Chapter – 17 : Theorems On Concurrence | Chapter Solution Class 9

Ganit Prakash Solution 9

Book Name	: Ganit Prakash
Subject	: Mathematics (Maths)
Class	: 9 (Madhyamik/WB)
Publisher	: Prof. Nabanita Chatterjee
Chapter Name	: Theorems On Concurrence (17th Chapter)

Let us do Yourself – 17.1

1. I draw an acute angled triangle PQR and let us prove that the perpendicular bisectors PQ, QR and RP are concurrent. Hence let us write where the circumcentre lies (inside/ out-side / on the side) of the acute angled triangle.

Solution :

Given:

Let, D, E, F are the mid-points of side PQ, QR and PR of a triangle PQR respectively. The two perpendiculars on the side PQ and QR respectively at the points D and E meet at the point O.( $\because$ P Q and Q R are not parallel) Join O, F

R.T.P: The three perpendicular drawn on $\mathrm{PQ}, \mathrm{QR} \text{ and } \mathrm{PR}$ at the piont D and F are concurrencies. Hence. it will be sufficient to prove that If OF perpendicular on PR. Then it will be proved that the three perpendicular bisectors are concurrent.

Construction : Join $\mathrm{O}, \mathrm{P} ; \mathrm{O}, \mathrm{Q} ; \mathrm{O}, \mathrm{R}$

Proof: In $\mathrm{\triangle P O D \text{ and } \triangle Q O D}$

$\mathrm{PD}=\mathrm{QB} \quad[\because \mathrm{D} \text{ is mid-point of } PQ]$

$\angle \mathrm{PDO}=\angle \mathrm{QDO}=90^{\circ} \quad[\because \mathrm{OD} \perp \mathrm{PQ}]$

OD is common side.

$\because \triangle \mathrm{POD} \cong \triangle \mathrm{QOD} [\text{Congruent} \ \mathrm{S}-\mathrm{A}-\mathrm{S} \ \text{hypothesis}]$

$\therefore \mathrm{OP = OQ}$ [Corresponding sides of congreuent triangle]…….(i)

Similarly congruent S-A-S hypothesis, $\triangle Q O E \cong \triangle R O E.$

$\therefore$ OQ = OR ………. (ii)

$\therefore$ (i) and (ii) we get OP = OR ……….. (iii)

Again, in $\triangle \mathrm{PFO} \text{ and } \triangle \mathrm{RFO}$

$\mathrm{OP}=\mathrm{OR} \\$

$\mathrm{PF}=\mathrm{RF}[\because F \text { is mid-point of } P R]$

OF is common side.

$\therefore \triangle \mathrm{QFO} \cong \triangle \mathrm{RFO}$ [ Corresponding angle of congruent triangle.]

If OF stands on line segment PR, the produced adjusted angles are equal.

OF, is perpendicular on PR.

So that three perpendicular bisectors on sides of $\triangle \mathrm{PQR}$ concurrent.

Hence, the circumcentre lies inside the triangle.

2. I draw an obtuse triangle and let us prove that the perpendicular bisectors of sides are concurrent. Let us write where the circumcentre lies (Inside/outside/on side of the triangular region.

Solution: Follow the same process in question (1). The position of the circumcentre outside of the triangular region.

3. Rita draws a right-angled triangle. Let us prove logically that the perpendicular bisectors of sides are concurrent and where the position of the circumcenter (inside/outside/on the side) is.

Solution: Follow the same process in question (1). The position of the circumcentre onside of the triangular region.

Let us do Yourself – 17.2

1. Let us write what is the length of circum radius of : triangle lengths of sides of which are 6 cm, 8 cm and 10 cm

Solution:

The triangle containing the sides 6 cm, 8 cm and 8 cm is a right angled triangle.

As. $6^2+8^2=10^2$

$\therefore$ The circum-centre of this triangle lies on the mid-point of hypotenuse,

So, the length of Circum-radius $=\frac{10}{2} cm . =5 cm.$ (Ans.)

2. If the length of circum-radius of a right-angled triangle is 10 cm, then let us write how much length of hypotenuse of triangle is.

Solution : In right angled triangle, the circum-centre lies on the mid-point of hypotenuse,

So, length of hypotenuse of triangle is = 2 × 10 cm

= 20 cm (Ans.)

