RS Aggarwal Solutions: Triangles- 3 | RS Aggarwal Solutions for Class 10 Mathematics PDF Download

Since the man goes C to A = 24 m west and then A to B = 10 m north, he is forming a right angle triangle with respect to starting point C.
His distance from the starting point can be calculated by using Pythagoras theorem
(AC)² = (AB)² + (BC)²
⇒ (AC)² = (24)² + (10)²
⇒ (AC)² = 576 + 100
⇒ (AC)² = 676
⇒ AC = 26

Let AB and CE be the two poles of the height 13 cm and 7 cm each which are perpendicular to the ground. The distance between them is 8 cm.
Now since CE and AB are ⊥ ground AE
BD ⊥ to CE and BD = 8 cm
Top of pole AB is B and top of pole CE is C
Now Δ BDC is right angled at D and BC, the hypotenuse is the distance between the top of the poles and CD = 13 – 7 = 6
(BC)² = (BD) ² + (CD)²
⇒ (BC)² = 64 + 36
⇒ (BC)² = 100
⇒ (BC)= 10 cm
The distance between the top of the poles is 10 cm

Let DF be the stick of 1.8 m height and AB be the pole of 6 m height.
AC and FE are the shadows of the pole and stick respectively.
FE = 45cm = .45 m

Since the shadows are formed at the same time, the two Δs are similar by AA similarity criterion
So 

⇒ x = 1.5 m

Let DE be the pole of 6 m length casting shadow of 3.6 m . Let AB be the tower x meter height casting shadow of 18mat the same time.
Since pole and tower stands vertical to the ground, they form right angled triangle with ground.
Δ ABC and Δ EDF are similar by AA similarity criterion
∴ x/6 = 18 /3.6
⇒ x = 30
The height of the tower is 30 m.

SINCE BOTH the tree and the stick are forming shadows at the same time the sides of the triangles so formed, would be in same ration
∵ of AA similarity criterian
12.5 /5 = x / 2
⇒ x = 5
Shadow of the tree would be 5 m long.

Let BC be the ladder placed against the wall AB. The distance of the ladder from the wall is the base of the right angled triangle as building stands vertically straight to the ground.
By Pythagoras theorem
(BC)² = (AB)² + (AC)²
(25)² = (24)² + (x)²
x = 7
the distance of ladder from the wall is 7m

The ΔMOP is right angled at O so MP is hypotenuse
(MP)² = (OM)² + (OP)²
(MP)² = (16) ² + (12)²
(MP)² = 400
MP = 20 cm
Δ NMP is right angled at M so NP is the hypotenuse so
(NP)² = (21)² + (20)²
NP = 29

Given (BC) = 25 cm
By Pythagoras theorem
(BC)² = (AB)² + (AC)²
(25) ² = (x + 5)² + x²
625 = x² + 25 + 10x + x²∵ (a + b)² = a² + b² + 2ab
x² + 5x – 300 = 0
x (x + 15) – 10(x + 15) = 0
Since x = -15 is not possible so side of the triangle is 15 cm and 20 cm

Since Δ ABC is an equilateral triangle so the altitude (Height = h) from the C is the median for AB dividing AB into two equal halves of 6 cm each
Now there are two right angled Δs
h² = a² - 1/2 (AB)²
h² = (12)² - 6²
h = 6 √3

The given triangle is isosceles so the altitude from the one of the vertex is median for the side opposite to it.
AB = AC = 13 cm
h = 5 cm (altitude)
ΔABX is a right angled triangle, right angled at X
(AB)² = h² + (BX)² (BX = 1/2 BC)
169 = 25 + (BX)²
BX = 12
⇒ BC = 24

By internal angle bisector theorem, the bisector of vertical angle of a triangle divides the base in the ratio of the other two sides.
Hence in ΔABC, we have

Here AB = 6 cm, AC = 8 cm
So 

By internal angle bisector theorem, the bisector of vertical angle of a triangle divides the base in the ratio of the other two sides”
Hence in ΔABC, we have


⇒ x = 7.5cm

By internal angle bisector theorem, the bisector of vertical angle of a triangle divides the base in the ratio of the other two sides”
Hence in Δ ABC


⇒ 10x -84 + 14x = 0
⇒ x = CD = 3.5 cm

In Δ ABC, AD bisects ∠ A and meets BC in D such that BD = DC
Extend AD to E and join C to E such that CE is ∥ to AB
∠ BAD = ∠ CAD
Now AB ∥ CE and AE is transversal
∠ BAD = ∠ CED (alternate interior ∠s)
But ∠ BAD = ∠ CED = ∠CAD
In Δ AEC
∠ CEA = ∠CAE
∴ AC = CE …. 1
In Δ ABD and Δ DCE
∠BAD = ∠CED (alternate interior ∠s)
∠ADB = ∠ CDE (vertically opposite ∠s)
BD = BC (given)
Δ ABD ≅ Δ DCE
AB = EC (CPCT)
AC = EC (from 1)
⇒AB = AC
⇒ ABC is an isosceles Δ with AB = AC

