| Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 |
Exercise 5.5
Question 1
Sides of some triangles are given below. Determine which of them are right triangles
(i) 8 cm, 15 cm, 17 cm(ii) (2a —1) cm, $2 \sqrt{2}$ cm and (2a + 1) cm
(iii) 7 cm, 24 cm, 25 cm
(iv) 1.4 cm, 4.8 cm, 5 cm
Sol :
(i) Using Pythagoras theorem, i.e. if the square of the hypotenuse is equal to the sum of the other two sides. Then, the given triangle is a right angled triangle, otherwise not.
Here, (8)2 + (15)2 = 64 + 225 = 289 = (17)2
∴ given sides 8cm, 15cm and 17cm make a right angled triangle.
(ii) Using Pythagoras theorem, i.e. if the square of the hypotenuse is equal to the sum of the other two sides. Then, the given triangle is a right angled triangle, otherwise not.
Here, (2a – 1)2 + (2√(2a))2
⇒ 4a2 + 1 – 4a + 8a
⇒ 4a2 + 1 + 4a
= (2a + 1)2
∴ given sides (2a —1) cm, $2 \sqrt{2 a}$ cm and (2a + 1) cm make a right angled triangle.
(iii) Using Pythagoras theorem, i.e. if the square of the hypotenuse is equal to the sum of the other two sides. Then, the given triangle is a right angled triangle, otherwise not.
Here, (7)2 + (24)2 = 49 + 576 = 625 = (25)2
∴ given sides 7cm, 24cm and 25cm make a right angled triangle.
(iv) Using Pythagoras theorem, i.e. if the square of the hypotenuse is equal to the sum of the other two sides. Then, the given triangle is a right angled triangle, otherwise not.
Here, (1.4)2 + (4.8)2 = 1.96 + 23.04 = 25 = (5)2
∴ given sides 1.4cm, 4.8cm and 5cm make a right angled triangle.
Here, (8)2 + (15)2 = 64 + 225 = 289 = (17)2
∴ given sides 8cm, 15cm and 17cm make a right angled triangle.
(ii) Using Pythagoras theorem, i.e. if the square of the hypotenuse is equal to the sum of the other two sides. Then, the given triangle is a right angled triangle, otherwise not.
Here, (2a – 1)2 + (2√(2a))2
⇒ 4a2 + 1 – 4a + 8a
⇒ 4a2 + 1 + 4a
= (2a + 1)2
∴ given sides (2a —1) cm, $2 \sqrt{2 a}$ cm and (2a + 1) cm make a right angled triangle.
(iii) Using Pythagoras theorem, i.e. if the square of the hypotenuse is equal to the sum of the other two sides. Then, the given triangle is a right angled triangle, otherwise not.
Here, (7)2 + (24)2 = 49 + 576 = 625 = (25)2
∴ given sides 7cm, 24cm and 25cm make a right angled triangle.
(iv) Using Pythagoras theorem, i.e. if the square of the hypotenuse is equal to the sum of the other two sides. Then, the given triangle is a right angled triangle, otherwise not.
Here, (1.4)2 + (4.8)2 = 1.96 + 23.04 = 25 = (5)2
∴ given sides 1.4cm, 4.8cm and 5cm make a right angled triangle.
Question 2
A ladder 26 m long reaches a window 24 m above the ground. Find the distance of the foot of the ladder from the base of the wall.
Sol :
Let AC be the position of a window from the ground and BC be the ladder, then the height of the window, AC =24m and length of the ladder, BC = 26m
Let AB = x m be the distance of the foot of the ladder from the base of the wall.
