The document RD Sharma Solutions Ex-12.2, (Part -2), Heron's Formula, Class 9, Maths Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.

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**Q8. Find the area of the quadrilateral ABCD in which AD = 24 cm, angle BAD = 90 ^{âˆ˜} and BCD forms an equilateral triangle whose each side is equal to 26 cm. [Take âˆš3 = 1.73]**

**Solution:**

Given that, in a quadrilateral ABCD in which AD = 24 cm,

Angle BAD = 90âˆ˜

BCD is an equilateral triangle and the sides BC = CD = BD = 26 cm

In triangle BAD, by applying Pythagoras theorem,

BA^{2} = BD^{2}âˆ’AD^{2}

BA^{2} = 26^{2}+24^{2}

Area of the triangle

Area of the triangle

Area of the triangle BAD = 120 cm^{2}

Area of the equilateral triangle

Area of the equilateral triangle QRS

Area of the equilateral triangle BCD = 292.37 cm^{2}

Therefore, the area of quadrilateral ABCD = area of triangle BAD + area of the triangle BCD

The area of quadrilateral ABCD = 120 + 292.37

= 412.37 cm^{2}

**Q9. Find the area of quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm and the diagonal BD = 20 cm.**

**Solution:**

Given

AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm, and the diagonal

BD = 20 cm.

Now, for the area of triangle ABD

Perimeter of triangle ABD 2s = AB + BD + DA

2s = 34 cm + 42 cm + 20 cm

s = 48 cm/p>

By using Heronâ€™s Formula,

Area of the triangle

= 336cm^{2}

Now, for the area of triangle BCD

Perimeter of triangle BCD 2s = BC + CD + BD

2s = 29cm + 21cm + 20cm

s = 35 cm

By using Heronâ€™s Formula,

Area of the triangle BCD

= 210cm^{2}

Therefore, Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD

Area of quadrilateral ABCD = 336 + 210

Area of quadrilateral ABCD = 546 cm^{2}

**Q10. Find the perimeter and the area of the quadrilateral ABCD in which AB = 17 cm, AD = 9 cm, CD = 12 cm, AC = 15 cm and angle ACB = 90 ^{âˆ˜}.**

**Solution:**

Given are the sides of the quadrilateral ABCD in which

AB = 17 cm, AD = 9 cm, CD = 12 cm, AC = 15 cm and an angle ACB = 90^{âˆ˜}

By using Pythagoras theorem

BC^{2} = AB^{2}âˆ’AC^{2}

BC^{2} = 17^{2}âˆ’15^{2}

BC = 8 cm

Now, area of triangle

area of triangle

area of triangle ABC = 60 cm^{2}

Now, for the area of triangle ACD

Perimeter of triangle ACD 2s = AC + CD + AD

2s = 15 + 12 +9

s = 18 cm

By using Heronâ€™s Formula,

Area of the triangle ACD

= 54cm^{2}

Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD

Area of quadrilateral ABCD = 60 cm^{2} + 54cm^{2}

Area of quadrilateral ABCD = 114 cm^{2}

**Q11. The adjacent sides of a parallelogram ABCD measure 34 cm and 20 cm, and the diagonal AC measures 42 cm. Find the area of parallelogram.**

**Solution:**

Given,

The adjacent sides of a parallelogram ABCD measures 34 cm and 20 cm, and the diagonal AC measures 42 cm.

Area of the parallelogram = Area of triangle ADC + Area of triangle ABC

Note: Diagonal of a parallelogram divides into two congruent triangles

Therefore,

Area of the parallelogram = 2 Ã— (Area of triangle ABC)

Now, for area of triangle ABC

Perimeter = 2s = AB + BC + CA

2s = 34 cm + 20 cm + 42 cm

s = 48 cm

By using Heronâ€™s Formula,

Area of the triangle ABC =

= 336cm^{2}

Therefore, area of parallelogram ABCD = 2 Ã— (Area of triangle ABC)

Area of parallelogram = 2 Ã— 336cm^{2}

Area of parallelogram ABCD = 672 cm^{2}

**Q12. Find the area of the blades of the magnetic compass shown in figure given below.**

**Solution:**

Area of the blades of magnetic compass = Area of triangle ADB + Area of triangle CDB

Now, for the area of triangle ADB

Perimeter = 2s = AD + DB + BA

2s = 5 cm + 1 cm + 5 cm

s = 5.5 cm

By using Heronâ€™s Formula,

Area of the triangle

= 2.49cm^{2}

Also, area of triangle ADB = Area of triangle CDB

Therefore area of the blades of the magnetic compass = 2 Ã— area of triangle ADB

Area of the blades of the magnetic compass = 2 Ã—2.49

Area of the blades of the magnetic compass = 4.98 cm^{2}

**Q13. A hand fan is made by sticking 10 equal size triangular strips of two different types of paper as shown in the figure. The dimensions of equal strips are 25 cm, 25 cm and 14 cm. Find the area of each type of paper needed to make the hand fan.**

**Solution:**

Given that,

The sides of AOB

AO = 25 cm

OB = 25 cm

BA = 14 cm

Area of each strip = Area of triangle AOB

Now, for the area of triangle AOB

Perimeter = AO + OB + BA

2s = 25 cm +25 cm + 14 cm

s = 32 cm

By using Heronâ€™s Formula,

Area of the triangle

= 168cm^{2}

Also, area of each type of paper needed to make a fan = 5 Ã— Area of triangle AOB

Area of each type of paper needed to make a fan = 5 Ã— 168 cm^{2}

Area of each type of paper needed to make a fan = 840cm^{2}

**Q14. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, find the height of a parallelogram.**

**Solution:**

The sides of the triangle DCE are

DC = 15 cm,

CE = 13 cm,

ED = 14 cm

Let the h be the height of parallelogram ABCD

Now, for the area of triangle DCE

Perimeter = DC + CE + ED

2s = 15 cm + 13 cm + 14 cm

s = 21 cm

By using Heronâ€™s Formula,

Area of the triangle

= 84cm^{2}

Also, area of triangle DCE = Area of parallelogram ABCD â‡’ 84cm^{2}

24 Ã— h = 84cm^{2}

h = 6 cm.

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