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Learn Exercise 6.5 with Free Lessons & Tips

Question 1:

A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

Solution 1:

Let OA be the wall and AB be the ladder.

Therefore, by Pythagoras theorem,

$AB^{2}=OA^{2}+BO^{2}$

$(10m^{2})=8m^{2}+OB^{2}$

OB=6m

Therefore, the distance of the foot of the ladder from the base of the wall is 6 m.

Comments

Tasneem

Question 2:

A guy wire attached to a vertical pole of height 18 m is 24 m 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?

Solution 2:

Let OB be the pole and AB be the wire.

By Pythagoras theorem,

$AB^{2}=OB^{2}+OA^{2}$

$(24 m)^{2}=(18m)^{2}+OA^{2}$

$OA=\sqrt{252}m=\sqrt{6\times 6\times 7}m=6\sqrt{7}m$

Therefore, the distance from the base is m.

Comments

Tasneem

Question 3:

An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after $1\tfrac{1}{2}$ hours?

Solution 3:

Distance travelled by the plane flying towards north in

Similarly, distance travelled by the plane flying towards west in

Let these distances be represented by OA and OB respectively.

Applying Pythagoras theorem,

Distance between these planes after, AB $=\sqrt{OA^{2}+OB^{2}}$

$\left ( \sqrt{(1500)^{2}+(1800)^{2}} \right ) km=\left ( \sqrt{2250000+3240000} \right )km$ = km

Therefore, the distance between these planes will be km after.

Comments

Tasneem

Question 4:

Tick the correct answer and justify :

In $\triangle ABC$ , AB = $6\sqrt{3}$ cm, AC = 12 cm and BC = 6 cm. The angle B is :

(A) $120^{\circ}$ (B) $60^{\circ}$ (C) $90^{\circ}$ (D) $45^{\circ}$

Solution 4:

Given : AB = 6√3cm

AC= 12 cm

BC=6 cm

Ans : check if (AC)²=(BC)²+(AB)²

12²=(6)²+(6√3)²

144= 36+108

144=144

Angle B is 90º

Option (C) is correct

Comments

Nitin Pathak

Question 5:

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

(i) 7 cm, 24 cm, 25 cm

(ii) 3 cm, 8 cm, 6 cm

(iii) 50 cm, 80 cm, 100 cm

(iv) 13 cm, 12 cm, 5 cm

Solution 5:

(i) The sides of the triangle are 7 cm, 24 cm, and 25 cm.

Squaring the lengths of these sides, we will obtain 49, 576, and 625.

49 + 576 = 625

Or,

Satisfies the pythagoras theorem

It is a right triangle.

The length of the hypotenuse is is 25 cm.

(ii) The sides of the triangle are 3 cm, 8 cm, and 6 cm.

Squaring the lengths of these sides, we will obtain 9, 64, and 36.

However, 9 + 36 ≠ 64

Or, 3² + 6² ≠ 8²

The sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.

Therefore, the given triangle is not satisfying Pythagoras theorem.

Hence, it is not a right triangle.

(iii)The sides are 50 cm, 80 cm, and 100 cm.

Squaring the lengths of these sides, we will obtain 2500, 6400, and 10000.

However, 2500 + 6400 ≠ 10000

Or, 50² + 80² ≠ 100²

The sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.

Therefore, the given triangle is not satisfying Pythagoras theorem.

Hence, it is not a right triangle.

(iv)Thesides are 13 cm, 12 cm, and 5 cm.

Squaring the lengths of these sides, we will obtain 169, 144, and 25.

144 +25 = 169

Or,

The sides of the given triangle are satisfying Pythagoras theorem.

Therefore, it is a right triangle.

The length of the hypotenuse of this triangle is 13 cm.

Comments

Tasneem

Question 6:

PQR is a triangle right angled at P and M is a point on QR such that $PM\perp QR$ . Show that $PM^{2}=QM.MR.$ .

Solution 6:

Ler $\angle MPR=x$

In $\Delta MPR, \angle MPR=180^{\circ}-90^{\circ}-x$

$\Delta MPR, \angle MPR=90^{\circ}-x$

Similarly, in $\Delta MPQ,$

$\angle MPQ=90^{\circ}-\angle MPR=90^{\circ}-x$

$\angle MQP=180^{\circ}-90^{\circ}-(90^{\circ}-x)$

$\angle MQP=x$

In $\Delta QMP and \Delta PMR$ ,

$\angle MPQ=\angle MPR$

$\angle PMQ=\angle RPM$

$\angle MQP=\angle MPR$

$\therefore \Delta QMP\sim \Delta PMR$ ( By AAA similarity)

$\Rightarrow \frac{QM}{PM}=\frac{MP}{MR}$

Therefore, $PM^{2}=QM\times MR$

Comments

Tasneem

Question 7:

In the figure, ABD is a triangle right angled at A and $AC\perp BD$ .

Show that:

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

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

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

Solution 7:

$(i) In \Delta ADB and \Delta CAB,$

$\angle DAB=\angle ACB$ (Each 90 degress)

$\angle ABD=\angle CBA$ (Common angle)

$\therefore \Delta ADB\sim \Delta CAB$ (AA similarity criteria)

$\Rightarrow \frac{AB}{CB}=\frac{BD}{AB}$

$AB^{2}=CB\times BD$

$(ii) Let \angle CAB=x$

$In \Delta CBA, \angle CBA=180^{\circ}-90^{\circ}-x$

$\angle CBA=90^{\circ}-x$

$Similarly, in \Delta CAD,$

$\angle CAD=90^{\circ}-\angle CAB=90^{\circ}-x$

$\angle CDA=180^{\circ}-90^{\circ}-(90^{\circ}-x)$

$\angle CDA=x$

$In \Delta CBA and \angle CAD$

$\angle CBA=\angle CAD$

$\angle CAB=\angle CDA$

$\angle ACB=\angle DCA$ (Each 90 degress)

$\therefore \Delta CBA\sim \Delta CAD$ ( By AAA similarity)

$\Rightarrow \frac{AC}{DC}=\frac{BC}{AC}$

$AC^{2}=DC\times BC$

$(iii) In \Delta DCA and \Delta DAB,$

∠DCA = ∠ DAB (Each 90º)

∠CDA = ∠ ADB (Common angle)

$\therefore \Delta DCA\sim \Delta DAB$ ( By AAA similarity)

$\Rightarrow \frac{DC}{DA}=\frac{DA}{DB}$

$AD^{2}=BD\times CD$

Comments

Tasneem

Question 8:

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

Solution 8:

Given that ΔABC is an isosceles triangle.

∴ AC = CB

Applying Pythagoras theorem in ΔABC (i.e., right-angled at point C), we obtain

$AC^{2}+CB^{2}=AB^{2}$

$AC^{2}+AC^{2}=AB^{2}$ (AC=CB)

$\Rightarrow 2AC^{2}=AB^{2}$

Comments

Tasneem

Question 9:

ABC is an isosceles triangle with AC = BC. If $AB^{2}=2AC^{2}$ , prove that ABC is a right triangle.

Solution 9:

$\Rightarrow AB^{2}=2AC^{2}$

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

$AB^{2}=AC^{2}+BC^{2}$ (As AC =BC)

The triangle satisfies Pythagoras theorem

Hence it is a right angled triangle

Comments

Tasneem

Question 10:

ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Solution 10:

Let AD be the altitude in the given equilateral triangle, ΔABC.

We know that altitude bisects the opposite side.

∴ BD = DC = a

In $\Delta ADB, \angle ADB=90^{\circ}$

Applying pythagoras theorem, we get

$AD^{2}+DB^{2}=AB^{2}$

$AD^{2}+a^{2}=(2a)^{2}$

$AD=a\sqrt{3}$

In an equilateral triangle, all the altitudes are equal in length.

Therefore, the length of each altitude will be.

Comments

Tasneem

Question 11:

Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Solution 11:

In ΔAOB, ΔBOC, ΔCOD, ΔAOD,

Applying Pythagoras theorem, we obtain

$AB^{2}=AO^{2}+OB^{2}..............(i)$

$CD^{2}=CO^{2}+OD^{2}..............(iii)$

$AD^{2}=AO^{2}+OD^{2}..............(iv)$

On addition of the above equations, we get

$AB^{2}+BC^{2}+CD^{2} +AD^{2}=2(AO^{2}+OB^{2}+OC^{2}+OD^{2})$

$=2\left ( \left ( \frac{AC}{2}^{2}\right ) +\left ( \frac{BD}{2} \right )^{2}+\left ( \frac{AC}{2} \right ) ^{2}+\left ( \frac{BD}{2}\right )^{2} \right )$

(diagonals bisect each other)

$=2\left ( \left ( \frac{AC}{2}^{2}\right ) +\left ( \frac{BD}{2} \right )^{2}\right )$

$=(AC)^{2}+(BD)^{2}$

Comments

Tasneem

Question 12:

In the figure, O is a point in the interior of a triangle ABC, $OD\perp BC,OE\perp AC$ and $OF\perp AB$ .

