Area of triangle questions with solutions are provided here for students to help them practice various kinds of questions based on the area of the triangle. In mensuration, the area of any figure is the region enclosed by a closed figure.

Learn more about the Area of shapes.

Following are some important formulae which we will use while solving questions:

Shape	Formula for Area	Variables
Triangle	½ × b × h	h = height of triangle b = base on which height is measured.
Triangle (whose length of sides are known)	√[s(s – a)(s – b)(s – c)] Heron’s Formula where s = (a + b + c)/2	a, b, c are the length of the sides of the triangle. s is the semi-perimeter of the triangle.
Equilateral Triangle	(√ 3/4 ) a²	a is the length of the sides.
Isosceles Triangle	¼ b√(4a² – b²)	a is the length of equal sides. b is the length of the unequal side.
Right Isosceles Triangle	a²/2	a is the length of equal sides.
Triangle (length of two sides and angle between them is given)	½ bc sin A ½ ab sin C ½ ca sin B	a, b, and c are the lengths of the sides. A, B, and C are measures of angles of the triangle.

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

Area of Triangle Questions with Solutions

Now, we will use the above formulae to solve the following area of triangle questions.

Question 1: Find the area of a triangle whose altitude and base are 12 cm and 7 cm, respectively.

Solution:

Base of triangle = 7 cm

Height of triangle = 12 cm

Area of triangle = ½ × base × height

= ½ × 7 × 12 = 42 cm².

Question 2: Find the area of the triangle whose height is 34 cm and the base is 1/4th the length of the height.

Solution:

Height of the triangle = 34 cm

Base of the triangle = 34 × ¼ = 8.5 cm

Area of the triangle = ½ × 34 × 8.5 = 144.5 cm².

Question 3: Find the height of the triangle whose base is two-thirds of its height and area is 225 cm².

Solution:

Let x be the height of the triangle, and then 2x/3 will be the base of the triangle.

Area of triangle = 225 cm²

⇒ ½ × x × 2x/3 = 225

⇒ x² = 225 × 3

⇒ x = √(225 × 3)

⇒ x = 15√3 cm.

Question 4: Find the area of a triangle whose length of each side is 3 inches, 5 inches and 4 inches.

Solution:

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

Let a = 3 inches, b = 5 inches, c = 4 inches

The semi perimeter of a triangle, s = (a + b + c)/2 = (3 + 4 + 5)/2 = 12/2 = 6 inches.

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

= √[6 × (6 – 3) × (6 – 4) × (6 – 5)]

= √[6 × 3 × 2 × 1]

= √36 = 6 sq. inches.

Question 5: A right triangle is inscribed inside a circle of radius 5 cm such that one the perpendicular sides of the triangle is the diameter of the circle and the other perpendicular side is 2/5th the radius of the circle. Find the area of the triangle.

Solution:

Radius of circle = 5 cm

Diameter of the triangle = 10 cm

Length of one perpendicular side of triangle = 10 cm

Length of the other perpendicular side = ⅖ × 5 = 2 cm

Area of the triangle = ½ × 10 × 2 = 10 cm².

Question 6: Find the area of the triangle whose perimeter is 170 cm and sides are in ratio 1:2:3.

Solution:

Let the length of the sides of triangle be x, 2x and 3x.

Perimeter of triangle = 150 cm

⇒ x + 2x + 3x = 150

⇒ 6x = 150

⇒ x = 25 cm

Then the sides of the triangle are 25 cm, 50 cm and 75 cm.

Let a = 25 cm, b = 50 cm, c = 75 cm and s = 150/2 = 75 cm

Area of triangle = √[s(s – a)(s – b)(s – c)] = √[75 × 50 × 25 × 0] = 0

Since the area is 0, therefore points on the plane are linear.

Question 7: Find the area of triangle whose length of two sides are 5 cm and 7 cm and angle between them is 30^o.

Solution:

Length of sides of triangle are 5 cm and 7 cm.

Angle between the sides = 30^o

Area of triangle = ½ × 5 × 7 × sin 30^o

= ½ × 35 × ½ = 8.75 cm²

Question 8: Find the sides of an equilateral triangle whose area is 9√3 cm².

Solution:

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

Area of an equilateral triangle = 9√3 cm²

⇒ (√3/4) a² = 9√3

⇒ a² = (9√3 × 4)/√3

⇒ a² = 36

⇒ a = 6 cm.

Question 9: A triangle is inscribed within a circle of diameter 17 cm. If one of the sides of the triangle is the diameter of the circle and another side is 15 cm, find the area of the triangle.

Solution:

According to the question, 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 17 cm, and one of the perpendicular sides is 15 cm. Let the length of the other perpendicular side be x, then by Pythagoras theorem,

x = √(17² – 15²) = √(289 – 225) = √64 = 8 cm.

Area of triangle = ½ × 8 × 15 = 60 cm².

Area of Similar Triangles:

If two triangles are similar, then the ratio of their areas is proportional to the square of ratios of their corresponding sides.

If △ABC is similar to △DEF, then

\(\begin{array}{l}\frac{Area \:of \:\Delta ABC}{Area \:of \:\Delta DEF}=\left ( \frac{AB}{DE} \right )^{2}=\left ( \frac{BC}{EF} \right )^{2}=\left ( \frac{CA}{FD} \right )^{2}\end{array} \)

Also Check:

Question 10: If △ABC ~ △PQR such that area of △ABC is 256 cm², area of △PQR is 169 cm² and BC = 32 cm, find the length of QR.

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.

∴

\(\begin{array}{l}\frac{Area \:of \:\Delta ABC}{Area \:of \:\Delta PQR}=\left ( \frac{BC}{QR} \right )^{2}\end{array} \)

\(\begin{array}{l}\Rightarrow\frac{256}{169}=\frac{32^{2}}{QR^{2}} \end{array} \)

\(\begin{array}{l}\Rightarrow\frac{16}{13}=\frac{32}{QR} \end{array} \)

\(\begin{array}{l}\Rightarrow QR=\frac{32 \times 13}{16} = 26\;cm\end{array} \)

Therefore, QR = 26 cm.

Question 11: If △ABC ~ △MNO such that Area of △ABC = 3△MNO and AB = 9 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,

\(\begin{array}{l}\frac{Area \:of \:\Delta ABC}{Area \:of \:\Delta MNO} = 3\end{array} \)

\(\begin{array}{l}\Rightarrow 3=\frac{AB^{2}}{MN^{2}}\end{array} \)

\(\begin{array}{l}\Rightarrow \sqrt{3}=\frac{AB}{MN}=\frac{9}{MN}\end{array} \)

\(\begin{array}{l}\Rightarrow MN=\frac{9}{\sqrt{3}}=3\sqrt{3} \end{array} \)

∴ MN = 3√3 cm.

Video Lesson on Area of Similar Triangles

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Practice Questions

1. Find the area of right isosceles triangle whose hypotenuse is 13√2 cm.

2. Find the base of the triangle whose height is 16 cm and area is 64 cm².

3. Find the area of isosceles triangles whose unequal side is one more than the length of equal sides and whose perimeter is 22 cm.

4. Find the area of triangle whose length of two sides are 5 cm and 6 cm, respectively and angle between them is 30^o.

5. If △POQ and △XYZ are similar to each other such that ratio of their areas is 9:16 and OQ = 4.5 cm. Find the length of YZ.

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