(i) ∠DAC = ∠BAC ...(1) [∵ Given] ∠DAC = ∠BCA ...(2) [∵ Alternate angle] ∠BAC = ∠ACD ...(3) [∵Alternate angles] From the equations (1), (2) and (3), we have ∠ACD = ∠BCA ...(4) Hence, Diagonal AC bisects angle C also. (ii) From the equation (2) and (4), we have ∠ACD = ∠DAC In ΔADC, ∠ACD = ∠DAC [∵ ProvRead more

(i) ∠DAC = ∠BAC …(1) [∵ Given]
∠DAC = ∠BCA …(2) [∵ Alternate angle]
∠BAC = ∠ACD …(3) [∵Alternate angles]
From the equations (1), (2) and (3), we have
∠ACD = ∠BCA …(4)
Hence, Diagonal AC bisects angle C also.
(ii) From the equation (2) and (4), we have
∠ACD = ∠DAC
In ΔADC,
∠ACD = ∠DAC [∵ Prove above]
AD = DC [∵ In a triangle, the sides opposite to equal angle are equal]
A parallelogram whose adjacent sides are equal, is a rhombus. Hence, ABCD is a rhombus.

(i) In ΔAPD and ΔCQB, DP = BQ [∵ Given] ∠ADP = ∠CBQ [∵ Alternate angle] AD = BC [∵ Opposite sides of a parallelogram] Hence, ΔAPD ≅ ΔCQB [∵ SAS Congruency rule] (ii) ΔAPD ≅ CQB [∵ Prove above] AP = CQ ...(1) [∵ CPCT] (iii) In ΔAQB and ΔCPD, QB = DB [∵ Given] ∠ABQ = ∠CDP [∵ Alternate angle] AB = CD [Read more

(i) In ΔAPD and ΔCQB,
DP = BQ [∵ Given]
∠ADP = ∠CBQ [∵ Alternate angle]
AD = BC [∵ Opposite sides of a parallelogram]
Hence, ΔAPD ≅ ΔCQB [∵ SAS Congruency rule]

(ii) ΔAPD ≅ CQB [∵ Prove above]
AP = CQ …(1) [∵ CPCT]

(iii) In ΔAQB and ΔCPD,
QB = DB [∵ Given]
∠ABQ = ∠CDP [∵ Alternate angle]
AB = CD [∵ Opposite sides of a parallelogram]
Hence, ΔAQB ≅ ΔCPD [ SAS Congruency rule]

(i) In ΔAPB and ΔCQD, ∠APB = ∠CQD [∵ Each 90°] ∠ABP = ∠CDQ [∵ Alternate angles] AB = CD [∵ Opposite sides of a parallelogram] Hence, ΔAPB ≅ ΔCQD [∵ SAS Congruency rule] (ii) ΔAPB ≅ ΔCQD [∵ Prove above] AP = CQ [∵ CPCT]

(i) In ΔAPB and ΔCQD,
∠APB = ∠CQD [∵ Each 90°]
∠ABP = ∠CDQ [∵ Alternate angles]
AB = CD [∵ Opposite sides of a parallelogram]
Hence, ΔAPB ≅ ΔCQD [∵ SAS Congruency rule]

(i) In ABED, AB = DE [∵ Given] AB ∥ DE [∵ Given] Hence, ABED is a parallelogram. (ii) In BEFC, BC = EF [∵ Given] BC ∥ EF [∵ Given] Hence, BEFC is a parallelogram. (iii) In ABED, AD = BE ...(1) [∵ ABED is a parallelogram] AD ∥ BE ...(2) [∵ ABED is a parallelogram] In BEFC, BE = CF ...(3) [∵ ABED is aRead more

(i) In ABED, AB = DE [∵ Given]
AB ∥ DE [∵ Given]
Hence, ABED is a parallelogram.

(ii) In BEFC, BC = EF [∵ Given]
BC ∥ EF [∵ Given]
Hence, BEFC is a parallelogram.