Show that:

(i) $OA^{2}+OC^{2}-OD^{2}-OE^{2}-OF^{2}=AF^{2}+BD^{2}+CE^{2}$ ,

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

Solution 12:

Join OA, OB, and OC.

(i) Applying Pythagoras theorem in ΔAOF, we obtain

Similarly, in ΔBOD,

Similarly, in ΔCOE,

$OC^{2}=OE^{2}+EC^{2}$

On addition of these equations,

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

$OA^{2}+OB^{2}+OC^{2}$ $-OD^{2}-OE^{2}-OF^{2}$ $=AF^{2}+BD^{2}+EC^{2}$

(ii) From the above result,

$AF^{2}+BD^{2}+EC^{2}(OA^{2}-OE^{2})+(OC^{2}-OD^{2})+(OB^{2}-OF^{2})$

$\therefore$ AF² + BD² + CE²= AE² + CD² + BF²

Comments

Tasneem

Question 13:

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

Solution 13:

Given : distance btween two poles = 12 m

Small pole height = 6 m

Big pole height = 11 m

To find : distnace btween two poles

Small to top pole distance can be find by pythgores therom top (hypotenuse)²=(base)²+(height)²

Base =12 m

Height = 11-6= 5 m

(Hypotenuse)²=(12)²+(5)²

Hypotenuse =√(12)²+(5)²m

Hypotenuse=√144+25= √169m

distance between their tops is13 m.

Comments

Nitin Pathak

Question 14:

D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that $AE^{2}+BD^{2}=AB^{2}+DE^{2}$ .

Solution 14:

$AC^{2}+CE^{2}=AE^{2}$ $............(i)$

On applying the Pythagoras theorem in $\Delta BCD$

$BC^{2}+CD^{2}=BD^{2}.............(ii)$

Using (i) and (ii) we obtain

$AC^{2}+CE^{2}+BC^{2}+CD^{2}=AE^{2}+BD^{2}............(iii)$

Applying the pythagoras theorem in $\Delta CDE$ we get

$DE^{2}=CD^{2}+CE^{2}$

Applying the pythagoras theorem in $\Delta ABC$ we get

$AB^{2}=AC^{2}+CB^{2}$

Placing the values in (iii)

AE²+ BD² = AB² + DE²

Comments

Tasneem

Question 15:

The perpendicular from A on side BC of a â?? ABC intersects BC at D such that DB = 3 CD (see the figure). Prove that $2AB^{2}=2AC^{2}+BC^{2}$ .

Solution 15:

Applying Pythagoras theorem for ΔACD, we obtain

$AC^{2}=AD^{2}+DC^{2}$

$AD^{2}=AC^{2}-DC^{2}...........(i)$

Applying Pythagoras theorem in ΔABD, we obtain

$AB^{2}=AD^{2}+DB^{2}$

$AD^{2}=AB^{2}-DB^{2}...........(ii)$

From (i) and (ii)

$AC^{2}-DC^{2}=AB^{2}-DB^{2}.........(iii)$

It is given that 3DC=DB

$\therefore DC=\frac{BC}{4}$ and $DB= \frac{3BC}{4}$

Placing these values in (iii)

$AC^{2}-\left ( \frac{BC}{4} \right )^{2}=AB^{2}-\left ( \frac{3BC}{4} \right )^{2}$

$AC^{2}-\frac{BC^{2}}{16}=AB^{2}-\frac{9BC^{2}}{16}$

$16AC^{2}-BC^{2}=16AB^{2}-9BC^{2}$

On solving, it is proved that 2 AB² = 2 AC² + BC²

Comments

Tasneem

Question 16:

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

Solution 16:

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.

∴ BE = EC = =

And, AE =

Given that, BD = BC

∴ BD =

DE = BE − BD =

Applying Pythagoras theorem in ΔADE, we obtain

AD² = AE² + DE²

⇒ 9 AD² = 7 AB²

Comments

Tasneem

Question 17:

In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Solution 17:

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.

∴ BE = EC = =

Applying Pythagoras theorem in ΔABE, we obtain

AB² = AE² + BE²

$a^{2}=AE^{2}+\left ( \frac{a}{2} \right )^{2}$

$AE^{2}=a^{2}-\frac{a^{2}}{4}$

$AE^{2}=3\frac{a^{2}}{4}$

4AE² = 3a²

⇒ 4 × (Square of altitude) = 3 × (Square of one side)

Comments

Tasneem

All Exercises - Chapter 6 - Triangles

Exercise 6.5

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Learn Exercise 6.5 with Free Lessons & Tips

All Exercises - Chapter 6 - Triangles

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