Rs Aggarwal 2017 for Class 9 Math Chapter 1 - Real Numbers

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Page No 351:

Question 1:

In the adjoining figure, show that ABCD is a parallelogram.
Calculate the area of || gm ABCD.

Answer:

Given: A quadrilateral ABCD and BD is a diagonal.
To prove: ABCD is a parallelogram.
Construction: Draw AM ⊥ DC and CL ⊥ AB (extend DC and AB). Join AC, the other diagonal of ABCD.

Proof: ar(quad. ABCD) = ar(∆ABD) + ar(∆DCB)
= 2 ar(∆ABD) [∵ ar(∆ABD) = ar(∆DCB)]
∴ ar(∆ABD) = $\frac{1}{2}$ ar(quad. ABCD) ...(i)

Again, ar(quad. ABCD) = ar(∆ABC) + ar(∆CDA)
= 2 ar(∆ ABC) [∵ ar(∆ABC) = ar(∆CDA)]
∴ ar(∆ABC) = $\frac{1}{2}$ ar(quad. ABCD) ...(ii)
From (i) and (ii), we have:
ar(∆ABD) = ar(∆ABC) = $\frac{1}{2}$ AB ⨯ BD = $\frac{1}{2}$ AB ⨯ CL
⇒ CL = BD
⇒ DC || AB
Similarly, AD || BC.
Hence, ABCD is a paralleogram.
∴ ar(|| gm ABCD) = base ⨯ height = 5 ⨯ 7 = 35 cm²

Page No 351:

Question 2:

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

Answer:

ar(parallelogram ABCD) = base ⨯ height
⇒ AB ⨯DL = AD ⨯ BM
⇒ 10 ⨯ 6 = AD ⨯ BM
⇒ AD ⨯ 8 = 60 cm²
⇒ AD = 7.5 cm
∴ AD = 7.5 cm

Page No 351:

Question 3:

Find the area of rhombus, the lengths of whose diagonals are 16 cm and 24 cm respectively.

Answer:

ar(rhombus) = $\frac{1}{2}$ ⨯ product of diagonals
= $\frac{1}{2}$ ⨯16 ⨯ 24
= 192 cm²
Hence, the area of the rhombus is 192 cm².

Page No 351:

Question 4:

Find the area of a trapezium whose parallel sides are 9 cm and 6 cm respectively and the distance between these sides is 8 cm.

Answer:

ar(trapezium) = $\frac{1}{2}$ ⨯ (sum of parallel sides) ⨯ (distance between them)
= $\frac{1}{2}$ ⨯ (9 + 6) ⨯ 8
= 60 cm²
Hence, the area of the trapezium is 60 cm².

Page No 351:

Question 5:

(i) Calculate the area of quad. ABCD, given in Fig. (i).
(ii) Calculate the area of trap. PQRS, given in Fig. (ii).

Answer:

(i)
ar(quad. ABCD) = ar(∆ ABD) + ar(∆ BDC)
∆ABD and ∆BDC are right angle triangles.
So, BD = $\sqrt{(17^{2} - 8^{2})} = \sqrt{225} = 15 cm$

and AB =

\sqrt{(15^{2} - 9^{2})} = \sqrt{144} = 12 cm

Now, ar(∆ABD) =

\frac{1}{2}

⨯AD ⨯ AB
=

\frac{1}{2}

⨯9 ⨯ 12
= 54 cm²

Also, ar(∆BDC) =

\frac{1}{2}

⨯BD ⨯ BC
=

\frac{1}{2}

⨯15 ⨯ 8
= 60 cm²

Hence, area of quad. ABCD = 54 + 60 = 114 cm²

(ii)
ar(quad. PQRS) = ar(rectangle PTRS) + ar(∆TRQ)
∆TRQ is a right angle triangle.
So, TR =

\sqrt{(17^{2} - 8^{2})} = \sqrt{225} = 15 cm

Now, ar(rectangle PQRS) = length ⨯ height
= PT ⨯ TR
= 8 ⨯ 15
= 120 cm²
Also, ar(∆ TRQ) =

\frac{1}{2}

⨯ TQ ⨯ TR
=

\frac{1}{2}

⨯8 ⨯ 15
= 60 cm²
∴ ar(quad. PQRS) = 120 +60 = 180 cm²

Page No 351:

Question 6:

In the adjoining figure, ABCD is a trapezium in which AB || DC; AB = 7 cm; AD = BC = 5 cm and the distance between AB and DC is 4 cm. Find the length of DC and hence, find the area of trap. ABCD.

Answer:

∆ADL is a right angle triangle.
So, DL = $\sqrt{(5^{2} - 4^{2})} = \sqrt{9} = 3 cm$
Similarly, in ∆BMC, we have:
MC = $\sqrt{(5^{2} - 4^{2})} = \sqrt{9} = 3 cm$
∴ DC = DL + LM + MC = 3 + 7 + 3 = 13 cm
Now, ar(trapezium. ABCD) = $\frac{1}{2}$ ⨯ (sum of parallel sides) ⨯ (distance between them)
= $\frac{1}{2}$ ⨯ (7 + 13) ⨯ 4
= 40 cm²
Hence, DC = 13 cm and area of trapezium = 40 cm²

Page No 351:

Question 7:

BD is one of the diagonals of a quad. ABCD. If AL ⊥ BD and CM ⊥ BD, show that $ar (quad . A B C D) = \frac{1}{2} \times B D \times (A L + C M)$ .

Answer:

ar(quad. ABCD) = ar(∆ABD) + ar (∆DBC)
ar(∆ABD) = $\frac{1}{2}$ ⨯ base ⨯ height = $\frac{1}{2}$ ⨯ BD ⨯ AL ...(i)
ar(∆DBC) = $\frac{1}{2}$ ⨯ BD ⨯ CL ...(ii)
From (i) and (ii), we get:
ar(quad ABCD) = $\frac{1}{2}$ ⨯BD ⨯ AL + $\frac{1}{2}$ ⨯ BD ⨯ CL
$\Rightarrow$ ar(quad ABCD) = $\frac{1}{2}$ ⨯ BD ⨯ (AL + CL)
Hence, proved.

Page No 352:

Question 8:

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

Answer:

ar(quad ABCD) = ar( $△$ ABD) + ar( $△$ BDC)
= $\frac{1}{2}$ ⨯BD ⨯ AL + $\frac{1}{2}$ ⨯BD ⨯ CM
= $\frac{1}{2}$ ⨯BD ⨯ ( AL + CM)
By substituting the values, we have;
ar(quad ABCD) = $\frac{1}{2}$ ⨯ 14 ⨯ ( 8 + 6)
= 7 ⨯14
= 98 cm²

Page No 352:

Question 9:

In the adjoining figure, ABCD is a trapezium in which AB || DC and its diagonals AC and BD intersect at O. Prove that ar(∆AOD) = ar(∆BOC).

Answer:

∆CDA and ∆CBD lies on the same base and between the same parallel lines.
So, ar(∆CDA) = ar(CDB) ...(i)
Subtracting ar(∆OCD) from both sides of equation (i), we get:
ar(∆CDA) $-$ ar(∆OCD) = ar(∆CDB) $-$ ar (∆OCD)
⇒ ar(∆AOD) = ar(∆BOC)

Page No 352:

Question 10:

In the adjoining figure, DE || BC. Prove that
(i) ar(∆ACD) = ar(∆ABE),
(ii) ar(∆OCE) = ar(∆OBD),

Answer:

∆DEC and ∆DEB lies on the same base and between the same parallel lines.
So, ar(∆DEC) = ar(∆DEB)                      ...(1)

(i) On adding ar(∆ADE) in both sides of equation (1), we get:
ar(∆DEC) + ar(∆ADE) = ar(∆DEB) + ar(∆ADE)
⇒ ar(∆ACD) = ar(∆ABE)

(ii) On subtracting ar(ODE) from both sides of equation (1), we get:
   ar(∆DEC) $-$ ar(∆ODE) = ar(∆DEB) $-$ ar(∆ODE)
⇒ ar(∆OCE) = ar(∆OBD)

Page No 352:

Question 11:

In the adjoining figure, D and E are points on the sides AB and AC of ∆ABC such that ar(∆BCE) = ar(∆BCD). Show that DE || BC.