(iii) In ABED,
AD = BE …(1) [∵ ABED is a parallelogram]
AD ∥ BE …(2) [∵ ABED is a parallelogram]
In BEFC,
BE = CF …(3) [∵ ABED is a parallelogram]
BE ∥ CF …(4) [∵ ABED is a parallelogram]
From (2) and (4), we have
AD ∥ CF …(5)
From (1) and (3), we have
AD = CF …(6)
(iv) In ACFD,
AD = CF [∵ From (6)]
AD ∥ CF [∵ From (5)]
Hence, ACFD is a parallelogram
(v) In ACFD,
AC = DF . [∵ ACFD is a parallelogram]

(vi) In ΔABC and ΔDEF
AB = DE [∵ Given]
AC = DF [∵ Proved above]
BC = EF [∵ Given]
Hence, ΔABC ≅ ΔDEF [∵ SSS Congruency rule]

Triangle ABP and parallelogram ABCD lie on the same base AB and between the same parallels, AB ∥ CD Hence, ar (APB) = (1/2) ar (ABCD) ....(1) [∵ If a parallelogram and a triangle are on the same base and between the same parallels, than area of the triangle is half the area of the parallelogram.] SiRead more

Triangle ABP and parallelogram ABCD lie on the same base AB and between the same parallels, AB ∥ CD
Hence, ar (APB) = (1/2) ar (ABCD) ….(1)
[∵ If a parallelogram and a triangle are on the same base and between the same parallels, than area of the triangle is half the area of the parallelogram.]
Similarly,
Triangle BQC and Parallelogram ABCD lie on the same base AB and between same parallels, AD ∥ BC.
Hence, ar(BQC = (1/2)ar(ABCD) …(2)
From the equation (1) and (2), ar (APB) = ar(BQC).

Inner radius of wood in pencil r = 1/2 = 0.5 mm = 0.05 cm, Outer radius R = 7/2 = 3.5 mm = 0.35 cm and length h = 14 cm Volume of wood used in pencil = π(R² - r²)h = 22/7 × [(0.35)² - (0.05)²] × 14 = 22 × (0.1225 - 0.0025) × 2 = 22 × 0.12 × 2 = 5.28 cm³ Radius of graphite inside the wood r = 1/2 = 0Read more

Inner radius of wood in pencil r = 1/2 = 0.5 mm = 0.05 cm,
Outer radius R = 7/2 = 3.5 mm = 0.35 cm and length h = 14 cm
Volume of wood used in pencil = π(R² – r²)h
= 22/7 × [(0.35)² – (0.05)²] × 14
= 22 × (0.1225 – 0.0025) × 2
= 22 × 0.12 × 2
= 5.28 cm³
Radius of graphite inside the wood r = 1/2 = 0.5 mm = 0.05 cm and height h = 14 cm.
Volume of graphite in pencil = πr²h
= 22/7 × (0.05)² × 14
= 22 × 0.0025 × 2
= 0.11 cm³
Hence, in pencil, the volume of wood is 5.28 cm³ and that of graphite is 0.11 cm³.

(I) Radius of cone r = 6 cm and height h = 7 cm Volume of cone = 1/3πr²h = 1/3 × 22/7 × 6 × 6 × 7 = 264 cm³ Hence, the volume of right circular cone is 264 cm³. (II) Radius of cone r = 3.5 cm and height h = 12 cm Volume of cone = 1/3πr²h = 1/3 × 22/7 × 3.5 × 3.5 × 12 = 154 cm³ Hence, the volume of rRead more

(I) Radius of cone r = 6 cm and height h = 7 cm
Volume of cone = 1/3πr²h
= 1/3 × 22/7 × 6 × 6 × 7 = 264 cm³
Hence, the volume of right circular cone is 264 cm³.
(II) Radius of cone r = 3.5 cm and height h = 12 cm
Volume of cone = 1/3πr²h
= 1/3 × 22/7 × 3.5 × 3.5 × 12 = 154 cm³
Hence, the volume of right circular cone is 154 cm³.

Radis of sphere r = 14 cm Surface area of sphere = 4πr² = 4 × 22/7 × 14 × 14 = 4 × 22 × 2 × 14 = 2464 cm² Hence, the surface area of sphere is 2464 cm².

Radis of sphere r = 14 cm
Surface area of sphere = 4πr²
= 4 × 22/7 × 14 × 14 = 4 × 22 × 2 × 14 = 2464 cm²
Hence, the surface area of sphere is 2464 cm².

Circumference of the base of a cylindrical vessel C = 132 cm height h = 25 cm Let, the radius of cylindrical vessel = r cm Circumference of base of cylindrical vessel = 2πr ⇒ 132 = 2πr ⇒ 132 = 2 × 22/7 × r ⇒ r = (132 × 7/22 × 2) ⇒ r = 21 cm Volume of cylindrical vessel = πr²h = 22/7 × 21 × 21 × 25 =Read more

Circumference of the base of a cylindrical vessel C = 132 cm height h = 25 cm
Let, the radius of cylindrical vessel = r cm
Circumference of base of cylindrical vessel = 2πr
⇒ 132 = 2πr ⇒ 132 = 2 × 22/7 × r
⇒ r = (132 × 7/22 × 2) ⇒ r = 21 cm
Volume of cylindrical vessel = πr²h
= 22/7 × 21 × 21 × 25
= 34650 cm³
= 34650/1000 = 34.65 litres [∵ 1000 cm³ = 1 litres]
Hence, the cylindrical vessel can hold 34.65 litres of water.

## Diagonal AC of a parallelogram ABCD bisects ∠A see Figure. Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus.

(i) ∠DAC = ∠BAC ...(1) [∵ Given] ∠DAC = ∠BCA ...(2) [∵ Alternate angle] ∠BAC = ∠ACD ...(3) [∵Alternate angles] From the equations (1), (2) and (3), we have ∠ACD = ∠BCA ...(4) Hence, Diagonal AC bisects angle C also. (ii) From the equation (2) and (4), we have ∠ACD = ∠DAC In ΔADC, ∠ACD = ∠DAC [∵ ProvRead more

(i) ∠DAC = ∠BAC …(1) [∵ Given]

See less∠DAC = ∠BCA …(2) [∵ Alternate angle]

∠BAC = ∠ACD …(3) [∵Alternate angles]

From the equations (1), (2) and (3), we have

∠ACD = ∠BCA …(4)

Hence, Diagonal AC bisects angle C also.

(ii) From the equation (2) and (4), we have

∠ACD = ∠DAC

In ΔADC,

∠ACD = ∠DAC [∵ Prove above]

AD = DC [∵ In a triangle, the sides opposite to equal angle are equal]

A parallelogram whose adjacent sides are equal, is a rhombus. Hence, ABCD is a rhombus.

## In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ see Figure. Show that: (i) ∆ APD ≅ ∆ CQB (ii) AP = CQ (iii) ∆ AQB ≅∆ CPD (iv) AQ = CP (v) APCQ is a parallelogram

(i) In ΔAPD and ΔCQB, DP = BQ [∵ Given] ∠ADP = ∠CBQ [∵ Alternate angle] AD = BC [∵ Opposite sides of a parallelogram] Hence, ΔAPD ≅ ΔCQB [∵ SAS Congruency rule] (ii) ΔAPD ≅ CQB [∵ Prove above] AP = CQ ...(1) [∵ CPCT] (iii) In ΔAQB and ΔCPD, QB = DB [∵ Given] ∠ABQ = ∠CDP [∵ Alternate angle] AB = CD [Read more

(i) In ΔAPD and ΔCQB,

DP = BQ [∵ Given]

∠ADP = ∠CBQ [∵ Alternate angle]

AD = BC [∵ Opposite sides of a parallelogram]

Hence, ΔAPD ≅ ΔCQB [∵ SAS Congruency rule]

(ii) ΔAPD ≅ CQB [∵ Prove above]

AP = CQ …(1) [∵ CPCT]

(iii) In ΔAQB and ΔCPD,

QB = DB [∵ Given]

∠ABQ = ∠CDP [∵ Alternate angle]

AB = CD [∵ Opposite sides of a parallelogram]

Hence, ΔAQB ≅ ΔCPD [ SAS Congruency rule]

(iv) ΔAQB ≅ ΔCPD [∵ Prove above]

AQ = CP …(2) [∵ CPCT]

(v) In APCQ,

See lessAP = CQ [∵ From (1)]

AQ = CP [∵ From (2)]

The opposite sides of quadrilateral APCQ are equal.

Hence, APCQ is a parallelogram.

## ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD see Figure. Show that (i) ∆ APB ≅ ∆ CQD (ii) AP = CQ

(i) In ΔAPB and ΔCQD, ∠APB = ∠CQD [∵ Each 90°] ∠ABP = ∠CDQ [∵ Alternate angles] AB = CD [∵ Opposite sides of a parallelogram] Hence, ΔAPB ≅ ΔCQD [∵ SAS Congruency rule] (ii) ΔAPB ≅ ΔCQD [∵ Prove above] AP = CQ [∵ CPCT]

(i) In ΔAPB and ΔCQD,

∠APB = ∠CQD [∵ Each 90°]

∠ABP = ∠CDQ [∵ Alternate angles]

AB = CD [∵ Opposite sides of a parallelogram]

Hence, ΔAPB ≅ ΔCQD [∵ SAS Congruency rule]

(ii) ΔAPB ≅ ΔCQD [∵ Prove above]

See lessAP = CQ [∵ CPCT]

## In ∆ ABC and ∆ DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram (ii) quadrilateral BEFC is a parallelogram (iii) AD || CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ∆ ABC ≅ ∆ DEF.

(i) In ABED, AB = DE [∵ Given] AB ∥ DE [∵ Given] Hence, ABED is a parallelogram. (ii) In BEFC, BC = EF [∵ Given] BC ∥ EF [∵ Given] Hence, BEFC is a parallelogram. (iii) In ABED, AD = BE ...(1) [∵ ABED is a parallelogram] AD ∥ BE ...(2) [∵ ABED is a parallelogram] In BEFC, BE = CF ...(3) [∵ ABED is aRead more

(i) In ABED, AB = DE [∵ Given]

AB ∥ DE [∵ Given]

Hence, ABED is a parallelogram.

(ii) In BEFC, BC = EF [∵ Given]

BC ∥ EF [∵ Given]

Hence, BEFC is a parallelogram.

(iii) In ABED,

AD = BE …(1) [∵ ABED is a parallelogram]

AD ∥ BE …(2) [∵ ABED is a parallelogram]

In BEFC,

BE = CF …(3) [∵ ABED is a parallelogram]

BE ∥ CF …(4) [∵ ABED is a parallelogram]

From (2) and (4), we have

AD ∥ CF …(5)

From (1) and (3), we have

AD = CF …(6)

(iv) In ACFD,

AD = CF [∵ From (6)]

AD ∥ CF [∵ From (5)]

Hence, ACFD is a parallelogram

(v) In ACFD,

AC = DF . [∵ ACFD is a parallelogram]

(vi) In ΔABC and ΔDEF

See lessAB = DE [∵ Given]

AC = DF [∵ Proved above]

BC = EF [∵ Given]

Hence, ΔABC ≅ ΔDEF [∵ SSS Congruency rule]

## P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).

Triangle ABP and parallelogram ABCD lie on the same base AB and between the same parallels, AB ∥ CD Hence, ar (APB) = (1/2) ar (ABCD) ....(1) [∵ If a parallelogram and a triangle are on the same base and between the same parallels, than area of the triangle is half the area of the parallelogram.] SiRead more

Triangle ABP and parallelogram ABCD lie on the same base AB and between the same parallels, AB ∥ CD

See lessHence, ar (APB) = (1/2) ar (ABCD) ….(1)

[∵ If a parallelogram and a triangle are on the same base and between the same parallels, than area of the triangle is half the area of the parallelogram.]

Similarly,

Triangle BQC and Parallelogram ABCD lie on the same base AB and between same parallels, AD ∥ BC.

Hence, ar(BQC = (1/2)ar(ABCD) …(2)

From the equation (1) and (2), ar (APB) = ar(BQC).

## A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

Inner radius of wood in pencil r = 1/2 = 0.5 mm = 0.05 cm, Outer radius R = 7/2 = 3.5 mm = 0.35 cm and length h = 14 cm Volume of wood used in pencil = π(R² - r²)h = 22/7 × [(0.35)² - (0.05)²] × 14 = 22 × (0.1225 - 0.0025) × 2 = 22 × 0.12 × 2 = 5.28 cm³ Radius of graphite inside the wood r = 1/2 = 0Read more

Inner radius of wood in pencil r = 1/2 = 0.5 mm = 0.05 cm,

See lessOuter radius R = 7/2 = 3.5 mm = 0.35 cm and length h = 14 cm

Volume of wood used in pencil = π(R² – r²)h

= 22/7 × [(0.35)² – (0.05)²] × 14

= 22 × (0.1225 – 0.0025) × 2

= 22 × 0.12 × 2

= 5.28 cm³

Radius of graphite inside the wood r = 1/2 = 0.5 mm = 0.05 cm and height h = 14 cm.

