5 Mark Question - Maths STD 9 Questions - Vidyadip

Questions

5 Mark Question

🎯

Test yourself on this topic

29 questions · timed · auto-graded

Question 15 Marks

PQRS is a rectangle inscribed in a quadrant of a circle of radius 13cm. A is any point on PQ. If PS = 5cm, then
find $\text{ar}(\triangle\text{RAS}).$

Answer

Given: Here from the given figure we get

PQRS is a rectangle inscribed in a quadrant of a circle with radius 10cm,
PS = 5cm
PR = 13cm (radius of the quadrant)

To find: Area of $\triangle\text{RAS}.$
Calculation: In right $\triangle\text{PSR},$
(Using Pythagoras Theorem)$ PR^2$
$= PS^2 + SR^2 13^2$
$= 5^2 + SR^2 SR^2$
$= 13^2 - 5^2 SR^2$
$= 169 - 25 SR^{2}$
$ = 144 SR = 12cm$

Area of Rectangle $\triangle=\frac{1}{2}\times\text{base}\times\text{height}$$\text{Area of }\triangle\text{RAS}=\frac{1}{2}\times\text{base}\times\text{height}$
$=\frac{1}{2}\times\text{RS}\times\text{SP}$
$=\frac{1}{2}\times12\times5$
$=\frac{1}{2}\times60$
$\text{Area of }\triangle\text{RAS}=30\text{cm}^2$
Hence we get the Area of $\triangle\text{RAS}=30\text{cm}^2$

View full question & answer→

Question 25 Marks

In figure, ABCD and AEFD are two parallelograms. Prove that:

$\text{PE}=\text{FQ}$
$\text{ar}(\triangle\text{APE}) : \text{ar}(\triangle\text{PFA})\\=\text{ar}(\triangle\text{QFD}): \text{ar}(\triangle\text{PFD})$
$\text{ar}(\triangle\text{PEA})=\text{ar}(\triangle\text{QFD})$

Answer

Given that, ABCD and AEFD are two parallelograms

In triangles, EPA and FQD

$\angle\text{PEA}=\angle\text{QFD}$ [corresponding angles]

$\angle\text{EPA}=\angle\text{FQD}$ [corresponding angles]

PA = QD [opposite sides of parallelogram]

Then, $\triangle\text{EPA}\cong\triangle\text{FQD}$ [By AAS condition]

Therefore, EP = FQ [C.P.C.T]

Since triangles, PEA and QFD stand on equal bases PE and FQ lies between the same parallels EQ and AD

Therefore, $\text{ar}(\triangle\text{PEA})=\text{ar}(\triangle\text{QFD})\ ...(1)$

Since, triangles PEA and PFD stand on the same base PF and between same parallels PF and AD

Therefore, $\text{ar}(\triangle\text{PFA})=\text{ar}(\triangle\text{PFD})\ ...(2)$

Divide the equation 1 by equation 2

$\text{ar}(\triangle\text{PEA})\text{ ar}(\triangle\text{PFA})=\text{ar}(\triangle\text{QFD})\text{ ar}(\triangle\text{PFD})$

From part (i),

$\triangle\text{EPA}\cong\triangle\text{FQD}$

Then, $\text{ar}(\triangle\text{PEA})=\text{ar}(\triangle\text{QFD}).$.

View full question & answer→

Question 35 Marks

In figure, find the area of $\triangle\text{GEF}.$

Answer

Given:

ABCD is a rectangle.
CD = 6cm
AD = 8cm

To find: Area of rectangle $\triangle\text{GEF}.$ Calculation: We know that, Area of parallelogram = base × height If a triangle and a parallelogram are on the same base and between the same parallels , the area of the triangle is equal to half of the parallelogram

Here we can see that rectangle ABCD and Parallelogram GEF are between the same base and same parallels. Hence, Area of Rectangle $\triangle\text{GEF}=\frac{1}{2}\text{Area of parallelogram ABCD}$$=\frac{1}{2}\times\text{AD}\times\text{CD}$
$=\frac{1}{2}\times8\times6$
$=24\text{cm}^2$
Hence we get the result as Area of $\triangle\text{GEF}=24\text{cm}^2$

View full question & answer→

Question 45 Marks

In figure, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that $\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ABD}).$

Answer

Given that CD is bisected by AB at O To prove: $\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ABD}).$ Construction: Draw $\text{CP}\perp\text{AB}$ and $\text{DQ}\perp\text{AB}.$ Proof:$\text{ar}(\triangle\text{ABC})=\frac{1}{2}\times\text{AB}\times\text{CP}\ ⋅⋅⋅⋅⋅ (1)$
$\text{ar}(\triangle\text{ABD})=\frac{1}{2}\times\text{AB}\times\text{DQ}\ ⋅⋅⋅⋅⋅ (2)$
In $\triangle\text{CPO}$ and $\triangle\text{DQO}$$\angle\text{CPO}=\angle\text{DQO}$ [Each 90°]
Given that, CO = OD$\angle\text{CPO}=\angle\text{DQO}$ [Vertically opposite angles are equal]
Then, $\triangle\text{CPO}\cong\text{DQO}$ [By AAS condition]$\therefore$ CP = DQ (3) [C.P.C.T]
Compare equation (1), (2) and (3)$\therefore\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ABD}).$

View full question & answer→

Question 55 Marks

In figure, ABCD is a trapezium in which $AB = 7cm, AD = BC = 5cm, DC = x cm$, and distance between AB and DC is $4cm$. Find the value of x and area of trapezium ABCD.