Answer:

∆BEC and ∆BCD lies on the same base BC and have equal areas.
Darw EM as the altitude of ∆BEC and DL as the altitude of ∆BCD.
ar(∆BCE) = $\frac{1}{2}$ ⨯ base ⨯ height = $\frac{1}{2}$ ⨯ BC ⨯EM
ar(∆BCD) = $\frac{1}{2}$ ⨯ base ⨯ height = $\frac{1}{2}$ ⨯ BC ⨯DL
However, ar(∆BCE) = ar(∆BCD)
⇒ $\frac{1}{2}$ ⨯ BC ⨯EM = $\frac{1}{2}$ ⨯ BC ⨯DL
⇒ EM = DL
i.e., the corresponding altitudes of both the triangles are equal.
∴ DE || BC

Page No 352:

Question 12:

In the adjoining figure, O is any point inside a parallelogram ABCD. Prove that
(i) $ar (∆ O A B) + ar (∆ O C D) = \frac{1}{2} ar (| | gm A B C D),$
(ii) $ar (∆ O A D) + ar (∆ O B C) = \frac{1}{2} ar (| | gm A B C D) .$

Answer:

Constructions: Draw EOF || AB and GOH || AD.
Proof: EOF || AB and DA cuts them.
∴ ∆OAB and parallelogram EABF being on the same base and between the same parallels AB and EF, we have:
ar(∆OAB) = $\frac{1}{2}$ ar(parallelogram EABF)                      ...(i)
Similarly, ar(∆OCD) = $\frac{1}{2}$ ar(parallelogram EFCD)             ...(ii)
On adding (i) and (ii), we get:
ar(∆OAB) + ar(∆OCD) = $\frac{1}{2}$ ⨯ ar(parallelogram EABF) + $\frac{1}{2}$ ar(parallelogram EFCD)
⇒ ar(∆OAB) + ar(∆OCD) = $\frac{1}{2}$ ⨯ ar(parallelogram ABCD)

(ii)
∴ ∆OAD and parallelogram AHGD being on the same base and between the same parallels AD and GH, we have:
   ar(∆OAD) = $\frac{1}{2}$ ar(parallelogram AHGD)                ...(iii)
Similarly, ar(∆OBC) = $\frac{1}{2}$ ar(parallelogram BCGH)             ...(iv)

On adding (iii) and (iv), we get:
ar(∆OAD) + ar(∆OBC) = $\frac{1}{2}$ ⨯ ar(parallelogram AHGD) + $\frac{1}{2}$ ar(parallelogram BCGH)
⇒ ar(∆OAD) + ar(∆OBC) = $\frac{1}{2}$ ⨯ ar(parallelogram ABCD)

Page No 352:

Question 13:

In the adjoining figure, ABCD is a quadrilateral. A line through D, parallel to AC, meets BC produced in P. Prove that ar(∆ABP) = ar(quad. ABCD).

Answer:

We have:
ar(quad. ABCD) = ar(∆ACD) + ar(∆ABC)
ar(∆ABP) = ar(∆ACP) + ar(∆ABC)

∆ACD and ∆ACP are on the same base and between the same parallels AC and DP.
∴ ar(∆ACD) = ar(∆ ACP)
By adding ar(∆ABC) on both sides, we get:
ar(∆ACD) + ar(∆ABC) = ar(∆ACP) + ar(∆ABC)
⇒ ar (quad. ABCD) = ar(∆ABP)
Hence, proved.

Page No 352:

Question 14:

In the adjoining figure, ∆ABC and ∆DBC are on the same base BC with A and D on opposite sides of BC such that ar(∆ABC) = ar(∆DBC). Show that BC bisects AD.

Answer:

Given: ∆ABC and ∆DBC are on the same base BC.
ar(∆ABC) = ar(∆DBC)
To prove: BC bisects AD
Construction: Draw AL ⊥ BC and DM ⊥ BC.
Proof:
Since ∆ABC and ∆DBC are on the same base BC and they have equal areas, their altitudes must be equal.
i.e., AL = DM
Let AD and BC intersect at O.
Now, in ∆ALO and ∆DMO, we have:
AL = DM
∠ALO = ∠DMO = 90^o
∠AOL = ∠DOM (Vertically opposite angles)
i.e., ∆ ALO ≅ ∆ DMO
∴ OA = OD
Hence, BC bisects AD.

Page No 353:

Question 15:

In the adjoining figure, AD is one of the medians of a ∆ABC and P is a point on AD.
Prove that
(i) ar(∆BDP) = ar(∆CDP),
(ii) ar(∆ABP) = ar(∆ACP),

Answer:

A median of a triangle divides it into two triangles of equal areas.
AD is a median of ∆ABC.
i.e., ar(∆ABD) = ar(∆ACD) ...(i)

(i) Now, PD is also a median of ∆PBC.
So ar(∆BDP) = ar(∆CDP) ...(ii)

(ii) Now, from (i) and (ii), we have:
ar(∆ABD) $-$ ar(∆BDP) = ar(∆ACD) $-$ ar(∆CDP)
∴ ar(∆ABP) = ar(∆ACP)

Page No 353:

Question 16:

In the adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O.
If BO = OD, prove that
ar(∆ABC) = ar(∆ADC),

Answer:

Given: BO = OD
To prove: ar(∆ABC) = ar(∆ADC)
Proof:
Since BO = OD, O is the mid point of BD.
We know that a median of a triangle divides it into two triangles of equal areas.
CO is a median of ∆BCD.
i.e., ar(∆COD) = ar (∆COB) ...(i)

AO is a median of ∆ABD.
i.e., ar(∆AOD) = ar(∆AOB) ...(ii)

From (i) and (ii), we have:
ar(∆COD) + ar(∆AOD) = ar(∆COB) + ar(∆AOB)
∴ ar(∆ADC ) = ar(∆ABC)

Page No 353:

Question 17:

ABC is a triangle in which D is the midpoint of BC and E is the midpoint of AD.
Prove that $ar (∆ B E D) = \frac{1}{4} ar (∆ A B C)$ .

Answer:

Given: D is the midpoint of BC and E is the midpoint of AD.
To prove: ar(∆BED) = $\frac{1}{4}$ ar(∆ABC)
Proof:
Since D is the mid point of BC, AD is median of ∆ABC and BE is the median of ∆ABD.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(∆ABD) = $\frac{1}{2}$ ar(∆ABC) ...(i)

Also, ar(∆BED) = $\frac{1}{2}$ ar(∆ABD) ...(ii)

From (i) and (ii), we have:
ar(∆BED) = $\frac{1}{2}$ ⨯ $\frac{1}{2}$ ⨯ ar(∆ABC)
∴ ar(∆BED) = $\frac{1}{4}$ ⨯ ar(∆ABC)

Page No 353:

Question 18:

The vertex A of ∆ABC is joined to a point D on the side BC. The midpoint of AD is E.
Prove that $ar (∆ B E C) = \frac{1}{2} ar (∆ A B C)$ .

Answer:

Given: D is the midpoint of BC and E is the midpoint of AD.
To prove: $ar (∆ B E C) = \frac{1}{2} ar (∆ A B C)$
Proof:
Since E is the midpoint of AD, BE is the median of ∆ABD.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(∆BED ) = $\frac{1}{2}$ ar(∆ABD)                 ...(i)
Also, ar(∆CDE ) = $\frac{1}{2}$ ar(∆ADC)             ...(ii)

From (i) and (ii), we have:
ar(∆BED) + ar(∆CDE) = $\frac{1}{2}$ ⨯ ar(∆ABD) + $\frac{1}{2}$ ⨯ ar(∆ADC)
⇒ ar(∆BEC ) = $\frac{1}{2}$ ⨯ [ar(∆ABD) + ar(∆ADC)]
⇒ ar(∆BEC ) = $\frac{1}{2}$ ⨯ ar(∆ABC)

Page No 353:

Question 19:

D is the midpoint of side BC of ∆ABC and E is the midpoint of BD. If O is the midpoint of AE, prove that $ar (∆ B O E) = \frac{1}{8} ar (∆ A B C)$ .