In ∆CAB, using Pythagoras Theorm,
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AC)2 + (AB)2 = (BC)2
⇒ (24)2 + (AB)2 = (26)2
⇒ (AB)2 = (26)2 – (24)2
⇒ (AB)2 = (26 – 24)(26+24)
[∵ (a2 – b2)=(a+b)(a – b)]
⇒ (AB)2 = (2)(50)
⇒ (AB)2 = 100
⇒ AB = ±10
⇒ AB = 10 [taking positive square root]
Hence, the distance of the foot of the ladder from base of the wall is 10m
Question 3
A man goes 15 m due west and then 8 m due north. How far is he from the starting point?
Sol :
Let AB = 15m and AC = 8m
In ∆CAB, using Pythagoras Theorm,
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AC)2 + (AB)2 = (BC)2
⇒ (8)2 + (15)2 = (BC)2
⇒ (BC)2 = 64 + 225
⇒ (BC)2 = 289
⇒ BC = ±17
⇒ BC = 17 [taking positive square root]
Hence, the man is 17m far from the starting point.
In ∆CAB, using Pythagoras Theorm,
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AC)2 + (AB)2 = (BC)2
⇒ (8)2 + (15)2 = (BC)2
⇒ (BC)2 = 64 + 225
⇒ (BC)2 = 289
⇒ BC = ±17
⇒ BC = 17 [taking positive square root]
Hence, the man is 17m far from the starting point.
Question 4
A ladder 10 m long just reaches the top of a building 8 m high from the ground. Find the distance of the foot of the ladder from the building.
Sol :
Let AC be the top of the building from the ground and BC be the ladder, then the height of the building, AC = 8m and length of the ladder, BC = 10m
Let AB = x m be the distance of the foot of the ladder from the building.
In ∆CAB, using Pythagoras Theorm,
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AC)2 + (AB)2 = (BC)2
⇒ (8)2 + (AB)2 = (10)2
⇒ (AB)2 = (10)2 – (8)2
⇒ (AB)2 = (10 – 8)(10+8)
[∵ (a2 – b2)=(a+b)(a – b)]
⇒ (AB)2 = (2)(18)
⇒ (AB)2 = 36
⇒ AB = ±6
⇒ AB = 6 [taking positive square root]
Hence, the distance of the foot of the ladder from building is 6m
Question 5
Find the length of a diagonal of a rectangle whose adjacent sides are 30 cm and 16 cm.
Sol :
Let ABCD be a rectangle and AB and BC are the adjacent sides of length 30cm and 16cm respectively.
Let AC be the diagonal.
In ∆CBA, using Pythagoras Theorm,
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AB)2 + (BC)2 = (AC)2
⇒ (30)2 + (16)2 = (AC)2
⇒ (AC)2 = 900 + 256
⇒ (AC)2 = 1156
⇒ AB = ±34
⇒ AB = 34 [taking positive square root]
Hence, the length of a diagonal of a rectangle is 34cm
Question 6
A 13 m-long ladder reaches a window of a building 12 m above the ground. Determine the distance of the foot of the ladder from the building.
Sol :
Let AC be the position of a window from the ground and BC be the ladder, then the height of the window, AC =12m and length of the ladder, BC = 13m
Let AB = x m be the distance of the foot of the ladder from the base of the wall.
In ∆CAB, using Pythagoras Theorem,
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AC)2 + (AB)2 = (BC)2
⇒ (12)2 + (AB)2 = (13)2
⇒ (AB)2 = (13)2 – (12)2
⇒ (AB)2 = (13 – 12)(13+12)
[∵ (a2 – b2)=(a+b)(a – b)]
⇒ (AB)2 = (1)(25)
⇒ (AB)2 = 25
⇒ AB = ±5
⇒ AB = 5 [taking positive square root]
Hence, the distance of the foot of the ladder from base of the wall is 5m
Question 7
Two vertical poles of height 9 m and 14 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.
Sol :
Let BC and AD be the two poles of height 14m and 9m respectively. Again, let CD be the distance between tops of the poles.
