Maths M.C.Q

2. $\frac{1}{2}\text{ar}(\triangle\text{ABC})$

Solution:

AQRP is a required parallelogram by joining the mid-points.

All 4 triangles formed are congruent and are equal in area.

So area of any one $\triangle{}=\frac{1}{4}\text{AR}(\triangle\text{ABC})$

$\text{Ar}(\triangle\text{APQ})+\text{Ar}(\triangle\text{PQR})=\frac{1}{2}\text{Ar}(\triangle\text{ABC})$

$\Rightarrow\text{Ar}(\text{AQRP})=\frac{1}{2}\text{Ar}(\triangle\text{ABC})$

Hence, correct option is (b).

4. Triangles of equal areas.

Solution:

A median divides the base in two equal parts but height of a triangle remains the same.

Now, since bases and heights are equal, areas of both $\triangle\text{s}$ are equal.

Hence, correct option is (d).

1. A rhombus of area 24cm²

Solution:

Figure obtained by joining the mid-points of adjacent sides of rectangle ABCD is a rhombus PQRS.

AB = 8cm, AD = 6cm

QS and PR are diagonals of Rhombus PQRS.

QS = AB = 8cm

PR = AD = 6cm

Ar(Rhombus PQRS) = 4 × Area of $\triangle\text{POR}$

$=4\times\frac{1}{2}\times\text{OQ}\times\text{OP}$ $\Big(\triangle\text{POQ}$ is a Right $\triangle\text{OQ}=\frac{\text{QS}}{2},\ \text{OP}=\frac{\text{PR}}{2}\Big)$

$=\frac{{4}^2}{2}\times4\times3$

⇒ Ar(Rhombus PQRS) = 24cm²

Hence, correct option is (a).

2. 48cm²

Solution:

AC = 12cm and BD = 16cm (given)

Now, consider $\triangle\text{ABC}.$

P and Q are mid-points of sides AB and BC

So line joining them will be parallel to the third side AC and equal to $\frac{1}{2}\text{AC}.$

$\Rightarrow\text{PQ}=\frac{1}{2}\text{AC}=6\text{cm}$

Similiary, in $\triangle\text{ABD},$ PS || BD

$\Rightarrow\text{PS}=\frac{1}{2}\text{BD}=8\text{cm}$

Now we know that by joining mid-points of adjacent sidea of a Rhombus.

We get a Rectangle whose sides are PQ and PS.

⇒ Area = PQ × PS = 6 × 8 = 48cm²

Hence, correct option is (b).

2. 9cm²

Solution:

AQ, CP and RB are medians of $\triangle\text{ABC}.$

Consider $\triangle\text{ACP}\ \&\ \triangle\text{ACQ}$

$\text{Ar}(\triangle\text{ACP})=\frac{27}{2}\text{cm}^2$ $\big[$Median divides a $\triangle$ into two equal Area$\big]$

$\text{Ar}(\triangle\text{ACQ})=\frac{27}{2}\text{cm}^2$ $\big[$Median divides a $\triangle$ into two equal Area$\big]$

$\Rightarrow\text{Ar}(\triangle\text{ACP})=\text{Ar}(\text{ACQ})$

$\text{Ar}(\triangle\text{AGC})$ is common in both the triangles.

$\Rightarrow\text{Ar}(\triangle\text{CGQ})=\text{Ar}(\triangle\text{AGP})\ ...(1)$

Similiarly $\text{Ar}\big(\triangle\text{ABR}\big)=\frac{27}{2}\text{cm}^2=\text{Ar}\big(\triangle\text{AQB}\big)$

$\text{Ar}\big(\triangle\text{ARG}\big)$ is commpn in both the triangle.

$\Rightarrow\text{Ar}\big(\triangle\text{ARG}\big)=\text{Ar}\big(\triangle\text{GQB}\big)\dots(2)$

From figure GR, GP, GQ are also medians for $\triangle\text{AGC},\triangle\text{AGB }\&\triangle\text{CGB}$ respectively.

$\Rightarrow\text{Ar}\big(\triangle\text{AGC}\big)+\text{Ar}\big(\triangle\text{AGB}\big)+\text{Ar}\big(\triangle\text{CGB}\big)=27\text{cm}^2$

$\Rightarrow2\text{Ar}\big(\triangle\text{ARG}\big)+2\text{Ar}\big(\triangle\text{AGP}\big)+\text{Ar}\big(\triangle\text{BGC}\big)=27\text{cm}^2$

$\Rightarrow\big(2\text{Ar}\big(\triangle\text{ARG}\big)+\text{Ar}\big(\triangle\text{AGP}\big)+\text{Ar}\big(\triangle\text{BGC}\big)=27\text{cm}^2$

From equauations (1) and (2).

