)

(i.e„) AB = BC = CA, AD ⊥ BC

In ∆^{les} ADB and ADC

∠ADB = ∠ADC = 90°

Hyp. AB = Hyp. AC

AD is common.

∴ ∆ ADB = ∆ADC

∴ BD = DC

In ∆ADB, ∠ADB = 90°

By Pythagoras Theorem,

AB^{2} = AD^{2} + BD^{2}

= AD^{2} + \(\left(\frac{\mathrm{BC}}{2}\right)^2\) (∵ BD = CD)

= AD^{2} + \(\frac{\mathrm{BC}^2}{4}\) = \(\frac{4 \mathrm{AD}^2+\mathrm{BC}^2}{4}\)

= \(\frac{4 \mathrm{AD}^2+\mathrm{AB}^2}{4}\) (∵ BC = AB)

⇒ 4AB^{2} = 4AD^{2} + AB^{2}

⇒ 4AB^{2} – AB^{2} = 4AD^{2}

⇒ 3AB^{2} = 4AD^{2}

Question 4.

PQR is a triangle right angled at P and M is a point on QR such that PM ⊥QR. Show that PM^{2} = QM . MR (AS_{2})

Solution:

∆PRM and ∆PQM are right angled triangles.

∴ PR^{2} = PM^{2} + RM^{2} ……………. (1)

and PQ^{2} = PM^{2} + QM^{2} …………… (2)

Adding (1) & (2), we get

PR^{2} + PQ^{2} = PM^{2} + RM^{2} + PM^{2} + QM^{2}

⇒ QR^{2} = 2PM^{2} + RM^{2} + QM^{2}

(∵ In ∆PQR, ∠P = 90° ∴ QR^{2} = PR^{2} + PQ^{2})

⇒ (QM + MR)^{2} = 2PM^{2} + RM^{2} + QM^{2}

⇒ QM^{2} + RM^{2} + 2QM. MR = 2PM^{2} + RM^{2} + QM^{2}

⇒ 2QM. MR = 2PM^{2}

⇒ PM^{2} = QM. MR

Question 5.

ABD is a triangle right angled at A and AC ⊥ BD. Show that (AS_{2})

(i) AB^{2} = BC. BD

(ii) AC^{2} = BC . DC

(iii) AD^{2} = BD . CD

Solution:

(i) In ∆^{les} BAD and BCA,

∠BAD = ∠BCA = 90°

∠B is common.

∴ ∆BAD ~ ∆BCA

Hence, their corresponding sides are in proportion.

⇒ \(\frac{\mathrm{BD}}{\mathrm{AB}}\) = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)

⇒ AB^{2} = BC . BD ……………. (1)

(ii) In ∆^{les} ACB and DCA,

∠ACB = ∠DCA = 90°

∠A is common.

∴ ∆ ACB ~ ∆DCA

Hence, their corresponding sides are in pro-portion.

⇒ \(\frac{\mathrm{AC}}{\mathrm{DC}}\) = \(\frac{\mathrm{BC}}{\mathrm{AC}}\)

⇒ AC^{2} = BC . DC ……………. (2)

iii) In ∆^{les} DAB and DCA,

∠DAB = ∠DCA = 90°

∠D is common.

∴ ∆DAB ~ ∆DCA

Hence, their corresponding sides are in proportion.

⇒ \(\frac{\mathrm{BD}}{\mathrm{AD}}\) = \(\frac{\mathrm{AD}}{\mathrm{CD}}\)

⇒ AD^{2} = BD . CD ……………. (3)

Question 6.

ABC is an isosceles triangle right angled at C. Prove that AB^{2} = 2AC^{2}. (AS_{2})

Solution:

In ∆ABC, ∠ACB = 90°

∴ AB is the hypotenuse, since it is an isosceles triangle, AC = BC

By Pythagoras Theorem,

AB^{2} = AC^{2} + BC^{2}

= AC^{2} + (AC)^{2} (∵ BC = AC)

= AC^{2} + AC^{2} = 2AC^{2}

Question 7.

‘O’ is any point in the interior of a triangle ABC. OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that

i) OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2} = AF^{2} + BD^{2} + CE^{2}

ii) AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}. (AS_{2})

Solution:

In ∆ABC, OD, OE and OF are drawn perpendicular to BC, CA and AB respectively. Join OB, OC and OA.

i) In ∆AFO, ∠AFO = 90°

∴ By Pythagoras Theorem,

OA^{2} = AF^{2} + OF^{2} …………….. (1)

In ABDO, ∠BDO = 90°

∴ OB^{2} = OD^{2} + BD^{2} ………………. (2)

In ACEO, ∠CEO = 90°

∴ OC^{2} = OE^{2} + CE^{2} ……………….. (3)

Adding (1), (2) & (3), we get

OA^{2} + OB^{2} + OC^{2} = AF^{2} + OF^{2} + OD^{2} + BD^{2} + OE^{2} + CE^{2}

⇒ OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2}

= AF^{2} + BD^{2} + CE^{2} …………….. (4)

ii) From (4), we get

AF^{2} + BD^{2} + CE^{2} = (OA^{2} – OE^{2}) + (OB^{2} – OF^{2}) + (OC^{2} – OD^{2})

= AE^{2} + BF^{2} + CD^{2} (∵ In the right triangle AEO,

∠AEO = 90° ∴ OA^{2} = OE^{2} + AE^{2}

⇒ OA^{2} – OE^{2} = AE^{2}

Similarly, in the right triangle OFB,

∠OFB = 90°; ∴ OB^{2} = OF^{2} + BF^{2}

⇒ OB^{2} – OF^{2} = BF^{2}

Lastly, in the right triangle ODC,

∠ODC = 90°,

OC^{2} = OD^{2} + CD^{2}

⇒ O^{2} – OD^{2} = CD^{2}

Question 8.