Then, CE = BC – AD = 14 – 9 = 5m [∵AD =BE]
Also, AB =12m
In ∆CED, using Pythagoras theorem, we get
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (CE)2 + (DE)2 = (CD)2
⇒ (5)2 + (12)2 = (CD)2
⇒ (CD)2 = 25 + 144
⇒ (CD)2 = 169
⇒ CD = √169
⇒ CD = ±13
⇒ CD = 13 [taking positive square root]
Hence, the distance between the tops of the poles is 13m
Question 8
A man goes 10 m due south and then 24 m due west. How far is he from the starting point?
Sol :
Let AB = 10m and AC = 24m
In ∆CAB, using Pythagoras Theorem,
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AC)2 + (AB)2 = (BC)2
⇒ (24)2 + (10)2 = (BC)2
⇒ (BC)2 = 576 + 100
⇒ (BC)2 = 676
⇒ BC = ±26
⇒ BC = 26 [taking positive square root]
Hence, the man is 26m far from the starting point.
Question 9
A man goes 80 m due east and then 150 m due north. How far is he from the starting point?
Sol :
Let AB = 80m and AC = 150m
In ∆CAB, using Pythagoras Theorem,
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AC)2 + (AB)2 = (BC)2
⇒ (150)2 + (80)2 = (BC)2
⇒ (BC)2 = 22500 + 6400
⇒ (BC)2 = 28900
⇒ BC = ±170
⇒ BC = 170 [taking positive square root]
Hence, the man is 170 m far from the starting point.
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AC)2 + (AB)2 = (BC)2
⇒ (150)2 + (80)2 = (BC)2
⇒ (BC)2 = 22500 + 6400
⇒ (BC)2 = 28900
⇒ BC = ±170
⇒ BC = 170 [taking positive square root]
Hence, the man is 170 m far from the starting point.
Question 10
∆ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ΔABC is a right triangle.
Sol :
Given an isosceles triangle ABC with AC = BC, and AB2 = 2AC2
To Prove: ∆ABC is a right triangle
Proof: AB2 = 2AC2 (given)
⇒ AB2 = AC2 +AC2
⇒ AB2 = AC2 +BC2 [∵AC =BC]
⇒ ∆ABC is a right triangle right angled at C.
Question 11
Find the length of each side of a rhombus whose diagonals are 24 cm and 10 cm long.
Sol :
Let ABCD be a rhombus where AC = 10cm and BD =24cm
Let AC and BD intersect each other at O.
Now, we know that diagonals of rhombus bisect each other at 90°
Thus, we have
$\mathrm{AO}=\frac{1}{2} \times \mathrm{AC}$
$\Rightarrow \frac{1}{2} \times 10=5 \mathrm{cm}$
$\mathrm{BO}=\frac{1}{2} \times \mathrm{BD}$
$=\frac{1}{2} \times 24=12 \mathrm{cm}$
Since, AOB is a right angled triangle
So, by Pythagoras theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AO)2 + (BO)2 = (AB)2
⇒ (5)2 + (12)2 = (AB)2
⇒ (AB)2 = 25 + 144
⇒ (AB)2 = 169
⇒ AB = √169
⇒ AB = ±13
⇒ AB = 13 [taking positive square root]
Hence, AB = 13cm
Thus, length of each side of rhombus is 13cm
Since, AOB is a right angled triangle
So, by Pythagoras theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AO)2 + (BO)2 = (AB)2
⇒ (5)2 + (12)2 = (AB)2
⇒ (AB)2 = 25 + 144
⇒ (AB)2 = 169
⇒ AB = √169
⇒ AB = ±13
⇒ AB = 13 [taking positive square root]
Hence, AB = 13cm
Thus, length of each side of rhombus is 13cm
Question 12
ΔABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.
Sol :
Given: ABC is an isosceles triangle right angled at C.