$2\big[\text{Ar}\big(\triangle\text{GQB}\big)+\text{Ar}\big(\triangle\text{AGB}\big)\big]+\text{Ar}\big(\triangle\text{CGB}\big)=27\text{cm}^2$

$\Rightarrow2\big[\text{Ar}\big(\triangle\text{BGC}\big)\big]+\text{Ar}\big(\triangle\text{BGC}\big)=27\text{cm}^2$

$\Rightarrow\text{Ar}\big(\triangle\text{BGC}\big)=9\text{cm}^2$

Hence, Correct option is (b).

2. 6sq. units

Solution:

When a triangle is formed by joining the mid-points of sides of a triangle, the triangle formed is Congruent to triangles formed around that.

i.e. $\triangle\text{PQR}$ is congruent to $\triangle\text{RPA},\ \triangle\text{QBP}\ \&\ \triangle\text{CQR}$

Hence, Area of all four triangles formed inside $\triangle\text{ABC}$ is same.

So $($4 × Area of any one $\triangle)$ = Area of $\triangle\text{ABC}$

4 × $($Area of $\triangle\text{PQR})$ = 24sq. units

Area of $\triangle\text{PQR}$ = 6sq. units

Hence, correct option is (b).

4. 1 : 1

Solution:

Diagonal SQ divides ||^gm in two equal areas.

Hence $\text{Ar}(\triangle\text{QSR})=\frac{1}{2}\text{Ar}(\text{PQRS})$

Also XY divides the ||^gm into two equal parts.

Hence, Ratio of $\text{Area}(||^{\text{gm}}\text{XQRY})=\frac{1}{2}\text{Ar}(\text{PQRS})$

Thus, Ratio of $\text{Area}(||^{\text{gm}}\text{XQRY}):\text{Ar}(\triangle\text{QSR})=1:1$

Hence, correct option is (d).

3. 24cm²

Solution:

$\text{Ar}(||^{\text{gm}}\text{ABCD})=\text{Ar}(||^{\text{gm}}\text{FECG})$

Ar of $(||^{\text{gm}}\text{AQED})$ is common in both,

$\Rightarrow\text{Ar}(||^{\text{gm}}\text{AQED})=\text{Ar}(||^{\text{gm}}\text{FGBQ})\ ...(1)$

Now AE is diagonal of AQED.

$\Rightarrow\text{Ar}(\triangle\text{AQE})=\frac{1}{2}\text{Ar}(||^{\text{gm}}\text{AQED})$

$\Rightarrow12\text{cm}^2=\frac{1}{2}\text{Ar}(||^{\text{gm}}\text{AQED})$

$\Rightarrow\text{Ar}(||^{\text{gm}}\text{AQED})=2\times12\text{cm}=24\text{cm}^2$

$\Rightarrow\text{Ar}(||^{\text{gm}}\text{FGBQ})=24\text{cm}^2$ [From (1)]

Hence, correct option is (c).

3. 8cm²

Solution:

E is the mid-point of AD and CE is median of $\triangle\text{ACD}.$

Hence $\text{Ar}(\triangle\text{AEC})=\text{Ar}(\triangle\text{CED})=4\text{cm}^2\ ...(1)$

(Median divides a $\triangle$ in two equal Areas)

Also AD is median of $\triangle\text{ABC}$ and ED is median of $\triangle\text{BEC}.$

So $\text{Ar}(\triangle\text{BED})=\text{Ar}(\triangle\text{CED})=4\text{cm}^2$ [From eq (1)]

So $\text{Ar}(\triangle\text{BEC})=\text{Ar}(\triangle\text{BED})\\+\text{Ar}(\triangle\text{CED})=4+4=8\text{cm}^2$

Hence, correct option is (c).