Volume of graphite in pencil = πr²h

= 22/7 × (0.05)² × 14

= 22 × 0.0025 × 2

= 0.11 cm³

Hence, in pencil, the volume of wood is 5.28 cm³ and that of graphite is 0.11 cm³.

## A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?

Radius of cylindrical bowl r = 7/2 = 3.5 and height of soup inside the cylindrical bowl h = 4 cm Volume of cylindrical bowl = πr²h = 22/7 × (3.5)² × 4 = 22/7 × 3.5 × 3.5 × 4 = 22 × 0.5 × 3.5 × 4 = 154 cm³ Therefore, the volume of soup per day for 250 patient = 250 × 154 = 38500 cm³ Hence, hospital hRead more

Radius of cylindrical bowl r = 7/2 = 3.5 and height of soup inside the cylindrical bowl h = 4 cm

See lessVolume of cylindrical bowl = πr²h

= 22/7 × (3.5)² × 4

= 22/7 × 3.5 × 3.5 × 4

= 22 × 0.5 × 3.5 × 4

= 154 cm³

Therefore, the volume of soup per day for 250 patient = 250 × 154 = 38500 cm³

Hence, hospital has to prepare 38500 cm³ soup daily to serve 250 patients.

## Find the volume of the right circular cone with

(I) Radius of cone r = 6 cm and height h = 7 cm Volume of cone = 1/3πr²h = 1/3 × 22/7 × 6 × 6 × 7 = 264 cm³ Hence, the volume of right circular cone is 264 cm³. (II) Radius of cone r = 3.5 cm and height h = 12 cm Volume of cone = 1/3πr²h = 1/3 × 22/7 × 3.5 × 3.5 × 12 = 154 cm³ Hence, the volume of rRead more

(I) Radius of cone r = 6 cm and height h = 7 cm

See lessVolume of cone = 1/3πr²h

= 1/3 × 22/7 × 6 × 6 × 7 = 264 cm³

Hence, the volume of right circular cone is 264 cm³.

(II) Radius of cone r = 3.5 cm and height h = 12 cm

Volume of cone = 1/3πr²h

= 1/3 × 22/7 × 3.5 × 3.5 × 12 = 154 cm³

Hence, the volume of right circular cone is 154 cm³.

## Find the surface area of a sphere of radius: 14 cm.

Radis of sphere r = 14 cm Surface area of sphere = 4πr² = 4 × 22/7 × 14 × 14 = 4 × 22 × 2 × 14 = 2464 cm² Hence, the surface area of sphere is 2464 cm².

Radis of sphere r = 14 cm

See lessSurface area of sphere = 4πr²

= 4 × 22/7 × 14 × 14 = 4 × 22 × 2 × 14 = 2464 cm²

Hence, the surface area of sphere is 2464 cm².

## The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm³= 1l)

Circumference of the base of a cylindrical vessel C = 132 cm height h = 25 cm Let, the radius of cylindrical vessel = r cm Circumference of base of cylindrical vessel = 2πr ⇒ 132 = 2πr ⇒ 132 = 2 × 22/7 × r ⇒ r = (132 × 7/22 × 2) ⇒ r = 21 cm Volume of cylindrical vessel = πr²h = 22/7 × 21 × 21 × 25 =Read more

Circumference of the base of a cylindrical vessel C = 132 cm height h = 25 cm

See lessLet, the radius of cylindrical vessel = r cm

Circumference of base of cylindrical vessel = 2πr

⇒ 132 = 2πr ⇒ 132 = 2 × 22/7 × r

⇒ r = (132 × 7/22 × 2) ⇒ r = 21 cm

Volume of cylindrical vessel = πr²h

= 22/7 × 21 × 21 × 25

= 34650 cm³

= 34650/1000 = 34.65 litres [∵ 1000 cm³ = 1 litres]

Hence, the cylindrical vessel can hold 34.65 litres of water.