Let AC = BC
In ∆ACB, using Pythagoras theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AC)2 + (BC)2 = (AB)2
⇒ (AC)2 + (AC)2 = (AB)2
[∵ABC is an isosceles triangle, AC =BC]
⇒ 2(AC)2 = (AB)2
Hence Proved
Question 13
ΔABC is an isosceles triangle with AB = AC = 13 cm. The length of altitude from A on BC is 5 cm. Find BC.
Sol :
Given: ΔABC is an isosceles triangle with AB = AC = 13 cm
Suppose the altitude from A on Bc meets BC at M.
∴ M is the midpoint of BC. AM = 5 cm
In ∆AMB, using Pythagoras theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AM)2 + (BM)2 = (AB)2
⇒ (5)2 + (BM)2 = (13)2
⇒ (BM)2 = (13)2 – (5)2
⇒ (BM)2 = (13 – 5)(13+5)
[∵ (a2 – b2) = (a + b)(a – b)]
⇒ (BM)2 = (8)(18)
⇒ (BM)2 = 144
⇒ BM = ±12
⇒ BM = 12 [taking positive square root]
∴ BC = 2BM or 2MC = 2×12 = 24cm
Question 14
In an equilateral triangle ABC, AD is drawn perpendicular to BC, meeting BC in D. Prove that AD2 = 3BD2.
Sol :
Given: ABC is an equilateral triangle
∴ AB = AC = BC
and AD ⊥ BC
Now, In ∆ADB, using Pythagoras theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AD)2 + (BD)2 = (AB)2
⇒ (AD)2 + (BD)2 = (BC)2 [∵ AB = BC]
⇒ (AD)2 + (BD)2 = (2BD)2 [ as AD⊥BC]
⇒ (AD)2 + (BD)2 = 4BD2
⇒ AD2 = 3BD2
Question 15
Find the length of altitude AD of an isosceles Δ ABC in which AB = AC = 2a units and BC = a units.
Sol :
Given: ABC is an isosceles triangle
∴ AB = AC = 2a and BC = a
and AD is the altitude on BC. Therefore, BC = 2BD
Now, In ∆ADB, using Pythagoras theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AD)2 + (BD)2 = (AB)2
$\Rightarrow(\mathrm{AD})^{2}+\left(\frac{\mathrm{a}}{2}\right)^{2}=(2 \mathrm{a})^{2}$
$\Rightarrow(\mathrm{AD})^{2}=(2 \mathrm{a})^{2}-\left(\frac{\mathrm{a}}{2}\right)^{2}$
$\Rightarrow(\mathrm{AD})^{2}=4 \mathrm{a}^{2}-\frac{\mathrm{a}^{2}}{4}$
$\Rightarrow(\mathrm{AD})^{2}=\frac{16 \mathrm{a}^{2}-\mathrm{a}^{2}}{4}$
$\Rightarrow(\mathrm{AD})^{2}=\frac{15 \mathrm{a}^{2}}{4}$
$\Rightarrow A D=\sqrt{\frac{15 a^{2}}{4}}$
$\Rightarrow A D=\frac{\sqrt{15}}{2} a$ [taking positive square root]
Question 16
Δ ABC is an equilateral triangle of side 2a units. Find each of its altitudes.
Sol :
Given: ABC is an equilateral triangle
∴ AB = AC = BC = 2a
And let AD is an altitude on BC.
Therefore, $B D=\frac{1}{2} \times B C=a$
Now, In ∆ADB, using Pythagoras theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AD)2 + (BD)2 = (AB)2
⇒ (AD)2 + (a)2 = (2a)2
⇒ (AD)2 = 4a2 – a2
⇒ (AD)2 = 3a2
⇒ AD = a√3 units
Now, In ∆ADB, using Pythagoras theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AD)2 + (BD)2 = (AB)2
⇒ (AD)2 + (a)2 = (2a)2
⇒ (AD)2 = 4a2 – a2
⇒ (AD)2 = 3a2
⇒ AD = a√3 units
Question 17
Find the height of an equilateral triangle of side 12 cm.