Δ ABC is an equilateral triangle
By Pythagoras theorem in triangle ABD
AB² = AD² + BD²
but BD = 1/2 BC (∵ In a triangle, the perpendicular from the vertex to the base bisects the base)
thus AB² = AD² + {1/2 BC}²
AB² = AD² + 1/4 BC²
4 AB² = 4AD² + BC²
4 AB² - BC² = 4 AD²
(as AB = BC we can subtract them )
Thus 3AB² = 4AD²

Since the diagonals of the rhombus bisects each other at 90°
∴ DO = OB = 6 cm
∠ AOD = ∠ DOC = ∠ COB = ∠BOA = 90°
Δ AOD is right angled Δ with
AD = 10 cm (given)
OD = 6 cm
∠ AOD = 90°
So DA = 10 = hypotenuse
(DA)² = (DO)² + (AO)²
100 – 36 = (AO)²
8 = AO
∴ AC = 16

In a rhombus the diagonals bisect each other at 90°
AC = 24cm (given)
∴ AD = 12cm
BD = 10cm (given)
∴ BO = 5cm
In right angled Δ AOB
AB² = AO² + OB²
AB² = (12)² + 5²
AB² = 144 + 25
AB² = 169
AB = 13cm
Hence the length of the sides of the rhombus is 13 cm

Given that ABCD is a quadrilateral and diagonals AC and BD intersect at O such that 
IN Δ AOD and ΔBOC

∠ AOD = ∠COB
Thus Δ AOC ∼ ΔBOC (SAS similarity criterion)
⇒ ∠OAD = ∠ OCB  ..1
Now transversal AC intersect AD and BC such the ∠CAD = ∠ACB (from …..1 ) (alternate opposite angles)
So AD ∥ BC
Hence ABCD is a trapezium

ABCD is a Trapezium with AC and BD as diagonals and AB ∥ DC
In Δ AOB and Δ DOC
∠AOB = ∠DOC (vertically opposite angles)
∠CDO = ∠ OBA (alternate interior angles) (AB ∥ DC and BD is transversal)
Δ AOB ∼ Δ DOC (AA similarity criterion)


(3x - 1)(6x - 5) = (5x - 3)(2x + 1)
18 x² – 21x + 5 = 10 x² – x – 3
8 x² – 20 x + 8 = 0
2x² – 5 x + 2 = 0
2x² – 4 x – x + 2 = 0
2x (x- 2) – 1(x – 2) = 0
(2x- 1) (x -2) = 0
x = 1/2, 2
x = 1/2 is not possible so x = 2 cm

In the given quadrilateral ABCD
P, Q, R, S are the midpoints of the sides AB, BC, CD and AD respectively.
Construction: - Join AC
In Δ ABC and Δ ADC
P and Q are midpoints of AB and CB
S and R are midpoints of AD and DC
So by Mid Point Theorem
PQ ∥ AC and PQ = 1/2 AC ….1
And SR ∥ AC and SR = 1/2 AC …2
From 1 and 2
PQ ∥ SR and PQ = SR
Since a pair of opposite side is equal (= ) and parallel (∥)
PQRS is a parallelogram

Given in Δ ABC, AD bisects the ∠A meeting BC at D
BD = DC and ∠BAD = ∠ CAD …. 1
Construction:- Extend BA to E and join C to E such CE ∥ AD … 4
∠BAD = ∠AEC (corresponding ∠s) … 2
∠CAD = ∠ ACE (alternate interior ∠s) ….. 3
From 1 , 2 and 3
∠ ACE = ∠AEC
In Δ AEC
∠ ACE = ∠AEC
∴ AC = AE (sides opposite to equal angles are equal) ….. 5
In Δ BEC
AD ∥ CE (From ….4)
And D is midpoint of BC (given)
By converse of midpoint theorem
A line drawn from the midpoint of a side, parallel to the opposite side of the triangle meets the third side in its middle and is half of it
∴ A is midpoint of BE
BA = AE … 6
From 5 and 6
AB = BC
⇒ ΔABC is an isosceles triangle

It is given that in Δ ABC,

∠B = 70° and ∠C = 50°
∠A = 180° – (70° + 50°) (∠ sum property of triangle)
= 180° - 120°
= 60°
Since, 
∴ AD is the bisector of ∠A
Hence, ∠BAD = 60°/2 = 30°

By Basic Proportionality Theorem
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
In Δ ABC, DE ∥ BC


DB = 3.6 cm
AB = AD + DB
AB = 6 cm

BY Basic Proportionality Theorem:
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Or 
Adding 1 to both sides



7.2/4.5 = 6.4/AE
AE = 4 cm

By Basic Proportionality Theorem:
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Or 
(7x - 4)/(3x + 4) = (5x - 2)/3x
3x (7x — 4) = (5x — 2) (3x + 4)
21 x² – 12x = 15x² + 14x – 8
6 x² – 26x + 8 = 0
3 x² – 13x + 4 = 0
3 x² – 12x – x + 4 = 0
3x(x - 4) – 1(x - 4) = 0
X = 1/3, 4
Since x cannot be 1/3 so x = 4

BY Basic Proportionality Theorem:
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.



