CBSE Class 9 Mathematics Area Of Triangles Herons Formula Worksheet Set 01

Question. The perimeter of an equilateral triangle is 60 m, then its area is:
(a) 10√3m²
(b) 15√3m²
(c) 20√3m²
(d) 100√3m²

Answer: D

Question. The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then the area of the triangle is
(A) 1322 cm²
(B) 1311 cm²
(C) 1344 cm²
(D) 1392 cm²

Answer: C

Question. An isosceles right triangle has area 8 cm². The length of its hypotenuse is
(a) √32cm
(b) √16cm
(c) √48cm
(d) √24cm

Answer: A

Question. The length of each side of an equilateral triangle having an area of 9√3cm² is
(A) 8 cm
(B) 36 cm
(C) 4 cm
(D) 6 cm

Answer: D

Question. If the area of an equilateral triangle is 16√3cm² then the perimeter of the triangle is
(A) 48 cm
(B) 24 cm
(C) 12 cm
(D) 36 cm

Answer: B

Question. The area of an equilateral triangle with side 2√3cm is
(a) 5.196 cm²
(b) 0.886 cm²
(c) 3.496 cm²
(d) 1.732 cm²

Answer: A

Question. The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude
(a) 16√5cm
(b) 10√5cm
(c) 24√5cm
(d) 28 cm

Answer: C

Question. The edges of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it at the rate of 9 paise per cm² is
(A) Rs2.00
(B) Rs2.16
(C) Rs2.48
(D) Rs3.00

Answer: B

Question. The base and the corresponding altitude of a parallelogram are 10 cm and 3.5 cm, respectively. The area of the parallelogram is 30 cm².
Answer: The base of the parallelogram is 10 cm and the corresponding altitude is 3.5 cm.
Area of || gm = base × corresponding altitude.
= 10 cm × 3.5 cm = 35 cm².
Hence, the given statement is false.

Question. The area of a triangle with base 4 cm and height 6 cm is 24 cm².
Answer: Area of Δ = 1/2 X base X height = 1/2 X 4X 6 = 12cm²
Hence, the statement is false.

Question. The area of the equilateral triangle is 20√3cm² whose each side is 8 cm.
Answer: Area of equilateral = √3/4 (side)²
= √3/4 (8)² = √3/4 X 64 = 16 √3cm²
Hence, the given statement is false.

Question. The area of a trapezium is 475 cm2 and the height is 19 cm. Find the lengths of its two parallel sides if one side is 4 cm greater than the other.
Answer: Area of trapezium = 1/2 × (Sum of the parallel side) × height
⇒ 475 = 1/2 x (x + x + 4) x 19cm
⇒ 2x + 4 = 950/19 = 50
⇒ 2x + 50 - 4 = 46, x = 46 ÷ 2 = 23
Hence, the length of two parallel sides are 23 cm and (23 + 4) cm i.e., 23 cm and 27 cm.

Question. A rectangular plot is given for constructing a house, having a measurement of 40 m long and 15 m in the front. According to the laws, a minimum of 3 m, wide space should be left in the front and back each and 2 m wide space on each of other sides. Find the largest area where house can be constructed.
Answer: The length of the rectangular plot = 40 m
And the breath of the plot = 15 m.
As a minimum of 3 m wide space should be left in the front and back 2 m wide space each
of other side, so the largest area where the house can be constructed.
= [40 − 2(3)] [15− 2(2)] = 34 × 11 = 374m²

