A triangle is a polygon with the three edges and three vertices. The sum of the interior angles of the triangle is always 180 degrees. The Triangles can be classified based on their sides and angles. Understanding the properties and types of the triangles is crucial for the solving various geometric problems.
Pythagoras Theorem
In a right-angled triangle the square of the hypotenuse is equal to sum of the squares of the other two sides.
c2 = a2 + b2
where c is the hypotenuse and a and b are the other two sides.
Area of a Triangle
The area of a triangle is given by:
Area = \frac{1}{2} \times \text{base} \times \text{height}
This formula is used to find the area of the triangle when the base and height are known.
Similarity of Triangles
Two triangles are similar if:
- Corresponding angles are equal.
- Corresponding sides are in proportion.
Basic Proportionality Theorem (Thales' Theorem)
If a line is drawn parallel to the one side of a triangle to the intersect the other two sides in distinct points the other two sides are divided in the same ratio.
Heron’s Formula
To find the area of a triangle when all three sides are known:
Area = \sqrt{s(s-a)(s-b)(s-c)}
where s = \frac{a+b+c}{2} is the semi-perimeter of the triangle.
Let us consider ∆ABC and ∆DEF, AL and DM are the medians of ∆ABC and ∆DEF
It is given that the area of ∆ABC = 121 cm2 and area of ∆DEF = 64 cm2
AL = 12.1 cm
Let us assume DM = x cm
Given that, ∆ABC ~ ∆DEF
So,
ar(∆ABC)/ar(∆DEF) = AL2/DM2
= 121/64 = (12.1)2/x2
11/8 = 12.1/x
⇒ x = (8 × 12.1)/11 = 8.8
Hence, the median of the second triangle is 8.8cm
Given that,
area (∆ABC) = 20 cm²
area (∆DEF) = 45 cm²
AB = 5 cm
Let us consider DE = x cm
Also, given that ∆ABC ~ ∆DEF
ar(∆ABC)/ar(∆DEF) = AB2/DE2
⇒20/45 = (5)2/x2
⇒20/45 = 25/x2
⇒x2 = (25 × 45)/20 = 225/4 = (15/2)2
x = 15/2 = 7.5
DE = 7.5cm
It is given that, in ∆ABC, PQ || BC and line PQ divide the ∆ABC into two parts
∆APQ and trap. BPQC equally
i.e., area ∆APQ = area BPQC
Now we have to find BP/AB.
As we know that PQ||BC
So, ∆APQ ∼ ∆ABC
⇒ ar.(∆APQ)/ar.(∆ABC) = AP2/AB2
⇒ ar.(∆ABC)/ar.(∆APQ) = AB2/AP2
2/1 = AB2/AP2
{area ∆APQ = area trap. BPQC
Area ∆ABC = 2area (∆APQ)}
⇒ AB/AP = √2/1
⇒√2 AP = AB = AP + PB
⇒√2AP - AP = PB
⇒(√2 - 1)AP = PB
BP/AP = (√2 - 1)/1
Given that, area (∆ABC) : area (∆PQR) = 9 : 16
∆ABC ~ ∆PQR
and BC = 4.5 cm
Let us considered QR = x cm
As we know that ∆ABC ~ ∆PQR
ar.(∆ABC)/ar.(∆PQR) = BC2/QR2 ⇒ 9/16 = (4.5)2/x2
⇒ (3/4)2 = (4.5/x)2 ⇒ 4.5/x = 3/4
x = (4.5 × 4)/3 = 60
Hence, the length of QR is 6cm
Given that, in ∆ABC, P and Q are two points on line AB and AC
AP = 1 cm, PB = 3 cm, AQ = 1.5 cm and QC = 4.5 cm
Now, AP/PB = 1/3 and AQ/QC = 1.5/4.5 = 1/3
In ∆APQ and ∆ABC
AP/PB = AQ/QC
PQ||BC
Hence, ∆APQ ∼ ∆ABC
So, ar.(∆APQ)/ar.(∆ABC) = AP2/PB2 = AP2/(AP + PB)2
ar.(∆APQ)/ar.(∆ABC) = 12/(1 + 3)2 = 1/16
Hence, area of ∆APQ = 1/16 of area of ∆ABC
Given that in ∆ABC, D is a point on AB such that AD : DB = 3 : 2
DE||AC
In ∆BDE and ∆ABC
∠BDE = ∠A
∠DBE = ∠ABC
So, by AA, ∆BED ∼ ∆ABC
Therefore, ar.(∆ABC)/ar.(∆BDE) = AB2/BD2 = (BD + AD)2/BD2
= (2 + 3)2/22 = 52/22 = 25/4
Hence, the ratio of areas of ∆ABC and ∆BDE is 25:4
Given that ∆ABC and ∆DBE are equilateral triangles, where D is mid point of BC
So, BD = 1/2BC
Now area of ∆ABC
√3/4(side)2 = √3/4BC2
and area of ∆DBE
√3/4(side)2 = √3/4BD2
√3/4(side)2 = √3/4(1/2BC)2
√3/4(side)2 = √3/16(BC)2
So, the ratio between areas is
= area(∆ABC)/area(∆DBE) =\frac{\frac{\sqrt{3}}{4}BC^2}{\frac{\sqrt{3}}{16}BC^2}\\ =\frac{\sqrt{3}}{4}×\frac{16}{\sqrt{3}}=\frac{4}{1}
Hence, the ratio of areas of ∆ABC and ∆BDE is 4:1
Let us consider two triangles, ∆ABC and ∆XYZ and these triangles have equal vertical angle, i.e., ∠A and ∠X
And AD and XO is the heights of these triangles.
