The document RD Sharma Solutions (Part - 1) - Ex-20.3, Mensuration - I, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.

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**Find the area of a parallogram with base 8 cm and altitude 4.5 cm.**

We have,

Base = 8 cm and altitude = 4.5 cm

Thus,

Area of the parallelogram = Base x Altitude

= 8 cm x 4.5 cm

= 36 cm

Find the area in square metres of the parallelogram whose base and altitudes are as under:

(i) Base = 15 dm, altitude = 6.4 dm

(ii) Base = 1 m 40 cm, altitude = 60 cm

We have,

(i) Base = 15 dm = (15 x 10) cm = 150 cm = 1.5 m [Since 100 cm = 1 m]

Altitude = 6.4 dm = (6.4 x 10) cm = 64 cm = 0.64 m

Thus,

Area of the parallelogram = Base x Altitude

= 1.5 m x 0.64 m

= 0.96 m^{2}

(ii) Base = 1 m 40 cm = 1.4 m [Since 100 cm = 1 m]

Altitude = 60 cm = 0.6 m

Thus,

Area of the parallelogram = Base x Altitude

= 1.4 m x 0.6 m

= 0.84 m^{2}

**Find the altitude of a parallelogram whose area is 54 dm ^{2} and base is 12 dm.**

We have,

Area of the given parallelogram = 54 dm^{2}

Base of the given parallelogram = 12 dm

∴ Altitude of the given parallelogram = Area/Base=54/12dm = 4.5 dm

**The area of a rhombus is 28 m ^{2}. If its perimeter be 28 m, find its altitude.**

We have,

Perimeter of a rhombus = 28 m

∴ 4(Side) = 28 m [Since perimeter = 4(Side)]

⇒ Side = 28 m/4 =7 m

Now,

Area of the rhombus = 28 m^{2}

⇒ (Side x Altitude) = 28 m^{2}

⇒ (7 m x Altitude) = 28 m^{2}

⇒ Altitude = 28 m^{2}/ 7m =4 m

**In Fig. 20, ABCD is a parallelogram, DL ⊥ AB and DM ⊥ BC. If AB = 18 cm, BC = 12 cm and DM = 9.3 cm, find DL.**

We have,

Taking *BC* as the base,*BC* = 12 cm and altitude* DM* = 9.3 cm

∴ Area of parallelogram *ABCD* = Base x Altitude

= (12 cm x 9.3 cm) = 111.6 cm^{2} ......... (i)

Now,

Taking *AB* as the base, we have,

Area of the parallelogram *ABCD* = Base x Altitude = (18 cm x *DL*).................(ii)

From (i) and (ii), we have

18 cm x *DL* = 111.6 cm^{2}

**The longer side of a parallelogram is 54 cm and the corresponding altitude is 16 cm. If the altitude corresponding to the shorter side is 24 cm, find the length of the shorter side.**

We have,*ABCD* is a parallelogram with the longer side* AB* = 54 cm and corresponding altitude *AE* = 16 cm.

The shorter side is *BC* and the corresponding altitude is* CF *= 24 cm.

Area of a parallelogram = base × height.We have two altitudes and two corresponding bases. So,

Hence, the length of the shorter side *BC *= *AD* = 36 cm.

**In Fig. 21, ABCD is a parallelogram, DL ⊥ AB. If AB = 20 cm, AD = 13 cm and area of the parallelogram is 100 cm^{2}, find AL.**

We have,*ABCD* is a parallelogram with base *AB* = 20 cm and corresponding altitude *DL.*

It is given that the area of the parallelogram *ABCD* = 100 cm^{2}

Now,

Area of a parallelogram = Base x Height

100 cm^{2} = *AB* x *DL*

100 cm^{2} = 20 cm x *DL*

Again by Pythagoras theorem, we have,

(*AD*)^{2} = (*AL*)^{2 }+ (*DL*)^{2}

⇒ (13)^{2} = (*AL*)^{2} + (5)^{2}

⇒ (*AL*)^{2 }= (13)^{2 }- (5)^{2 } = 169 − 25 = 144

⇒ (*AL*)^{2} = (12)^{2}

⇒ *AL* = 12 cm

Hence. length of *AL* is 12 cm.

**In Fig. 21, if AB = 35 cm, AD = 20 cm and area of the parallelogram is 560 cm^{2}, find LB.**

We have,*ABCD *is a parallelogram with base *AB *= 35 cm and corresponding altitude *DL*. The adjacent side of the parallelogram *AD* = 20 cm.

It is given that the area of the parallelogram* ABCD* = 560 cm^{2}

Now,

Area of the parallelogram = Base x Height

560 cm^{2} = *AB* x *DL*

560 cm^{2} = 35 cm x *DL*

Again by Pythagoras theorem, we have,

(*AD*)^{2} = (*AL*)^{2 }+ (*DL)*^{2}

⇒ (20)^{2} = (*AL*)^{2} + (16)^{2}

⇒ (*AL*)^{2 }= (20)^{2 }− (16)^{2 } = 400 − 256 = 144

⇒ (*AL)*^{2} = (12)^{2}

⇒ *AL *= 12 cm

From the figure,*AB* =* AL* + *LB*

35 cm = 12 cm +* LB*

∴* LB* = 35 cm − 12 cm

= 23 cm

Hence, length of* LB* is 23 cm.

**The adjacent sides of a parallelogram are 10 m and 8 m. If the distance between the longer sides is 4 m, find the distance between the shorter sides.**

We have,*ABCD* is a parallelogram with side *AB* = 10 m and corresponding altitude *AE* = 4 m.

The adjacent side *AD* = 8 m and the corresponding altitude is* CF.*

Area of a parallelogram = Base × Height

We have two altitudes and two corresponding bases. So,*AD* x *CF* = *AB* x *AE*

⇒ 8 m x *CF* = 10 m x 4 m

Hence, the distance between the shorter sides is 5 m.

**The base of a parallelogram is twice its height. If the area of the parallelogram is 512 cm ^{2}, find the base and height.**

Let the height of the parallelogram be *x* cm.

Then the base of the parallelogram is *2x* cm.

It is given that the area of the parallelogram = 512 cm^{2}

So,

Area of a parallelogram = Base x Height

512 cm^{2} = 2*x* x *x*

512 cm^{2} = *2x ^{2}*

Hence, base = *2x* *= *2 x 16 = 32 cm and height = *x *= 16 cm.

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