Let us do Yourself – 17.3

1. I draw a triangle PQR and let us prove logically that the medians of $\triangle \mathrm{PQR}$ are concurrent:

Solution :

Given: Let $\triangle P Q R$ is triangle, two medians of which QE and RF intersect at G. Join P, G and produced AG meets QR at D.

R.T.P.: Medians of triangle are concurrent. Hence, the medians of a triangle will be concurrent if it is shown that the point of side QR.

Construction: PD is produced to H in such a way that $\mathrm{PG}=\mathrm{GH}. \mathrm{B}, \mathrm{H} \text{ and } \mathrm{C}, \mathrm{H}$ are joined

Proof: In $\triangle \mathrm{PQH}, \mathrm{F} \text { is the mid-point of } \mathrm{PQ}$ [Hypotenuse]

G is the mid-point of PH [Construction]

$\therefore \mathrm{FG} \| \mathrm{QH} \quad$ [ $\because$ line joining two mid-points of any two sides of a triangle is parallel to third sides of a triangle]

Again, similarly, in $\triangle$ PRH, E is the mid-point of PR [Hypotenuse] and G is the mid-point of PH [by construction]

$\therefore \mathrm{GE} \| \mathrm{HR}$

$\therefore$ We get in quadrilateral $GR \| \mathrm{QH} \text{ and } \mathrm{QG} \| \mathrm{HR}.$

$\therefore \mathrm{QGRH}$ is a parallelogram, diagonals of which are QR and GH.

$\therefore \quad$ D is the mid-point of QR [ $\therefore$ diagonal of parallelogram bisect each other]

$\therefore$ The three medians of a triangle are concurrent.

2. I draw acute angle-triangle, right angled triangle and obtuse angled triangle seperately and let us find what is the position of centroid of triangles.

Solution :

Acute angle-triangle

Position of centroid inside the triangle

Right angle-triangle

Position of centroid inside the triangle.

Obtuse angle-triangle

Position of centroid inside the triangle.

Let us do yourself – 17

1. The bisector of <B and <C of $\triangle A B C$ intersect at the point I. Let us prove that $<B I C=90^{\circ}+ <BAC$

Solution :

Given:

Let ABC be a triangle, and the bisectors of <B and <C intersect at the point I.

R.T.P: $<\mathrm{BIC}=90^{\circ}+\frac{<\mathrm{BAC}}{2}$

Proof: $\because$ <B is the bisector of ABC.

$\therefore<\mathrm{IBC}= <\mathrm{ABI} \text{ and } \therefore < \mathrm{ABC}=2<\mathrm{IBC} \text{ and } < \mathrm{C} \text{ is the bisector of } < A B C$

$\therefore<\mathrm{ICB}=<\mathrm{ACl} \text { and } \therefore<\mathrm{ACB}=2<\mathrm{ICB} \\$

$\therefore<\mathrm{ABC}+<\mathrm{ACB}=2(<\mathrm{IBC}+<\mathrm{ICB})$

We have,

$<\mathrm{BAC} =180^{\circ}-(<\mathrm{ABC}+<\mathrm{ACB}) \\$

$=180^{\circ}-2(<\mathrm{IBC}+<\mathrm{BIC}) \quad[\text { by }(\mathrm{i})] \\$

$=180^{\circ}-2\left(180^{\circ}-<\mathrm{BIC}\right) \\$

$=180^{\circ}-360^{\circ}+2<\mathrm{BIC} \\$

$=2<\mathrm{BIC}-180^{\circ}$

or, $2<\mathrm{BIC}=180^{\circ}+<\mathrm{BAC}$

$\text { or, } <\mathrm{BIC}=\frac{180^{\circ}+ <\mathrm{BAC}}{2} \\$

$\therefore <\mathrm{BIC}=90^{\circ}+\frac{180^{\circ} + <\mathrm{BAC}}{2}$ (Proved)

2. If the length of two medians of a triangle are equal, Let us prove that the triangle is an issoceles triangle.

Solution :

Given:

Let, the medians BE and CF of the triangle ABC are equal.

R.T.P : The triangle ABC is an Isosceles triangle.