Question 1. The sides of a triangle are 8cm, 15cm and 17cm. find its perimeter and its area. Also find the length of the altitude drawn on the side with length 17cm.
Answer: Let the sides of the triangle be \(a = 8\text{ cm}\), \(b = 15\text{ cm}\), and \(c = 17\text{ cm}\).
Perimeter of the triangle = \(a + b + c\)
Perimeter = \(8 + 15 + 17 = 40\text{ cm}\)
Semi-perimeter (\(s\)) = \(\frac{40}{2} = 20\text{ cm}\)
Using Heron's formula, Area = \(\sqrt{s(s - a)(s - b)(s - c)}\)
Area = \(\sqrt{20(20 - 8)(20 - 15)(20 - 17)}\)
Area = \(\sqrt{20 \times 12 \times 5 \times 3}\)
Area = \(\sqrt{(4 \times 5) \times (4 \times 3) \times 5 \times 3}\)
Area = \(\sqrt{4^2 \times 5^2 \times 3^2} = 4 \times 5 \times 3 = 60\text{ cm}^2\)
To find the length of the altitude drawn on the side with length 17 cm:
Area = \(\frac{1}{2} \times \text{base} \times \text{height}\)
\(60 = \frac{1}{2} \times 17 \times h\)
\(h = \frac{120}{17}\text{ cm} \approx 7.06\text{ cm}\)
In simple words: The perimeter is 40 cm, the area is 60 square centimeters, and the altitude on the 17 cm side is 120/17 cm.

Exam Tip: Always check if the given sides form a right-angled triangle (\(8^2 + 15^2 = 64 + 225 = 289 = 17^2\)). Since it is a right-angled triangle, you can also calculate the area directly as \(\frac{1}{2} \times 8 \times 15 = 60\text{ cm}^2\), which verifies your Heron's formula calculation.

Question 2. The perimeter of the triangle is 44cm. if its sides are in the ratio 9:7:6 find its area.
Answer: Let the sides of the triangle be \(9x\), \(7x\), and \(6x\).
Perimeter = \(9x + 7x + 6x = 44\)
\(22x = 44\)
\(x = 2\)
Therefore, the sides are:
\(a = 9(2) = 18\text{ cm}\)
\(b = 7(2) = 14\text{ cm}\)
\(c = 6(2) = 12\text{ cm}\)
Semi-perimeter (\(s\)) = \(\frac{44}{2} = 22\text{ cm}\)
Using Heron's formula, Area = \(\sqrt{s(s - a)(s - b)(s - c)}\)
Area = \(\sqrt{22(22 - 18)(22 - 14)(22 - 12)}\)
Area = \(\sqrt{22 \times 4 \times 8 \times 10}\)
Area = \(\sqrt{(2 \times 11) \times 4 \times (4 \times 2) \times (2 \times 5)}\)
Area = \(\sqrt{4 \times 11 \times 4 \times 4 \times 2 \times 2 \times 5}\)
Area = \(\sqrt{4^2 \times 2^2 \times 4 \times 11 \times 5}\)
Area = \(4 \times 2 \times \sqrt{220} = 8\sqrt{55}\text{ cm}^2 \approx 59.33\text{ cm}^2\)
In simple words: First find the actual side lengths using the given ratio and perimeter, then apply Heron's formula to find the area.

Exam Tip: When sides are given in a ratio, always introduce a common variable \(x\) and sum them to equate with the given perimeter.

Question 3. The base of a right angled triangle is 5cm and hypotenuse is 13cm. Find its area.
Answer: Let the base be \(b = 5\text{ cm}\) and hypotenuse be \(h = 13\text{ cm}\).
Using Pythagoras theorem to find the perpendicular (\(p\)):
\(p^2 + b^2 = h^2\)
\(p^2 + 5^2 = 13^2\)
\(p^2 + 25 = 169\)
\(p^2 = 169 - 25 = 144\)
\(p = 12\text{ cm}\)
Area of the right-angled triangle = \(\frac{1}{2} \times \text{base} \times \text{perpendicular}\)
Area = \(\frac{1}{2} \times 5 \times 12 = 30\text{ cm}^2\)
In simple words: Find the missing height using Pythagoras theorem, then use the standard area formula for a right triangle.

Exam Tip: Memorize common Pythagorean triples like \((5, 12, 13)\) to save time during exams.