Answer:

Given: D is the midpoint of BC; E is the midpoint of BD; O is the mid point of AE.
To prove: ar(∆BOE) = $\frac{1}{8}$ ⨯ ar(∆ABC)
Proof:
D is the midpoint of BC, so AD is the median of ∆ABC.
E is the midpoint of BD, so AE is the median of ∆ABD.
O is the mid point of AE, so BO is median of ∆ABE.
We know that a median of a triangle divides it into two triangles of equal areas. So, we have:
ar(∆ABD ) = $\frac{1}{2}$ ar(∆ABC)               ...(i)
ar(∆ABE ) = $\frac{1}{2}$ ar (∆ ABD)                ...(ii)
ar(∆BOE ) = $\frac{1}{2}$ ar (∆ ABE)              ...(iii)

From (i), (ii) and (iii), we have:
ar(∆BOE) = $\frac{1}{2}$ ar(∆ABE)
ar(∆BOE) = $\frac{1}{2}$ ⨯ $\frac{1}{2}$ ⨯ $\frac{1}{2}$ ⨯ ar(∆ABC)
∴ ar(∆BOE) = $(\frac{1}{8})$ ⨯ ar(∆ABC)

Page No 353:

Question 20:

In the adjoining figure, ABCD is a parallelogram and O is any point on the diagonal AC. Show that ar(∆AOB) = ar(∆AOD).

Answer:

Given: ABCD is a parallelogram and O is any point on AC.
To prove: ar(∆AOB) =  ar(∆AOD)
Proof:
Since the diagonals of a parallelogram bisect each other, P is the midpoint of BD.
Now, AP is a median of ∆BAD.
i.e., ar(∆PAB ) = ar(∆PAD)                     ...(i)
OP is median of ∆ODB
i.e., ar(∆OPB ) = ar(∆ODP)                       ...(ii)

Adding (i) and (ii), we have:
ar(∆PAB) + ar(∆OPB) = ar(∆PAD) + ar(∆ODP)
∴ ar(∆AOB) = ar(∆AOD)

Page No 353:

Question 21:

P, Q, R, S are respectively the midpoints of the sides AB, BC, CD and DA of || gm ABCD. Show that PQRS is a parallelogram and also show that
$ar (| | gm P Q R S) = \frac{1}{2} \times ar (| | gm A B C D)$ .

Answer:

Given: ABCD is a parallelogram and P, Q, R and S are the midpoints of sides AB, BC, CD and DA, respectively.
To prove: ar(parallelogram PQRS ) = $\frac{1}{2}$ × ar(parallelogram ABCD )
Proof:
In ∆ABC, PQ || AC and PQ = $\frac{1}{2}$ × AC [ By midpoint theorem]
Again, in ∆DAC, the points S and R are the mid points of AD and DC, respectively.
∴ SR || AC and SR = $\frac{1}{2}$ × AC [ By midpoint theorem]
Now, PQ || AC and SR || AC
⇒ PQ || SR
Also, PQ = SR = $\frac{1}{2}$ × AC
∴ PQ || SR and PQ = SR
Hence, PQRS is a parallelogram.

Now, ar(parallelogram PQRS) = ar(∆PSQ) + ar(∆SRQ) ...(i)
also, ar(parallelogram ABCD) = ar(parallelogram ABQS) + ar(parallelogram SQCD)            ...(ii)

∆PSQ and parallelogram ABQS are on the same base and between the same parallel lines.
So, ar(∆PSQ ) = $\frac{1}{2}$ × ar(parallelogram ABQS)                         ...(iii)
Similarly, ∆SRQ and parallelogram SQCD are on the same base and between the same parallel lines.
So, ar(∆SRQ ) = $\frac{1}{2}$ × ar(parallelogram SQCD)                        ...(iv)
Putting the values from (iii) and (iv) in (i), we get:
  ar(parallelogram PQRS) = $\frac{1}{2}$ × ar(parallelogram ABQS) + $\frac{1}{2}$ × ar(parallelogram SQCD)
From (ii), we get:
ar(parallelogram PQRS) = $\frac{1}{2}$ × ar(parallelogram ABCD)

Page No 353:

Question 22:

The given figure shows a pentagon ABCDE. EG, drawn parallel to DA, meets BA produced at G, and CF, drawn parallel to DB, meets AB produced at F. Show that ar(pentagon ABCDE) = ar(∆DGF).

Answer:

Given: ABCDE is a pentagon. EG || DA and CF || DB.
To prove: ar(pentagon ABCDE ) = ar( ∆DGF)
Proof:
ar(pentagon ABCDE ) = ar(∆DBC) + ar(∆ADE ) + ar(∆ABD) ...(i)
Also, ar(∆DGF) = ar(∆DBF) + ar(∆ADG) + ar(∆ABD )                ...(ii)

Now, ∆DBC and ∆DBF lie on the same base and between the same parallel lines.
∴ ar(∆DBC) = ar(∆DBF)                         ...(iii)
Similarly, ∆ADE and ∆ADG lie on same base and between the same parallel lines.
∴ ar(∆ADE) = ar(∆ADG)                       ...(iv)

From (iii) and (iv), we have:
ar(∆DBC) + ar(∆ADE) = ar(∆DBF) + ar(∆ADG)
Adding ar(∆ABD) on both sides, we get:
ar(∆DBC) + ar(∆ADE) + ar(∆ABD) = ar (∆DBF) + ar(∆ADG) + ar(∆ABD)
By substituting the values from (i) and (ii), we get:
ar(pentagon ABCDE) = ar(∆DGF)

Page No 354:

Question 23:

Prove that a median divides a triangle into two triangles of equal area.

Answer:

Let AD is a median of ∆ABC and D is the midpoint of BC. AD divides ∆ABC in two triangles: ∆ABD and ∆ADC.
To prove: ar(∆ABD) = ar(∆ADC)
Construction: Draw AL ⊥ BC.
Proof:
Since D is the midpoint of BC, we have:
BD = DC
Multiplying with $\frac{1}{2}$ AL on both sides, we get:
$\frac{1}{2}$ × BD × AL = $\frac{1}{2}$ × DC × AL
⇒ ar(∆ABD) = ar(∆ADC)

Page No 354:

Question 24:

Show that a diagonal divides a parallelogram into two triangles of equal area.

Answer:

Let ABCD be a parallelogram and BD be its diagonal.
To prove: ar(∆ABD) = ar(∆CDB)

Proof:
In ∆ABD and ∆CDB, we have:
AB = CD [Opposite sides of a parallelogram]
AD = CB [Opposite sides of a parallelogram]

BD = DB [Common]

i.e., ∆ABD ≅ ∆CDB [ SSS criteria]
∴ ar(∆ABD) = ar(∆CDB)

Page No 354:

Question 25:

The base BC of ∆ABC is divided at D such that $B D = \frac{1}{2} D C$ . Prove that $ar (∆ A B D) = \frac{1}{3} \times ar (∆ A B C)$ .

Answer:

Given: D is a point on BC of ∆ABC, such that BD = $\frac{1}{2}$ DC
To prove: ar(∆ABD) = $\frac{1}{3}$ ar(∆ABC)
Construction: Draw AL ⊥ BC.
Proof:
In ∆ABC, we have:
BC = BD + DC
⇒ BD + 2 BD = 3 × BD
Now, we have:
ar(∆ABD) = $\frac{1}{2}$ × BD × AL
ar(∆ABC) = $\frac{1}{2}$ × BC × AL
⇒ ar(∆ABC) = $\frac{1}{2}$ × 3BD × AL = 3 × $(\frac{1}{2} \times B D \times A L)$
⇒ ar(∆ABC) = 3 × ar(∆ABD)
∴ ar(∆ABD) = $\frac{1}{3}$ ar(∆ABC)

Page No 354:

Question 26:

In the adjoining figure, the point D divides the side BC of ∆ABC in the ratio m : n. Prove that ar(∆ABD) : ar(∆ADC) = m : n.

Answer:

Given: D is a point on BC of ∆ ABC, such that BD : DC = m : n
To prove: ar(∆ABD) : ar(∆ADC) = m : n
Construction: Draw AL ⊥ BC.
Proof:
ar(∆ABD) = $\frac{1}{2}$ × BD × AL ...(i)
ar(∆ADC) = $\frac{1}{2}$ × DC × AL ...(ii)
Dividing (i) by (ii), we get:
$\frac{ar (△ A B D)}{ar (∆ A D C} = \frac{\frac{1}{2} \times B D \times A L}{\frac{1}{2} \times D C \times A L} = \frac{B D}{D C} = \frac{m}{n}$

∴ ar(∆ABD) : ar(∆ADC) = m : n

Page No 356:

Question 1:

Out of the following given figures which are on the same base but not between the same parallels?