Proof: Let the point of intersection of the medians BE and CF be G. Hence, G is the centroid of the triangle ABC (because the third median of the triangle ABC must pass through G).

Again, the centroid of triangle divide each median internally in the ratio 2 : 1.

$\therefore \mathrm{GE}=\frac{1}{3} \mathrm{BE} \text { and } \mathrm{GF}=\frac{1}{3} \mathrm{CF}$

But it is given that, $\mathrm{BE}=\mathrm{CF}$

$\therefore \frac{1}{3} \mathrm{BE}=\frac{1}{3} \mathrm{CF} \therefore \mathrm{BE}=\mathrm{CF}$

Also, $\mathrm{BG}=\mathrm{BE}-\mathrm{GE}=\mathrm{BE}-\frac{1}{3} \mathrm{BE}=\frac{2}{3} \mathrm{BE}$

and $\mathrm{CG}=\mathrm{CF}-\mathrm{GF}=\mathrm{CF}-\frac{1}{3} \mathrm{CF}=2 \mathrm{CF}$

Since, $\mathrm{BE}=\mathrm{CF} \therefore \mathrm{BG}=\mathrm{CG}.$

Now, in the triangles GEC and GFB

$\mathrm{GE}=\mathrm{GF}, \mathrm{CG}=\mathrm{BG}$

and included $<\mathrm{CGE}= \text { included } <\mathrm{BGF}$ (vertically opposite angles)

$\therefore \triangle \mathrm{GEC} \cong \triangle \mathrm{GFB}$ (by side – angle – side congruency)

$\therefore \quad \mathrm{BF}=\mathrm{CE}$

or, $\frac{1}{2} \mathrm{AB}=\frac{1}{2} \mathrm{AC}$

or, $\mathrm{AB}=\mathrm{BC}$

Hence, $\triangle A B C$ is an isosceles triangle. (Proved)

3. Let us prove that in an equilateral triangle, circumcentre, incentre, centriod, ortho-centre will coincide.

Solution :

Given:

Let, ABC be an equilateral triangle

R.T.P: In the equilateral triangle ABC, circumcentre, incentre, centroid and orthocentre are the same point.

Construction: AD is drawn perpendicular from A on BC, BE is drawn perpendicualar from B on CA and CF is drawn perpendicular from AB. Let, the point of intersection of the three perpendiculars be O.

Proof: According to the construction, O is the ortho-centre of the $\triangle A B C$

Now, in the $\triangle A B D$ and the $\triangle A C D$

(Since each of them is a right angle)

$\therefore \triangle A B D \cong \triangle A C D \\$

$\therefore \quad < B A D = < C A D$

$\therefore \quad$ AD is the bisector of <BAC

Similarly, BE and CF are respectively the bisector of <ABC and <ACB

$\therefore$ O is incentre of the triangle.

$\therefore \triangle A B D \cong \triangle A C D \\$

$\therefore$ BD = CD

Therefore, AD is a median of the triangle. Similarly, BE and CF are the other two medians. Therefore O is the centroid of the triangle. Again in $\triangle B O D$ and $\triangle COD$ , BD = DC, OD is the common side and <ODB = <ODC.

$\therefore \triangle B O D \cong \triangle C O D \\$

$\therefore O B=O C \text {. similarly } O C=O A$

4. AD, BE and CF are three medians of a triangle ABC. Let us prove that the cetroid of ABC and DEF are the same point.

Solution :

Given:

Let AD, BE and CF are three medians of a triangle ABC. Let G be the centroid of $\triangle A B C$ , BE cuts FD at Q, DE cuts FC at R and FE cuts AD at P respectively.

R.T.P : The Centroid of $\triangle \mathrm{ABC}$ and $\triangle D E F$ are the same point.

Proof: E and F are the mid-points of AC and AB. Then, $EF \| BC$

$\therefore \quad E F=\frac{1}{2} \mathrm{BC}$

$\therefore \quad \mathrm{EF}=\mathrm{BD}$ [ $\because$ D is mid-point of BC]

$\therefore \mathrm{EF}\|\mathrm{BC}\| \mathrm{BD} \text{ and } \mathrm{EF}=\mathrm{BD}.$

$\therefore \quad$ BDEF is a parallelogram.

Similarly,

CDFE is a parallelogram and AFDE is a parallelogram.