So, ∆ABC/∆XYZ = AB/AC = XY/XZ
In ∆ABC and ∆XYZ
∠A = ∠X
AB/AC = XY/XZ
So, by SAS
∆ABC ~ ∆XYZ
So, ar(∆ABC)/ar(∆XYZ) = AD2/XO2
As we know that ar(∆ABC)/ar(∆XYZ) = 36/25
So,
36/25 = AD2/XO2
6/5 = AD/XO
Hence, the ratio of their corresponding heights is 6:5
Given that two ∆ABC and ∆DBC are on the same base BC as shown in the given figure
AC and BD intersect each other at O
Now, Draw AL ⊥ BC and DM ⊥BC
Prove: \frac{ar.(∆ABC)}{ar.(∆DBC)}=\frac{AO}{DO}
Proof:
In ∆ALO and ∆DMO,
∠L =∠M = 90°
∠AOL = ∠DOM [Vertically opposite angles]
So, by AA, ∆ALO ∼ ∆DMO
So, AL/DM = AO/DO
Now \frac{ar(∆ABC)}{ar(∆DBC)}=\frac{\frac{1}{2}BC×AL}{\frac{1}{2}BC×DM}=\frac{AL}{DM}
But AL/DM = AO/DO (Proved above)
So, \frac{ar.(∆ABC)}{ar.(∆DBC)}=\frac{AO}{DO}
Hence proved
Given that ABCD is a trapezium in which AB || CD and the diagonals AC and BD intersect at O
Now, in the figure from point D, draw DL⊥AC
(i) In ∆AOB and ∆COD
∠AOB =∠COD [Vertically opposite angles]
∠OAB =∠OCD [Alternate angles]
So, by AA criterion
∆AOB ~ ∆COD
(ii) Given that OA = 6 cm, OC = 8 cm
As we know that ∆AOB ~ ∆COD
So, OA/OC = OB/OD = AB/CD
(a) ar(∆AOB)/ar(∆COD) = AO2/OC2
= 62/82 = 36/64 = 9/16
Therefore, ar(∆AOB)/ar(∆COD) = 9/16
(b) As we know that ∆AOD and ∆COD have their bases on the same line and their vertex A is common
Therefore, ar(∆AOD)/ar(∆COD) = AO/OC = 6/8 = 3/4
Given that, ABC is a triangle, in which P divides the side AB such that
AP : PB = 1 : 2. Q is a point in AC such that PQ || BC
In ∆APQ and ∆ABC
∠APQ = ∠B
∠PAQ = ∠BAC
So, by AA criterion
∆APQ ∼ ∆ABC
So,
ar(∆APQ)/ar(∆ABC) = (AP)2/(AB)2
ar(∆APQ)/ar(∆ABC) = (1)2/(1 + 2)2 = (1)2/(3)2 = 1/9
9 ar(∆APQ) = ar(∆ABC)
9 ar(∆APQ) = ar(∆APQ) + ar(trap. BPQC)
9 ar(∆APQ) = ar(trap BPQC)
ar(∆APQ)/ar(trap BPQC) = 1/9
Hence, the ratio of the areas of ∆APQ and trapezium BPQC is 1:9
Given that AD is an altitude of an equilateral triangle ABC.
On AD as base, another equilateral triangle ADE is constructed
Prove: Area (∆ADE) : Area (∆ABC) = 3 : 4
Proof:
Area of ∆ABC = √3/4 BC2
and AD = √3/2 BC
Area of ∆ADE = √3/4 AD2
= √3/4 (√3/2 BC)2 = 3√3/16 BC2
So, the ratio of area (∆ADE):area(∆ABC) = 3√3/16 BC2 : √3/4 BC2
= 3/4:1 = 3:4