3. 12cm²

Solution:

Area of $\triangle\text{ABC}$ = Area of $\triangle\text{AEF}$ + Area of trapazium FBCE

We know that any triangle formed by joining the mid-point of sides of triangle.

has area $=\frac{1}{4}\times(\text{parent}\triangle)$

$\Rightarrow\text{Area of }\triangle\text{AEF}=\frac{1}{4}\times\text{Ar}(\triangle\text{ABC})\\=\frac{1}{4}\times16\text{cm}^2=4\text{cm}^2$

⇒ Area of Trapezium = (16 - 4)cm² = 12cm²

Hence, correct option is (c).

3. 1 : 6

Solution:

A median divides a triangle in two equal triangles.

$\Rightarrow\text{Ar}(\triangle\text{ABD})=\text{Ar}(\triangle\text{ADC})$

$\text{Ar}(\triangle\text{PDC})=\text{Ar}(\triangle\text{ADC})-\text{Ar}(\triangle\text{ADP})$

$\Rightarrow\text{Ar}(\triangle\text{PDC})=\text{Ar}(\triangle\text{ABD})-\text{Ar}(\triangle\text{ADP})\ ...(1)$

Also,

$\text{Ar}(\triangle\text{ABC})=2\times\text{Ar}(\triangle\text{ABD})$

Dividing equation (1) by $\text{Ar}(\triangle\text{ABC}),$ we get

$\frac{\text{Ar}(\triangle\text{PDC})}{\text{Ar}(\triangle\text{ABC})}=\frac{\text{Ar}(\triangle\text{ABD})}{\text{Ar}(\triangle\text{ABC})}-\frac{\text{Ar}(\triangle\text{ADP})}{\text{Ar}(\triangle\text{ABC})}$

$\Rightarrow\frac{\text{Ar}(\triangle\text{PDC})}{\text{Ar}(\triangle\text{ABC})}=\frac{\text{Ar}(\triangle\text{ABD})}{2\text{Ar}(\triangle\text{ABD})}-\frac{\text{Ar}(\triangle\text{ADP})}{2\text{Ar}(\triangle\text{ABD})}$

$\Rightarrow\frac{1}{2}-\frac{1}{2}\times\frac{2}{3}$

$=\frac{1}{6}$

$=1:6$

Hence, correct option is (c).

2. $\triangle\text{BOC}$

Solution:

$\triangle\text{ABD}\ \&\ \triangle\text{ABC}$ have same base AB and are between same parallels.

Then,

$\text{Ar}(\triangle\text{ABD})=\text{Ar}(\triangle\text{ABC})$

But $\text{Ar}(\triangle\text{AOB})$ is common in both.

Thus, $\text{Ar}(\triangle\text{AOD})=\text{Ar}(\triangle\text{BOC})$

Hence, correct option is (b).

2. 1 : 2

Solution:

Area of a triangle DFC $=\frac{1}{2}\times\text{base}\times\text{height}\\=\frac{1}{2}\times\text{DC}\times\text{FG}=\frac{1}{2}\times\text{DC}\times\text{h}$

Area of parallelogram ABCD = base × height = DC × AE = DC × h

Required Ratio $=\frac{\frac{1}{2}\times\text{DC}\times\text{h}}{\text{DC}\times\text{h}}=1:2$

Hence, correct option is (b).

3. (3a + b) : (a + 3b)

Solution:

AP is drawn parallel to BC.

ABCP is a parallelogram.

AB = PC = a

DP = DC - PC = b - a

$\text{Area}(\text{ABFE})=\text{Ar}(\triangle\text{AQE})+\text{Ar}(||^{\text{gm}}\text{ABFQ})$

$=\frac{1}{4}(\text{Ar}(\triangle\text{ADP}))+\frac{1}{2}\text{Ar}(||^{\text{gm}}\text{ABCP})$

$=\frac{1}{4}\times\frac{1}{2}\times(\text{b}-\text{a})\text{h}+\frac{1}{2}\times\text{a}\times\text{h}$

$=\frac{(3\text{a}+\text{b})\text{h}}{8}$

Similiarly, Area of trapazium EFCD

$=\text{Ar}(\text{EQPD})+\text{Ar}(||^{\text{gm}}\text{QFCP})$

$=\frac{3}{4}(\text{Ar}(\triangle\text{ADP}))+\frac{1}{2}\text{Ar}(||^{\text{gm}}\text{ABCP})$

$=\frac{3}{4}\times\frac{1}{2}\times(\text{b}-\text{a})\times\text{h}+\frac{1}{2}\times\text{a}\times\text{h}$

$=\frac{(3\text{b}+\text{a})\text{h}}{8}$

⇒ Ratio of Ar(Quad ABFE) : Ar(Quad EFCD) $=\frac{(3\text{a}+\text{b})\text{h}}{8}:\frac{(3\text{b}+\text{a})\text{h}}{8}$

$=(3\text{a}+\text{b})(\text{a}+3\text{b})$

Hence, correct option is (c).

3. 6cm

Solution:

$\triangle\text{ABD}\ \&\ \triangle\text{ABC}$ are on same base AB and are between same parallels.

$\Rightarrow\text{Ar}(\triangle\text{ABD})=\text{Ar}(\triangle\text{ABC})$

$\text{Ar}(\triangle\text{ABD})=\frac{1}{2}\times8\times\text{h}=24\text{cm}^2$

$\Rightarrow\text{h}=6\text{cm}$

Now, $\text{Ar}(\triangle\text{ABC})=\frac{1}{2}\times8\times\text{h}=24\text{cm}^2$

$\Rightarrow\text{h}=6\text{cm}$

Hence, correct option is (c).

4. 18cm²

Solution:

A line PQ is drawn fromn AB parallel to AD & BC.

Now, $\triangle\text{AOD}$ has height = AP

And, $\triangle\text{BOC}$ has height = BP

Area of $\triangle\text{AOD}=\frac{1}{2}\times\text{AD}\times\text{AP}=3\text{cm}^2$

$\Rightarrow\text{AD}\times\text{AP}=6\text{cm}^2\ ...(1)$

$​​\text{Ar}(\triangle\text{BOC})=\frac{1}{2}\times\text{BC}\times\text{BP}=6\text{cm}^2$

$\Rightarrow\text{BC}\times\text{BP}=12\text{cm}^2\ ...(2)$

Adding equation (1) and (2), we get

$\text{AD}\times\text{AP}+\text{BC}\times\text{BP}=18\text{cm}^2$

$\Rightarrow\text{AD}\times\text{AP}+\text{AD}\times\text{BP}=18\text{cm}^2$ (AD = BC)

$\Rightarrow\text{AD}(\text{AP}+\text{BP})=18\text{cm}^2$

$\Rightarrow\text{AD}\times\text{AB}=18\text{cm}^2$ = Area of rectangle ABCD

Hence, correct option is (d).

3. 35cm²

Solution:

Area of trapazium ABCP = Area of $\triangle\text{APB}$ + ar of $\triangle\text{BPC}\ ...(1)$

$\triangle\text{APC}$ and $\triangle\text{BPC}$ have same base PC and are batween same parallels.

⇒ Area of $\triangle\text{APC}$ = Area of $\triangle\text{BPC}$ = 20cm² ...(2)

From figure, $\text{Ar}(\triangle\text{ADP})+\text{Ar}(\triangle\text{APC})=\frac{1}{2}\text{Ar}(||^{\text{gm}}\text{ABCD})$

$\Rightarrow\text{Ar}(||^{\text{gm}}\text{ABCD})=2(20+15)=70\text{cm}^2$

Area of trapazium ABCP $=\text{Ar}(||^{\text{gm}}\text{ABCD})-\text{Ar}(\triangle\text{ADP)}=70-15=55\text{cm}^2$

⇒ Area of $\triangle\text{APB}$ Area of trapezium ABCP - Area of $\triangle\text{BPC}$

$=(55-20)\text{cm}^2$ [From (1)]

$=35\text{cm}^2$

Hence, correct option is (c).

2. 18cm²

Solution:

D and B are joined.

DB || PQ || RS

Now, consider parallelogram PQBD and parallelogram DBRS.

In PQBD, $\triangle\text{ABD}$ has same base and same height as parallelogram PQBD.

So, Area of $\triangle\text{ABD}=\frac{1}{2}\times\text{Ar}(\text{PQBD})$

Similarly, Area of $\triangle\text{CDB}=\frac{1}{2}\times\text{Ar}(\text{RSDB})$

Area of (ABCD) = $\text{Area of }\triangle\text{ABD}+\text{Area of }\triangle\text{CDB}$

$=\frac{1}{2}[\text{Ar }(\text{PQBD})+\text{Ar}(\text{RSDB})]$

$=\frac{1}{2}\text{Area of PQRS}$

$=\frac{1}{2}\times36$

$=18\text{cm}^2$

Hence, correct option is (b).