In parallelogram BDEF, FQ = QD

In Parallelogram CDFE, DR = RE

In parallelogram AFDE, FP = PE

Hence, $\quad$ FQ = QD, Q is mid-point of FD

$\therefore \quad \mathrm{QE}$ is a median of $\triangle \mathrm{DEF}$

DR = RE, R is mid-point of DE

FR is a median of $\triangle D E F$

$\mathrm{FP}=\mathrm{PE}, \mathrm{P}$ is mid-point EF.

DP is a median of $\triangle D E F$

Hence, The three medians QE, FR and DP meets at the point G

$\therefore$ The centroid of $\triangle \mathrm{ABC} \triangle \mathrm{DEF}$ are the same point. (Proved)

5. Let us prove that two medians of triangle are together greater than the third median.

Solution :

Given:

Let, the three medians of the triangle ABC be AD, BE and CF the point of intersection of these three medians is G.

R.T.P: The sum of the any two medians of the triangle ABC is greater than the third median.

Construction : AD is produced upto H in such a way that $\mathrm{GD}=\mathrm{DH}. \mathrm{CH}$ is joined.

Proof: In triangles BGD and DHC

$\mathrm{GD}=\mathrm{DH}$ (By construction)

BD = DC ( $\because$ D is the mid-point of BC)

and included $<\mathrm{CDH}$ (Vertically opposite angle)

$\therefore \triangle B D G \cong \triangle D H C \\$

$\therefore \quad$ BG = CH

Now, in $\triangle \mathrm{GHC}, \mathrm{GC}+\mathrm{CH}>\mathrm{GH}$

or, $\mathrm{GC}+\mathrm{BG}>\mathrm{GH}$ [ $\because$ it has been proved that BG = CH]

or, $\mathrm{GC}+\mathrm{BG}>\mathrm{AG}[\because \mathrm{GH}=2 \mathrm{GD}=2 . \frac{1}{2} \mathrm{AG}=\mathrm{AG}]$

or, $\frac{2}{3} \mathrm{CF}+ \frac{2}{3} \mathrm{BE}>{ } \frac{2}{3} \mathrm{AD}$

or, $\mathrm{CF}+\mathrm{BE}>\mathrm{AD}$

$\text { i.e, } \mathrm{BE}+\mathrm{CF}>\mathrm{AD}$ …………. (i)

$\mathrm{BE}+\mathrm{AD}>\mathrm{CF}$ …………. (ii)

and $\mathrm{CF}+\mathrm{AD}>\mathrm{BE}$ …………. (iii)

From (i), (ii) and (iii) It may be concluded that, the sum of two medians of a triangle is greater thatn the third median. (Proved)

6. AD, BE and CF are three medians of $\triangle A B C.$

Let us prove that (i) $4(\mathrm{AD}+\mathrm{BE}+\mathrm{CF})>3(\mathrm{AB}+\mathrm{BC}+\mathrm{CA})$

(ii) $3(\mathrm{AB}+\mathrm{BC}+\mathrm{CA})>2(\mathrm{AD}+\mathrm{BE}+\mathrm{CF})$

Solution :

Proof: Let, the point of intersection of the three medians be G. Then G is the centroid of the triangle ABC.

Now, In $\triangle \mathrm{ABG}, \mathrm{AG}+\mathrm{BG}>\mathrm{AB}$

In $\triangle \mathrm{BCG}, \mathrm{BG}+\mathrm{CG}>\mathrm{BC}$

$\text { In } \triangle \mathrm{CAG}, \mathrm{CG}+\mathrm{AG}>\mathrm{CA}$

Adding

$2(\mathrm{AG}+\mathrm{BG}+\mathrm{CG})>\mathrm{AB}+\mathrm{BC}+\mathrm{CA}$

Again, Since the median of a triangle is divided at the centroid in the ratio 2 : 1 Therefore,

$\mathrm{AG}= \frac{2}{3} \mathrm{AD}, \mathrm{BG}=\frac{2}{3} \mathrm{BE} \text { and } \mathrm{CG}=\frac{2}{3}{\mathrm{CF}}$