Sol :
Given: ABC is an equilateral triangle
∴ AB = AC = BC = 12cm
And let AD is an altitude on BC.
Therefore, $\mathrm{BD}=\frac{1}{2} \times \mathrm{BC}=6 \mathrm{cm}$
Now, In ∆ADB, using Pythagoras theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AD)2 + (BD)2 = (AB)2
⇒ (AD)2 + (6)2 = (12)2
⇒ (AD)2 = 144 – 36
⇒ (AD)2 = 108
⇒ AD = √108
⇒ AD = 6√3
Hence, the height of an equilateral triangle is 6√3 cm
Now, In ∆ADB, using Pythagoras theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AD)2 + (BD)2 = (AB)2
⇒ (AD)2 + (6)2 = (12)2
⇒ (AD)2 = 144 – 36
⇒ (AD)2 = 108
⇒ AD = √108
⇒ AD = 6√3
Hence, the height of an equilateral triangle is 6√3 cm
Question 18
L and M are the mid-points of AB and BC respectively of ΔABC, right-angled at B. Prove that 4LC2 = AB2 + 4BC2
Sol :
Given: ABC is a right triangle right angled at B
and L and M are the mid-points of AB and BC respectively.
⇒ AL = LB and BM = MC
In ∆LBC, using Pythagoras theorem we have,
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (LB)2 + (BC)2 = (LC)2
$\Rightarrow\left(\frac{A B}{2}\right)^{2}+(B C)^{2}=(L C)^{2}$
⇒ (AB)2 + 4(BC)2 = 4(LC)2
Hence Proved
Question 19
Find the length of the second diagonal of a rhombus, whose side is 5 cm and one of the diagonals is 6 cm.
Sol :
Let ABCD be a rhombus having AD = 5cm and AC = 6cm
Now, we know that diagonals of rhombus bisect each other at 90°
Thus, we have
$A O=\frac{1}{2} \times A C \Rightarrow \frac{1}{2} \times 6=3 \mathrm{cm}$
Since, AOD is a right angled triangle
So, by Pythagoras theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AO)2 + (BO)2 = (AD)2
⇒ (3)2 + (BO)2 = (5)2
⇒ (BO)2 = 25 – 9
⇒ (BO)2 = 16
⇒ BO = √16
⇒ BO = ±4
⇒ BO = 4 [taking positive square root]
Hence, BO = 4cm
⇒ BC = 2BO = 2 × 4 = 8cm
Thus, length of each side of rhombus is 13cm.
So, by Pythagoras theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AO)2 + (BO)2 = (AD)2
⇒ (3)2 + (BO)2 = (5)2
⇒ (BO)2 = 25 – 9
⇒ (BO)2 = 16
⇒ BO = √16
⇒ BO = ±4
⇒ BO = 4 [taking positive square root]
Hence, BO = 4cm
⇒ BC = 2BO = 2 × 4 = 8cm
Thus, length of each side of rhombus is 13cm.
Question 20
In ΔABC, ∠B = 90° and D is the midpoint of BC. Prove that AC2 = AD2 + 3CD2.
Sol :
Given: ∠B = 90° and D is the midpoint of BC .i.e. BD = DC
To Prove: AC2 = AD2 + 3CD2
In ∆ABC, using Pythagoras theorem we have,
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AB)2 + (BC)2 = (AC)2
⇒ (AB)2 + (2CD)2 =(AC)2
⇒ (AB)2 + 4(CD)2 =(AC)2
⇒ (AD2 – BD2) + 4(CD2) = AC2
[∵ In right triangle ∆ABD, AD2 =AB2 + BD2 ]
⇒ AD2 – BD2 + 4CD2 = AC2
⇒ AD2 – CD2 + 4CD2 = AC2
[∵ D is the midpoint of BC, BD = DC]
⇒ AD2 +3CD2 = AC2
or AC2 = AD2 + 3CD2
Hence Proved
Question 21
In ∆ABC, ∠C = 90° and D is the midpoint of BC. Prove that AB2 = 4AD2 — 3AC2.
