**RD Sharma Class 9 Solutions Chapter 12 Exercise 12.2: **We have the perfect Maths guide for all the Class 9 students, i.e., RD Sharma Solutions Class 9 Maths. You can practice the questions and clear all your doubts. RD Sharma Class 9 Solutions Chapter 12 Exercise 12.2 solutions are designed as per the current CBSE Syllabus by our subject matter experts. To know more, read the whole blog.

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RD Sharma Class 9 Solutions Chapter 12 Exercise 12.2

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**Question 1: Find the area of the quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.**

**Solution:**

Area of the quadrilateral ABCD = Area of △ABC + Area of △ADC ….(1)

△ABC is a right-angled triangle, which is right-angled at B.

Area of △ABC = 1/2 x Base x Height

= 1/2×AB×BC

= 1/2×3×4

= 6

Area of △ABC = 6 cm2 ……(2)

Now, In △CAD,

Sides are given, apply Heron’s Formula.

Perimeter = 2s = AC + CD + DA

2s = 5 cm + 4 cm + 5 cm

2s = 14 cm

s = 7 cm

Area of the △CAD = 9.16 cm2 …(3)

Using equations (2) and (3) in (1), we get

Area of quadrilateral ABCD = (6 + 9.16) cm2

= 15.16 cm2.

**Question 2: The sides of a quadrilateral field, taken in order, are 26 m, 27 m, 7 m, and 24 m, respectively. The angle contained by the last two sides is a right angle. Find its area.**

**Solution:**

Here,

AB = 26 m, BC = 27 m, CD = 7 m, DA = 24 m

AC is the diagonal joined at A to C point.

Now, in △ADC,

From the Pythagoras theorem,

AC2 = AD2 + CD2

AC2 = 142 + 72

AC = 25

Now, the area of △ABC

All the sides are known, Apply Heron’s Formula.

Perimeter of △ABC= 2s = AB + BC + CA

2s = 26 m + 27 m + 25 m

s = 39 m

= 291.84

Area of a triangle ABC = 291.84 m2

Now, for the area of △ADC, (Right angle triangle)

Area = 1/2 x Base X Height

= 1/2 x 7 x 24

= 84

Thus, the area of a △ADC is 84 m2

Therefore, the area of rectangular field ABCD = Area of △ABC + Area of △ADC

= 291.84 m2 + 84 m2

= 375.8 m2

**Question 3: The sides of a quadrilateral, taken in order as 5, 12, 14, and 15 meters, respectively, and the angle contained by the first two sides is a right angle. Find its area.**

**Solution:**

Here, AB = 5 m, BC = 12 m, CD =14 m and DA = 15 m

Join the diagonal AC.

Now, the area of △ABC = 1/2 ×AB×BC

= 1/2×5×12 = 30

The area of △ABC is 30 m2

In △ABC (right triangle),

From the Pythagoras theorem,

AC2 = AB2 + BC2

AC2 = 52 + 122

AC2 = 25 + 144 = 169

or AC = 13

Now in △ADC,

All sides are known, apply Heron’s Formula:

Perimeter of △ADC = 2s = AD + DC + AC

2s = 15 m +14 m +13 m

s = 21 m

= 84

Area of △ADC = 84 m2

Area of quadrilateral ABCD = Area of △ABC + Area of △ADC

= (30 + 84) m2

= 114 m2

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