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MATHS 3 Marks Question

Areas Of Parallelograms And Triangles · Rajasthan Board (RBSE) STD 9 (Rajasthan - English Medium). 21 questions.

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Question 13 Marks
Let $\text{ABCD}$ be a parallelogram of area $124\ cm^2.$ If $E$ and $F$ are the mid$-$points of sides $AB$ and $CD$ respectively, then find the area of parallelogram $\text{AEFD}.$
Answer


Given,
Area of parallelogram $\text{ABCD} = 124\ cm^2$
Comstruction: Draw $\text{AP}\perp\text{DC}$
Proof:
Area of parallelogram $\text{AEFD} = DE \times AP ...(1)$
And area of parallelogram $\text{EBCF} = FC \times AP ...(2)$
And $DF = FC ...(3) [F$ is the mid$-$point of $DC]$
Compare equation $(1),(2)$ and $(3)$
Area of parallelogram $\text{AEFD} =$ area of parallelogram $\text{EBCF}$
Therefore, Area of parallelogram $\text{AEFD} =\frac{\text{Area of parallelogram ABCD}}{2}=\frac{124}{2}=62\text{ cm}^2$
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Question 23 Marks
In figure, ABCD, ABFE and CDEF are parallelograms. Prove that $\text{ar}(\triangle\text{ADE})=\text{ar}(\triangle\text{BCF}).$
Answer
Given that:
ABCD is parallelogram ⇒ AD = BC
CDEF is parallelogram ⇒ DE = CF
ABFE is parallelogram ⇒ AE = BF
Thus, in $\triangle\text{ADE}$ and $\triangle\text{BCF},$ we have
AD = BC, DE = CF and AE = BF
So, by SSS criterion of congruence, we have
$\triangle\text{ADE}\cong\triangle\text{BCF}$
$\text{ar}(\triangle\text{ADE})=\text{ar}(\triangle\text{BCF})$
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Question 33 Marks
In figure, ABC is a right angled triangle at A, BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment $\text{AX}\perp\text{DE}$ meets BC at Y. Show that:

$\text{ar}(\text{BYXD})=\text{ar}(\text{ABMN})$
Answer
Since triangles MBC and square MBAN are on the same base Mb and between the same parallels MB and NC.
$\therefore2\text{ar}(\triangle\text{MBC})=\text{ar}(\text{MBAN})$
From equ. 2 and 3, we have
$\text{ar}(\text{sq. }\text{MBAN})=\text{ar}(\text{rect }\text{BYXD})$
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Question 43 Marks
In figure, ABC is a right angled triangle at A, BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment $\text{AX}\perp\text{DE}$ meets BC at Y. Show that:

$\text{ar}(\text{CYXE})=\text{ar}(\text{ACFG})$
Answer
Clearly, triangle FCb and rectangle FCAG are on the same base FC and between the same parallels FC and BG.
$\therefore2\text{ar}(\triangle\text{FCB})=\text{ar}(\text{FCAG})$
From 4 and 5, we get
$\text{ar}(\text{CYXE})=\text{ar}(\text{ACFG})$
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Question 53 Marks
In figure, D and E are two points on BC such that BD = DE = EC. Show that $\text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{ADE})=\text{ar}(\triangle\text{AEC}).$
Answer
Draw a line l through A parallel to BC.
Given that, BD = DE = EC
We observed that the triangles ABD and AEC are on the equal bases and between the same parallels l and BC. Therefore, their areas are equal.
Hence, $\text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{ADE})=\text{ar}(\triangle\text{AEC}).$
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Question 63 Marks
If $P$ is any point in the interior of a parallelogram $\text{ABCD},$ then prove that area of the triangle $\text{APB}$ is less than half the area of parallelogram.
Answer
Draw $\text{DN}\perp\text{AB}$ and $\text{PM}\perp\text{AB}$
Now,
ar $(||^{ gm}\text{ABCD}) =\text{AB}\times\text{DN},\ \text{ar}(\triangle\text{APB})=\Big(\frac{1}{2}\Big)(\text{AB}\times\text{PM})$
Now$, PM < DN$
$\Rightarrow AB \times PM < AB \times DN$
$\Rightarrow\Big(\frac{1}{2}\Big)(\text{AB}\times\text{PM})<\Big(\frac{1}{2}\Big)(\text{AB}\times\text{DN})$
$\Rightarrow\text{ar}(\triangle\text{APB})<\frac{1}{2}\text{ar}(||^{\text{gm}}\text{ABCD})$
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Question 73 Marks
In figure, ABC is a right angled triangle at A, BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment $\text{AX}\perp\text{DE}$ meets BC at Y. Show that:

$\text{ar}(\text{BYXD})=2\text{ar}(\triangle\text{MBC})$
Answer
Clearly, triangle ABC and rectangle BYXD are on the same base BD and between the same parallels AX and BD.
$\therefore\text{ar}(\triangle\text{ABD})=\Big(\frac{1}{2}\Big)\text{ar}(\text{rect }\text{BYXD})$
$\Rightarrow\text{ar}(\text{rect }\text{BYXD})=2\text{ar}(\triangle\text{ABD})$
$\Rightarrow\text{ar}(\text{rect }\text{BYXD})=2\text{ar}(\triangle\text{MBC})$ [From equ. 1]
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Question 83 Marks
In figure, PQRS is a square and T and U are, respectively, the mid-points of PS and QR. Find the area of $\triangle\text{OTS}$ if PQ = 8cm.
Answer
From the figure,
T and U are mid points of PS and QR respectively
$\therefore$ TU || PQ
⇒ TU || PQ
Thus, in $\triangle\text{PQS},$ T is the midpoint of PS and TO || PQ
$\therefore\text{TO}=\Big(\frac{1}{2}\Big)\text{PQ}=4\text{cm}$
Also, $\text{TS}=\Big(\frac{1}{2}\Big)\text{PS}=4\text{cm}$
$\therefore(\triangle\text{OTS)}=\Big(\frac{1}{2}\Big)=(\text{TO}\times\text{TS})\\ \ =\Big(\frac{1}{2}\Big)(4\times4)\text{cm}^2=8\text{cm}^2$
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Question 93 Marks
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that: $\text{ar}(\triangle\text{APB})\times\text{ar}(\triangle\text{CPD})\\=\text{ar}(\triangle\text{APD})\times\text{ar}(\triangle\text{BPC})$
Answer
Construction: - Draw $\text{BQ}\perp\text{AC}$ and $\text{DR}\perp\text{AC}$
Proof:-
$\text{L.H.S}$
$=\text{ar}(\triangle\text{APB})\times\text{ar}(\triangle\text{CDP})$
$=\Big(\frac{1}{2}\Big)\big[(\text{AP}\times\text{BQ})\big]\times\Big(\frac{1}{2}\Big)\times\text{PC}\times\text{DR}$
$=\Big(\frac{1}{2}\times\text{PC}\times\text{BQ}\Big)\times\Big(\frac{1}{2}\times\text{AP}\times\text{DR}\Big)$
$=\text{ar}(\triangle\text{APD})\times\text{ar}(\triangle\text{BPC}).$
$\text{= R.H.S}$
Hence proved.
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Question 103 Marks
ABCD is a parallelogram whose diagonals AC and BD intersect at O. A line through O intersects AB at P and DC at Q. Prove that $\text{ar}(\triangle\text{POA})=\text{ar}(\triangle\text{QOC}).$
Answer
In triangles POA and QOC, we have
$\angle\text{AOP}=\angle\text{COQ}$
$\text{AO}=\text{OC}$
$\angle\text{PAC}=\angle\text{QCA}$
So, by ASA congruence criterion, we have
$\triangle\text{POA}\cong\triangle\text{QOC}$
$\Rightarrow\text{ar}(\triangle\text{POA})=\text{ar}(\triangle\text{QOC}).$
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Question 113 Marks
In figure, ABC is a right angled triangle at A, BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment $\text{AX}\perp\text{DE}$ meets BC at Y. Show that:

$\triangle\text{FCB}\cong\triangle\text{ACE}$
Answer
In triangles FCB and ACE, we have
FC = AC
CB = CE
And, $\angle\text{FCB}=\angle\text{ACE}$ $\Big[$since, $\angle\text{FCB}$ and $\angle\text{ACE}$ are obtained by adding $\angle\text{ACB}$ to a right angle.$\Big]$
So, by SAS congruence criterion, we have
$\triangle\text{FCB}\cong\triangle\text{ACE}$
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Question 123 Marks
In figure ABC and BDE are two equilateral triangles such that D is the midpoint of BC. AE intersects BC in F. Prove that:

$\text{ar}(\triangle\text{BDE})=\frac{1}{2}\text{ar}(\triangle\text{BAE})$
Answer
Given that ABC and BDE are two equilateral triangles.
Let AB = BC = CA = x. Then, $\text{BD}=\frac{\text{x}}{2}=\text{DE}=\text{BE}$
It is given that triangles, ABC and BED are equilateral triangles
$\angle\text{ACB}=\angle\text{DBE}=60^\circ$
⇒ BE || AC (Since, Alternative angles are equal)
Trinagles BAF and BEC are on the same base BE and between same parallels BF and AC.
Therefore, $\text{ar}(\triangle\text{BAE})=\text{ar}(\triangle\text{BEC})$
$\Rightarrow\text{ar}(\triangle\text{BAE})=2\text{ar}(\triangle\text{BDE})$ $[$Since , ED is a median of triangle EBC; $\text{ar}(\triangle\text{BEC})=2\text{ar}(\triangle\text{BDE})]$
$\therefore\text{ar}(\triangle\text{BDE})=\frac{1}{2}\text{ar}(\triangle\text{BAE}).$
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Question 133 Marks
If fig. 14.26, ABCD is a parallelogram, $\text{AE}\perp\text{DC}$ and $\text{CF}\perp\text{AD}.$ If AB = 16cm. AE = 8cm and CF = 10cm, find AD.
Answer
In parallelogram ABCD, CD = AB = 16cm [Opposite sides of a parallelogram are equal]
We know that,
Area of parallelogram = Base × corresponding attitude
Area of parallelogram ABCD = CD × AE = AD × CF
16cm × 8cm = AD × 10cm $\text{AD}=\frac{16\times8}{10}\text{cm}= 12.8\text{cm}.$
Thus, the length of AD is 12.8cm.
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Question 143 Marks
In figure, ABC is a right angled triangle at A, BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment $\text{AX}\perp\text{DE}$ meets BC at Y. Show that:

$\triangle\text{MBC}\cong\triangle\text{ABD}$
Answer
In $\triangle\text{MBC}$ and $\triangle\text{ABD},$ we have
MB = AB
BC = BD
And $\angle\text{MBC}=\angle\text{ABD}$ $\Big[$since, $\angle\text{MBC}$ and $\angle\text{ABC}$ are obtained by adding $\angle\text{ABC}$ to a right angle.$\Big]$
So, by SAS congruence criterion, we have
$\triangle\text{MBC}\cong\triangle\text{ABD}$
$\Rightarrow\text{ar}(\triangle\text{MBC})=\text{ar}(\triangle\text{ABD})$
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Question 153 Marks
In figure ABC and BDE are two equilateral triangles such that D is the midpoint of BC. AE intersects BC in F. Prove that:

$\text{ar}(\triangle\text{FED})=\frac{1}{8}\text{ar}(\triangle\text{AFC})$
Answer
Given that ABC and BDE are two equilateral triangles.
Let AB = BC = CA = x. Then, $\text{BD}=\frac{\text{x}}{2}=\text{DE}=\text{BE}$
$\text{ar}(\triangle\text{AFC})=\text{ar}(\triangle\text{AFD})+\text{ar}(\triangle\text{ADC})$
$\Rightarrow\text{ar}(\triangle\text{BFE})+\Big(\frac{1}{2}\Big)\text{ar}(\triangle\text{ABC})$ [Using part (iii), and AD is the median of triangle ABC]
$=\text{ar}(\triangle\text{BFE})+\Big(\frac{1}{2}\Big)\times4\text{ar}(\triangle\text{BDE})$ (Using part (i))
$=\text{ar}(\triangle\text{BFE})=2\text{ar}(\triangle\text{FED})\ ...(3)$
$\text{ar}(\triangle\text{BDE})=\text{ar}(\triangle\text{BFE})+\text{ar}(\triangle\text{FED})$
$\Rightarrow2\text{ar}(\triangle\text{FED})+\text{ar}(\triangle\text{FED})$
$\Rightarrow3\text{ar}(\triangle\text{FED})\ ...(4)$
From 2, 3 and 4 we get
$\text{ar}(\triangle\text{AFC})=2\text{ar}(\triangle\text{FED})\\ \ +2\times3\text{ar}(\triangle\text{FED})=8\text{ar}(\triangle\text{FED})$
$\text{ar}(\triangle\text{FED})=\Big(\frac{1}{8}\Big)\text{ar}(\triangle\text{AFC})$
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Question 163 Marks
In figure ABC and BDE are two equilateral triangles such that D is the midpoint of BC. AE intersects BC in F. Prove that:

$\text{ar}(\triangle\text{BFE})=\text{ar}(\triangle\text{AFD})$
Answer
Given that ABC and BDE are two equilateral triangles.
Let AB = BC = CA = x. Then, $\text{BD}=\frac{\text{x}}{2}=\text{DE}=\text{BE}$
Since, triangles ABC and BDE are equilateral triangles
$\therefore\angle\text{ABC}=60^\circ$ and $\angle\text{BDE}=60^\circ$
$\angle\text{ABC}=\angle\text{BDE}$
⇒ AB || DE (Since, Alternative angles are equal)
Trinagles BED and AED are on the same base ED and between same parallels AB and DE.
Therefore, $\text{ar}(\triangle\text{BED})=\text{ar}(\triangle\text{AED})$
$\Rightarrow\text{ar}(\triangle\text{BED})-\text{ar}(\triangle\text{EFD})\\=\text{ar}(\triangle\text{AED})-\text{ar}(\triangle\text{EFD})$
$\Rightarrow\text{ar}(\triangle\text{BEF})=2\text{ar}(\triangle\text{AFD})]$
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Question 173 Marks
In figure, ABC is a right angled triangle at A, BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment $\text{AX}\perp\text{DE}$ meets BC at Y. Show that:

$\text{ar}(\text{CYXE})=2\text{ar}(\triangle\text{FCB})$
Answer
We have,
$\triangle\text{FCB}\cong\triangle\text{ACE}$
$\Rightarrow\text{ar}(\triangle\text{FCB})=\text{ar}(\triangle\text{ACE})$
Clearly, triangle ACE and rectangle CYXE are on the same base CE and between same parallels CE and AX.
$\therefore2\text{ar}(\triangle\text{ACE})=\text{ar}(\text{CYXE})$
$\Rightarrow2\text{ar}(\triangle\text{FCB})=\text{ar}(\triangle\text{CYXE})$
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Question 183 Marks
In figure ABC and BDE are two equilateral triangles such that D is the midpoint of BC. AE intersects BC in F. Prove that:

$\text{ar}(\triangle\text{BFE})=2\text{ar}(\triangle\text{EFD})$
Answer
Given that ABC and BDE are two equilateral triangles.
Let AB = BC = CA = x. Then, $\text{BD}=\frac{\text{x}}{2}=\text{DE}=\text{BE}$
Let h be the height of vertex E, corresponding to the side BD in triangle BDE.
Let H be the height of vertex A, corresponding to the side BC in triangle ABC
From part (i)
$\Rightarrow\text{ar}(\triangle\text{BDE})=\Big(\frac{1}{4}\Big)\text{ar}(\triangle\text{ABC})$
$\Rightarrow\Big(\frac{1}{2}\Big)\times\text{BD}\times\text{h}=\Big(\frac{1}{4}\Big)\Big(\frac{1}{2}\times\text{BC}\times\text{h}\Big)$
$\Rightarrow\text{BD}\times\text{h}=\Big(\frac{1}{4}\Big)(2\text{BD}\times\text{H})$
$\Rightarrow\text{h}=\Big(\frac{1}{2}\Big)\text{H}\ ...(1)$
From part (iii)
$\text{ar}(\triangle\text{BFE})=\text{ar}(\triangle\text{AFD})$
$\text{ar}(\triangle\text{BFE})=\Big(\frac{1}{2}\Big)\times\text{FD}\times\text{H}$
$\text{ar}(\triangle\text{BFE})=\Big(\frac{1}{2}\Big)\times\text{FD}\times2\text{h}$
$\text{ar}(\triangle\text{BFE})=2\Big(\Big(\frac{1}{2}\Big)\times\text{FD}\times\text{h}\Big)$
$\text{ar}(\triangle\text{BFE})=2\text{ar}(\triangle\text{EFD})$
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Question 193 Marks
In figure ABC and BDE are two equilateral triangles such that D is the midpoint of BC. AE intersects BC in F. Prove that:

$\text{ar}(\triangle\text{BDE})=\frac{1}{4}\text{ar}(\triangle\text{ABC})$
Answer
Given that ABC and BDE are two equilateral triangles.
Let AB = BC = CA = x. Then, $\text{BD}=\frac{\text{x}}{2}=\text{DE}=\text{BE}$
We have,
$\text{ar}(\triangle\text{ABC})=\frac{\sqrt3}{4}\text{x}^2$ and $\text{ar}(\triangle\text{BDE})=\frac{\sqrt3}{4}\Big(\frac{\text{x}}{2}\Big)^2=\frac{1}{4}\times\frac{\sqrt3}{4}\text{x}^2$
Therefore, $\text{ar}(\triangle\text{BDE})=\frac{1}{4}\text{ar}(\triangle\text{ABC}).$
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Question 203 Marks
In figure ABC and BDE are two equilateral triangles such that D is the midpoint of BC. AE intersects BC in F. Prove that:

$\text{ar}(\triangle\text{ABC})=2\text{ar}(\triangle\text{BEC})$
Answer
Given that ABC and BDE are two equilateral triangles.
Let AB = BC = CA = x. Then, $\text{BD}=\frac{\text{x}}{2}=\text{DE}=\text{BE}$
Since ED is the median of triangle BEC
Therefore, $\text{ar}(\triangle\text{BEC})=2\text{ar}(\triangle\text{BDE})$
$\Rightarrow\text{ar}(\triangle\text{BEC})=2\times\Big(\frac{1}{2}\Big)\text{ar}(\triangle\text{ABC})$ $\Big[$From 1, $\text{ar}(\triangle\text{BDE})=\Big(\frac{1}{2}\Big)\text{ar}(\triangle\text{ABC})\Big]$
$\Rightarrow\text{ar}(\triangle\text{BEC})=\Big(\frac{1}{2}\Big)\text{ar}(\triangle\text{ABC})$
$\Rightarrow\text{ar}(\triangle\text{ABC})=2\text{ar}(\triangle\text{BEC})]$
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Question 213 Marks
In figure, ABCD is a trapezium in which AB || DC. Prove that $\text{ar}(\triangle\text{AOD}) =\text{ar}(\triangle\text{BOC}).$
Answer
Given:  ABCD is a trapezium in which AB || DC
To prove: $\text{ar}(\triangle\text{AOD}) =\text{ar}(\triangle\text{BOC})$
Proof: Since, $\triangle\text{ADC}$ and $\triangle\text{BDC}$ are on the same base DC and between same parallels AB and DC
Then, $\text{ar}(\triangle\text{ADC}) =\text{ar}(\triangle\text{BDC})$
$\Rightarrow\text{ar}(\triangle\text{AOD})+\text{ar}(\triangle\text{DOC})\\ \ =\text{ar}(\triangle\text{BOC})+\text{ar}(\triangle\text{DOC})$
$\Rightarrow\text{ar}(\triangle\text{AOD}) =\text{ar}(\triangle\text{BOC})$
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