Answer

Draw $\text{AL}\perp\text{DC},\ \text{BM}\perp\text{DC}$ then, $AL = BM = 4cm$ and $LM = 7cm$. In
$\triangle\text{ADL},$ we have $AD^2 = AL^2 + DL^{2}$
$ \Rightarrow 25 = 16 + DL^2$
$\Rightarrow DL = 3cm$ Similarly,
$\text{MC}=\sqrt{\text{BC}^2-\text{BM}^2}=\sqrt{25-16}=3\text{cm}$
$\therefore$ $X = CD = CM + ML + LD = (3 + 7 + 3)cm = 13cm$
ar(trap. ABCD) $=\frac{1}{2}(\text{AB}+\text{CD})\times\text{AL}\\ =\frac{1}{2}(7+3)\times4\text{cm}^2=40\text{cm}^2$

View full question & answer→

Question 65 Marks

In a triangle ABC, if L and M are points on AB and AC respectively such that LM || BC. Prove that:

$\text{ar}(\triangle\text{LCM})=\text{ar}(\triangle\text{LBM})$
$\text{ar}(\triangle\text{LBC})=\text{ar}(\triangle\text{MBC})$
$\text{ar}(\triangle\text{ABM})=\text{ar}(\triangle\text{ACL})$
$\text{ar}(\triangle\text{LOB})=\text{ar}(\triangle\text{MOC})$

Answer

Clearly triangles LMB and LMC are on the same base LM and between the same parallels LM and BC.$\therefore\text{ar}(\triangle\text{LMB})=\text{ar}(\triangle\text{LMC})\ ...(1)$
We observe that triangles LBC and MBC are on the same base BC and between same parallels LM and BC.$\therefore\text{ar}(\triangle\text{LBC})=\text{ar}(\triangle\text{MBC})\ ...(2)$
We have,$\text{ar}(\triangle\text{LMB})=\text{ar}(\triangle\text{LMC})$ [From 1]
$\Rightarrow\text{ar}(\triangle\text{ALM})+\text{ar}(\triangle\text{LMB})\\=\text{ar}(\triangle\text{ALM})+\text{ar}(\triangle\text{LMC})$.
$\Rightarrow\text{ar}(\triangle\text{ABM})=\text{ar}(\triangle\text{ACL})$
We have,$\text{ar}(\triangle\text{LBC})=\text{ar}(\triangle\text{MBC})$ [From 1]
$\Rightarrow\text{ar}(\triangle\text{LBC})-\text{ar}(\triangle\text{BOC})\\=\text{ar}(\triangle\text{MBC})-\text{ar}(\triangle\text{BOC})$
$\Rightarrow\text{ar}(\triangle\text{LOB})=\text{ar}(\triangle\text{MOC}).$

View full question & answer→

Question 75 Marks

In figure, $\angle\text{AOB}=90^\circ,$ $AC = BC, OA = 12cm$ and $OC = 6.5cm$.
Find the area of $\triangle\text{AOB}.$

Answer

Since, the midpoint of the hypotenuse of a right triangle is equidistant from the vertices
$\therefore$ $CA = CB = OC$
$\Rightarrow CA = CB = 6.5cm$
$\Rightarrow AB = 13cm$ In right angled triangle OAB,
we have $AB^2 = OB^2 + OA^2$
$\Rightarrow 13^2 = OB^2 + 12^{2} $
$\Rightarrow OB^2 = 13^2 - 12^2= 169 - 144 = 25$
$\Rightarrow OB $= $5$
$\therefore\text{ar}(\triangle\text{AOB})=\Big(\frac{1}{2}\Big)(12\times5)=30\text{cm}^2$

View full question & answer→

Question 85 Marks

ABCD is a parallelogram. G is a point on AB such that AG = 2GB and E is point on DC such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:

$\text{ar}(\text{ADEG})=\text{ar}(\text{GBCE})$
$\text{ar}{(\triangle\text{EGB}})=\frac{1}{6}\text{ar}(\text{ABCD})$
$\text{ar}(\triangle\text{EFC})=\frac{1}{2}\text{ar}(\triangle\text{EBF})$
$\text{ar}(\triangle\text{EBG})=\text{ar}(\triangle\text{EFC})$
Find what portion of the parallelogram is the area of $\triangle\text{EFG}.$

