Exercise 13.3

1. Find the area of each of the following triangles.

2. In ΔPQR, PQ = 4 cm, PR = 8 cm and RT = 6 cm. Find (i) the area of ΔPQR (ii) the length of QS.

Answer:

Area of triangle = 1/2 × ×

Area = 1/2 × ×

Area = 1/2 × 24 = cm²

(ii) In ΔPRS, by Pythagoras theorem: QS² = PR² - RT²

QS = 2−2

QS = 64−36 =

QS = cm (rounded to two decimals)

3. ΔABC is right-angled at A. AD is perpendicular to BC, AB = 5 cm, BC = 13 cm and AC = 12 cm. Find the area of ΔABC. Also, find the length of AD.

Answer:

Area of triangle = 1/2 × ×

Area = 1/2 × ×

Area = 1/2 × 60 = cm²

Let AD = h. Area = 1/2 × ×

= 1/2 × 13 × h

30 = × h

h = 30 ÷ 6.5 = cm (rounded to two decimals)

4. ΔPQR is isosceles with PQ = PR = 7.5 cm and QR = 9 cm. The height PS from P to QR is 6 cm. Find the area of ΔPQR. What will be the height from R to PQ i.e. RT?

Answer:

Area of triangle = 1/2 × ×

Area = 1/2 × ×

Area = 1/2 × 54 = cm²

Let RT = h. Area = 1/2 × ×

= 1/2 × 7.5 × h

27 = × h

h = 27 ÷ 3.75 = cm

5. ABCD rectangle with AB = 8 cm, BC = 16 cm and AE = 4 cm. Find the area of ABCE. Is the area of ABEC equal to the sum of the area of ABAE and ACDE. Why?

Answer:

Area of rectangle ABCD = ×

Area = ×

Area = cm²

Area of ABCE = ×

Area = ×

Area = cm²

Area of EBCE = × ( - )

Area = 8 × 12 = cm²

Sum of areas = + = cm²

Yes, the area of ABEC is equal to the sum of the area of ABAE and ACDE because the sum covers the entire rectangle ABCD.

6. Ramu says that the area of ΔPQR is, A = 12 × 7 × 5 cm².
Gopi says that it is, A = 12 × 8 × 5 cm². Who is correct? Why?

Answer:

Area of triangle = 12 × ×

Area = 12 × ×

Area = 12 × 40 = cm²

Gopi is correct because the base and the corresponding height must be perpendicular. Here, QR = 8 cm is the base and the perpendicular from P to QR is 5 cm.

7. Find the base of a triangle whose area is 220 cm² and height is 11 cm.

Answer:

Area = 12 × ×

220 = 12 × ×

220 = 12 × 11 × base

220 = × base

base = 220 ÷ 5.5 = cm

8. In a triangle the height is double the base and the area is 400 cm². Find the length of the base and height.

Answer:

Let base = cm, height = cm

Area = 12 × ×

Area = 12 × 2x² = x

x² =

x = 400 = cm

Base = cm, Height = cm

9. The area of triangle is equal to the area of a rectangle whose length and breadth are 20 cm and 15 cm respectively. Calculate the height of the triangle if its base measures 30 cm.

Answer:

Area of rectangle = ×

Area = cm²

Area of triangle = 12 × ×

300 = 12 × 30 × height

300 = × height

height = 300 ÷ 15 = cm

10. In Figure ABCD find the area of the shaded region.

Answer:

Area of square = ×

Area = cm²

Area of central triangle = 12 × ×

Area = 12 × 1600 = cm²

Area of shaded region = 1600 - 800 = cm²

11. In Figure ABCD, find the area of the shaded region.

Answer:

Area of square ABCD = ×

Area = cm²

Area of triangle ADE = 12 × ×

Area = 12 × 160 = cm²

Area of triangle FBC = 12 × ×

Area = 12 × 200 = cm²

Area of triangle AFE = 12 × ×

Area = 12 × 120 = cm²

Area of shaded region = Area of square ABCD - (Area of triangle ADE + Area of triangle FBC + Area of triangle AFE)

Area of shaded region = - ( + + )

Area of shaded region = 400 -240 = cm²

12. Find the area of a parallelogram PQRS, if PR = 24 cm and QU = ST = 8 cm.

Answer:

Area of parallelogram PQRS = Area of triangle PSR + Area of triangle PQR

Area of triangle PSR = 12 × ×

Area = 12 × ×

Area = 12 × 192 = cm²

Area of triangle PQR = 12 × ×

Area = 12 × ×

Area = 12 × 192 = cm²

Area of parallelogram PQRS = +

Area = cm²

13. The base and height of the triangle are in the ratio 3:2 and its area is 108 cm². Find its base and height.

Answer:

Let base = cm, height = cm

Area of triangle = 12 × ×

108 = 12 × ×

108 = 12 × x2

108 = x2

x2 = 108 ÷

x2 =

x = 36 = cm

Base = 3x = 3 × = cm

Height = 2x = 2 × = cm