Hence, $\quad 2( \frac{2}{3} \mathrm{AD}+\frac{2}{3} \mathrm{BE}+\frac{2}{3} \mathrm{CF})>\mathrm{AB}+\mathrm{BC}+\mathrm{CA}$

$\text { or, } \frac{4}{3} (\mathrm{AD}+\mathrm{BE}+\mathrm{CF})>A B+B C+C A \\$

$\text { or, } 4(\mathrm{AD}+\mathrm{BE}+\mathrm{CF})>3(A B+B C+C A)$ ………. (i) Proved

Again,

$\text { In } \triangle A B D, A B+B D>A D \\$

$\text { In } \triangle B C D, B C+C E>B E \\$

$\text { In } \triangle C A F, C A+C E>C F$

Adding, $\mathrm{AB}+\mathrm{BC}+\mathrm{CA}+\mathrm{BD}+\mathrm{CE}+\mathrm{AF}>\mathrm{AD}+\mathrm{BE}+\mathrm{CF}$

or, $A B+B C+C A+\frac{B C}{2}+\frac{C A}{2}+\frac{A B}{2}>A D+B E+C F$

or, $\frac{3}{2} \mathrm{AB}+\frac{3}{2} \mathrm{BC}+\frac{3}{2} \mathrm{CA}>\mathrm{AD}+\mathrm{BE}+\mathrm{CF}$

or, $\frac{3}{2}(\mathrm{AB}+\mathrm{BC}+\mathrm{CA})>\mathrm{AD}+\mathrm{BE}+\mathrm{CF}$

or, $3(\mathrm{AB}+\mathrm{BC}+\mathrm{CA})>2(\mathrm{AD}+\mathrm{BE}+\mathrm{CF})$ ……… (ii) Proved.

7. Three medians AD, BE and CF of $\triangle \mathrm{ABC}$ intersect each other at the point G. If area of $\triangle A B C$ is 36 sq. cm. Let us calculate (i) Area of $\triangle A G B$ (ii) Area of $\triangle C G E$ (iii) Area of quadrilateral BDGF.

Solution :

Given, Let ABC is a triangle whose medians AD. BE and CF intersect each other at the point G.

$\triangle A B C=36 \text { sq. cm}$

Prove: $\mathrm{AD} \text{ is median of }\triangle \mathrm{ABC}$

$\therefore \quad \mathrm{GD} \text { is median of } \triangle B G C \text {. }$

$\therefore \triangle B G D=\triangle C G D$

$\because \quad A D \text{ is median of } \triangle A B C$

$\therefore \triangle \mathrm{ABD}=\triangle \mathrm{ACD}$

$\text { or, } \triangle \mathrm{AGB}+\triangle \mathrm{BGD}= \triangle \mathrm{AGC}+\triangle \mathrm{CGD}$

$\text { or, } \triangle \mathrm{AGB}+\triangle \mathrm{CGD}=\triangle \mathrm{AGC}+\triangle \mathrm{CGD} [\because \triangle \mathrm{BGD}=\triangle \mathrm{CGD}]$

$\therefore \triangle A G B=\triangle A G C$ ……….. (i)

$\triangle A F G+\triangle A G C=\triangle B F G+\triangle B G C$

$\because \quad \mathrm{CF}$ is median of $\triangle \mathrm{ABC}$

$\therefore \triangle \mathrm{ACF}=\triangle \mathrm{BCF}$ ……….. (i)

or, $\triangle \mathrm{BGF}+\triangle \mathrm{AGC}=\triangle \mathrm{BFG}+\triangle \mathrm{BGC}$

$\therefore \triangle \mathrm{AGC}=\triangle \mathrm{BGC}$

$[\because \triangle A F G=\triangle B F G]$

By (i) and (ii) we get,

$\triangle \mathrm{AGB}=\triangle \mathrm{AGC}=\triangle \mathrm{BGC} \\$

$\text { But, } \triangle \mathrm{AGB}+\triangle \mathrm{AGC}+\triangle \mathrm{BGC}=\triangle \mathrm{ABC} \\$

$\therefore \triangle \mathrm{AGR}=-\mathrm{AGC}=\triangle \mathrm{BGC}=\frac{1}{3} \triangle \mathrm{ABC} \\$

