Area Exercise 10.1 |

**Q.1: In the adjoining figure, show that ABCD is parallelogram .Calculate the area parallelogram ABCD.**

**Sol:**

Since the diagonal BD divides ABCD into two triangles of equal area.

Therefore,

Therefore, Area of parallelogram = 35 cm^{2}

**Q.2: In a parallelogram ABCD, it is being given that AB = 10cm and the altitudes corresponding to the sides AB and AD are DL = 6cm and BM = 8cm, respectively. Find AD.**

**Sol:**

Since, ABCD is a parallelogram and DL Is perpendicular to AB

So, its area =AB x DL = (10 x 6) cm^{2} = 60cm^{2}

Also in parallelogram ABCD,

Therefore, Area of parallelogram ABCD= AD x BM

60 = AD x 8cm

AD =

Therefore, AD = 7.5 cm

**Q.3: Find the area of a rhombus, the lengths whose diagonals are 16cm and 24cm respectively.**

**Sol:**

ABCD is a rhombus in which diagonal Ac = 24cm and BD = 16cm

These diagonals intersect at O:

Since, diagonals of a rhombus are perpendicular to each other. So, in

OD is its altitude and AC is its base

So, area of

= 1/2 x 24 x BD/2 = ^{2}

i.e. Area of

=

Now, the Area of rhombus = Area of ^{2} = 192 cm^{2}

**Q.4: Find the area of a rhombus, the lengths of whose diagonals are 16 cm and 24 cm respectively.**

**Sol:**

ABCD is a trapezium in which, AB || CD

AB = 9cm and CD = 6cm

CE is a perpendicular drawn to AB through C and CE = 8cm

Area of trapezium = 1/2 (sum of parallel sides) x distance between them = [1/2 (9+6) x 8] cm^{2 }= [1/2 x 15 x 8] cm^{2} = 60 cm^{2}

Therefore, the Area of trapezium = 60 cm^{2}

**Q.5: Calculate the area of quad ABCD, given in Fig.**

**Sol:**

ABCD is a quadrilateral:

Now in right angled

DB^{2 }= DC^{2} – CB^{2} = 17^{2} – 8^{2} = 289 – 64 = 225 cm^{2}

Therefore,

So, area of ^{2} = 60 cm^{2}

Again, in right angled

AB^{2} = DB^{2} – AD^{2} = 15^{2} – 9^{2} = 225 – 81 = 144cm^{2}

Therefore,

Therefore, the area of ^{2} = 54 cm^{2}

So, area of quadrilateral ABCD = Area of ^{2} =114 cm^{2}

Therefore, the area of quadrilateral ABCD = 114 cm^{2}

**Q.6: Calculate the area of trap PQRS, given in Fig. (ii)**

**Sol:**

In right angled

RT^{2} = RQ^{2} – TQ^{2} = 17^{2} – 8^{2} = 289 – 64 = 225 cm^{2}

Therefore,

Therefore, Area of trapezium = 1/2(sum of parallel sides) x distance between them

= 1/2 x (PQ + SR) x RT =1/2 x (16 +8) x 15 = (1/2 x 24 x 15) cm^{2} = 180 cm^{2}

Therefore, the Area of trapezium = 180 cm^{2}

**Q.7: BD is one of the diagonals of a diagonal of a quad. ABCD. If AL⊥BD and CM⊥BD , show that**

**Area (quad. ABCD) = 1/2 * BD * (AL + CM)**

**Sol:**

**Given:** ABCD is a quadrilateral and BD is one of its diagonals

**To prove:** area (quad. ABCD) = 1/2 x BD x (AL + CM)

**Proof:**

Area of

Area of

Therefore, Area of quadrilateral ABCD= Area of

Therefore, the Area of quadrilateral ABCD = 1/2 x BD [AL + CM]

**Q.8: In the adjoining figure, ABCD is a quadrilateral in which diag. BD = 14cm. If AL⊥BD and CM⊥BD such that AL = 8cm and CM = 6cm, find the area of quad ABCD.**

**Sol:**

Area of ^{2} = 56cm^{2}

Area of ^{2} = 42cm^{2}

Therefore, the Area of quadrilateral ABCD= Area of ^{2} = 98 cm^{2}

**Q.9: In the adjoining figure, ABCD is a trapezium in which AB∥BD and its diagonals AC and BD intersect at O.**

**Prove that area of triangle AOD = area of the triangle of BOC.**

**Sol:**

Consider

Triangles on the same base and between the same parallels are equal in area.

So

Therefore,

Now,

And,

=

Therefore, Area

**Q.10: In the adjoining figure, DE∥BC. Prove that:**

**(i) Area of triangle ACD = area ABE**

**(ii) Area of triangle OCE = area of triangle OBD**

**Sol:**

**(i)**

So, area

Adding area

Therefore,

**(ii)** Since

Subtracting

Therefore,

**Q.11: In the adjoining figure, D and E are points on the sides AB and AC of triangle ABC such that area of triangle BCE = area of triangle BCD.**

**Show that: DE∥BC**

**Sol:**

**Given:** A

Such that

**To prove:**

**Proof:**

As

This means that they lie between two parallel lines. Therefore,

**Question 12: In the adjoining figure, O is any point inside a parallelogram ABCD. Prove that:**

**(i) ar.(**

**(ii) ar.( △OAD)+ ar.( △OBC)=**

**Sol:**

**Given:** A parallelogram ABCD in which O is a point inside it

**To prove: (i)**

**(ii)**

**Construction:** Though O draw

**Proof: (i)**

If a triangle and a parallelogram are on the same base, and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram.

Therefore,

Similarly,

So,

=

=

=

**(ii)**

Thererfore,

Similarly,

Therefore,

=

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