Answer:

In this figure, both the triangles are on the same base (QR) but not on the same parallels.

Page No 357:

Question 2:

In which of the following figures, you find polynomials on the same base and between the same parallels?

Answer:

In this figure, the following polygons lie on the same base and between the same parallel lines:
a) Parallelogram ABCD
b) Parallelogram ABPQ

Page No 357:

Question 3:

The median of a triangle divides it into two
(a) triangles of equal areas
(b) congruent triangles
(c) isosceles triangles
(d) right triangles

Answer:

(a) triangles of equal areas

Page No 357:

Question 4:

The area of quadrilateral ABCD in the given figure is
(a) 57 cm²
(b) 108 cm²
(c) 114 cm²
(d) 195 cm²

Answer:

(c)114 cm²

ar (quad. ABCD) = ar (∆ ABC) + ar (∆ ACD)
In right angle triangle ACD, we have:
AC = $\sqrt{(17^{2} - 8^{2})} = \sqrt{225} = 15 cm$
In right angle triangle ABC, we have:
BC = $\sqrt{(15^{2} - 9^{2})} = \sqrt{144} = 12 cm$
Now, we have the following:
ar(∆ABC) = $\frac{1}{2}$ × 12 × 9 = 54 cm²
ar(∆ADC) = $\frac{1}{2}$ × 15 × 8 = 60 cm²
ar(quad. ABCD) = 54 + 60 = 114 cm²

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Question 5:

The area of trapezium ABCD in the given figure is
(a) 62 cm²
(b) 93 cm²
(c) 124 cm²
(d) 155 cm²

Answer:

(c)124 cm²

In the right angle triangle BEC, we have:
EC = $\sqrt{17^{2} - 15^{2}} = \sqrt{289 - 225} = \sqrt{64} = 8 cm$
ar(trapez. ABCD) = $\frac{1}{2} \times (sum of parallel sides) \times distance between them = \frac{1}{2} \times 31 \times 8 = 124$ cm²

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Question 6:

In the given figure, ABCD is a || gm in which AB = CD = 5 cm and BD ⊥ DC such that BD = 6.8 cm. Then, the area of || gm ABCD = ?
(a) 17 cm²
(b) 25 cm²
(c) 34 cm²
(d) 68 cm²

Answer:

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Question 7:

In the given figure, ABCD is a || gm in which diagonals AC and BD intersect at O. If ar(||gm ABCD) is 52 cm², then the ar(∆OAB) = ?
(a) 26 cm²
(b) 18.5 cm²
(c) 39 cm²
(d) 13 cm²

Answer:

(d) 13 cm²
The diagonals of a parallelogram divides it into four triangles of equal areas.
∴ Area of ∆OAB = $\frac{1}{4}$ ⨯ ar(||gm ABCD)
⇒ ar(∆OAB) = $\frac{1}{4}$ ⨯ 52 = 13 cm²

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Question 8:

In the given figure, ABCD is a || gm in which DL ⊥ AB. If AB = 10 cm and DL = 4 cm, then the ar(||gm ABCD) = ?
(a) 40 cm²
(b) 80 cm²
(c) 20 cm²
(d) 196 cm²

Answer:

(a) 40 cm²
ar(||gm ABCD) = base × height = 10 × 4 = 40 cm²

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Question 9:

In || gm ABCD, it is given that AB = 10 cm, DL ⊥ AB and BM ⊥ AD such that DL = 6 cm and BM = 8 cm. Then, AD = ?
(a) 7.5 cm
(b) 8 cm
(c) 12 cm
(d) 14 cm

Answer:

(a) 7.5 cm

ar (||gm ABCD) = base × height
⇒ AB × DL = AD × BM
⇒ 10 × 6 = AD × 8
⇒ AD = 60 ÷ 8 = 7.5 cm

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Question 10:

The lengths of the diagonals of a rhombus are 12 cm and 16 cm. The area of the rhombus is
(a) 192 cm²
(b) 96 cm²
(c) 64 cm²
(d) 80 cm²

Answer:

(b) 96 cm²
Area of the rhombus = $\frac{1}{2}$ × product of diagonals = $\frac{1}{2}$ × 12 × 16 = 96 cm²

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Question 11:

Two parallel sides of a trapezium are 12 cm and 8 cm long and the distance between them is 6.5 cm. The area of the trapezium is
(a) 74 cm²
(b) 32.5 cm²
(c) 65 cm²
(d) 130 cm²
Figure

Answer:

(c) 65 cm²
Area of the trapezium = $\frac{1}{2}$ × (sum of parallel sides) × distance between them
= $\frac{1}{2}$ × ( 12 + 8) × 6.5
= 65 cm²

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Question 12:

In the given figure ABCD is a trapezium such that AL ⊥ DC and BM ⊥ DC. If AB = 7 cm, BC = AD = 5 cm and AL = BM = 4 cm, then ar(trap. ABCD) = ?
(a) 24 cm²
(b) 40 cm²
(c) 55 cm²
(d) 27.5 cm²

Answer:

(b) 40 cm²

In right angled triangle MBC, we have:
MC = $\sqrt{5^{2} - 4^{2}} = \sqrt{9} = 3 cm$
In right angled triangle ADL, we have:
DL = $\sqrt{5^{2} - 4^{2}} = \sqrt{9} = 3 cm$

Now, CD = ML + MC + LD = 7 + 3 + 3 = 13 cm
∴ Area of the trapezium = $\frac{1}{2}$ × (sum of parallel sides) × distance between them
= $\frac{1}{2}$ × ( 13 + 7) × 4
= 40 cm²

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Question 13:

In a quadrilateral ABCD, it is given that BD = 16 cm. If AL ⊥ BD and CM ⊥ BD such that AL = 9 cm and CM = 7 cm, then ar(quad. ABCD) = ?
(a) 256 cm²
(b) 128 cm²
(c) 64 cm²
(d) 96 cm²

Answer:

(b)128 cm²

ar(quad ABCD) = ar (∆ ABD) + ar (∆ DBC)

We have the following:
ar(∆ABD) = $\frac{1}{2}$ × base × height = $\frac{1}{2}$ × 16 × 9 = 72 cm²
ar(∆DBC) = $\frac{1}{2}$ × base × height = $\frac{1}{2}$ × 16 × 7 = 56 cm²

∴ ar(quad ABCD) = 72 + 56 = 128 cm²

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Question 14:

ABCD is a rhombus in which ∠C = 60°. Then, AC : BD = ?
(a) $\sqrt{3} : 1$
(b) $\sqrt{3} : \sqrt{2}$
(c) 3 : 1
(d) 3 : 2

Answer:

(a) $\sqrt{3} : 1$

ABCD is a rhombus. So all of its sides are equal.
Now, BC = DC
⇒ ∠BDC = ∠DBC = x^o (Angles opposite to equal sides are equal)
Also, ∠BCD = 60^o
∴ x^o + x^o + 60^o = 180^o
⇒ 2x^o = 120^o
⇒ x^o = 60^o
i.e., ∠BCD = ∠BDC = ∠DBC = 60^o
So, ∆BCD is an equilateral triangle.
∴ BD = BC = a
Also, OB = a /2
Now, in ∆OAB, we have:
$A B^{2} = O A^{2} + O B^{2}$
$\Rightarrow O A^{2} = A B^{2} - O B^{2} = a^{2} - {(\frac{a}{2})}^{2} = a^{2} - \frac{a^{2}}{4} = \frac{3 a^{2}}{4} \Rightarrow O A^{2} = \frac{3 a^{2}}{4} \Rightarrow O A = \sqrt{\frac{3 a^{2}}{4}} \Rightarrow O A = \frac{\sqrt{3} a}{2} Now, A C = 2 \times O A = 2 \times \frac{\sqrt{3} a}{2} = \sqrt{3} a ∴ A C : B D = \sqrt{3} a : a = \sqrt{3} : 1$

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Question 15:

In the given figure ABCD and ABFE are parallelograms such that ar(quad. EABC) = 17 cm² and ar(||gm ABCD) = 25 cm². Then, ar(∆BCF) = ?
(a) 4 cm²
(b) 4.8 cm²
(c) 6 cm²
(d) 8 cm²

Answer:

(d) 8 cm²

Since ||gm ABCD and ||gm ABFE are on the same base and between the same parallel lines, we have:
ar(||gm ABFE) = ar(||gm ABCD) = 25 cm²
⇒ ar(∆BCF ) = ar(||gm ABFE) $-$ ar(quad EABC) = ( 25 $-$ 17) = 8 cm²

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Question 16:

∆ABC and ∆BDE are two equilateral triangles such that D is the midpoint of BC. Then, ar(∆BDE) : ar(∆ABC) = ?
(a) 1 : 2
(b) 1 : 4
(c) $\sqrt{3} : 2$
(d) 3 : 4

Answer:

(b) 1:4

∆ABC and ∆BDE are two equilateral triangles and D is the midpoint of BC.
Let AB = BC = AC = a
Then BD = BE = ED = $\frac{a}{2}$
∴ $\frac{ar (∆ B D E)}{ar (∆ A B C)} = \frac{\frac{\sqrt{3}}{4} A B^{2}}{\frac{\sqrt{3}}{4} B E^{2}} = \frac{{(\frac{a}{2})}^{2}}{a^{2}} = \frac{1}{4}$
So, required ratio = 1 : 4

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Question 17:

In a || gm ABCD, if P and Q are midpoints of AB and CD respectively and ar(|| gm ABCD) = 16 cm², then ar(|| gm APQD) = ?
(a) 8 cm²
(b) 12 cm²
(c) 6 cm²
(d) 9 cm²

Answer:

(a) 8 cm²

Let the distance between AB and CD be h cm.
Then ar(||gm APQD) = AP × h
= $\frac{1}{2}$ × AB ×h (AP = $\frac{1}{2}$ AB )
= $\frac{1}{2}$ × ar(||gm ABCD) [ ar(|| gm ABCD) = AB ×h )
∴ ar (||gm APQD) = $\frac{1}{2}$ × 16 = 8 cm²

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Question 18:

The figure formed by joining the midpoints of the adjacent sides of a rectangle of sides 8 cm and 6 cm is a
(a) rectangle of area 24 cm²
(b) square of area 24 cm²
(c) trapezium of area 24 cm²
(d) rhombus of area 24 cm²

Answer:

(d) rhombus of 24 cm²

We know that the figure formed by joining the midpoints of the adjacent sides of a rectangle is a rhombus.
So, PQRS is a rhombus and SQ and PR are its diagonals.
i.e., SQ = 8 cm and PR = 6 cm
∴ ar(rhombus PQRS) = $\frac{1}{2}$ × product of diagonals = $\frac{1}{2}$ × 8 × 6 = 24 cm²

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Question 19:

In ∆ABC, if D is the midpoint of BC and E is the midpoint of AD, then ar(∆BED) = ?
(a) $\frac{1}{2} ar (∆ A B C)$
(b) $\frac{1}{3} ar (∆ A B C)$
(c) $\frac{1}{4} ar (∆ A B C)$
(d) $\frac{2}{3} ar (∆ A B C)$

Answer:

(c) $\frac{1}{4}$ ar (∆ ABC )

Since D is the mid point of BC, AD is a median of ∆ABC and BE is the median of ∆ABD.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(∆ABD ) = $\frac{1}{2}$ ar(∆ABC) ...(i)

⇒ ar(∆BED) = $\frac{1}{2}$ ar(∆ABD) ...(ii)

From (i) and (ii), we have:
ar(∆BED) = $\frac{1}{2}$ ⨯ $\frac{1}{2}$ ⨯ ar(∆ABC)
∴ ar(∆BED) = $\frac{1}{4}$ ⨯ ar(∆ABC)
ar(∆ABC)

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Question 20:

The vertex A of ∆ABC is joined to a point D on BC. If E is the midpoint of AD, then ar(∆BEC) = ?
(a) $\frac{1}{2} ar (∆ A B C)$
(b) $\frac{1}{3} ar (∆ A B C)$
(c) $\frac{1}{4} ar (∆ A B C)$
(d) $\frac{1}{6} ar (∆ A B C)$

Answer:

(a) $\frac{1}{2} ar (∆ A B C)$

Since E is the midpoint of AD, BE is a median of ∆ABD.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(∆BED) = $\frac{1}{2}$ ⨯ ar(∆ABD)                ...(i)
Since E is the midpoint of AD, CE is a median of ∆ADC.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(∆CED ) = $\frac{1}{2}$ ⨯ ar(∆ADC)               ...(ii)

Adding (i) and (ii), we have:
ar(∆BED ) + ar(∆CED ) = $\frac{1}{2}$ ⨯ ar(∆ABD) + $\frac{1}{2}$ ⨯ ar(∆ADC)
⇒ ar (∆ BEC ) = $\frac{1}{2} (∆ A B D + A D C) = \frac{1}{2} ∆ A B C$

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Question 21:

In ∆ABC, it is given that D is the midpoint of BC; E is the midpoint of BD and O is the midpoint of AE. Then, ar(∆BOE) = ?
(a) $\frac{1}{3} ar (∆ A B C)$
(b) $\frac{1}{4} ar (∆ A B C)$
(c) $\frac{1}{6} ar (∆ A B C)$
(d) $\frac{1}{8} ar (∆ A B C)$

Answer:

(d) $\frac{1}{8}$ ar (∆ ABC)

Given: D is the midpoint of BC, E is the midpoint of BD and O is the mid point of AE.
Since D is the midpoint of BC, AD is the median of ∆ABC.
E is the midpoint of BC, so AE is the median of ∆ABD. O is the midpoint of AE, so BO is median of ∆ABE.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(∆ABD ) = $\frac{1}{2}$ ⨯ ar(∆ABC)                ...(i)
ar(∆ABE ) = $\frac{1}{2}$ ⨯ ar(∆ABD)                        ...(ii)
ar(∆BOE) = $\frac{1}{2}$ ⨯ ar(∆ABE)                       ...(iii)

From (i), (ii) and (iii), we have:
ar(∆BOE ) = $\frac{1}{2}$ ar(∆ABE)
ar(∆BOE ) = $\frac{1}{2}$ ⨯ $\frac{1}{2}$ ⨯ $\frac{1}{2}$ ⨯ ar(∆ABC)
∴ ar(∆BOE ) = $\frac{1}{8}$ ar(∆ABC)18ar (∆ ABC)

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Question 22:

If a triangle and a parallelogram are on the same base and between the same parallels, then the ratio of the area of the triangle to the area of the parallelogram is
(a) 1 : 2
(b) 1 : 3
(c) 1 : 4
(d) 3 : 4
Figure

Answer:

(a) 1:2

If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangles is half the area of the parallelogram.
i.e., area of triangle = $\frac{1}{2}$ × area of parallelogram
∴ Required ratio = area of triangle : area of parallelogram = $\frac{1}{2}$ : 1 = 1 : 2

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Question 23:

In the given figure ABCD is a trapezium in which AB || DC such that AB = a cm and DC = b cm. If E and F are the midpoints of AD and BC respectively. Then, ar(ABFE) : ar(EFCD) = ?
(a) a : b
(b) (a + 3b) : (3a + b)
(c) (3a + b) : (a + 3b)
(d) (2a + b) : (3a + b)

Answer:

(c) (3a +b) : (a +3b)

Clearly, EF = $\frac{1}{2}$ (a + b) [Mid point theorem]
Let d be the distance between AB and EF.
Then d is the distance between DC and EF.
$Now, we have : ar (trap A B E F) = \frac{1}{2} (a + \frac{a + b}{2}) d = \frac{(3 a + b) d}{4} ar (trap E F C D) = \frac{1}{2} (b + \frac{a + b}{2}) d = \frac{(a + 3 b) d}{4} ∴ Required ratio = \frac{(3 a + b) d}{4} : \frac{(a + 3 b) d}{4} = (3 a + b) : (a + 3 b)$

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Question 24:

ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD is
(a) a rectangle
(b) a || gm
(c) a rhombus
(d) all of these

Answer:

(d) all of these
In all the mentioned quadrilaterals, a diagonal divides them into two triangles of equal areas.