$=\frac{1}{3} \times 36 \text { sq. cm }=12 \text { Sq.cm }$

$\therefore \text{ (i) } \triangle \mathrm{AGB}=12 \mathrm{sq . cm.}$ (Ans)

(ii) GE is median of $\triangle A G C.$

$\therefore \quad \triangle \mathrm{CGE} =\frac{1}{2} \triangle \mathrm{AGC} \\$

$=\frac{1}{2} \times 12 \mathrm{sq} . cm . \\$

$=6 \mathrm{sq.} cm . \quad \text { (Ans.) }$

(iii) $\text { Quadrilateral BDGF } =\triangle B F G+\triangle B G D \\$

$=\frac{1}{2} \triangle A G B+\frac{1}{2} \triangle B G C \\$

$= (\frac{1}{2} \times 12+\frac{1}{2} \times 12) \text { Sq.cm. } \\$

$=(6+6)=12 \text { sq. cm. Ans. }$

8. AD, BE and CF are the medians of $\triangle A B C.$

If 2AD = BC let us prove that the angle between two medians is $90^{\circ}.$

Solution :

Given: AD, BE and CF are the Medians of $\triangle A B C$ and $\frac{2}{3} \mathrm{AD}=\mathrm{BC}$

R.T.P : <CGB = $90^{\circ}.$

Proof: BC = $\frac{2}{3}AD$

$\text { Also, } A G=\frac{2}{3} A D \ G D=\frac{1}{3} A D \\$

$\therefore B C=A G, G D=\frac{1}{3} \times \frac{3}{2} B C=\frac{1}{2} B C \\$

$\therefore B D=C D=\frac{1}{2} B C \\$

$\therefore B D=G D \text { and } G D=C D$

Hence, $\triangle B G D \text{ and } \triangle C G D \text{ are isosceles triangles. }$

We take, $<\mathrm{GBD}=< \mathrm{BGD}= x \text { and } <\mathrm{GCD}=<\mathrm{CGD}=\mathrm{y}$

$\therefore \mathrm{BDG} =180^{\circ}-(<\mathrm{GBD}+\angle \mathrm{BGD}) \\$

$=180^{\circ}-(\mathrm{x}+\mathrm{x}) \\$

$=180^{\circ}-(2 \mathrm{x}) \\$

$\text { and } < \mathrm{CDG} =180^{\circ}-(<\mathrm{CGD}+\angle \mathrm{GCD}) \\$

$=180^{\circ}-(\mathrm{y}+\mathrm{y}) \\$

$=180^{\circ}-(2 \mathrm{y})$

$\because \quad$ BC is line segment.

$\therefore 180^{\circ}-2 x+180^{\circ}-2 y=180^{\circ}$

or, $x+y=90^{\circ}.$

$\therefore \quad < \mathrm{BGD}+< \mathrm{CGB}=90^{\circ}$

or, $< \mathrm{CGD}=90^{\circ}.$

Hence,

The angle between two medians is $90^{\circ}$ (Proved)

9. P and Q are the mid-points of sides BC and CD of a parallelogram ABCD respectively; the A P and A Q cuts diagonal BD at the points K and L. Let us prove that, BK = KL = LD

Solution:

Given:

ABCD is a parallelogram in which P and Q are the mid-points of Sides BC and CD.

The line AP and AQ cuts diagonals BD at the points K and L

R.T.P : BK = KL = LD.

Construction : We draw $KC \| L \mathrm{LQ} \text{ and } CL\|PK$

Proof: $\triangle \mathrm{In} \mathrm{LBC}, \mathrm{CL} \| \mathrm{PK} \text{ and } \mathrm{P}$ is mid point of BC

$\therefore$ K is mid point of BL.

$\therefore \quad$ BK = KL ……….. (i)

In $\triangle K C D, K C \| L Q \text{ and } Q$ is mid point of CD.

$\therefore$ L is mid point of DK.