Answer

ABCD is a parallelogram AG = 2GB, CE = 2DE, BF = 2FC To prove: ABCD is ||gm ar(EBG) = ar(EFC) AB || CD (AB = CD)$\text{BG}=\frac{1}{3}\text{AB},\ \text{DE}=\frac{1}{3}\text{CD}$
$\therefore\ \text{BG}=\text{DE}$
$\therefore\ \text{ADEH}$ is ||gm
ar(||gm ADEH) = ar(||gm BCIG) ...(i) ar$(\triangle\text{HEG})$ = ar$(\triangle\text{EGI})$ ..(ii) ($\because$ diagonal of ||gm divided into 2 equal areas) (i) and (ii)$\therefore\text{ar}(||\text{gm}\text{ ADEG})=\text{ar}(||\text{gm}\text{ GBCE})$
Height, h of ||gm ABCD and $\triangle\text{EGB}$ is Its same Base of $\triangle\text{EGB}=\frac{1}{3}\text{AB}$ area of ABCD = h × AB$\text{ar}(\triangle\text{EGB})=\frac{1}{6}\times\text{h}\times\frac{1}{3}\text{AB}=\frac{1}{6}\text{h}\times\text{AB}$
$\text{ar}(\text{EGB})=\frac{1}{6}\text{Area}(\text{ABCD})$
let dislaver between EH and CB = x$\text{ar}(\text{EBF})=\frac{1}{2}\times\text{BF}\times\text{x}=\frac{1}{2}\times\frac{2}{3}\text{BC}\times\text{x}=\frac{1}{3}\times\text{BC}\times\text{x}$
$\text{ar}(\text{EFC})=\frac{1}{2}\times\text{CF}\times\text{x}=\frac{1}{2}\times\frac{1}{3}\text{BC}\times\text{x}=\frac{1}{2}\times\text{ar}\times(\text{EBF})$
$\Rightarrow\text{ar}(\text{EFC})=\frac{1}{2}\times\text{area of EBF}$
If g = altitute from AD to BC$\text{ar}(\text{EFC})=\frac{1}{2}\times\frac{2}{3}\text{g}\times\frac{1}{3}\text{BC}=\frac{1}{9}\text{ar}(\text{ABCD})$
$\Rightarrow\text{ar}(\text{EFC})=\frac{2}{3}\text{ar}\text{ EBG}$
$\text{ar}(\text{EFG})=\text{ar}(\text{EGB})+\text{ar}(\text{FBF})+\text{ar}(\text{EFC})$
$=\frac{1}{6}\text{ABCD}+2\text{ar}(\text{EFC})+\text{ar}(\text{EFC})$
$=\frac{1}{6}\text{ABCD}+3\text{ar}(\text{EFC})$
$=\frac{1}{6}\text{ABCD}+\not3\times\frac{1}{\not9}\times\text{ABCD}$
$=\Big(\frac{1}{6}+\frac{1}{3}\Big)\text{ABCD}$
$=\Big(\frac{1+2}{6}\Big)\text{ABCD}$
$=\frac{1}{2}\text{ABCD}$

View full question & answer→

Question 95 Marks

If ABC and BDE are two equilateral triangles such that D is the mid-point of BC, then find $\text{ar}(\triangle\text{ABC}) : \text{ar}(\triangle\text{BDE}).$

Answer

Given:$\triangle\text{ABC}$ is equilateral triangle.
$\triangle\text{BDE}$ is equilateral triangle.
D is the midpoint of BC. To find: $\text{ar}(\triangle\text{ABC}):\text{ar}(\triangle\text{BDE})$ Proof: Let us draw the figure as per the instruction given in the question.

We know that area of equilateral triangle $=\frac{\sqrt3}{4}\times\text{a}^2,$ where a is the side of the triangle. Let us assume that length of BC is a cm. This means that length of BD is $\frac{\text{a}}{2}\text{cm},$ Since D is the midpoint of BC.$\therefore$ Area of equilateral $\triangle\text{ABC}=\frac{\sqrt3}{4}\times\text{a}^2\ ...(1)$
Area of equilateral $\triangle\text{BDE}=\frac{\sqrt3}{4}\times\Big(\frac{\text{a}^2}{2}\Big)\ ...(2)$ Now, $\triangle\text{ABC}:\triangle\text{BDE}=\frac{\sqrt3}{4}\times\text{a}^2:\frac{\sqrt3}{4}\times\Big(\frac{\text{a}^2}{2}\Big)$ [From 1 and 2]$=4:1$
Hence we get the result $\text{ar}(\triangle\text{ABC}):\text{ar}(\triangle\text{BDE})=4:1$

View full question & answer→

Question 105 Marks

In figure, ABCD is a rectangle in which CD = 6cm, AD = 8cm. Find the area of parallelogram CDEF.

Answer

Given:

ABCD is a rectangle.
CD = 6cm
AD = 8cm

To find: Area of rectangle CDEF.
Calculation: We know that,
Area of parallelogram = $base \times height$
The Area of parallelogram and a rectangle on the same base and between the same parallels are equal in area.

Here we can see that rectangle ABCD and Parallelogram CDEF are between the same base and same parallels.
Hence,
Area of Rectangle CDEF = Area of parallelogram ABCD
$= AD \times CD$
$= 8 \times 6$
$= 48cm^2$
Hence we get the result as Area of Rectangle CDEF $= 48cm^2$

View full question & answer→

Question 115 Marks

A point D is taken on the side BC of a $\triangle\text{ABC},$ such that BD = 2DC. Prove that $\text{ar}(\triangle\text{ABD})=2\text{ar}(\triangle\text{ADC}).$

Answer

Given that, In $\triangle\text{ABC},$ BD = 2DC To prove: $\text{ar}(\triangle\text{ADB})=2\text{ar}(\triangle\text{ADC}).$ Construction: Take a point E on BD such that BE = ED Proof: Since, BE = ED and BD = 2 DC Then, BE = ED = DC We know that median of triangle divides it into two equal triangles.$\therefore$ In $\triangle\text{ABD},$ AE is the median.
Then, $\text{ar}(\triangle\text{ABD})=2\text{ar}(\triangle\text{AED})\ ...(1)$ In $\triangle\text{AEC},$ AD is the median. Then, $\text{ar}(\triangle\text{ADE})=2\text{ar}(\triangle\text{ADC})\ ...(2)$ Compare equation 1 and 2$\text{ar}(\triangle\text{ABD})=2\text{ar}(\triangle\text{ADC}).$

