The document NCERT Solutions(Part- 2)- Mensuration Class 8 Notes | EduRev is a part of the Class 8 Course Class 8 Mathematics by VP Classes.

All you need of Class 8 at this link: Class 8

**Question: **We know that parallelogram is also a quadrilateral. Let us also split such a quadrilateral into two triangles, find their areas and hence that of the parallelogram. Does this agree with the formula that you know already?

**Solution: **The diagonal BD of quadrilateral ABCD is joined and it divides the quadrilateral into two triangles.

Now,

Area of quadrilateral ABCD = Area of D ABD + Area of D BCD

Infact ABCD is a parallelogram.

âˆ´ Area of a parallelogram ABCD = b * h

Area of a parallelogram = Base * Height

We know that a parallelogram can also be a trapezium. We already know that

Area of trapezium ABCD = 1/2 (Sum of parallel sides) * [Perpendicular distance between the parallel sides]

or Area of parallelogram ABCD = bh.

Yes, the above relation agrees with formula that we know already.

**Question: **A parallelogram is divided into two congruent triangles by drawing a diagonal across it. Can we divide a trapezium into two congruent triangles?

**Solution: **No, a trapezium cannot be divided into two congruent triangles.

**Area of Special Quadrilaterals**

Let ABCD be a rhombus. Therefore, its diagonals are perpendicular to each other.

Area of rhombus ABCD = Area of Î” ACD + Area of Î” ABC

where d_{1} and d_{2} are the diagonals of the rhombus.

Thus, the area of a rhombus =1/2 * The product of its diagonals

Note: Splitting a quadrilateral into triangles is called triangulation.

**Example 2. **Find the area of a rhombus whose diagonals are 12 cm and 9.2 cm.

**Solution:** Let d_{1} and d_{2} be the diagonals of the rhombus.

âˆ´ d_{1} = 12 cm and d_{2} = 9.2 cm

âˆµ Area of rhombus =1/2 * d_{1} * d_{2}

âˆ´ Area of the given rhombus =1/2 * 12 * 9.2 cm^{2}

= 6 * 9.2 cm^{2} = 55.2 cm^{2}

**Question:** Find the area of these quadrilaterals:

**Solution:**

**(i)** Area of quadrilateral ABCD

= 1/2 * AC * [Sum of perpendiculars on AC from opposite vertices]

= 1/2 * 6 cm * [3 cm + 5 cm]

= 1/2 * 6 cm * 8 cm = 3 cm * 8 cm = 24 cm^{2}

**(ii)** The given figure is a rhombus having d_{1} = 7 cm and d_{2} = 6 cm.

âˆ´ Area of the given rhombus =1/2 * Product of diagonals

= 1/2 * d_{1} * d_{2}

= 1/2 *7 cm * 6 cm

= 7 cm * 3 cm = 21 cm^{2}

**(iii)** The given figure is a parallelogram. Its diagonal divides it in totwo congruent triangles.

âˆ´ Area of the parallelogram = 2 * [Area of one of the triangles]

**Area of a Polygon**

To find area of polygons, we divide them into shapes, for which we have a formula for the area. First we find the areas of various parts and then add them to get the area of given polygon.

**Question 1.** Divide the following polygons into parts (triangles and trapezium) to find out its area.

**Solution:**

**(a) **We draw perpendiculars from opposite vertices on FI, i.e. GL âŠ¥ FI, HM âŠ¥ FI and EN âŠ¥ FI

Area of the polygon EFGHI

= ar (Î” GFL) + ar (trapezium GLMH) + ar (Î” HMI) + ar (Î” NEI) + ar (Î” EFN)

**(b)** NQ is a diagonal. Draw OA âŠ¥ NQ, MB âŠ¥ NQ, PC âŠ¥ NQ and RD âŠ¥ NQ

âˆ´ Area of polygon OPQRMN = ar (Î” OAN) + ar (trap. CPOA) + ar (Î” PCQ) + ar (Î” RDQ)+ ar (trap. MBDR) + ar (Î” MBN)

**Question 2.** Fill in the blanks.

Polygon ABCDE is divided into parts as shown below. Find its area if AD = 8 cm, AH = 6 cm, AG = 4 cm, AF = 3 cm and perpendiculars BF = 2 cm, CH = 3 cm, EG = 2.5 cm.

Area of polygon ABCDE = Area of Î”AFB + â€¦

Area of Î” AFB =1/2 * AF * BF = 1/2 * 3 * 2 = â€¦

Area of trapezium FBCH =

So, the area of polygon ABCDE = â€¦

**Solution:** Area of polygon ABCDE = Area of Î” AFB + Area of trapezium FBCH + Area of Î” CHD + Area of Î” ADE

Area of D AFB = 1/2 * AF * BF

= 1/2 * 3 * 2 = 3 cm^{2}

Area of trapezium FBCH

So, the area of polygon ABCD = 3 cm^{2} + 7.5 cm^{2} + 3 cm^{2} + 10 cm^{2} = 23.5 cm^{2}

**Question 3.** Find the area of polygon MNOPQR if MP = 9 cm, MD = 7 cm MC = 6 cm, MB = 4 cm, MA = 2 cm. NA, OC, QD and RB are perpendiculars to diagonal MP.

**Solution: **Area of polygon MNOPQR = ar (Î” MAN) + ar (trap. ACON) + ar (Î” OCP) + ar (Î” PDQ) + ar (trap. DBRQ) + ar Î” RBM)

âˆµ

âˆ´ Area of polygon MNOPQR

93 docs|16 tests

### NCERT Solutions(Part- 3)- Mensuration

- Doc | 5 pages
### NCERT Solutions(Part- 4)- Mensuration

- Doc | 5 pages
### NCERT Solutions(Part- 5)- Mensuration

- Doc | 4 pages
### NCERT Solutions(Part- 6)- Mensuration

- Doc | 6 pages
### Extra Questions- Mensuration

- Doc | 5 pages

- NCERT Solutions(Part- 1)- Mensuration
- Doc | 7 pages
- Mensuration - MCQ
- Test | 10 ques | 10 min