# Triangles : Exercise - 6.5 (Mathematics NCERT Class 10th)

Q.1    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
Sol.

(i)      Let a = 7 cm, b = 24 cm and c = 25 cm
Here the larger side is c = 25 cm
We have, ${a^2} + {b^2} = {7^2} + {24^2} = 49 + 576 = 625 = {c^2}$
So, the triangle with the given sides is a right triangle. Its hypotenuse = 25 cm.
(ii)     Let a = 3 cm, b = 8 cm and c = 6 cm
Here the larger side is b = 8 cm
We have, ${a^2} + {c^2} = {3^2} + {6^2} = 9 + 36 = 45 \ne {b^2}$
So, the triangle with the given sides is not a right triangle.
(iii)    Let a = 50 cm, b = 80 cm and c = 100 cm
Here the larger side is c = 100 cm
We have, ${a^2} + {b^2} = {50^2} + {80^2} = 2500 + 6400$ = 8900 $\ne$ ${c^2}$
So, the triangle with the given sides is not a right triangle.
(iv)     Let a = 13 cm, b = 12 cm and c = 5 cm
Here the larger side is a = 13 cm
We have, ${b^2} + {c^2} = {12^2} + {5^2} = 144 + 25 = 169 = {a^2}$
So, the triangle with the given sides is a right triangle. Its hypotenuse = 13 cm.

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

Given : PQR is a triangle right angled at P and $PM \bot QR$.
To prove : $P{M^2} = QM.MR$
Proof : Since $PM \bot QR$ Therefore $\Delta PQM \sim \Delta PRM$
$\Rightarrow$ ${{PM} \over {QM}} = {{MR} \over {PM}}$
$\Rightarrow$ $P{M^2} = QM.MR$

Q.3    In figure, ABD is a triangle right angled at A and $AC \bot BD$. SHow that
(i) $A{B^2} = BC.BD$
(ii) $A{C^2} = BC.DC$
(iii) $A{D^2} = BD.CD$ Sol.

Given : ABD is a triangle right angled at A and $AC \bot BD$.
To prove :
(i) $A{B^2} = BC.BD$

(ii) $A{C^2} = BC.DC$
(iii) $A{D^2} = BD.CD$
Proof :
(i) Since $AC \bot BD$

Therefore $\Delta ABC \sim \Delta ADC$ and each triangle is similar to $\Delta ABD$
Because $\Delta ABC \sim \Delta ABD$
Therefore ${{AB} \over {BD}} = {{BC} \over {AB}}$ $\Rightarrow$ $A{B^2} = BC.BD$
(ii) Since, $\Delta ABC \sim \Delta ADC$
Therefore ${{AC} \over {BC}} = {{DC} \over {AC}}$ $\Rightarrow$ $A{C^2} = BC.DC$
(iii) Since, $\Delta ACD \sim \Delta ABD$
${{AD} \over {CD}} = {{BD} \over {AD}}$ $\Rightarrow$ $A{D^2} = BD.CD$

Q.4     ABC is an isosceles triangle right angled at C. Prove that $A{B^2} = 2A{C^2}$.
Sol.

Since, ABC is an isosceles right triangle, right angled at C. Therefore $A{B^2} = A{C^2} + B{C^2}$
$\Rightarrow$ $A{B^2} = A{C^2} + A{C^2}$ [Because BC = AC, given]
$\Rightarrow$ $A{B^2} = 2A{C^2}$

Q.5     ABC is an isosceles triangle with AC = BC. If $A{B^2} = 2A{C^2}$, prove that ABC is a right triangle.
Sol.

Since, ABC is an isosceles triangle with AC = BC and $A{B^2} = 2A{C^2}$
$\Rightarrow$ $A{B^2} = A{C^2} + A{C^2}$
$\Rightarrow$ $A{B^2} = A{C^2} + B{C^2}$ [Because AC = BC, given]
Therefore $\Delta$ABC is right angled at AC.

Q.6      ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Sol.

Let ABC be an equilateral triangle of side 2a units.
Draw $AD \bot BC$. Then, D is the mid-point of BC. $\Rightarrow$ $BD = {1 \over 2}BC = {1 \over 2} \times 2a = a$
Since, ABD is a right triangle, right angled at D.
Therefore $A{B^2} = A{D^2} + B{D^2}$
$\Rightarrow$ ${\left( {2a} \right)^2} = A{D^2} + {\left( a \right)^2}$
$\Rightarrow$ $A{D^2} = 4{a^2} - {a^2} = 3{a^2}$
Therefore each of its altitude = $\sqrt 3 a$.

Q.7      Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Sol.

Let the diagonals AC and BD of rhombus ABCD interesect each other at O. Since the diagonals of a rhombus bisect each other at right angles. Therefore $\angle AOB = \angle BOC = \angle COD$ = $\angle DOA = 90^\circ$
and, AO = CO, BO = OD
Since, AOB is a right triangle, right angled at O.
Therefore $A{B^2} = O{A^2} + O{B^2}$
$\Rightarrow$ $A{B^2} = {\left( {{1 \over 2}AC} \right)^2} + {\left( {{1 \over 2}BD} \right)^2}$ [Because OA = OC and OB = OD]
$\Rightarrow$ $4A{B^2} = A{C^2} + B{D^2}$ ................ (1)
Siumilarly, we have
$4B{C^2} = A{C^2} + B{D^2}$ ............... (2)
$4C{D^2} = A{C^2} + B{D^2}$ ............... (3)
and $4A{D^2} = A{C^2} + B{D^2}$ ................(4)
Adding all these results, we get
$4{\left( {A{B^2} + B{C^2} + C{D^2} + AD} \right)^2} = 4\left( {A{C^2} + B{D^2}} \right)$
$\Rightarrow$ $A{B^2} + B{C^2} + C{D^2} + D{A^2} = A{C^2} + B{D^2}$

Q.8     In figure, O is a point in the interior of a triangle ABC, $OD \bot BC$, $OE \bot AC$ and $OF \bot AB$. Show that
(i) $O{A^2} + O{B^2} + O{C^2} - O{D^2} - O{E^2} - O{F^2} = A{F^2} + B{D^2} + C{E^2}$

(ii) $A{F^2} + B{D^2} + C{E^2} = A{E^2} + C{D^2} + B{F^2}$ Sol.

Join AO, BO and CO
(i)    In right $\Delta s$ OFA, ODB and OEC, we have
$O{A^2} = A{F^2} + O{F^2}$
$O{B^2} = B{D^2} + O{D^2}$
and, $O{C^2} = C{E^2} + O{E^2}$ Adding all these, we get
$O{A^2} + O{B^2} + O{C^2} = A{F^2} + B{D^2} + C{E^2} + O{F^2} + O{D^2} + O{E^2}$
$\Rightarrow$ $O{A^2} + O{B^2} + O{C^2} - O{D^2} - O{E^2} - O{F^2}$

= $A{F^2} + B{D^2} + C{E^2}$
(ii)      In right $\Delta s$ ODB and ODC, we have
$O{B^2} = O{D^2} + B{D^2}$
and, $O{C^2} = O{D^2} + C{D^2}$
$\Rightarrow$ $O{B^2} - O{C^2} = B{D^2} - C{D^2}$ ............... (1)
Suimilarly, we have
$O{C^2} - O{A^2} = C{E^2} - A{E^2}$ ................ (2)
and, $O{A^2} - O{B^2} = A{F^2} - B{F^2}$ ................. (3)
Adding equations (1), (2) and (3), we get
$\left( {O{B^2} - O{C^2}} \right) + \left( {O{C^2} - O{A^2}} \right) + \left( {O{A^2} - O{B^2}}\right)$

= $\left( {B{D^2} - C{D^2}} \right) + \left( {C{E^2} - A{E^2}} \right) + \left( {A{F^2} - B{F^2}}\right)$
$\Rightarrow$ $\left( {B{D^2} + C{E^2} + A{F^2}} \right) - \left( {A{E^2} + C{D^2} + B{F^2}}\right)$ = 0
$\Rightarrow$ $A{F^2} + B{D^2} + C{E^2}$
= $A{E^2} + B{F^2} + C{D^2}$.

Q.9    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.
Sol.

Let AB be the ladder, B be the window and CB be the wall. Then, ABC is a right triangle, right angled at C.
Therefore $A{B^2} = A{C^2} + B{C^2}$ $\Rightarrow$ ${10^2} = A{C^2} + {8^2}$
$\Rightarrow$ $A{C^2} = 100 - 64$
$\Rightarrow$ $A{C^2} = 36$
$\Rightarrow$ $AC = 6$
Hence, the foot of the ladder is at a distance 6 m from the base of the wall.

Q.10    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 ?
Sol.

Let AB (= 24 m) be a guy wire attached to a vertical pole. BC of height 18 m. To keep the wire taut, let it be fixed to a stake at A. Then, ABC is a right triangle, right angled at C.
Because $A{B^2} = A{C^2} + B{C^2}$ $\Rightarrow$ ${24^2} = A{C^2} + {18^2}$
$\Rightarrow$ $A{C^2} = 576 - 324$
$\Rightarrow$ $A{C^2} = 252$
$\Rightarrow$ $AC = \sqrt {252} = 6\sqrt 7$
Hence, the stake may be placed at distance of $6\sqrt 7$ m from the base of the pole.

Q.11    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 1½ hours ?
Sol.

Let the first aeroplane starts from O and goes upto A towards north where $OA = \left( {1000 \times {3\over 2}} \right)$ km = 1500 km. Let the second aeroplane starts from O at the same time and goes upto B towards west where $OB = \left({1200 \times {3 \over 2}} \right)$ km = 1800 km.
According to the problem the required distane = BA.
In right angled $\Delta$ABC, by Pythagoras theorem, we have,
$A{B^2} = O{A^2} + O{B^2}$
= ${\left( {1500} \right)^2} + {\left( {1800} \right)^2}$
= $2250000 + 3240000$
= $5490000 = 9 \times 61 \times 100 \times 100$
$\Rightarrow$ $AB = 3 \times 100\sqrt {61} = 300\sqrt {61}$ km.

Q.12   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.
Sol.

Let AB = 11 m and CD = 6 m be the two poles such that BD = 12 m.
Draw $CE \bot AB$ and join AC Therefore CE = DB = 12 m,
AE = AB – BE = AB – CD = (11–6) m = 5 m.
In right angled $\Delta ACE$, by Pythagoras theorem, we have
$A{C^2} = C{E^2} + A{E^2}$
= ${\left( {12} \right)^2} + {\left( 5 \right)^2}$
= $144 + 25 = 169$
$\Rightarrow$ $AC = \sqrt {169} = 13$
Hence, the distance between the tops of the two poles is 13 m.

Q.13    D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that $A{E^2} + B{D^2} = A{B^2} + D{E^2}$.
Sol.

In right angled $\Delta s$ ACE and DCB, we have
$A{E^2} = A{C^2} + C{E^2}$
and, $B{D^2} = D{C^2} + B{C^2}$
$\Rightarrow$ $A{E^2} + B{D^2} = \left( {A{C^2} + C{E^2}} \right) + \left( {D{C^2} + B{C^2}}\right)$
$\Rightarrow$ $A{E^2} + B{D^2} = \left( {A{C^2} + B{C^2}} \right) + \left( {D{C^2} + C{E^2}}\right)$
$\Rightarrow$ $A{E^2} + B{D^2} = A{B^2} + D{E^2}$
[By Pythagoras theorem, $A{C^2} + B{C^2} = A{B^2}$ and $D{C^2} + C{E^2} = D{E^2}$] Q.14   The perpendicular from A on side BC of a $\Delta ABC$ intersects BC at D such that DB = 3 CD (see figure). Prove that $2A{B^2} = 2A{C^2} + B{C^2}$. Sol.

We have, DB = 3CD
Now, BC = DB + CD
$\Rightarrow$ BC = 3CD + CD [Because BD = 3CD]
$\Rightarrow$ BC = 4CD
Therefore $CD = {1 \over 4}BC$ and,
$DB = 3CD = {3 \over 4}BC$ ............. (1)
Since, $\Delta ABD$ is a right triangle, right angled at D. Therefore, by Pythagoras theorem, we have
$A{B^2} = A{D^2} + D{B^2}$ ............... (1)
Similarly, from $\Delta ACD$, we have
$A{C^2} = A{D^2} + C{D^2}$ ........... (2)
(1) – (2) gives,
$A{B^2} - A{C^2} = D{B^2} - C{D^2}$
$\Rightarrow$ $A{B^2} - A{C^2} = {\left( {{3 \over 4}BC} \right)^2} - {\left( {{1 \over 4}BC}\right)^2}$ [Using (1)]
$\Rightarrow$ $A{B^2} - A{C^2} = \left( {{9 \over {16}} - {1 \over {16}}} \right)B{C^2}$
$\Rightarrow$ $A{B^2} - A{C^2} = {1 \over 2}B{C^2}$
$\Rightarrow$ $2A{B^2} - 2A{C^2} = B{C^2}$
$\Rightarrow$ $2A{B^2} = 2A{C^2} + B{C^2}$
Which is the required result.

Q.15   In an equilateral triangle ABC, D is a point on side BC such that $BD = {1 \over 3}BC$. Prove that $9A{D^2} = 7A{B^2}$
Sol.

Let ABC be an equilateral triangle and let D be a point on BC such that
$BD = {1 \over 3}BC$
Draw $AE \bot BC$. Join AD.
In $\Delta s$ AEB and AEC, we have AB = AC [Because $\Delta ABC$ is equilateral]
$\angle AEB = \angle AEC$ [Because each = 90°]
and, AE = AE
Therefore by SAS-criterion of similarity, we have
$\Delta AEB \sim \Delta AEC$
$\Rightarrow$ BE = EC
Thus, we have
$BD = {1 \over 3}BC,DC = {2 \over 3}BC$
and, $BE = EC = {1 \over 2}BC$ ................. (1)
Since, $\angle C = 60^\circ$
Therefore $\Delta ADC$ is an acute triangle.
Therefore $A{D^2} = A{C^2} + D{C^2} - 2DC \times EC$
= $A{C^2} + {\left( {{2 \over 3}BC} \right)^2} - 2 \times {2 \over 3}BC \times {1 \over 2}BC$ [Using (1)]
= $A{C^2} + {4 \over 9}B{C^2} - {2 \over 3}B{C^2}$
= $A{B^2} + {4 \over 9}A{B^2} - {2 \over 3}A{B^2}$ [Because AB = BC = AC]
= ${{\left( {9 + 4 - 6} \right)A{B^2}} \over 9} = {7 \over 9}A{B^2}$
$\Rightarrow$ $9A{D^2} = 7A{B^2}$, which is the required result.

Q.16    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.
Sol.

Let ABC be an equilateral triangle and let $AD \bot BC$.
In $\Delta ADB$ and $\Delta ADC$, we have
AB = AC [Given]
$\angle B = \angle C$ [Each = 60°]
and, $\angle ADB = \angle ADC$ [Each = 90°]
By RHS criterion of congruence, we have
$\Delta ADB \cong \Delta ADC$
$\Rightarrow$ $BD = DC$
$\Rightarrow$ $BD = DC = {1 \over 2}BC$
Since $\Delta ADB$ is a right triangle, right angled at D, by Pythagoras theorem, we have $A{B^2} = A{D^2} + B{D^2}$
$\Rightarrow$ $A{B^2} = A{D^2} + {\left( {{1 \over 2}BC} \right)^2}$
$\Rightarrow$ $A{B^2} = A{D^2} + {1 \over 4}B{C^2}$
$\Rightarrow$ $A{B^2} = A{D^2} + {{A{B^2}} \over 4}$ [Because BC = AB]
$\Rightarrow$ ${3 \over 4}A{B^2} = A{D^2}$
$\Rightarrow$ $3A{B^2} = 4A{D^2}$
Hence, the result.

Q.17    Tick the correct answer and justify : In $\Delta ABC$, $AB = 6\sqrt 3$ cm, AC = 12 cm and BC = 6 cm. The angles A and B are respectively :
(a) 90° and 30°
(b) 90° and 60°
(c) 30° and 90°
(d) 60° and 90°
Sol.

In $\Delta ABC$, we have AB = $6\sqrt 3$ cm,
AC = 12 cm
and BC = 6 cm ............... (1)
Now, $A{B^2} + B{C^2} = {\left( {6\sqrt 3 } \right)^2} + {\left( 6 \right)^2}$
= 36 × 3 + 36 = 108 + 36
= $144 = {\left( {AC} \right)^2}$
Thus, $\Delta ABC$ is a right triangle, right angled at B.
Therefore $\angle B = 90^\circ$
Let D be the mid-point of AC. We know that the mid-point of the hypotenuse of a right triangle is equidistant from the vertices.
$\Rightarrow$ CD = BD = 6 cm [Because $CD = {1 \over 2}AC$]
Also BC = 6 cm
Therefore in $\Delta BDC$, we have
BD = CD = BC
$\Rightarrow$ $\Delta BDC$ is equilateral $\Rightarrow$ $\angle ACB = 60^\circ$
Therefore $\angle A = 180^\circ - \left( {\angle B + \angle C} \right)$
= 180° – (90° + 60°) = 30°
Thus, $\angle A = 30^\circ$ and $\angle B = 90^\circ$
Therefore (C) is the correct choice.