View full question & answer→

Question 125 Marks

In the given figure, ABCD is a rectangle with sides AB = 10cm and AD = 5cm. Find the area of $\triangle\text{EFG}.$

Answer

Given:

ABCD is a rectangle.
AB = 10cm
AD = 5cm

To find: Area of rectangle $\triangle\text{EGF}.$ Calculation: We know that, Area of Rectangle = base × height If a triangle and a parallelogram are on the same base and between the same parallels , the area of the triangle is equal to half of the parallelogram

Here we can see that Rectangle ABCD and triangle GEF are between the same base and same parallels. Hence, Area of Rectangle $\triangle\text{GEF}=\frac{1}{2}$ Area of parallelogram ABCD$=\frac{1}{2}\times\text{AD}\times\text{CD}$
$=\frac{1}{2}\times10\times5$
$=25\text{cm}^2$
Hence we get the result as Area of $\triangle\text{GEF}=25\text{cm}^2$

View full question & answer→

Question 135 Marks

In square ABCD, P and Q are mid-point of AB and CD respectively. If AB = 8cm and PQand BD intersect at O, then find area of $(\triangle\text{OPB}).$

Answer

Given: Here from the given question we get

ABCD is a square,
P is the midpoint of AB
Q is the midpoint of CD
PQ and BD intersect at O.
AB = 8cm

To find: Area of $\triangle\text{OPB}.$ Calculation: Since P is the midpoint of AB,$\text{BP}=\frac{1}{2}(\text{AB})$
$=\frac{1}{2}(8)$
$=4\text{cm}$
$\text{BP}=4\text{cm}\ ...(1)$

Area of triangle $=\frac{1}{2}\times\text{base}\times\text{height}$$\text{Area of }\triangle\text{OPB}=\frac{1}{2}\times\text{BP}\times\text{PO}$ [From 1]
$=\frac{1}{2}\times4\times4$ $\Big(\text{PO}=\frac{1}{2}\text{AD},\text{APQD}$ is a rectangle$\Big)$
$=\frac{1}{2}\times16$
$\text{Area of }\triangle\text{OBP}=8\text{cm}^2$
Hence we get the Area of $\triangle\text{OBP}=8\text{cm}^2$

View full question & answer→

Question 145 Marks

$A B C D$ is a parallelogram. $E$ is a point on $B A$ such that $B E=2 E A$ and $F$ is point on $D C$ such that $D F=2 F C$. Prove that AECF is a parallelogram whose area is one third of the area of parallelogram $A B C D$.

Answer

Draw $FG \perp AB$
We have, $B E=2 E A$ and $D F=2 F C$
$\Rightarrow A B-A E=2 A E$ and $D C-F C=2 F C$
$\Rightarrow A B=3 A E$ and $D C=3 F C$
$\Rightarrow AE=\left(\frac{1}{3}\right) AB \text { and } FC=\left(\frac{1}{3}\right) DC \ldots$
But $A B=D C$ Then, $A E=F C$ [opposite sides of $\left.\|^{g m}\right]$ Thus, $A E=F C$ and $A E \| F C$ Then, $A E C F$ is a parallelogram Now, area of parallelogram $AECF = AE \times FG \Rightarrow \operatorname{ar}\left(| |^{ gm } AECF \right)=\frac{1}{3} AB \times FG$ from (1)
$\Rightarrow 3 a r\left(\|{ }^{g m} A E C F\right)=A B \times F G \ldots$...(2) And $\operatorname{ar}(\| g gm A B C D)=A B \times F G \cdots(3)$
Compare equation 2 and 3
$\Rightarrow 3 \operatorname{ar}\left(\|^{gm} AECF\right)=\operatorname{ar}\left(\| \|^{gm} ABCD\right)$
$\Rightarrow \operatorname{ar}\left(\|^{gm} AECF\right)=\frac{1}{3} \operatorname{ar}\left(\|^{gm} ABCD\right)$

View full question & answer→

Question 155 Marks

In figure, ABCD is a parallelogram. O is any point on AC. PQ || AB and LM || AD. Prove that: $\text{ar}(||^{\text{gm}}\text{ DLOP})=\text{ar}(||^{\text{gm}}\text{ BMOQ}).$

Answer

Since a diagonal of a parallelogram divides it into two triangles of equal area Therefore, $\text{ar}(\triangle\text{ADC})=\text{ar}(\triangle\text{ABC})$$\Rightarrow\text{ar}(\triangle\text{APO})+\text{ar}(||^{\text{gm}}\text{ DLOP})+\text{ar}(\triangle\text{OLC})$
$\Rightarrow\text{ar}(\triangle\text{AOM})+\text{ar}(||^{\text{gm}}\text{ BMOQ})+\text{ar}(\triangle\text{OQC})\ ...(1)$
Since AO and Oc are diagonals of parallelograms AMOP and OQCL respectively.$\therefore\text{ar}(\triangle\text{APO})=\text{ar}(\triangle\text{AMO})\ ...(2)$
And $\text{ar}(\triangle\text{OLC})=\text{ar}(\triangle\text{OQC})\ ...(3)$ Subtracting 2 and 3 from 1, we get$\text{ar}(||^{\text{gm}}\text{ DLOP})=\text{ar}(||^{\text{gm}}\text{ BMOQ}).$

View full question & answer→

Question 165 Marks

In figure, PSDA is a parallelogram in which PQ = QR = RS and AP || BQ || CR. Prove that $\text{ar}(\triangle\text{PQE})=\text{ar}(\triangle\text{CFD}).$