3 (5.6-AE) = 5 AE
16.8 = 8AE
AE = 2.1 cm

Since the ΔABC ∼ ΔDEF
Their sides will be same ratios. Let the ratio be K

AB + BC + AC = K (DE + EF + DF)
30/18 = K
1.67 = k
From…… 1
BC/EF = 1.67
EF = 5.4 cm

Since the ΔABC ∼ ΔDEF
So the sides of the triangles are in the same ratio be k


K = 1.4


Perimeter of ΔABC = 1.4 × 25
Perimeter of ΔABC = 35 cm

Perimeter of Δ ABC = AB + BC + CA
= 9 + 6 + 7.5
= 22.5 cm
Since the Δ ABC ∼ Δ DEF
(ratio of perimeter of triangles is equal to the ratio of the sides of the triangle )

Δ ABC and Δ BDE are two equilateral triangles
Let a be the side of Δ ABC
Since D is midpoint of BC
So the side of equilateral ΔBDE = a/2
Area of equilateral Δ =  √3/4 (side)2
Area of Δ ABC = √3/4 (a²) ………….1
Area of Δ BDE = 
= 
Putting value of   …………from 1
Area of Δ BDE = 1/4 Area of Δ ABC

In ΔABC,
∠ A + ∠B + ∠ C = 180°
30° + ∠B + 50° = 180°
∠ B = 100°
Given that Δ ABCΔDEF
∠D = ∠A = 30°
∠ E = ∠ B = 100°
∠ F = ∠ C = 50°
Also, 

DE = 4.6875
And as neither BC nor EF is given we can not find either of them. So, none of the given options is correct.
Now,
Δ ABC~ΔDFE,
Then,
∠D = ∠ A = 30°
∠ F = ∠ B = 100°
∠ E = ∠ C = 50°

Therefore, the correct option is (b).

∠ABD + ∠ BAD = 90°
∠ ABD + (90-∠CAD) = 90°
∠ ABD = ∠ DAC
In ΔBDA and Δ ADC,
∠ ABD = ∠ CAD
∠ BDA = ∠ ADC = 90°
Therefore, ΔBDA and Δ ADC are similar by AAA.

Therefore the correct option is (c).

AB = 6√3cm.
In ΔABC,
AB² + BC² = AC²

Since the square of the longest side is equal to the sum of the squares of the remaining two sides of Δ ABC. Therefore ABC is right angled at B.
∠ B = 90°.

With the ratio given, we can observe that Δ ABC ~Δ EDF,
∠ A = ∠E,
∠ B = ∠D,
∠ C = ∠ F

Given ∠ D = ∠ Q and ∠E = ∠R
By AA similarity, ΔDEF~ΔQRP

We have to find the option which is not true.

ΔABC~ΔEDF,
Therefore, 
We have to find the option which is not true.
∴ The correct option is (c) .

Δ ABC~Δ DEF
By AA similarity, the triangles are similar
For triangles to be congruent, AB = DE, but given that AB = 3DE.

Given 

Therefore ΔABC~ΔQRP
or ΔPQR~ΔCAB.

In ΔAPB and Δ DPC,
∠APB = ∠DPC = 50°


By SAS property, Δ APB~ΔDPC
∠PBA = ∠DPC
In Δ DPC,
∠D + ∠ P + ∠C = 180°
∠ C = 100°
∴ ∠PBA = ∠DPC = 100°

If two triangles are similar, the ratio of the area of triangle is equal to the square of the ratio of the sides.
Ratio of Area = (Ratio of Side)2 = (4/9)² = 16:81
∴ The correct option is (d).

If two triangles are similar, the ratio of the area of triangle is equal to the square of the ratio of the sides.

Given that D and E are of AB and AC respectively,
Therefore,

Δ ABC~Δ ADE
If two triangles are similar, the ratio of the area of triangle is equal to the square of the ratio of the sides.

∴ The correct option is (b).

Therefore Δ ABC~Δ DEF
If two triangles are similar, the ratio of the area of triangle is equal to the square of the ratio of the sides.

It is given that the corresponding angles are equal, that implies that the triangles are similar.

In this figure,
As given that the inner triangle is formed by joining the midpoints of the sides.
Therefore the outer three triangles are similar to bigger triangle.
By Basic Proportionality Theorem,
The inner triangle is also similar to the bigger triangle.

ΔABC~ΔQRP

PR = 2/3 BC
PR =  = 10cm

In Δ DOB and Δ AOC,
∠DOB = ∠AOC = 45° (vertically opposite angle)
∠OAC = ∠ODB (angles in the same segment)
∠OCA = ∠OBD (angles in the same segment)
Therefore, Δ DOB ~ Δ AOC by AA similarity,

Therefore, OC = OA.

AB² = 2AC²
AB² = AC² + AC²
AB² = AC² + BC²
Therefore, it is an isosceles triangle right angled at C. ∠ C = 90°

AB² + BC² = 16² + 12² = 256 + 144 = 400 = 20² = AC²
Therefore, ABC is a right angled triangle.

If two triangles ABC and PQR are similar,

That is their corresponding sides are proportional.

The ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.