Sol :
Given: ∠C = 90° and D is the midpoint of BC .i.e. BC = 2CD
To Prove: AB2 = 4AD2 — 3AC2
In ∆ABC, using Pythagoras theorem we have,
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AC)2 + (BC)2 = (AB)2
⇒ (AC)2 + (2CD)2 =(AB)2
⇒ (AC)2 + 4(CD)2 =(AB)2
⇒ (AC)2 + 4(AD2 – AC2) = AB2
[∵ In right triangle ∆ACD, AD2 =AC2 + CD2 ]
⇒ AC2 +4AD2 – 4AC2 = AB2
⇒ 4AD2 – 3AC2 = AB2
or AB2 = 4AD2 — 3AC2
Hence Proved
Question 22
In an isosceles ΔABC, AB = AC and BD ⊥ AC. Prove that BD2 — CD2 = 2CD AD.
Sol :
Given: AB = AC and BD
ACTo Prove: BD2 – CD2 = 2CD × AD
In ∆BDC, using Pythagoras theorem we have,
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (BD)2 + (CD)2 = (BC)2 …(i)
In ∆BDA, using Pythagoras theorem we have,
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (BD)2 + (AD)2 = (AB)2
⇒ (BD)2 + (AD)2 = (AC)2 [∵ AB = AC]
Multiply this eq. by 2, we get
⇒ 2(BD)2 + 2(AD)2 = 2(AC)2 …(ii)
Subtracting Eq. (ii) from (i), we get
⇒ CD2 – BD2 = BC2 – 2 AC2 + 2 AD2
= BC2 – 2 (AD +CD)2 + 2 AD2
= BC2 – 2 CD2 – 4 AD × CD
= BD2 + CD2 – 2 CD2 – 4 AD × CD
= BD2 – CD2 – 4 AD × CD
⇒ CD2 – BD2 –BD2 +CD2 = –4AD × CD
⇒ –2(BD2 – CD2) = –4AD × CD
⇒ BD2 – CD2 = 2CD × AD
Hence Proved
Question 23
In a quadrilateral, ΔBCD, ∠B = 90°. If AD2= AB2 + BC2 + CD2, prove that ∠ACD = 90°.
Sol :
Given: ABCD is a quadrilateral and ∠B = 90°
and AD2= AB2 + BC2 + CD2
To Prove: ∠ACD = 90°
In right triangle ∆ABC, using Pythagoras theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AB)2 + (BC)2 = (AC)2 …(i)
Given: AD2= AB2 + BC2 + CD2
⇒ AD2= AC2 + CD2 [from (i)]
In ∆ACD
AD2= AC2 + CD2
∴ ∠ACD = 90° [converse of Pythagoras theorem]
Hence Proved
Question 24
In a rhombus ABCD, prove that: AB2 + BC2 + CD2 + DA2 = AC2 + BD2
Sol :
In rhombus ABCD, AB = BC = CD = DA
We know that diagonals bisect each other at 90°
And $\mathrm{OA}=\mathrm{OC}=\frac{1}{2} \times \mathrm{AC}$,$\mathrm{OB}=\mathrm{OD}=\frac{1}{2} \times \mathrm{BC}$
Consider right triangle AOB
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (OA)2 + (OB)2 = (AB)2
$\Rightarrow\left(\frac{A C}{2}\right)^{2}+\left(\frac{B D}{2}\right)^{2}=A B^{2}$
⇒ AC2 + BD2 = 4AB2
⇒ AC2 + BD2 = AB2 + AB2 + AB2 + AB2
⇒ AC2 + BD2 = AB2 + BC2 + CD2 + DA2
Hence Proved
Question 25
In an equilateral triangle ABC, AD is the altitude drawn from A on side BC. Prove that 3AB2= 4AD2.