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Question 25:

In the given figure, a || gm ABCD and a rectangle ABEF are of equal area. Then,
(a) perimeter of ABCD = perimeter of ABEF
(b) perimeter of ABCD < perimeter of ABEF
(c) perimeter of ABCD > perimeter of ABEF
(d) perimeter of $A B C D = \frac{1}{2} (perimeter of A B E F)$

Answer:

(c) perimeter of ABCD > perimeter of ABEF

Parallelogram ABCD and rectangle ABEF lie on the same base AB, i.e., one side is common in both the figures.
In ||gm ABCD, we have:
AD is the hypotenuse of right angled triangle ADF.
So, AD > AF
∴ Perimeter of ABCD > perimeter of ABEF

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Question 26:

In the given figure, ABCD is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If $A D = 2 \sqrt{5}$ cm, then area of the rectangle is
(a) 32 cm²
(b) 40 cm²
(c) 44 cm²
(d) 48 cm²

Answer:

(b) 40 cm²

Radius of the circle, AC = 10 cm
Diagonal of the rectangle, AC = 10 cm

$Now, A B = \sqrt{A C^{2} - B C^{2}} = \sqrt{10^{2} - {(2 \sqrt{5})}^{2}} = \sqrt{80} = 4 \sqrt{5} cm ∴ Area of the rectangle = A B \times A D = 2 \sqrt{5} \times 4 \sqrt{5} = 40 {cm}^{2}$

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Question 27:

Look at the statements given below:
I. A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
II. In a || gm ABCD, it is given that AB = 10 cm. The altitudes DE on AB and BF on AD being 6 cm and 8 cm respectively, then AD = 7.5 cm.
III. Area of a || gm $= \frac{1}{2} \times base \times altitude .$
Which is true?
(a) I only
(b) II only
(c) I and II
(d) II and III

Answer:

(c) I and II

Statement I is true, because if a parallelogram and a rectangle lie on the same base and between the same parallel lines, then they have the same altitude and therefore equal areas.
Statement II is also true as area of a parallelogram = base × height
$\Rightarrow$ AB × DE = AD × BF
$\Rightarrow$ 10 × 6 = 8 × AD
$\Rightarrow$ AD = 60 ÷ 8 = 7.5 cm
Hence, statements I and II are true.

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Question 28:

Assertion: In a trapezium ABCD we have AB || DC and the diagonals AC and BD intersect at O. Then, ar(∆AOD) = ar(∆BOC)

Reason: Triangles on the same base and between the same parallels are equal in areas.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

In trapezium ABCD, ∆ABC and ∆ABD are on the same base and between the same parallel lines.
∴ ar(∆ABC) = ar(∆ABD)
⇒ ar(∆ABC) $-$ ar(∆AOB) = ar(∆ABD) $-$ ar(∆AOB)
⇒ ar(∆BOC) = ar(∆AOD)
∴ Assertion (A) is true and, clearly, reason (R) gives (A).

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Question 29:

Assertion: If ABCD is a rhombus whose one angle is 60°, then the ratio of the lengths of its diagonals is $\sqrt{3} : 1$ .
Reason: Median of triangle divides it into two triangles of equal area.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.

Reason (R) is clearly true.
The explanation of assertion (A) is as follows:
ABCD is a rhombus. So, all of its sides are equal.
Now, BC = DC
⇒∠BDC = ∠DBC = x $°$
Also, ∠BCD = 60 $°$
∴ x $°$ + x $°$ + 60 $°$ = 180 $°$
⇒2x $°$ = 120 $°$
⇒ x $°$ = 60 $°$
∴ ∠BCD = ∠BDC = ∠DBC = 60 $°$
So, ∆BCD is an equilateral triangle.
i.e., BD = BC = a
∴ OB = $\frac{a}{2}$
Now, in ∆ OAB, we have:
$O A^{2} = A B^{2} - O B^{2} = a^{2} - {(\frac{a}{2})}^{2} = \frac{3 a^{2}}{4} \Rightarrow O A = \frac{\sqrt{3} a}{2} \Rightarrow A C = 2 \times \frac{\sqrt{3} a}{2} = \sqrt{3} a$
$∴ A C : B D = \sqrt{3} a : a = \sqrt{3} : 1$
Thus, assertion (A) is also true, but reason (R) does not give (A).
Hence, the correct answer is (b).

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Question 30:

Assertion: The diagonals of a || gm divide it into four triangles of equal area.
Reason: A diagonal of a || gm divides it into two triangles of equal area.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

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Question 31:

Assertion: The area of a trapezium whose parallel sides measure 25 cm and 15 cm respectively and the distance between them is 6 cm, is 120 cm².
Reason: The area of an equilateral triangle of side 8 cm is $16 \sqrt{3} {cm}^{2}$ .
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not a correct explanation of Assertion (A).

Explanation:
Reason (R):
∴ ar(∆ABC ) = $\frac{\sqrt{3}}{4} \times (side)^{2} = (\frac{\sqrt{3}}{4} \times 8 \times 8) = 16 \sqrt{3} {cm}^{2}$
Thus, reason (R) is true.

Assertion (A):
Area of trapezium = $\frac{1}{2} \times (25 + 15) \times 6 = 120 {cm}^{2}$
Thus, assertion (A) is true, but reason (R) does not give assertion (A).

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Question 32:

Assertion: In the given figure, ABCD is a || gm in which DE ⊥ AB and BF ⊥ AD. If AB = 16 cm, DE = 8 cm and BF = 10 cm, then AD is 12 cm.

Reason: Area of a || gm = base × height.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(d) Assertion is false and Reason is true.

Clearly, reason (R) is true.
Assertion: Area of a parallelogram = base × height
$\Rightarrow$ AB × DE = AD × BF
$\Rightarrow$ AD = (16 × 8) ÷ 10 = 12.8 cm

So, the assertion is false.

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Question 33:

Which of the following is a false statement?
(a) A median of a triangle divides it into two triangles of equal areas.
(b) The diagonals of a || gm divide it into four triangles of equal areas.
(c) In a ∆ABC, if E is the midpoint of median AD, then $ar (∆ B E D) = \frac{1}{4} ar (∆ A B C)$ .

(d) In a trap. ABCD, it is given that AB || DC and the diagonals AC and BD intersect at O. Then, ar(∆AOB) = ar(∆COD).

Answer:

(d) In a trap. ABCD, it is given that AB || DC and the diagonals AC and BD intersect at O. Then ar(∆AOB) = ar(∆COD).

Consider ∆ADB and ∆ADC, which do not lie on the same base but lie between same parallel lines.
i.e., ar(∆ADB) $\neq$ ar(∆ADC)
Subtracting ar(∆AOD) from both sides, we get:
ar(∆ADB) $-$ ar(∆AOD) $\neq$ ar(∆ADC) $-$ ar(∆AOD)
Or ar(∆ AOB) $\neq$ ar(∆ COD)

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Question 34:

Which of the following is a false statement?
(a) If the diagonals of a rhombus are 18 cm and 14 cm, then its area is 126 cm².
(b) Area of $a | | gm = \frac{1}{2} \times base \times corresp o nding height .$
(c) A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
(d) If the area of a || gm with one side 24 cm and corresponding height h cm is 192 cm², then h = 8 cm.

Answer:

(b) $Area of a | | gm = \frac{1}{2} \times base \times corresponding height$

Area of a parallelogram = base × corresponding height

Page No 367:

Question 1:

The area of || gm ABCD is
(a) AB × BM
(b) BC × BN
(c) DC × DL
(d) AD × DL

Answer:

(c) DC × DL

Area of a parallelogram ABCD = base × corresponding height
= AB × DL
But AB = DC
∴ Area of ABCD = DC × DL

Page No 367:

Question 2:

Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is
(a) 1 : 2
(b) 1 : 1
(c) 2 : 1
(d) 3 : 1

Answer:

(b) 1:1

Two parallelograms are on equal bases and between the same parallels.
i.e., their corresponding heights must be the same.
Thus, areas of both parallelograms are equal. Hence, the ratio of their areas is 1:1.

Page No 367:

Question 3:

ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area. Then, ABCD
(a) is a rectangle
(b) is a rhombus
(c) is a parallelogram
(d) need not be any of (A), (B), (C)

Answer:

(d) need not be any of (A), (B), (c)

A quadrilateral whose diagonal AC divides it into two parts of equal areas can be a parallelogram or a rectangle or a rhombus or a square.

Page No 367:

Question 4:

In the given figure, ABCD and ABPQ are two parallelograms and M is a point on AQ and BMP is a triangle.
Then, $ar (∆ B M P) = \frac{1}{2} ar (| | gm A B C D)$ .
(a) True
(b) False

Answer:

(a) True

ABCD and ABPQ are on thesame base and between the same parallel lines.
So, ar(parallelogram ABCD) = ar(parallelogram ABPQ) ...(i)
Now, parallelogram ABPQ and ∆BMP lie on the same base (BP) and between the same parallel lines.
So, ar(∆BMP) = $\frac{1}{2}$ ar(|| ABPQ)
From equation(i), we have:
ar(∆BMP) = $\frac{1}{2}$ ar(||ABCD)

Page No 367:

Question 5:

The midpoints of the sides of a triangle along with any of the vertices as the fourth point makes a parallelogram of area equal to
(a) $\frac{1}{2} (ar ∆ A B C)$
(b) $\frac{1}{3} (ar ∆ A B C)$
(c) $\frac{1}{4} (ar ∆ A B C)$
(d) $ar (∆ A B C)$

Answer:

(a) $\frac{1}{2}$ ( ar ∆ ABC)

Join, FE.
∆ABC has been divided into 4 triangles of equal areas.
So, ar(∆AFE) = $\frac{1}{4}$ × (ar∆ABC)
∴ ar(∣∣gm AFDE) = ar (∆AFE) + ar(∆FED)
= 2 × ar(∆AFE) = 2 × $\frac{1}{4}$ × (ar∆ABC) = $\frac{1}{2}$ (ar∆ABC)
Hence, ar(∣∣gm AFDE) = $\frac{1}{2}$ (ar∆ABC)

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Question 6:

Let ABCD be a || gm in which DL ⊥ AB and BM ⊥ AD such that AD = 6 cm, BM = 10 and DL = 8 cm. Find AB.

Answer:

We have ar(∣∣gm ABCD) = base × corresponding height
⇒ AB × DL = AD × BM
⇒ AB × 8 = 6 × 10
⇒ AB = 60 ÷ 8 = 7.5 cm
∴ AB = 7.5 cm

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Question 7:

Find the area of the trapezium whose parallel sides are 14 cm and 10 cm and whose height is 6 cm.

Answer:

Area of the trapezium = $\frac{1}{2}$ × (sum of the parallel sides) × (distance between them)
= $\frac{1}{2}$ × ( 14 + 10) × 6
= 72 cm²
∴ The area of the trapezium is 72 cm².

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Question 8:

Show that the median of a triangle divides it into two triangles of equal area.

Answer:

Suppose that AD is a median of ∆ABC and D is the mid point of BC. AD divides ∆ABC in two triangles, i.e., ∆ABD and ∆ADC.
To prove: ar(∆ABD) = ar(∆ADC)
Construction: Draw AL ⊥ BC.
Proof: ar(∆ABD) = $\frac{1}{2}$ × base × height
⇒ ar(∆ABD) = $\frac{1}{2}$ × BD × AL
⇒ ar(∆ABD) = $\frac{1}{2}$ ×DC × AL ...(i) (BD = DC)
Also, ar(∆ADC) = $\frac{1}{2}$ × DC × AL ...(ii)
From equation (i) and (ii), we have:
ar(∆ABD) = ar(∆ADC)
Hence, the median of the triangle divides it into two triangles of equal areas.

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Question 9:

Prove that area of a $triangle = \frac{1}{2} \times base \times altitude$ .

Answer:

In ∆ABC, BC is the base and AL is the corresponding height.
Construction: Through A and C, draw AD|| BC and CD || BA, intersecting each other at D.

AD || BC and CD || BA
So, BCDA is a parallelogram. Its diagonal AC divides it into two triangles of equal areas.
∴ ar(∆ABC ) = $\frac{1}{2}$ × ar(||gm BCDA)
= $\frac{1}{2}$ × BC × AL [ ∵ ar (||gm BCDA) = BC× AL]
∴ Area of a triangle = $\frac{1}{2}$ × base × height
Hence, proved.

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Question 10:

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

Answer:

ar(quad ABCD) = ar(∆ABD) + ar(∆BDC)
                          = $\frac{1}{2}$ ⨯ BD ⨯ AL + $\frac{1}{2}$ ⨯ BD ⨯ CM
                          = $\frac{1}{2}$ ⨯ BD ⨯ (AL + CM)
By substituting the values, we have:
ar(quad ABCD) = $\frac{1}{2}$ ⨯ 14 ⨯ (8 + 6)
                         = 7 ⨯14
                         = 98 cm²
∴ Area of quad. ABCD = 98 cm²

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Question 11:

In the adjoining figure, ABCD is a quadrilateral. A line through D, parallel to AC, meets BC produced in P. Prove that ar(∆ABP) = ar(quad. ABCD).

Answer:

We have:
ar(quad. ABCD) = ar(∆ACD) + ar(∆ABC)
ar(∆ABP) = ar (∆ACP) + ar(∆ABC)

∆ACD and ∆ACP are on the same base and between the same parallels AC and DP.
∴ ar(∆ACD) = ar(∆ACP)
Adding ar(∆ABC) on both sides, we get:
⇒ ar(∆ACD) + ar(∆ABC) = ar(∆ACP) + ar(∆ABC)
⇒ ar(quad. ABCD) = ar(∆ABP)
Hence, proved.

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Question 12:

In the given figure, ABCD is a quadrilateral and BE || AC and also BE meets DC produced at E. Show that the area of ∆ADE is equal to the area of quad. ABCD.

Answer:

∆BAC and ∆EAC lie on the same base AC and between the same parallel lines.
So, ar(∆BAC) = ar(∆EAC)
Adding ar(∆ADC) on both sides, we get:
⇒ ar(∆BAC) + ar(∆ADC) = ar(∆EAC) + ar(∆ADC)
⇒ ar(quad. ABCD) = ar(∆ADE)
Hence, proved.

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Question 13:

In the given figure, area of || gm ABCD is 80 cm². Find (i) ar(||gm ABEF) (ii) ar(∆ABD) and (iii) ar(∆BEF).

Answer:

||gm ABCD, ||gm ABEF, ∆ABD and ∆BEF lie on the same base and between the same parallel lines.

(i) ||gm ABCD and ||gm ABEF are on the same base and between the same parallels.
∴ ar(||gm ABEF) = ar(||gm ABCD) = 80 cm²

(ii) ||gm ABCD and ∆ABD lie on the same base and between the same parallels.
∴ ar(∆ABD) = $\frac{1}{2}$ × ar (||gm ABCD) = $\frac{1}{2}$ × 80 = 40 cm²

(iii) ||gm ABEF and ∆BEF lie on the same base and between the same parallel lines.
∴ ar(∆BEF) = $\frac{1}{2}$ × ar (||gm ABEF) = $\frac{1}{2}$ × 80 = 40 cm²

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Question 14:

In trapezium ABCD, AB || DC and L is the midpoint of BD. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. Prove that ar(trap. ABCD) = ar(||gm APQD).

Answer:

Given: A trapezium ABCD, in which AB || DC and L is the mid point of BC (i.e., BL = CL).
Now, in ∆LPB and ∆LQC, we have:
BL = CL     (Given)
∠BLP = ∠CLQ            (Vertically opposite angles)
∠PBL = ∠QCL           (Alternate angles)
∴ ∆LPB ≅ ∆LQC    (ASA congruency)

Now, ar(trapz.ABCD) = ar (APLCD ) + ar(∆ LPB)
⇒ ar(trapzABCD) = ar(APLCD ) + ar(∆ LQC) [∴ (ar (∆ LPB) = ar(∆ LQC)]
⇒ ar(trapzABCD) = ar( $∥ gm$ APQD)
Hence proved.

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Question 15:

In the adjoining figure, ABCD is a || gm and O is a point on the diagonal AC. Prove that ar(∆AOB) = ar(∆AOD).

Answer:

Given: ABCD is a parallelogram and O is the point on diagonal AC.

To prove: ar(∆AOB) = ar(∆AOD)
Construction: Join DB to intersect AC at P.
Proof: Join DB to intersect AC at P.
Since the diagonals of a parallelogram bisect each other, P is the mid point of AC as well as that of BD.
We know that the median of a triangle divides it into two triangles of equal areas.
In ∆ABD, AP is the median.
i.e., ar(∆PAB) = ar(∆PAD) ...(i)
In ∆OBD, OP is the median.
i.e., ar(∆OPB) = ar(∆OPD) ...(ii)
Adding (i) and (ii), we get:
ar(∆PAB) + ar(∆OPB) = ar(∆PAD) + ar(∆OPD)
⇒ ar(∆AOB) = ar(∆AOD)
Hence, proved.

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Question 16:

∆ABC and ∆BDE are two equilateral triangles such that D is the midpoint of BC. Then, prove that $ar (∆ B D E) = \frac{1}{4} ar (∆ A B C)$ .

Answer:

∆ABC and ∆BDE are two equilateral triangles and D is the mid point of BC.
Let AB = BC = AC = a
Then BD = BE = ED = $\frac{a}{2}$
We know that the area of an equilateral triangle is given by $\frac{\sqrt{3}}{4} (side)^{2}$ .
So, ar(∆ABC ) = $\frac{\sqrt{3}}{4} A B^{2} = \frac{\sqrt{3}}{4} a^{2}$ ...(i)
Also, ar(∆BDE ) = $\frac{\sqrt{3}}{4} \times {(B D)}^{2} = \frac{\sqrt{3}}{4} \times {(\frac{a}{2})}^{2} = \frac{\sqrt{3}}{4} \times \frac{a^{2}}{4} = \frac{1}{4} (\frac{\sqrt{3}}{4} a^{2})$
From (i), we have:
ar(∆BDE ) = ar(∆ABC)
Hence, proved.

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Question 17:

In ∆ABC, D is the midpoint of AB and P is any point on BC. If CQ || PD meets AB in Q, then prove that $ar (∆ B P Q) = \frac{1}{2} ar (∆ A B C)$ .

Answer:

Given: ∆ABC, D is the midpoint of AB and P is any point on BC.
To prove: ar(∆BPQ) = $\frac{1}{2}$ × ar(∆ABC)
Construction: Join CD.
Proof: Now, in ∆ABC, CD is a median.
∴ ar(∆BDC) = $\frac{1}{2}$ × ar(∆ABC)
⇒ ar(∆BPD) + ar(∆PDC) = $\frac{1}{2}$ × ar (∆ABC)
∆PDC and ∆PQD are on the same base PD and between the same parallel lines.
Then ar(∆PDC) = ar(∆PQD)

Now, ar(∆BPD) + ar(∆ PQD) = $\frac{1}{2}$ × ar(∆ABC)
∴ ar ( ∆BPQ) = $\frac{1}{2}$ × ar ( ∆ABC) [ ∵ ar(∆BPQ) = ar(∆BPD) + ar(∆PQD)]
Hence, proved.

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Question 18:

Show that the diagonals of a || gm divide into four triangles of equal area.
Figure

Answer:

Given: ABCD is a parallelogram. Its diagonals AC and BD intersect at O.
To prove:
ar(∆OAB) = ar(∆OBC) = ar(∆OCD) = ar (∆OAD)

Proof: Since the diagonals of a parallelogram bisects each other, we have:
OA = OC and OB = OD
Also, a median of a triangle divides it into two triangles of equal areas.
Now, in ∆ABC, BO is the median.
∴ ar(∆OAB) = ar(∆OBC) ...(i)
In ∆ABD, AO is the median.
∴ ar(∆OAB) = ar(∆OAD) ...(ii)

Now, in ∆ ADC, DO is the median.
∴ ar(∆OCD) = ar(∆OAD) ...(iii)
Similarly, in ∆BDC, OC is the median.
∴ ar(∆OBC) = ar(∆ODC) ...(iv)
From (i), (ii), (iii) and (iv), we get:
ar(∆OAB) = ar(∆OBC) = ar(∆OCD) = ar(∆OAD)
Hence, the diagonals of a parallelogram divide it into four triangles of equal areas.

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Question 19:

In the given figure, BD || CA, E is the midpoint of CA and $B D = \frac{1}{2} C A$ . Prove that ar(∆ABC) = 2 × ar(∆DBC).

Answer:

Given: BD || CA and E is the midpoint of CA.
To prove: ar(∆ABC) = 2 × ar(∆DBC)
Construction: Join DE.
Proof:
Now, BD || CE and BD = CE   [ E is the mid point of AC]
∴ BCED is a parallelogram.

So, ar(∆ EBC) = ar(∆DBC)              ...(i)    [On the same base and between the same parallel lines]
∵  ar(∆EBC) = $\frac{1}{2}$ × ar(∆ABC)             ...(ii)    [ BE is the median of ∆ABC]
From equation (i) and (ii), we get:
ar(∆DBC) = $\frac{1}{2}$ × ar(∆ABC)
⇒ ar(∆ABC) = 2 × ar(∆DBC)
Hence, proved.

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Question 20:

The given figure shows a pentagon ABCDE in which EG, drawn parallel to DA, meets BA produced at G and CF drawn parallel to DB meets AB produced at F.
Show that ar(pentagon ABCDE) = ar(∆DGF).

Answer:

Given: ABCDE is a pentagon. EG || DA and CF || DB.
To prove: ar(pentagon ABCDE ) = ar( ∆DGF)
Proof:
ar(pentagon ABCDE ) = ar(∆DBC ) + ar(∆ADE ) + ar(∆ABD)              ...(i)
ar(∆DGF) = ar(∆DBF) + ar(∆ADG) + ar(∆ABD)                           ...(ii)

Now, ∆DBC and ∆DBF lie on the same base and between the same parallel lines.
∴ ar(∆DBC ) = ar(∆DBF)    ...(iii)
Similarly, ∆ADE and ∆ADG lie on the same base and between the same parallel lines.
∴ ar(∆ADE) = ar(∆ADG)                                ...(iv)

From (iii) and (iv), we have:
ar(∆DBC) + ar(∆ADE) = ar(∆DBF) + ar(∆ADG)
Adding ar(∆ABD) on both sides, we get:
ar(∆DBC) + ar(∆ADE) + ar(∆ABD) = ar(∆DBF) + ar(∆ADG) + ar(∆ABD)
By substituting the values from (i) and (ii), we get:
ar(pentagon ABCDE ) = ar(∆DGF)
Hence, proved.

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Question 21:

In the adjoining figure, the point D divides the side BC of ∆ABC in the ratio m : n. Prove that ar(∆ABD) : ar(∆ADC) = m : n.

Answer:

Given: D is a point on BC of ∆ABC such that BD : DC = m : n.
To prove: ar(∆ABD) : ar(∆ADC) = m : n
Construction: Draw AL ⊥ BC
Proof:
Area of a triangle = $\frac{1}{2}$ × base × height
ar(∆ABD) = $\frac{1}{2}$ × BD × AL ...(i)
ar (∆ ADC) = $\frac{1}{2}$ × DC × AL ...(ii)
Dividing equation (i) by (ii), we get:
$\frac{ar (∆ A B D)}{ar (∆ A D C)} = \frac{\frac{1}{2} \times B D \times A L}{\frac{1}{2} \times D C \times A L} = \frac{B D}{D C} = \frac{m}{n} (∵ \frac{B D}{D C} = \frac{m}{n})$
∴ ar(∆ABD) : ar(∆ADC) = m : n

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Question 22:

In the give figure, X and Y are the midpoints of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar(∆ABP) = ar(∆ACQ).

Answer:

Given: Y and X are midpoints of AB and AC, respectively. QP || BC and CYQ and BXP are straight lines.
To prove: ar(∆ABP) = ar(∆ACQ)
Proof:
X and Y are the mid points of AC and AB, respectively.
So, XY || BC
In ∆BYC and ∆AYQ, we have:

$B Y = A Y (Y is the midpoint of A B) ∠ Q A Y = ∠ C B Y (Alternate angles) ∠ A Y Q = ∠ B Y C (Vertically opposite angles) S o, △ B Y C ≅ △ A Y Q (By ASA congruency)$
∴ BC = AQ                         ...(ii)    (CPCT)

In ∆BXC and ∆AXQ, we have:

$C X = X A (X is the mid point of A C) ∠ P A X = ∠ B C X (Alternate angles) ∠ A X P = ∠ C X A (Vertically opposite angles) S o, △ B X C ≅ △ A X Q (By ASA congruency)$
∴ BC = AP                         ...(ii)          (CPCT)

From (i) and (ii), we get:
AQ = AP
Now, ∆ABP and ∆ACQ are on the equal base (AQ = AP) and between the same parallels.
∴ ar( ∆ABP) = ar(∆ACQ)

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