Answer

Given that PADA is a parallelogram Since, AP || BQ || CR || DS and AD || PS Therefore, PQ = CD (equ. 1) In triangle BED, C is the midpoint of BD and CF || BE Therefore, F is the midpoint of ED ⇒ EF = PE Similarly, EF = PE Therefore, PE = FD (equ. 2) In triangle PQE and CFD, we have PE = FD THerefore, $\angle\text{EPQ}=\angle\text{FDC}$ [Alternate angles] So, by SAS criterion, we have$\triangle\text{PQE}\cong\triangle\text{DCF}$
$\Rightarrow\text{ar}(\triangle\text{PQE})=\text{ar}(\triangle\text{DCF})$

View full question & answer→

Question 175 Marks

Compute the area of trapezium PQRS in figure.

Answer

We have, $ar(trap. PQRS)$= $ar(rect. PSRT)$ + $\text{ar}(\triangle\text{QRT})$
$\Rightarrow$ ar(trap. PQRS) $=\text{PT}×\text{RT}+\frac{1}{2}(\text{QT}\times\text{RT})$
$=8\times\text{RT}+\frac{1}{2}(8\times\text{RT})=12\times\text{RT}$
In $\triangle\text{QRT},$ we have$\text{QR}^2=\text{QT}^2+\text{RT}^2$
$\Rightarrow\text{RT}^2=\text{QR}^2-\text{QT}^2$
$\Rightarrow\text{RT}^2=17^2-8^2=225$
$\Rightarrow\text{RT}=15$
Hence, Area of trapezium =$12 \times 15cm^2 = 180cm^2$

View full question & answer→

Question 185 Marks

ABC is a triangle in which D is the mid-point of BC. E and F are mid-points of DC and AE respectively. IF area of $\triangle\text{ABC}$ is $16cm^2,$ find the area of $\triangle\text{DEF}.$

Answer

Given: Here from the given question we get

ABC is a triangle
D is the midpoint of BC
E is the midpoint of CD
F is the midpoint of A

Area of $\triangle\text{ABC}=16\text{cm}^2$ To find:
Area of $\triangle\text{DEF}$ Calculation:
We know that , The median divides a triangle in two triangles of equal area.

For $\triangle\text{ABC},$ AD is the median$\text{Area of }\triangle\text{ADC}=\frac{1}{2}(\text{Area of }\triangle\text{ABC})$
$=\frac{1}{2}(16)$
$=8\text{cm}^2$
$\text{Area of }\triangle\text{ADE}=8\text{cm}^2$
For $\triangle\text{ADC},$ AE is the median.$\text{Area of }\triangle\text{AED}=\frac{1}{2}(\text{Area of }\triangle\text{ABC})$
$=\frac{1}{2}(8)$
$=4\text{cm}^2$
$\text{Area of }\triangle\text{AED}=4\text{cm}^2$
Similarly, For $\triangle\text{AED},$ DF is the median.$\text{Area of }\triangle\text{DEF}=\frac{1}{2}(\text{Area of }\triangle\text{AED})$
$=\frac{1}{2}(4)$
$=2\text{cm}^2$
$\text{Area of }\triangle\text{DEF}=2\text{cm}^2$
Hence we get Area of $\triangle\text{DEF}=2\text{cm}^2$

View full question & answer→

Question 195 Marks

If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area. If G is the mid-point of median AD, prove that $\text{ar}(\triangle\text{BGC})=\text{ar}(\triangle\text{AGC}).$

Answer

Draw $\text{AM}\perp\text{BC}$ Since, AD is the median of $\triangle\text{ABC}$$\therefore$ BD = DC
⇒ BD = AM = DC × AM$\Rightarrow\Big(\frac{1}{2}\Big)(\text{BD}\times\text{AM})=\Big(\frac{1}{2}\Big)(\text{DC}\times\text{AM)}$
$\Rightarrow\text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{ACD})\ ⋅⋅⋅ (1)$
In $\triangle\text{BGC},$ GD is the median$\Rightarrow\text{ar}(\triangle\text{BGD})=\text{ar}(\triangle\text{CGD})\ ⋅⋅⋅ (2)$
In $\triangle\text{ACD},$ CG is the median$\Rightarrow\text{ar}(\triangle\text{AGC})=\text{ar}(\triangle\text{CGD})\ ⋅⋅⋅ (3)$
From (2) and (3) we have,$\text{ar}(\triangle\text{BGD})=\text{ar}(\triangle\text{AGC}).$
But, $\text{ar}(\triangle\text{BGC})=\text{ar}(\triangle\text{AGD}).$$\text{ar}(\triangle\text{BGC})=\text{ar}(\triangle\text{AGC}).$

View full question & answer→

Question 205 Marks

In figure, compute the area of quadrilateral ABCD.