Sol :
Given: ABC is an equilateral triangle
and AD is the altitude on side BC
To Prove: 3AB2= 4AD2
In right triangle ∆ADB, using Pythagoras theorem
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ AD2 + BD2 = AB2
$\Rightarrow A D^{2}+\left(\frac{B C}{2}\right)^{2}=A B^{2}$
⇒ 4AD2 + BC2 = 4AB2
⇒ 4AD2 = 4AB2 – BC2
⇒ 4AD2 = 4AB2 – AB2 [∵ABC is an equilateral triangle]
⇒ 4AD2 = 3AB2
Hence Proved
Question 26
In ΔABC, AB = AC. Side BC is produced to D. Prove that (AD2 —AC2) = BD . CD
Sol :
Construction: Draw an altitude from A on BC and named it O.
Given: ABC is an isosceles triangle with AB = AC
To Prove: AD2 —AC2 = BD × CD
In right triangle ∆AOD, using Pythagoras theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ AO2 + OD2 = AD2 …(i)
Now, in right triangle ∆AOB, using Pythagoras theorem, we have
⇒ AO2 + BO2 = AB2 …(ii)
Subtracting eq (ii) from (i), we get
AD2 – AB2 = AO2 + OD2 – AO2 – BO2
⇒ AD2 – AB2 = OD2 – BO2
⇒ AD2 – AB2 = (OD + BO)(OD – OB)
[∵ (a2 – b2)= (a + b)(a – b)]
⇒ AD2 – AB2 = (BD)(OD – OC) [∵OB = OC]
⇒ AD2 – AB2 = (BD)(CD)
⇒ AD2 – AC2 = (BD)(CD) [∵AB =AC]
Hence Proved
Question 27
In ΔABC, D is the mid-point of BC and AE ⊥ BC . If AC > AB, show that AB2 = AD2 — BC .DE + 1/4 BC2
Sol :
Given: In ΔABC, D is the mid-point of BC and AE
BCand AC > AB
In right triangle ∆AEB, using Pythagoras theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AE)2 + (BE)2 = (AB)2
⇒ (AE)2 + (BD – ED)2 = (AB)2
⇒ (AE)2 + (ED)2 + (BD)2 – 2 (ED)(BD) = (AB)2
[∵ (a – b)2 = a2 + b2 – 2ab]
⇒ (AE2 + ED2) + (BD)2 – 2 (ED)(BD) = (AB)2
⇒ (AD)2 + (BD)2 – 2 (ED)(BD) = (AB)2
[∵ In right angled ∆AED, AE2 + ED2 =AD2]
$\Rightarrow(\mathrm{AD})^{2}+\left(\frac{\mathrm{BC}}{2}\right)^{2}-2 \mathrm{ED}\left(\frac{\mathrm{BC}}{2}\right)=(\mathrm{AB})^{2}$
[∵ D is the midpoint of BC, so 2DC = BC]
$\Rightarrow A B^{2}=A D^{2}-B C \times E D+\frac{B C^{2}}{4}$
Hence Proved
Question 28
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 ΔABE and ΔACD.
Sol :
Given ∆ABC is an isosceles triangle in which ∠B is right angled i.e. 90°
⇒ AB = BC
In right angled ∆ABC, by Pythagoras theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AB)2 + (BC)2 = (AC)2
⇒ (AB)2 + (AB)2 = (AC)2
[∵ABC is an isosceles triangle, AB =BC]
⇒ 2(AB)2 = (AC)2
⇒ (AC)2 = 2(AB)2 …(i)
It is also given that ∆ABE ~ ∆ADC
And we also know that, the ratio of similar triangles is equal to the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ABE})}{\operatorname{ar}(\Delta \mathrm{ADC})}=\frac{\mathrm{AB}^{2}}{\mathrm{AC}^{2}}$
$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{ABE})}{\operatorname{ar}(\Delta \mathrm{ADC})}=\frac{\mathrm{AB}^{2}}{2 \mathrm{AB}^{2}}$ [from (i)]
$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{ABE})}{\operatorname{ar}(\Delta \mathrm{ADC})}=\frac{1}{2}$
∴ ar(∆ABE) : ar(∆ADC) = 1 : 2
Sol :
Given: ∠POR = 90°, OP = 6 cm and OR= 8 cm
and PQ = 24 cm and QR = 26 cm
To Prove: ΔPQR is right angled at P
In ∆POR, using Pythagoras theorem, we get
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (PO)2 + (OR)2 = (PR)2
⇒ (6)2 + (8)2 = (PR)2
⇒ 36 +64 = (PR)2
⇒ (PR)2= 100
⇒ PR =√100
⇒ PR = 10 [taking positive square root]
In ∆PQR
Using Pythagoras theorem, i.e. if the square of the hypotenuse is equal to the sum of the other two sides. Then, the given triangle is a right angled triangle, otherwise not.
Here, (PR)2 + (PQ)2
⇒ (10)2 + (24)2
= 100 + 576
= 676
= (26)2 = (QR)2
∴ given sides 10cm, 24cm and 26cm make a right triangle right angled at P.
Hence Proved
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ABE})}{\operatorname{ar}(\Delta \mathrm{ADC})}=\frac{\mathrm{AB}^{2}}{\mathrm{AC}^{2}}$
$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{ABE})}{\operatorname{ar}(\Delta \mathrm{ADC})}=\frac{\mathrm{AB}^{2}}{2 \mathrm{AB}^{2}}$ [from (i)]
$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{ABE})}{\operatorname{ar}(\Delta \mathrm{ADC})}=\frac{1}{2}$
∴ ar(∆ABE) : ar(∆ADC) = 1 : 2
Question 29
In the given figure, 0 is a point inside a ∠PQR such that ∠POR = 90°, OP = 6 cm and OR= 8 cm. If PQ = 24 cm and QR = 26 cm, prove that ΔPQR is right angled. P
Given: ∠POR = 90°, OP = 6 cm and OR= 8 cm
and PQ = 24 cm and QR = 26 cm
To Prove: ΔPQR is right angled at P
In ∆POR, using Pythagoras theorem, we get
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (PO)2 + (OR)2 = (PR)2
⇒ (6)2 + (8)2 = (PR)2
⇒ 36 +64 = (PR)2
⇒ (PR)2= 100
⇒ PR =√100
⇒ PR = 10 [taking positive square root]
In ∆PQR
Using Pythagoras theorem, i.e. if the square of the hypotenuse is equal to the sum of the other two sides. Then, the given triangle is a right angled triangle, otherwise not.
Here, (PR)2 + (PQ)2
⇒ (10)2 + (24)2
= 100 + 576
= 676
= (26)2 = (QR)2
∴ given sides 10cm, 24cm and 26cm make a right triangle right angled at P.
Hence Proved
| S.no | Chapters | Links |
|---|---|---|
| 1 | Real numbers | Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 |
| 2 | Polynomials | Exercise 2.1 Exercise 2.2 Exercise 2.3 |
| 3 | Pairs of Linear Equations in Two Variables | Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
| 4 | Trigonometric Ratios and Identities | Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4 |
| 5 | Triangles | Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 |
| 6 | Statistics | Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4 |
| 7 | Quadratic Equations | Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5 |
| 8 | Arithmetic Progressions (AP) | Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4 |
| 9 | Some Applications of Trigonometry: Height and Distances | Exercise 9.1 |
| 10 | Coordinates Geometry | Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 |
| 11 | Circles | Exercise 11.1 Exercise 11.2 |
| 12 | Constructions | Exercise 12.1 |
| 13 | Area related to Circles | Exercise 13.1 |
| 14 | Surface Area and Volumes | Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 |
| 15 | Probability | Exercise 15.1 |
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