The DAV Maths Class 7 Solutions and **DAV Class 7 Maths Chapter 11 Worksheet 4** Solutions of Perimeter and Area offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 11 WS 4 Solutions

Question 1.

The area of a triangle is 90 cm^{2}. If its base is 15 cm, find its altitude.

Answer:

Area of triangle = \(\frac{1}{2}\) × base × altitude

90 = \(\frac{1}{2}\) × 15 × altitude

Altitude = \(\frac{90 \times 2}{15}\) = 12 cm

Hence the required altitude = 12 cm.

Question 2.

Find the base of a triangle whose area is 0.48 dm^{2} and altitude is 8 cm.

Answer:

Area = 0.48 dm^{2}

= 0.48 x 100 cm^{2} = 48 cm^{2}

Area of triangle = \(\frac{1}{2}\) × base × altitude

⇒ 48 = \(\frac{1}{2}\) × base × 8

⇒ 48 = base × 4

base = \(\frac{48}{4}\) = 12 cm

Hence the required base = 12 cm.

Question 3.

Find the area of a right triangle whose hypotenuse is 13 cm and one side is 5 cm.

Answer:

Let the other side be x cm From Pythagorous theorem,

x^{2} + (5)^{2} = (13)^{2}

x^{2} + 25 = 169

x^{2} = 169 – 25 = 144

x = 12 cm

Area = \(\frac{1}{2}\) × base × altitude

= \(\frac{1}{2}\) × 12 × 5

= 30 cm^{2}

Hence the required area = 30 cm^{2}.

Question 4.

Find the area of an isosceles right triangle whose equal sides are 15 cm each.

Answer:

Here base = height = 15 cm

Area of the triangle = \(\frac{1}{2}\) × base × height

= \(\frac{1}{2}\) × 15 × 15

= \(\frac{225}{2}\)

= 112.5 cm^{2}

Hence the required area = 112.5 cm^{2}.

Question 5.

Find the area of an isosceles right triangle having the length of each equal sides 5 cm.

Answer:

Here base = height = 5 cm

∴ Area = \(\frac{1}{2}\) × base × height

= \(\frac{1}{2}\) × 5 × 5 = \(\frac{25}{2}\) = 12.5 cm^{2}

Hence the required area = 12.5 cm^{2}.

Question 6.

The base of a triangular field is three times its height. If the cost of cultivating the field at ₹ 36 per hectare is ₹ 486, find its base and height (1 hectare = 10,000 m^{2}).

Answer:

Area of the triangular field = \(\frac{486}{36}=\frac{27}{2}\) hectare

= \(\frac{1}{2}\) × 10000 m^{2}

= 27 × 5000 m^{2}

= 135000 m^{2}

Let the height be x m

∴ Base = 3 x m

Area = \(\frac{1}{2}\) × base × height

135000 = \(\frac{1}{2}\) × 3x × x

135000 × 2 = 3x^{2}

270000 = 3x^{2}

x^{2} = \(\frac{270000}{3}\)

x^{2} = 90000

∴ x = 300

Hence the height is 300 m and base = 3 × 300

= 900 m.

Question 7.

In the given figure, ST = 5 cm, QR = 9 cm. The area of the larger triangle is 50 cm^{2} . What is the area of shaded region?

Answer:

Area of the smaller Δ SQT

= \(\frac{1}{2}\) × QR × ST

= \(\frac{1}{2}\) × 9 × 5

= \(\frac{45}{2}\) = 22.5 cm

Area of the bigger ∆ PQR = 50 cm^{2}

Area of the shaded region = 50 cm^{2} – 22.5 cm^{2} = 27.5 cm^{2}

Hence the required area = 27.5 cm^{2} .

Question 8.

The area of a rhombus is 98 m^{2} . If one of its diagonal is 14 m, find the other diagonal

Answer:

If d_{1} and d_{2} be the diagonals of the rhombus, then

Area of the rhombus = \(\frac{1}{2}\) × d_{1} × d_{2}

98 = \(\frac{1}{2}\) × 14 × d_{2}

\(\frac{98 \times 2}{14}\) = d_{2}

d_{2} = 14 cm

Hence the length of the required diagonal = 14 cm.

Question 9.

The diagonals of a rhombus are 8 cm and 6 cm. Find its area.

Answer:

Area of a rhombus = \(\frac{1}{2}\) × d_{1} × d_{2}

= \(\frac{1}{2}\) × 8 × 6 = 4 × 6 = 24 cm^{2}

Hence the required area = 24 cm^{2}.

Question 10.

Find the area of a rhombus whose side is 13 cm and altitude is 2 cm.

Answer:

Area of a rhombus = Base × Altitude

= 13 × 2 = 26 cm^{2}

Hence the required area = 26 cm^{2}.

Question 11.

The area of a rhombus is 220.5 cm^{2}. If its altitude is 17.5 cm, find the length of each side of the rhombus.

Answer:

Area of rhombus = Base × Altitude

220.5 = Base × 17.5

Hence the side of the rhombus = 12.5 cm.

Question 12.

A diagonal of a quadrilateral is 40 m long and the perpendiculars to it from the opposite corner are 8 m and 10 m respectively. Find its area.

Answer:

Area of □ ABCD = area of Δ ABC + area of Δ ADC

= \(\frac{1}{2}\) × AC × BE + y × AC × DF

= \(\frac{1}{2}\) × AC × (BE + DE)

= \(\frac{1}{2}\) × 40 × (8 + 10)

= \(\frac{1}{2}\) × 40 × 18

= 20 × 18 = 360 m^{2}

Hence the required area = 360 m^{2}.

Question 13.

Find the area of a rhombus whose side is 15 cm long while one of the diagonal is 24 cm long.

Answer:

Join BD

As the diagonals of a rhombus bisect each other at right angle,

∴ OA = \(\frac{24}{2}\) = 12 cm

In right ΔAOD

OA^{2} + OD^{2} = AD^{2} (By Pythagoras Theorem)

(12)^{2} + OD^{2} = (15)^{2}

144 + OD^{2} = 225

OD^{2} = 225 – 144

OD^{2} = 81

OD = 9 cm

BD = 9 × 2 = 18 cm

Area of the rhombus = \(\frac{1}{2}\) × d_{1} × d_{2}

= \(\frac{1}{2}\) × 24 × 18

= 12 × 18

= 216 cm^{2}

Hence the area of the rhombus = 216 cm^{2}.