Question 4. Find the area of quadrilateral ABCD whose sides are 9m, 40m, 28m and 15m. The angle between the first two sides is a right angle.
Answer: Let the sides be \(AB = 9\text{ m}\), \(BC = 40\text{ m}\), \(CD = 28\text{ m}\), and \(DA = 15\text{ m}\).
The angle between \(AB\) and \(BC\) is a right angle, meaning \(\angle B = 90^\circ\).
Join diagonal \(AC\). In right-angled triangle \(ABC\):
\(AC^2 = AB^2 + BC^2\)
\(AC^2 = 9^2 + 40^2 = 81 + 1600 = 1681\)
\(AC = \sqrt{1681} = 41\text{ m}\)
Area of right-angled triangle \(ABC\) = \(\frac{1}{2} \times 9 \times 40 = 180\text{ m}^2\)
Now, for triangle \(ACD\), the sides are \(a = 28\text{ m}\), \(b = 15\text{ m}\), and \(c = 41\text{ m}\).
Semi-perimeter (\(s\)) = \(\frac{28 + 15 + 41}{2} = \frac{84}{2} = 42\text{ m}\)
Area of triangle \(ACD\) = \(\sqrt{42(42 - 28)(42 - 15)(42 - 41)}\)
Area = \(\sqrt{42 \times 14 \times 27 \times 1}\)
Area = \(\sqrt{(14 \times 3) \times 14 \times (9 \times 3)}\)
Area = \(\sqrt{14^2 \times 3^2 \times 3^2} = 14 \times 3 \times 3 = 126\text{ m}^2\)
Total area of quadrilateral \(ABCD\) = Area of \(\triangle ABC\) + Area of \(\triangle ACD\)
Total Area = \(180 + 126 = 306\text{ m}^2\)
In simple words: Divide the quadrilateral into two triangles using a diagonal, find the area of each triangle separately, and add them up.

Exam Tip: Always draw a rough sketch and use the diagonal to split a non-standard quadrilateral into two triangles.

Question 5. Find the area of a trapezium whose parallel sides measure 60cm and 77cm and non-parallel sides are 25cm and 26cm.
Answer: Let the parallel sides be \(AB = 60\text{ cm}\) and \(CD = 77\text{ cm}\), and non-parallel sides be \(AD = 25\text{ cm}\) and \(BC = 26\text{ cm}\).
Draw \(BE \parallel AD\) such that \(ABED\) is a parallelogram.
Then, \(BE = AD = 25\text{ cm}\), \(ED = AB = 60\text{ cm}\), and \(EC = CD - ED = 77 - 60 = 17\text{ cm}\).
Now consider triangle \(BEC\) with sides \(25\text{ cm}\), \(26\text{ cm}\), and \(17\text{ cm}\).
Semi-perimeter (\(s\)) = \(\frac{25 + 26 + 17}{2} = \frac{68}{2} = 34\text{ cm}\)
Area of triangle \(BEC\) = \(\sqrt{34(34 - 25)(34 - 26)(34 - 17)}\)
Area = \(\sqrt{34 \times 9 \times 8 \times 17}\)
Area = \(\sqrt{(17 \times 2) \times 9 \times (4 \times 2) \times 17}\)
Area = \(\sqrt{17^2 \times 3^2 \times 4^2 \times 2^2} = 17 \times 3 \times 4 \times 2 = 408\text{ cm}^2\)
Also, Area of triangle \(BEC\) = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 17 \times h\)
\(\frac{1}{2} \times 17 \times h = 408\)
\(17h = 816\)
\(h = 48\text{ cm}\)
Area of trapezium = \(\frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{height}\)
Area = \(\frac{1}{2} \times (60 + 77) \times 48\)
Area = \(\frac{1}{2} \times 137 \times 48 = 137 \times 24 = 3288\text{ cm}^2\)
In simple words: Convert the trapezium into a parallelogram and a triangle, find the height of the triangle using Heron's formula, and then use the standard trapezium area formula.

Exam Tip: Constructing a parallelogram inside the trapezium reduces the problem to finding the area and altitude of a standard triangle.