Answer

Given that: $DC =17 cm A D=9 cm$ and $B C=8 cm$ In
$\triangle BCD$
we have $C D^2=B D^2+B C^2$
$\Rightarrow(17)^2=B D^2+(8)^2$
$\Rightarrow B D^2=289-64$
$\Rightarrow B D=15 \ln$
$\triangle ABD$,
we have $A B^2+A D^2=B D^2$
$\Rightarrow(15)^2=A B^2+(9)^2$
$\Rightarrow A B^2=225-81$
$\Rightarrow A B=123 \operatorname{ar}(\text { quad, } A B C D)$
$=\operatorname{ar}(\triangle ABD)+(\triangle BCD)$
$\Rightarrow \operatorname{ar}(\text { quad, } A B C D)=\frac{1}{2}(12 \times 9)+\frac{1}{2}(8 \times 17)=54+68$
$=112 cm^2$
$\Rightarrow \operatorname{ar}(\text { quad, } A B C D)=\frac{1}{2}(12 \times 9)+\frac{1}{2}(8 \times 17)$
$=54+60 cm^2$
$=114 cm^2$

View full question & answer→

Question 215 Marks

ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F.

Prove that $\text{ar}(\triangle\text{ADF})=\text{ar}(\triangle\text{ECF}).$
If the area of $\triangle\text{DFB}=3\text{cm}^2,$ find the area of $||^{gm}ABCD$.

Answer

In triangles ADF and ECF, we have$\angle\text{ADF}=\angle\text{ECF}$ [Alternate interior angles, Since AD || BE]
AD = EC [since AD = BC = CE] And $\angle\text{DFA}=\angle\text{CFA}$ [Vertically opposite angles] So, by AAS congruence criterion, we have$\triangle\text{ADF}\cong\triangle\text{ECF}$
$\Rightarrow\text{ar}(\triangle\text{ADF})=\text{ar}(\triangle\text{ECF})$ and DF = CF.
Now, DF = CF ⇒ BF is a median in $\triangle\text{BCD}.$
$\Rightarrow\text{ar}(\triangle\text{BCD})=2\text{ar}(\triangle\text{BDF})$
$\Rightarrow\text{ar}(\triangle\text{BCD})=2\times3\text{cm}^2=6\text{cm}^2$
Hence, area of a parallelogram $=2\text{ar}(\triangle\text{BCD})=2\times6\text{cm}^2=12\text{cm}$

View full question & answer→

Question 225 Marks

P is any point on base BC of $\triangle\text{ABC}$ and D is the mid-point of BC. DE is drawn parallel to PA to meet AC at E. If $\text{ar}(\triangle\text{ABC})=12\text{cm}^2,$ then find area of $\triangle\text{EPC}.$

Answer

Given: $A r e a(A B C)=12 cm^2, D$ is midpoint of $B C$ and $A P$ is parallel to $E D$.
We need to find area of the triangle EPC.

Since, AP || ED, and we know that the area of triangles between the same parallel and on the same base are equal.
$\text { So, Area(APE) }=\text { Area(APD) }$
$\Rightarrow \text { Area(APM) }+ \text { Area(AME })=\text { Area(APM })+ \text { Area(PMD) }$
$\Rightarrow \text { Area(AME })=\text { Area(PMD) } \ldots . .(1) \text { Since, median divide triangles into two equal parts. }$
$\text { So,Area }(ADC)=\frac{1}{2} \text { Area }(ABC)=\frac{12}{2}=6 cm^2$
$\Rightarrow \text { Area(ADC) }=\text { Area(MDCE })+ \text { Area(AME })$
$\Rightarrow \text { Area(ADC) }=\text { Area(MDCE })+ \text { Area(PMD) (from equation (1)) }$
$\Rightarrow \text { Area(ADC) }=\text { Area(PEC) Therefore, Area(PEC) }=6 cm^2 .$

View full question & answer→

Question 235 Marks

In a $\triangle\text{ABC},$ P and Q are respectively the mid points of AB and BC and R is the midpoint of AP. Prove that:

$\text{ar}(\triangle\text{PBQ})=\text{ar}(\triangle\text{ARC}).$
$\text{ar}(\triangle\text{PRQ})=\frac{1}{2}\text{ar}(\triangle\text{ARC}).$
$\text{ar}(\triangle\text{EQC})=\frac{3}{8}\text{ar}(\triangle\text{ABC}).$

Answer

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

Since CR is the median of $\triangle\text{CAP}$

$\therefore\text{ar}(\triangle\text{CRA})=\Big(\frac{1}{2}\Big)\text{ar}(\triangle\text{CAP})\ ...(1)$

Also, CP is the median of a $\triangle\text{CAB}$

$\therefore\text{ar}(\triangle\text{CAP})=\text{ar}(\triangle\text{CPB})\ ...(2)$

From 1 and 2, we get

$\therefore\text{ar}(\triangle\text{ARC})=\Big(\frac{1}{2}\Big)\text{ar}(\triangle\text{CPB})\ ...(3)$

PQ is the median of a $\triangle\text{PBC}$

$\therefore\text{ar}(\triangle\text{CPB})=2\text{ar}(\triangle\text{PBQ})\ ...(4)$

From 3 and 4, we get

$\therefore\text{ar}(\triangle\text{ARC})=\text{ar}(\triangle\text{PBQ})\ ...(5)$

Since QP and QR medians of triangles QAB and QAP respectively

$\therefore\text{ar}(\triangle\text{QAP})=\text{ar}(\triangle\text{QBP})\ ...(6)$

And $\therefore\text{ar}(\triangle\text{QAP})=2\text{ar}(\triangle\text{QRP})\ ...(7)$

From 6 and 7, we get

$\therefore\text{ar}(\triangle\text{PRQ})=\Big(\frac{1}{2}\Big)\text{ar}(\triangle\text{PBQ})\ ...(8)$

From 5 and 8, we get

$\therefore\text{ar}(\triangle\text{PRQ})=\Big(\frac{1}{2}\Big)\text{ar}(\triangle\text{ARC})$

