Practice problems on the area of a triangle are provided here with solutions to help students enhance their skills on calculating the area of triangles. In mensuration , the area of any figure is the region enclosed within its boundary.

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Here are some crucial formulae that we will use to solve the problems:

Click here to learn more formulas for the area of a triangle .

Now, let's apply the above formulae to solve the following practice problems on the area of a triangle.

Problem 1: Calculate the area of a triangle with an altitude of 14 cm and a base of 8 cm.

Solution:

Triangle's base = 8 cm

Triangle's height = 14 cm

Area of triangle = ½ × base × height

= ½ × 8 × 14 = 56 cm 2 .

Problem 2: Calculate the area of the triangle with a height of 36 cm and a base that is 1/3rd the length of the height.

Solution:

Triangle's height = 36 cm

Triangle's base = 36 × ⅓ = 12 cm

Area of the triangle = ½ × 36 × 12 = 216 cm 2 .

Problem 3: Find the height of the triangle with a base that is three-fourths of its height and an area of 240 cm².

Solution:

Let the height of the triangle be x, then the base = (3/4)x

Area of the triangle = ½ × base × height
= ½ × (3/4)x × x = (3x²)/8

Given area = 240 cm²
⇒ (3x²)/8 = 240
⇒ x² = (240 × 8)/3 = 640
⇒ x = √640 ≈ 25.30 cm

 Correct height: 25.30 cm

Problem 4: Calculate the area of a triangle with each side measuring 4 inches, 6 inches, and 5 inches.

Solution:

We will use Heron’s formula to find the area of the triangle.

Let a = 4 inches, b = 6 inches, c = 5 inches

The semi perimeter of a triangle, s = (a + b + c)/2 = (4 + 6 + 5)/2 = 15/2 = 7.5 inches.

Area of triangle = √[s(s – a)(s – b)(s – c)]

= √[7.5 × (7.5 – 4) × (7.5 – 6) × (7.5 – 5)]

= √[7.5 × 3.5 × 1.5 × 2.5]

= √98.44 = 9.92 sq. inches.

Problem 5: A right triangle is inscribed inside a circle of radius 6 cm such that one of the perpendicular sides of the triangle is the diameter of the circle and the other perpendicular side is 1/3rd the radius of the circle. Calculate the area of the triangle.

Solution:

Radius of circle = 6 cm

Diameter of the triangle = 12 cm

Length of one perpendicular side of triangle = 12 cm

Length of the other perpendicular side = ⅓ × 6 = 2 cm

Area of the triangle = ½ × 12 × 2 = 12 cm 2 .

Problem 6: Calculate the area of the triangle with a perimeter of 180 cm and sides in the ratio 2:3:4.

Solution:

Let the sides be 2x, 3x, 4x

Perimeter = 180 cm
⇒ 2x + 3x + 4x = 9x = 180
⇒ x = 20

Sides: a = 40 cm, b = 60 cm, c = 80 cm
Semi-perimeter, s = (40 + 60 + 80)/2 = 90 cm

Using Heron’s formula:
Area = √[s(s − a)(s − b)(s − c)]
= √[90 × (90 − 40) × (90 − 60) × (90 − 80)]
= √[90 × 50 × 30 × 10] = √1350000 ≈ 1161.90 cm²

 Correct area: 1161.90 cm²

Problem 7: Calculate the area of a triangle with two sides measuring 6 cm and 8 cm and an angle of 45 o between them.

Solution:

Lengths of the triangle's sides are 6 cm and 8 cm.

Angle between the sides = 45 o

Area of triangle = ½ × 6 × 8 × sin 45 o

= ½ × 48 × 1/√2 = 24/√2 = 16.97 cm 2

Problem 8: Find the sides of an equilateral triangle with an area of 12√3 cm 2 .

Solution:

Let ‘a’ be the side of the equilateral triangle.

Area of an equilateral triangle = 12√3 cm 2

⇒ (√3/4) a 2 = 12√3

⇒ a 2 = (12√3 × 4)/√3

⇒ a 2 = 48

⇒ a = 6.93 cm.

Problem 9: A triangle is inscribed within a circle of diameter 18 cm. If one of the sides of the triangle is the diameter of the circle and another side is 16 cm, find the area of the triangle.

Solution:

According to the problem, the triangle is made within the semi-circular part of the circle. We know that the angle of a semi-circle is always 90 o .

Hence the given triangle is a right-angled triangle whose hypotenuse is 18 cm, and one of the perpendicular sides is 16 cm. Let the length of the other perpendicular side be x, then by Pythagoras theorem,

x = √(18 2 – 16 2 ) = √(324 – 256) = √68 = 8.25 cm.

Area of triangle = ½ × 8.25 × 16 = 66 cm 2 .

Also Check:

• Similar Triangles
• Area of Similar Triangles
• Area of Scalene Triangle
• Congruence of Triangles

Problem 10: If △ABC ~ △PQR such that area of △ABC is 275 cm², area of △PQR is 184 cm², and BC = 35 cm, find the length of QR.

Solution:

From the property of similar triangles:
(Area of △ABC / Area of △PQR) = (BC / QR)²

⇒ 275 / 184 = (35 / QR)²
⇒ (35 / QR)² = 275 / 184
⇒ (35 / QR) = √(275 / 184) ≈ √1.4946 ≈ 1.2226
⇒ QR = 35 / 1.2226 ≈ 28.63 cm

 Correct length of QR: 28.63 cm

Problem 11: If △ABC ~ △MNO such that Area of △ABC = 4△MNO and AB = 12 cm, find the length of MN.

Solution:

We know that if two triangles are similar, then the ratio of their areas is proportional to the square of ratios of their corresponding sides.

∴

\(\begin{array}{l}\frac{Area \:of \:\Delta ABC}{Area \:of \:\Delta MNO}=\left ( \frac{AB}{MN} \right )^{2}\end{array} \)

Now,

\(\begin{array}{l}\frac{Area \:of \:\Delta ABC}{Area \:of \:\Delta MNO} = 4\end{array} \)

\(\begin{array}{l}\Rightarrow 4=\frac{AB^{2}}{MN^{2}}\end{array} \)

\(\begin{array}{l}\Rightarrow 2=\frac{AB}{MN}=\frac{12}{MN}\end{array} \)

\(\begin{array}{l}\Rightarrow MN=\frac{12}{2} = 6\;cm\end{array} \)

∴ MN = 6 cm.

Practice Questions

1. Calculate the area of a right isosceles triangle with a hypotenuse of 15√2 cm.
2. Find the base of the triangle with a height of 18 cm and an area of 72 cm 2 .
3. Calculate the area of an isosceles triangle whose unequal side is two more than the length of equal sides and whose perimeter is 24 cm.
4. Find the area of a triangle with two sides measuring 7 cm and 9 cm, respectively, and an angle of 60 o between them.
5. If △POQ and △XYZ are similar to each other such that the ratio of their areas is 4:9 and OQ = 6 cm, find the length of YZ.

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