A wire attached to a vertical pole of height 18m is 24m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut ? (AS_{4})

Solution:

In triangle ABC, ∠B = 90°

AB stands for the vertical pole AC represents the wire ‘B’ denotes ‘base’ and C denotes ‘stake’.

By Pythagoras Theorem,

AC^{2} = AB^{2} + BC^{2}

⇒ BC^{2} = AC^{2} – AB^{2}

= (24)^{2} – (18)^{2}

= 576 – 324 = 252

∴ BC = \(\sqrt{252}\)

= \(\sqrt{252}\) = \(\sqrt{2 2 3 3 7}\)

= \(\sqrt{2^2 \times 3^2}\) × 7

= 2 × 3 × \(\sqrt{7}\) = 6\(\sqrt{7}\) Therefore, the stake should be driven 6 \(\sqrt{7}\) meters for the base of the pole so that the wire will be taut.

Question 9.

Two poles of heights 6 m and 11 m stand on a plane ground. If the distance be¬tween the feet of the poles is 12m find the distance between their tops. (AS_{4})

Solution:

AB and CD represent two poles of heights 11m and 6m respectively. BC represents the distance between their feet. BC = 12m

Draw DE || CB intersecting AB at E. Join AD.

ED = BC = 12m.

EB = DC = 6m

AE = AB – CD = 11 – 6 = 5m

In ∆AED, ∠AED = 90°

∴ By Pythagoras theorem,

AD^{2} = AE^{2} + DE^{2}

= 5^{2} + 12^{2} = 25 + 144 = 169

∴ AD = \(\sqrt{169}\) = 13m

∴ The distance between the tops of the poles = 13 meters.

Question 10.

In an equilateral triangle ABC, D is a point on side BC such that BD = \(\frac{1}{3}\) BC. Prove that 9AD^{2} = 7 AB^{2}. (AS_{2})

Solution:

ABC is an equilateral triangle.

∴ AB = BC = CA

D is a point on BC, such that

BD = \(\frac{1}{3}\) BC; Draw AE ⊥ BC.

∴ E bisects BC.

⇒ BE = CE

In ∆ABD, ∠ADB is obtuse.

∴ AB^{2} = AD^{2} + BD^{2} + 2 BD . DE

⇒ AD^{2} = AB^{2} – BD^{2} – 2BD. DE

Question 11.

In the given figure, ABC Is a triangle right angled at B. D and E are points on BC trisect it. Prove that 8AE^{2} = 3AC^{2} + 5AD^{2}. (AS_{2})

Solution:

In ∆ADC, ∠ADC is obtuse.

∠B = 90° and BD = DE = EC.

Let BD = DE = EC = x

AC^{2} = AD^{2} + DC^{2} + 2 DC. DB

= AD^{2} + (2x)^{2} + 2. 2x. x

= AD^{2} + 4x^{2} + 4x^{2}

= AD^{2} + 8x^{2} ………………… (1)

In ∆ABD, ∠B = 90°

∴ AD^{2} = AB^{2} + BD^{2} (By Pythagoras Theorem)

⇒ AB^{2} = AD^{2} – BD^{2} = AD^{2} – x^{2} …………. (2)

In ∆ABE = ∠B = 90°

∴ AE^{2} = AB^{2} + BE^{2} = AB^{2} + (2x)^{2} = AB^{2} + 4x^{2}

∴ AB^{2} = AE^{2} – 4x^{2} ……………. (3)

From (2) & (3), we get

AE^{2} – 4x^{2} = AD^{2} – x^{2}

AE^{2} – 4x^{2} = AD^{2} – x^{2} + 4x^{2} = AD^{2} + 3x^{2}

8AE^{2} = 8(AD^{2} + 3x^{2})

= 8AD^{2} + 24x^{2} ………………. (4)

From (1),

3AC^{2} + 5 AD^{2}

= 3 (AD^{2} + 8x^{2}) + 5 AD^{2}

= 3 AD^{2} + 24x^{2} + 5AD^{2}

= 8AD^{2} + 24x^{2} ……………….. (5)

From (4) & (5), we have,

8AE^{2} = 3AC^{2} + 5 AD^{2}

Question 12.

ABC is an isosceles triangle right angled at B. Similar triangles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of ∆ABE and ∆ACD.

Solution:

In ∆ABC, ∠B = 90°

Let AB = BC = x

∴ By Pythagoras theorem,

AC^{2} = AB^{2} + BC^{2} = x^{2} + x^{2} = 2x^{2}

∴ AC = \(\sqrt{2 \mathrm{x}^2}\) = \(\sqrt{2}\)x

By problem ∆ACD ~ ∆ABE,

Therefore, \(\frac{\text { area of }(\triangle \mathrm{ABE})}{\text { area of }(\triangle \mathrm{ACD})}\) = \(\frac{\mathrm{AB}^2}{\mathrm{AC}^2}\)

= \(\frac{(x)^2}{(\sqrt{2} x)^2}\) (∵ AB = x and AC = \(\sqrt{2}\)x )

= \(\frac{x^2}{2 x^2}\) = \(\frac{1}{2}\)

Hence, the ratio between the areas of ∆ABE and ∆ACD = 1 : 2