$\therefore$ KL = LD ………… (ii)

From (i) and (ii),

$\mathrm{BK}=\mathrm{KL}=\mathrm{LD} \quad \text { (Proved) }$

10. (M.C.Q)

(i) O is the circumcentre of ABC; if < BOC = $80^{\circ}$ the <BAC is

(a) $40^{\circ}$

(b) $160^{\circ}$

(c) $130^{\circ}$

(d) $110^{\circ}$

Solution :

We have,

<BOC = $80^{\circ}$

$<\mathrm{BAC} =\frac{1}{2}<\mathrm{BOC} \\$

$=\frac{1}{2} 80^{\circ}=40^{\circ}$

$\therefore \quad$ (a) is correct option

(ii) O is the Ortho centre of ABC; If <BAC = $40^{\circ}$ the <BOC is

(a) $80^{\circ}$

(b) $140^{\circ}$

(c) $110^{\circ}$

(d) $40^{\circ}$

Solution :

In quadrilateral AFOE,

$< \mathrm{BOC}=< \mathrm{FOE}=360^{\circ}-(90^{\circ}+90^{\circ}+40^{\circ}) \\$

$=360^{\circ}-220^{\circ} \\$

$=140^{\circ} \\$

$\therefore$ (b) is correct option

(iii) O is incentre of ABC; if $< \mathrm{BAC}=40^{\circ} \text { then } < \mathrm{BOC}$ = 4T.

(a) $80^{\circ}$

(b) $110^{\circ}$

(c) $140^{\circ}$

(d) $40^{\circ}$

Solution:

We have,

$< B O C =90+\frac{1}{2}<\mathrm{BAC} \\$

$=90+\frac{1}{2} \times 40^{\circ} \\$

$=90^{\circ}+20^{\circ}=110^{\circ}$

$\therefore$ (b) is correct option

(iv) G is the Centroid of triangle ABC; if area of GBC 12 sq.cm., then the area of ABC is

(a) 24 sq.cm.

(b) 6 sq.cm.

(c) 36 sq.cm.

(d) none of them

Solution :

$\quad \because \text{ Area of } \triangle \mathrm{ABC}$

= 3 × area of $\triangle GCB$

= 3 × 12 sq.cm.

= 36 sq. cm.

$\therefore$ (c) is correct option

(v) If the length of circumcentre of a right-angled triangle is 5cm.then the length of hypotenuse is

(a) 2.5cm

(b) 10cm

(c) 5cm

(d) none of these

Solution :

$\because \text { Length } \\$

$=2 \times 5 \mathrm{~cm} . \\$

$=10 \mathrm{~cm} .$

$\because \text { Length of hypotenuse }$

$\therefore (\mathrm{b})$ is correct option

12. Short answer type.

(i) If the length of sides of triangle are 6 cm, 8 cm, then let us write where the circumcentre of this triangle lies.

Solution :

We have,

$=(6)^2+(8)^2 \\$

= 36 + 64

= 100

$=(10)^2$

$\therefore$ The length of sides of triangle are 6cm, 8cm and 10cm.

So, it is right angled triangle.

The circumcentre lies at mid-point of length of side 10cm.

(ii) AD is the median and G is the centriod of an equilateral triangle. If the length of side $3 \sqrt{3 \mathrm{~cm}}$ , then let us write the length of AG.

Solution :

We have,

$\mathrm{AB}=3 \sqrt{3} \mathrm{~cm} \\$

$\mathrm{AD}=\frac{\sqrt{3}}{2} \times 3 \sqrt{3}$

$=\frac{3 \times 3}{2} cm=\frac{9}{2} cm \\$

$\mathrm{AG} =\frac{2}{3} \mathrm{AD} \\$

$=\frac{3}{2} \times \frac{9}{2} cm \\$

$=3 cm \text { (Ans.) }$

(iii) Let us write how many points are equidistant from sides of a triangle.

Solution : Only one point.

(iv) DEF is a pedal triangle of an equilateral triangle ABC. Let us write the measure of <FDA.

Solution :

$\because$ DEF is a pedal triangle

$\therefore \triangle \mathrm{ABC}=4 \times \triangle \mathrm{DEF} \\$

$\therefore \quad < \mathrm{FDE}=60^{\circ}$

and AD is bisector

$\therefore < \mathrm{FDA}=\frac{60^{\circ}}{2}=30^{\circ}$ (Ans.)

(v) ABC is an isosceles triangle in which $< \mathrm{ABC}=< \mathrm{ACB}$ and Median AD = 1 / 2 BC. IF $AB \sqrt{2} cm$ , let us write the length of circumradius of this triangle.

Solution: