RD Sharma 2020 solution class 9 chapter 14 Area of Parallelograms and Triangles FBQS

FBQS

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Question 1:

The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm is _________.

Answer 1:

Given:
A rhombus ABCD with diagonals 12 cm and 16 cm
i.e., AC = 16 cm and BD = 12 cm
And a quadrilateral PQRS formed by joining the mid-points of the adjacent sides of ABCD.



Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.

In ∆ABC,
PQ || AC
PQ = $\frac{1}{2}$AC
⇒ PQ = $\frac{1}{2}$(16)
⇒ PQ = 8 cm

In ∆ADC,
RS || AC
RS = $\frac{1}{2}$AC
⇒ RS = $\frac{1}{2}$(16)
⇒ RS = 8 cm

In ∆BCD,
RQ || BD
RQ = $\frac{1}{2}$BD
⇒ RQ = $\frac{1}{2}$(12)
⇒ RQ = 6 cm

In ∆BAD,
SP || BD
SP = $\frac{1}{2}$BD
⇒ SP = $\frac{1}{2}$(12)
⇒ SP = 6 cm

Since, PQ = 8 cm = RS and RQ = 6 cm = SP
and Diagonals of a rhombus intersect at right angle.
⇒ angle between AC and BD is 90°
⇒ angle between PQ and QR is 90°
Therefore, PQRS is a rectangle
Thus, Area of rectangle = PQ × QR
                                       = 8 × 6
                                       = 48 cm²

Hence, the area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm is 48 cm².

Question 2:

The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is a ________ of area _________.

Answer 2:

Given: 
A rectangle ABCD of sides 8 cm and 6 cm
i.e., AB = 8 cm and AD = 6 cm
And a quadrilateral PQRS formed by joining the mid-points of the adjacent sides of ABCD.



We can see that QS || AB and PR || AD
Also, QS = AB =  8 cm and PR = AD = 6 cm
Thus, angle between QS and PR is 90°
Therefore, PQRS is a rhombus.

Area of rhombus = $\frac{1}{2}$× SQ × PR
                            = $\frac{1}{2}$× 8 × 6
                            = 4 × 6
                            = 24 cm²

Hence, the figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is a rhombus of area 24 cm².

Question 3:

If P is any point on the median AD of a ΔABC, then $\frac{\mathrm{ar} \left(∆ABP\right)}{\mathrm{ar} \left(∆ACP\right)}$= _________.

Answer 3:

Given: P is any point on the median AD of a ΔABC

We know, median of a triangle divides it into two triangles of equal area.

Therefore, ar (ΔADB) = ar (ΔADC)       ...(1)
Also, ar (ΔPDB) = ar (ΔPDC)               ...(2)

Subtracting (2) from (1), we get
ar (ΔADB) − ar (ΔPDB) = ar (ΔADC) − ar (ΔPDC)
⇒ ar (ΔABP) = ar (ΔACP)
⇒$\frac{\mathrm{ar} \left(∆ABP\right)}{\mathrm{ar} \left(∆ACP\right)}$= 1


Hence, $\frac{\mathrm{ar} \left(∆ABP\right)}{\mathrm{ar} \left(∆ACP\right)}$= 1.

Question 4:

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 _________.

Answer 4:

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

Area of triangle = $\frac{1}{2}$ Area of the parallelogram
$\Rightarrow \frac{\mathrm{Area} \mathrm{of} \mathrm{triangle}}{\mathrm{Area} \mathrm{of} \mathrm{parallelogram}}=\frac{1}{2}$


Hence, 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 1 : 2.

Question 5:

Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is _________.

Answer 5:

Parallelograms on equal bases and between the same parallels are equal in area.

Area of first parallelogram = Area of the second parallelogram
$\Rightarrow \frac{\mathrm{Area} \mathrm{of} \mathrm{first} \mathrm{parallelogram}}{\mathrm{Area} \mathrm{of} \mathrm{second} \mathrm{parallelogram}}=\frac{1}{1}$


Hence, the ratio of their areas is 1 : 1.

Question 6:

ABCD is a parallelogram and X is the mid-point of AB. If ar(AXCD) = 24 cm² , then ar(ΔABC) = ________.

Answer 6:

Given: 
ABCD is a parallelogram
X is the mid-point of AB
ar(AXCD) = 24 cm²

We know, if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.
Thus, ar(ΔABC) = $\frac{1}{2}$ar(ABCD)     ...(1)

Since, X is the mid-point of AB
Therefore, ar(ΔXBC) = $\frac{1}{2}$ar(ABC)
                                   = $\frac{1}{2}$×$\frac{1}{2}$ar(ABCD)    (From (1))
                                   = $\frac{1}{4}$ar(ABCD)      ...(2)

Thus, ar(AXCD) = ar(ABCD) − ar(ΔXBC)
⇒ 24 = ar(ABCD) − $\frac{1}{4}$ar(ABCD)         (From (2))
⇒ 24 = $\frac{3}{4}$ar(ABCD)
⇒ ar(ABCD) = $24\times \frac{4}{3}$
⇒ ar(ABCD) = 8 × 4
⇒ ar(ABCD) = 32 cm²

From (1)
ar(ΔABC) = $\frac{1}{2}$ar(ABCD)
                 = $\frac{1}{2}$× 32
                 = 16 cm²


Hence, ​ar(ΔABC) = 16 cm².

Question 7:

PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then ar(ΔPAS) = _______.

Answer 7:

Given: 
PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm
PS = 5 cm
A is any point on PQ


QS = radius of the circle = 13 cm     ...(1)

In ΔPQS
Using pythagoras theorem,
 QS² = PS² + PQ²
⇒ 13² = 5² + PQ²
⇒ PQ² = 169 − 25
⇒ PQ² = 144
⇒ PQ = 12 cm = SR            ...(2)


Thus,
ar(ΔRAS) = $\frac{1}{2}$× base × height
                 = $\frac{1}{2}$× SR × PS
                 = $\frac{1}{2}$× 12 × 5
                 = 30 cm²


Hence, ​ar(ΔRAS) = 30 cm².

Disclaimer: The question is to find the area of ΔRAS instead of the area of ΔPAS.

Question 8:

If ABC and BDE are two equilateral triangles such that D is the mid-point of BC then ar(ΔABC) : ar(BDE) = _________.

Answer 8:

Given: 
ABC and BDE are two equilateral triangles
D is the mid-point of BC


 $$\begin{gathered} \mathrm{ar}\left(∆ABC\right)=\frac{\sqrt{3}}{4}\times {\left(\mathrm{side}\right)}^2 \\ =\frac{\sqrt{3}}{4}\times {\left(BC\right)}^2 ...\left(1\right) \\ \mathrm{ar}\left(∆BDE\right)=\frac{\sqrt{3}}{4}\times {\left(\mathrm{side}\right)}^2 \\ =\frac{\sqrt{3}}{4}\times {\left(BD\right)}^2 \\ =\frac{\sqrt{3}}{4}\times {\left(\frac{1}{2}BC\right)}^2 \left(\because D \mathrm{is} \mathrm{the} \mathrm{mid}-\mathrm{point} \mathrm{of} BC\right) \\ =\frac{\sqrt{3}}{4}\times \frac{1}{4}\times {\left(BC\right)}^2 \\ =\frac{\sqrt{3}}{16}\times {\left(BC\right)}^2 ...\left(2\right) \\ \mathrm{Dividing} \left(1\right) \mathrm{and} \left(2\right), \mathrm{we} \mathrm{get} \\ \frac{\mathrm{ar}\left(∆ABC\right)}{\mathrm{ar}\left(∆BDE\right)}=\frac{{\displaystyle \frac{\sqrt{3}}{4}}{\left(BC\right)}^2}{{\displaystyle \frac{\sqrt{3}}{16}}{\left(BC\right)}^2} \\ =\frac{4}{1} \end{gathered}$$


Hence, ar(ΔABC) : ar(ΔBDE) = 4 : 1.
 

Question 9:

If PQRS is a parallelogram whose area is 180 cm² and A is any point on the diagonal QS, then ar(ΔASR) is _______ 90 cm²?

Answer 9:

Given: 
PQRS is a parallelogram with area 180 cm²
A is any point on the diagonal QS


 We know, the diagonal of the parallelogram bisects it into two triangles of equal area.
Thus, ar(ΔPQS) = ar(ΔQRS) = $\frac{1}{2}$ar(PQRS)
⇒ ar(ΔPQS) = ar(ΔQRS) = 90 cm²

Therefore, ar(ΔASR) is always less than 90 cm² unless or until the point A coincides with Q or S.

Hence, ar(ΔASR) is less than 90 cm²

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Question 10:

The mid points of the sides of a triangle ABC along with any of the vertices as the fourth point make a parallelogram of area equal to ________.

Answer 10:

Given: 
ABC is a triangle


Let D is the mid-point of AB, E is the mid-point of BC and F is the mid-point of AC.

ADEF is a parallelogram having 2 triangles of equal area i.e., ∆ADF and ∆DEF.

But the ∆ABC is divided in 4 triangles of equal area i.e., ∆ADF, ∆DEF, ∆BED and ∆CEF.

Thus, area of ∆ABC = 2 × area of the parallelogram ADEF.


Hence, the mid-points of the sides of a triangle ABC along with any of the vertices as the fourth point make a parallelogram of area equal to half the area of the triangle ABC.

Question 11:

A median of a triangle divides it into two ___________.

Answer 11:

Let ABC be a triangle with a mid-point D on BC.
Therefore, BD = DC
Let AE be the altitude from A on BC.

Now, ar(∆ABD) = $\frac{1}{2}$× base × height
                          = $\frac{1}{2}$× BD × AE

Also, ar(∆ACD) = $\frac{1}{2}$× base × height
                          = $\frac{1}{2}$× CD × AE
                          = $\frac{1}{2}$× BD × AE         (∵ BD = CD)
                          = ar(∆ABD) 


Hence, a median of a triangle divides it into two triangles of equal area.

Question 12:

If ABCD is a rectangle, E and F are the mid-points of BC and AD respectively and G is any point on EF. Then ar(ΔGAB) : ar (rectangle ABCD) = _________.

Answer 12:

Given: 
ABCD is a rectangle
E and F are the mid-points of BC and AD respectively
G is any point on EF


Since, E and F are the mid-points of BC and AD respectively.
Therefore, ar(BEFA) = ar(ECDF) = $\frac{1}{2}$× ar(ABCD)            ...(1)

We know, if a triangle and a rectangle are on the same base and between the same parallels, then the area of the triangle is half the area of the rectangle.
Thus, ar(ΔGAB) = $\frac{1}{2}$ar(BEFA)             ...(2)

From (1) and (2),
ar(ΔGAB) = $\frac{1}{2}$×$\frac{1}{2}$× ar(ABCD)
⇒ ar(ΔGAB) = $\frac{1}{4}$× ar(ABCD)
⇒ $\frac{\mathrm{ar}\left(∆GAB\right)}{\mathrm{ar}\left(\mathrm{rectangle} ABCD\right)}=\frac{1}{4}$
​

Hence, ar(ΔGAB) : ar(rectangle ABCD) = 1 : 4.

Question 13:

PQRS is a square of side 8 cm. T and U and  respectively the mid points of PS and QR. If TU and QS intersect at 0, then ar(ΔOTS) = ________.


 

Answer 13:

Given: 
PQRS is a square of side 8 cm.
T and U are respectively the mid-points of PS and QR
TU and QS intersect at O



In ΔQOU and ΔOTS,
∠QOU = ∠TOS  (vertically opposite angles)
∠OQU = ∠OST  (alternate angles)
QU = TS (mid-points of sides of a square)

By AAS property,
ΔQOU ≅ ΔOTS

Thus, OU = OT (by CPCT)
⇒ OU + OT = PQ = 8 cm
⇒ OU = OT = 4 cm             ...(1)

Also, TS = 4 cm (T is the mid-point of PS)       ...(2)

ar(ΔOTS) = $\frac{1}{2}$× base × height
                = $\frac{1}{2}$× TS × OT
                = $\frac{1}{2}$× 4 × 4    (From (1) and (2))
                = 8 cm²
​

Hence, ar(ΔOTS) = 8 cm².

Question 14:

In the given figure, ABCD and EFGD are two parallelograms and G is the mid point of CD. Then, ar(ΔDPC) : ar(EFGD) = ________.

Answer 14:

Given: 
ABCD and EFGD are two parallelograms.
G is the mid point of CD


Since, G is the mid point of CD
Therefore, DG = GC

Since, ΔDPG and ΔGPC have equal base and common height,
Thus, ar(ΔDPG) = ar(ΔGPC) 
⇒ ar(ΔDPC) = 2 ar(ΔDPG)       ...(1)

Also, if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.
Thus, ar(ΔDPG) = $\frac{1}{2}$ar(EFGD)     ...(2)

From (1) and (2),
ar(ΔDPC) = 2 × $\frac{1}{2}$ar(EFGD)
⇒ ar(ΔDPC) = ar(EFGD)
​

Hence, ar(ΔDPC) : ar(EFGD) = 1 : 1.

Question 15:

If the given figure, PQRS and EFRS are two parallelograms, then ar(||^gm PQRS) : ar(ΔMFR) = ________.

Answer 15:

Given: 
PQRS and EFRS are two parallelograms



PQRS and EFRS are two parallelograms lying on the same base SR and between the same parallels SR and PF.

We know, if two parallelograms are on the same base and between the same parallels, then the area of the parallelograms are equal.

Thus, ar(PQRS) = ar(EFRS)             ...(1)

Also, ΔMFR and parallelogram EFRS is lying on the same base FR and between the same parallels SR and EF.

We know, if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.

Thus, ar(ΔMFR) = $\frac{1}{2}$ar(EFRS)     ...(2)

From (1) and (2),
ar(ΔMFR) = $\frac{1}{2}$ar(PQRS)
⇒ ar(||^gm PQRS) = 2 ar(ΔMFR)
​

Hence, ar(||^gm PQRS) : ar(ΔMFR) = 2 : 1.

Question 16:

In the given figure, if ABCD is a parallelogram of area 90 cm². Then, ar(||^gm ABEF) = ________ ar(ΔABD) = _______ and ar(ΔBEF) = _________.

Answer 16:

Given: 
ABCD is a parallelogram of area 90 cm²


ABCD and ABEF are two parallelograms lying on the same base AB and between the same parallels AB and CF.

We know, if two parallelograms are on the same base and between the same parallels, then the area of the parallelograms are equal.

Thus, ar(ABCD) = ar(ABEF) = 90 cm²           ...(1)

Also, ΔABD and parallelogram ABEF is lying on the same base AB and between the same parallels AF and BE.

We know, if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.

Thus, ar(ΔABD) = $\frac{1}{2}$ar(ABEF)
⇒ ar(||^gm ABEF) = 2 ar(ΔABD)          ...(2)

From (1) and (2),
ar(||^gm ABEF) = 2 ar(ΔABD) = 90 cm²
​
Also, diagonal of a parallelogram divides it into two triangles of equal area.
Thus,  ar(ΔBEF) = $\frac{1}{2}$ar(ABEF)
                            = $\frac{1}{2}$(90)
                            = 45 cm²


Hence, ar(||^gm ABEF) = 2 ar(ΔABD) = 90 cm² and ar(ΔBEF) = 45 cm².