Since, LR is a median of $\triangle\text{CAP}$

$\therefore\text{ar}(\triangle\text{ARC})=\Big(\frac{1}{2}\Big)\text{ar}(\triangle\text{CAD})$

$=\frac{1}{2}\times\Big(\frac{1}{2}\Big)\text{ar}(\triangle\text{ABC})$

$=\Big(\frac{1}{4}\Big)\text{ar}(\triangle\text{ABC})$

Since RQ is the median of $\triangle\text{RBC}.$

$\therefore\text{ar}(\triangle\text{RQC})=\Big(\frac{1}{2}\Big)\text{ar}(\triangle\text{RBC})$

$=\Big(\frac{1}{2}\Big)\{\text{ar}(\triangle\text{ABC})−\text{ar}(\triangle\text{ARC})\}$

$=\Big(\frac{1}{2}\Big)\Big\{\text{ar}(\triangle\text{ABC})–\Big(\frac{1}{4}\Big)\text{ar}(\triangle\text{ABC)}\Big\}$

$=\Big(\frac{3}{8}\Big)\text{ar}(\triangle\text{ABC})$

View full question & answer→

Question 245 Marks

In Q.No. 1, if AD = 6cm, CF = 10cm, and AE = 8cm, find AB.

Answer

Area of parallelogram ABCD = AD × CF ...(1)
Again area of parallelogram ABCD = DC × AE ...(2)
Compare equation (1) and equation (2)
AD × CF = DC × AE
⇒ 6 × 10 = DC × 8
$\Rightarrow\text{DC}=\frac{6\times10}{8}=7.5\text{cm}$
$\therefore$ AB = DC = 7.5cm [Opposite sides of || gm]

View full question & answer→

Question 255 Marks

D is the midpoint of side BC of $\triangle\text{ABC}$ and E is the midpoint of BD. If O is the midpoint of AE, Prove that $\text{ar}(\triangle\text{BOE})=\frac{1}{8}\text{ar}(\triangle\text{ABC}).$

Answer

Given that D is the midpoint of sides BC of triangle ABC E is the midpoint of BD and O is the midpoint of AE Since AD and AE are the medians of triangles, ABC and ABD respectively$\therefore\text{ar}(\triangle\text{ABD})=\Big(\frac{1}{2}\Big)\text{ar}(\triangle\text{ABC})\ ...(1)$
$\therefore\text{ar}(\triangle\text{ABE})=\Big(\frac{1}{2}\Big)\text{ar}(\triangle\text{ABD})\ ...(2)$
OB is the median of triangle ABE Therefore,$\therefore\text{ar}(\triangle\text{BOE})=\Big(\frac{1}{2}\Big)\text{ar}(\triangle\text{ABE})$
From 1, 2 and 3, we have$\therefore\text{ar}(\triangle\text{BOE})=\Big(\frac{1}{8}\Big)\text{ar}(\triangle\text{ABC})$

View full question & answer→

Question 265 Marks

PQRS is a trapezium having PS and QR as parallel sides. A is any point on PQ and B is a point on SR such that AB || QR. If area of $\triangle\text{PBQ}$ is $17cm^2$,
find the area of $\triangle\text{ASR}.$

Answer

Here from the given figure we get

PQRS is a trapezium having PS || QR
A is any point on PQ
B is any point on SR
AB || QR
Area of $\triangle\text{BPQ}=17\text{cm}^2$

To find : Area of $\triangle\text{ASR}.$

Calculation: We know that ‘If a triangle and a parallelogram are on the same base and the same parallels,
the area of the triangle is equal to half the area of the parallelogram’
Here we can see that:$\text{Area}(\triangle\text{APB})=\text{Area}(\triangle\text{ABS})\ ...(1)$
And, $\text{Area}(\triangle\text{AQR})=\text{Area}(\triangle\text{ABR})\ ...(2)$
Therefore,$\text{Area}(\triangle\text{ASR})=\text{Area}(\triangle\text{ABS})+\text{Area}(\triangle\text{ABR})$
From equation (1) and (2),
we have,$\text{Area}(\triangle\text{ASR})=\text{Area}(\triangle\text{APB})+\text{Area}(\triangle\text{AQR})$
$\Rightarrow\text{Area}(\triangle\text{ASR})=\text{Area}(\triangle\text{BPQ})=17\text{cm}^2$
Hence, the area of the triangle $\triangle\text{ASR}\text{ is }17\text{cm}^2.$

View full question & answer→

Question 275 Marks

In the given figure, CD || AE and CY || BA.

Name a triangle equal in area of $\triangle\text{CBX}$
Prove that $\text{ar}(\triangle\text{ZDE})=\text{ar}(\triangle\text{CZA})$
Prove that $\text{ar}(\text{BCZY})=\text{ar}(\triangle\text{EDZ}).$

Answer

Given: CD || AE. CY || BA. To find: Name a triangle equal in area of $\triangle\text{CBX}.$$\text{ar}(\triangle\text{ZDE})=\text{ar}(\triangle\text{CZA}).$
$\text{ar}(\text{BCZY})=\text{ar}(\triangle\text{EDZ}).$
Proof:

Since triangle BCY and triangle YCA are on the same base and between same parallel, so their area should be equal.

Therefore
$\text{ar}(\triangle\text{BCY})=\text{ar}(\triangle\text{YCA})$
$\Rightarrow\text{ar}(\triangle\text{CBX})+\text{ar}(\triangle\text{XYC})\\=\text{ar}(\triangle\text{XYC})+\text{ar}(\triangle\text{AXY})$
$\Rightarrow\text{ar}(\triangle\text{CBX})=\text{ar}(\triangle\text{AXY})$
Therefore area of triangle CBX is equal to area of triangle AXY

Triangle ADE and triangle ACE are on the same base AE and between the same parallel AE and CD.

$\Rightarrow\text{ar}(\triangle\text{ADE})=\text{ar}(\triangle\text{ACE})$
$\Rightarrow\text{ar}(\triangle\text{ADE})-\text{ar}(\triangle\text{AZE})\\=\text{ar}(\triangle\text{ACE})-\text{ar}(\triangle\text{AZE})$
$\Rightarrow\text{ar}(\triangle\text{ZDE})=\text{ar}(\triangle\text{ACZ})$

Triangle ACY and BCY are on the same base CY and between same parallels CY and BA. So we have $\text{ar}(\triangle\text{ACY})=\text{ar}(\triangle\text{BCY})$

Now we know that
$\text{ar}(\triangle\text{ACZ})=\text{ar}(\triangle\text{ZDE})$
$\Rightarrow\text{ar}(\triangle\text{ACY})+\text{ar}(\triangle\text{CYZ})=\text{ar}(\triangle\text{EDZ})$
$\Rightarrow\text{ar}(\triangle\text{BCY})+\text{ar}(\triangle\text{CYZ})=\text{ar}(\triangle\text{EDZ})$
$\Rightarrow\text{ar}(\triangle\text{BCY})=\text{ar}(\triangle\text{EDZ})$

View full question & answer→

Question 285 Marks

In figure, X and Y are the mid points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that $\text{ar}(\triangle\text{ABP})=\text{ar}(\triangle\text{ACQ}).$

Answer

Since X and Y are the mid points of AC and AB respectively. Therefore, XY || BC Clearly, triangles BYC and BXC are on the same base BC and between the same parallels XY and BC$\therefore\text{ar}(\triangle\text{BYC})=\text{ar}(\triangle\text{BXC})$
$\Rightarrow\text{ar}(\triangle\text{BYC})-\text{ar}(\triangle\text{BOC})\\ \ =\text{ar}(\triangle\text{BXC})-\text{ar}(\triangle\text{BOC})$
$\Rightarrow\text{ar}(\triangle\text{BOY})=\text{ar}(\triangle\text{COX})$
$\Rightarrow\text{ar}(\triangle\text{BOY})+\text{ar}(\triangle\text{XOY})\\ \ =\text{ar}(\triangle\text{COX})+\text{ar}(\triangle\text{XOY})$
$\Rightarrow\text{ar}(\triangle\text{BXY})=\text{ar}(\triangle\text{CXY})\ ...(1)$
We observed that the quadrilaterals XYAP and XYAQ are on the same base XY and between same parallels XY and PQ.$\therefore\text{ar}(\text{quad}.\ \text{XYAP})=\text{ar}(\text{quad}\text{ XYQA})\ ...(2)$
Adding 1 and 2, we get$\therefore\text{ar}(\triangle\text{BXY})+\text{ar}(\text{quad}.\text{ XYAP})\\ \ =\text{ar}(\triangle\text{CXY})+\text{ar}(\text{quad}\text{ XYQA})$
$\Rightarrow\text{ar}(\triangle\text{ABP})=\text{ar}(\triangle\text{ACQ})$

View full question & answer→

Question 295 Marks

$ABCD$ is a parallelogram. $P$ is the mid-point of $AB. BD$ and $CP$ intersect at $Q$ such that $CQ : QP = 3 : 1.$
If$\text{ar}(\triangle\text{PBQ})=10\text{cm}^2,$ find the area of parallelogram $ABCD.$

Answer

Given: $A B C D$ is a parallelogram. $P$ is a mid point of $A B, B D$ and $C P$ intersect at $Q$ such that $C Q: Q P=3: 1$
To find: the area of parallelogram $A B C D$.
As we know
$CQ: QP=3: 1$
$\text { or } \frac{CQ}{QP}=\frac{3}{1}$
$\text { Or } 3 Q P=C Q \ldots \text { Eq } 1$
and also
$\Rightarrow \frac{1}{2} \times PB \times QP=10 cm^2$
$\Rightarrow \frac{1}{2} \times PB \times \frac{CQ}{3}=10 cm^2[\text { From eq 1] }$
$\Rightarrow \frac{1}{2} \times PB \times CQ=3 \times 10 cm^2$
$\Rightarrow \frac{1}{2} \times PB \times CQ=30 cm^2$
but Area(triangle $P B Q$ ) + Area(triangle $C B Q$ ) $=$ Area(triangle $P B C$ )
$\Rightarrow 10 cm^2+30 cm^2=40 cm^2$
therefore Area(triangle $P B C$ ) $=40 cm^2$
Since $P$ is the mid point of $A B$
therefore $C P$ is the median of triangle $A B C$
thus Area(triangle PBC ) = Area(triangle APC )
Or Area(triangle APC ) $=40 cm^2$
thus area of triangle $A B C=40+40=80 cm^2$
Area of parallelogram $A B C D=160 cm^2$

View full question & answer→

Generate Paper Free All Areas Of Parallelograms And Triangles topics

5 Mark Question - Maths STD 9 Questions - Vidyadip